Total Gadha - Geometry PDF

Lines and Angles Fundamental concepts of Geometry: Point: It is an exact location. It is a fine dot which has neither length nor breadth nor thickness but has position i.e., it has no magnitude. Line segment: The straight path joining two points A and B is called a line segment AB. It has and poi...

Lines and Angles Fundamental concepts of Geometry: Point: It is an exact location. It is a fine dot which has neither length nor breadth nor thickness but has position i.e., it has no magnitude. Line segment: The straight path joining two points A and B is called a line segment AB. It has and points and a definite length. Ray: A line segment which can be extended in only one direction is called a ray. Intersecting lines: Two lines having a common point are called intersecting lines. The common point is known as the point of intersection. Concurrent lines: If two or more lines intersect at the same point, then they are known as concurrent lines. Angles: When two straight lines meet at a point they form an angle. In the figure above, the angle is represented as ∠AOB. OA and OB are the arms of ∠AOB. Point O is the vertex of ∠AOB. The amount of turning from one arm (OA) to other (OB) is called the measure of the angle (ÐAOB). Right angle: An angle whose measure is 90o is called a right angle. Acute angle: An angle whose measure is less then one right angle (i.e., less than 90o), is called an acute angle. 1 http://www.totalgadha.com Obtuse angle: An angle whose measure is more than one right angle and less than two right angles (i.e., less than 180o and more than 90o) is called an obtuse angle. Reflex angle: An angle whose measure is more than 180o and less than 360o is called a reflex angle. Complementary angles: If the sum of the two angles is one right angle (i.e., 90o), they are called complementary angles. Therefore, the complement of an angle θ is equal to 90° − θ. Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180o. Example: Angles measuring 130o and 50o are supplementary angles. Two supplementary angles are the supplement of each other. Therefore, the supplement of an angle θ is equal to 180° − θ. 2 http://www.totalgadha.com Vertically opposite angles: When two straight lines intersect each other at a point, the pairs of opposite angles so formed are called vertically opposite angles. In the above figure, ∠1 and ∠3 and angles ∠2 and ∠4 are vertically opposite angles. Note: Vertically opposite angles are always equal. Bisector of an angle: If a ray or a straight line passing through the vertex of that angle, divides the angle into two angles of equal measurement, then that line is known as the Bisector of that angle. A point on an angle is equidistant from both the arms. In the figure above, Q and R are the feet of perpendiculars drawn from P to OB and OA. It follows that PQ = PR. 3 http://www.totalgadha.com Parallel lines: Two lines are parallel if they are coplanar and they do not intersect each other even if they are extended on either side. Transversal: A transversal is a line that intersects (or cuts) two or more coplanar lines at distinct points. In the above figure, a transversal t is intersecting two parallel lines, l and m, at A and B, respectively. Angles formed by a transversal of two parallel lines: In the above figure, l and m are two parallel lines intersected by a transversal PS. The following properties of the angles can be observed: ∠3 = ∠5 and ∠4 = ∠6 [Alternate angles] ∠1 = ∠5, ∠2 = ∠6, ∠4 = ∠8, ∠3 = ∠7 [Corresponding angles] ∠4 + ∠5 = ∠3 + ∠6 = 180° [Supplementary angles] In the figure given below, which of the lines are parallel to each other? Answer: As 67° + 113° = 180°, lines P and S, R and S, and S and U are parallel. Therefore, lines P, R, S and U are parallel to each other. Similarly, lines Q and T are parallel to each other. 4 http://www.totalgadha.com In the figure given below, PQ and RS are two parallel lines and AB is a transversal. AC and BC are angle bisectors of ∠BAQ and ∠ABS, respectively. If ∠BAC = 30°, find ∠ABC and ∠ACB. Answer: ∠BAQ + ∠ABS = 180° [Supplementary angles] ∠BAQ ∠ABS 180° ⇒ + = = 90° ⇒ ∠BAC + ∠ABC = 90° 2 2 2 Therefore, ∠ABC = 60° and ∠ACB = 90°. For what values of x in the figure given below are the lines P-A-Q and R–B-S parallel, given that AD and BD intersect at D? Answer: We draw a line DE, parallel to RS, as shown in the figure below: In the above figure, ∠CDE = ∠RBD = 10x + 5 ⇒ ∠CDA = 10x + 5 −30 = 10x − 25. Let the line PQ and RS be parallel. Therefore, PQ // DE. ⇒ ∠EDA = ∠CAD = 10x − 25 = 6x − 5 ⇒ x = 5. In the figure given below, lines AB and DE are parallel. What is the value of ∠CDE? Answer: We draw a line CF // DE at C, as shown in the figure below. 5 http://www.totalgadha.com ∠BCF = ∠ABC = 55° ⇒ ∠DCF = 30°. ⇒ CDE = 180° − 30° = 150°. TRIANGLES Triangles are closed figures containing three angles and three sides. General Properties of Triangles: 1. The sum of the two sides is greater than the third side: a + b > c, a + c > b, b + c > a The two sides of a triangle are 12 cm and 7 cm. If the third side is an integer, find the sum of all the values of the third side. Answer: Let the third side be of x cm. Then, x + 7 > 12 or x > 5. Therefore, minimum value of x is 6. Also, x < 12 + 7 or x < 19. Therefore, the highest value of x is 18. The sum of all the integer values from 6 to 18 is equal to 156. 2. The sum of the three angles of a triangle is equal to 180°: In the triangle below ∠A + ∠B + ∠C = 180° Also, the exterior angle α is equal to sum the two opposite interior angle A and B, i.e. α = ∠A + ∠B. Find the value of a + b in the figure given below: Answer: In the above figure, ∠CED = 180° − 125° = 55°. ∠ACD is the exterior angle of ∆ABC. Therefore, ∠ACD = a + 45°. In ∆CED, a + 45° + 55° + b = 180° ⇒ a + b = 80° 6 http://www.totalgadha.com 3. Area of a Triangle: 1 1 Area of a triangle = × base × height = × a × h 2 2 1 1 1 Area of a triangle = bc sin A = ab sin C = ac sin B 2 2 2 a+b+c Area of a triangle = s(s − a)(s − b)(s − c) where s = 2 abc Area of a triangle = where R = circumradius 4R a+b+c Area of a triangle = r × s where r = inradius and s = 2 4. More Rules: SinA SinB SinC Sine Rule: = = a b c b + c − a2 2 2 a2 + c2 − b2 b2 + a2 − c2 Cosine Rule: CosA = , CosB = , CosC = 2bc 2ac 2ab AD x AE l Let D and E be on sides AB and AC of triangle ABC such that = and =. Then, area DB y EC m 1 1 triangle ADE =lx sin A and area triangle ABC = (l + m)(x + y) sin A.Therefore, 2 2 Area ∆ADE lx = Area ∆ABC (x + y)(l + m) 7 http://www.totalgadha.com Points D, E and F divide the sides of triangle ABC in the ratio 1: 3, 1: 4, and 1: 1, as shown in the figure. What fraction of the area of triangle ABC is the area of triangle DEF? Area ∆ADE 1 × 3 3 Area ∆BDF 1×1 1 Area ∆CFE 4 ×1 2 Answer: = = , = = , = = Area ∆ABC 4 × 5 20 Area ∆ABC 4 × 2 8 Area ∆ABC 5 × 2 5 Area ∆DEF ⎛ 3 1 2 ⎞ 13 Therefore, =1−⎜ + + ⎟= Area ∆ABC ⎝ 20 8 5 ⎠ 40 5. Medians of a triangle: The medians of a triangle are lines joining a vertex to the midpoint of the opposite side. In the figure, AF, BD and CE are medians. The point where the three medians intersect is known as the centroid. O is the centroid in the figure. The medians divide the triangle into two equal areas. In the figure, area ∆ABF = area ∆AFC = Area ∆ABC area ∆BDC = area ∆BDA = area ∆CBE = area ∆CEA = 2 AO BO CO The centroid divides a median internally in the ratio 2: 1. In the figure, = = =2 OF OD OE 2 2 2 2 2 2 2 2 2 2 Apollonius Theorem: AB + AC = 2(AF + BF ) or BC + BA = 2(BD + DC ) or BC + AC = 2(EC2 + AE2) ABCD is a parallelogram with AB = 21 cm, BC = 13 cm and BD= 14 cm. Find the length of AC. Answer: The figure is shown below. Let AC and BD intersect at O. O bisects AC and BD. Therefore, OD is the median in triangle ADC. ⇒ AD2 + CD2 = 2(AO2 + DO2) ⇒ AO = 16. Therefore, AC = 32. 8 http://www.totalgadha.com 6. Altitudes of a Triangle: The altitudes are the perpendiculars dropped from a vertex to the opposite side. In the figure, AN, BF, and CE are the altitudes, and their point of intersection, H, is known as the orthocenter. Triangle ACE is a right-angled triangle. Therefore, ∠ECA = 90° − ∠A. Similarly in triangle CAN, ∠CAN = 90° − ∠C. In triangle AHC, ∠CHA = 180° − (∠HAC + ∠HCA) = 180° − (90° − ∠A + 90° − ∠C) = ∠A + ∠C = 180° − ∠B. Therefore, ∠AHC and ∠B are supplementary angles. 7. Internal Angle Bisectors of a Triangle: In the figure above, AD, BE and CF are the internal angle bisectors of triangle ABC. The point of intersection of these angle bisectors, I, is known as the incentre of the triangle ABC, i.e. centre of the circle touching all the sides of a triangle. ⎛B C⎞ ⎛B + C⎞ ⎛ 180 − A ⎞ A ∠BIC = 180° − (∠IBC + ∠ICB) = 180 − ⎜ + ⎟ = 180 − ⎜ ⎟ = 180 − ⎜ ⎟ = 90 + ⎝2 2⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 2 AB BD = (internal bisector theorem) AC CD 8. Perpendicular Side Bisectors of a Triangle: In the figure above, the perpendicular bisectors of the sides AB, BC and CA of triangle ABC meet at O, the circumcentre (centre of the circle passing through the three vertices) of triangle ABC. In figure above, O is the centre of the circle and BC is a chord. Therefore, the angle subtended at the centre by BC will be twice the angle subtended anywhere else in the same segment. Therefore, ∠BOC = 2∠BAC. 9 http://www.totalgadha.com 9. Line Joining the Midpoints: In the figure above, D, E and F are midpoints of the sides of triangle ABC. It can be proved that: FE // BC, DE // AB and DF // AC. BC AB AC FE = , DE = , FD = 2 2 2 Area ∆ABC Area ∆DEF = Area ∆AFE = Area ∆BDF = Area ∆DEC = 4 Corollary: If a line is parallel to the base and passes through midpoint of one side, it will pass through the midpoint of the other side also. In the figure given below: AG = GE and GF // ED, EF //BD and ED // BC. Find the ratio of the area of triangle EFG to trapezium BCDE. Answer: We know that line parallel to the base and passing through one midpoint passes through another midpoint also. Using this principle, we can see that G, F, E and D are midpoints of AE, AD, AB, and AC respectively. Therefore, GF, EF, ED, and BD are medians in triangles AFE, AED, ADB and ABC. We know that medians divide the triangle into two equal areas. Let the area of triangle AGF = a. Therefore, the areas of the rest of the figures are as shown above. The required ratio = a/12a = 1/12. 10 http://www.totalgadha.com Types of triangles: Scalene Triangle Equilateral Triangle Isosceles Triangle Right Triangle No side equal. All the general properties One of the angles is a Each side equal Two sides equal. right angle, i.e. 90°. of triangle apply Each angle = 60° The angles opposite to 1 Length of altitude the opposite sides are Area = AB × BC 2 3 equal. = a AC2 = AB2 + BC2 2 AB × BC Altitude BD = 3 2 AC Area = a 4 The midpoint of the a hypotenuse is equidistant Inradius = 2 3 from all the three vertices, i.e. EA = EB = a Circumradius = EC 3 In triangle DEF shown below, points A, B, and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. If angle D = 40° then what is angle ACB in degrees? (CAT 2001) 1. 140 2. 70 3. 100 4. None of these Answer: Let ∠AEC = ∠EAC = α and ∠CBF = ∠CFB = β. We know that α + β = 180° − ∠D = 140°. ∠ACB = 180° − (∠ECA + ∠BCF) = 180° − (180° − 2α + 180° − 2β) = 100°. In the figure (not drawn to scale) given below, if AD = CD = BC, and ∠BCE = 96°, how much is ∠DBC? (CAT 2003) 1. 32° 2. 84° 3. 64° 4. Cannot be determined. Answer: Let ∠DAC = ∠ACD = α and ∠CDB = ∠CBD = β. As ∠CDB is the exterior angle of triangle ACD, β = 2α. Now ∠ACD + ∠DCB + 96° = 180° ⇒ α + 180° − 2β + 96° = 180° ⇒ 3α = 96° ⇒ α = 32° ⇒ β = 64° 11 http://www.totalgadha.com Similarity of triangles: Two triangles are similar if their corresponding angles are equal or corresponding sides are in proportion. In the figure given above, triangle ABC is similar to triangle PQR. Then ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R and AB BC CA AI AJ = = = (altitudes) = (medians) PQ QR RP PK PL Therefore, if you need to prove two triangles similar, prove their corresponding angles to be equal or their corresponding sides to be in proportion. Ratio of Areas: If two triangles are similar, the ratio of their areas is the ratio of the squares of the length of their corresponding sides. Therefore, Area of triangle ABC AB2 BC 2 CA2 = = = Area of triangle PQR PQ2 QR2 RP2 DE 1 In triangle AC, shown above, DE // BC and =. If area of triangle ADE is 10, find the area BC 4 of the trapezium BCED and the area of the triangle CED. Answer: ∆ADE and ∆ABC are similar. Therefore, Area of triangle ABC BC2 = = 16 ⇒ Area of triangle ABC = 160 ⇒ Area of trapezium BCDE = Area Area of triangle ADE DE2 ∆ABC − Area ∆ADE = 160 − 10 = 150. 12 http://www.totalgadha.com To find the area of triangle CDE, we draw altitudes of triangle BDC and CDE, as shown above. Let the length of the altitudes be h. 1 1 Area of triangle BCD = × BC × h and area of triangle CDE = × DE × h 2 2. Area of triangle BCD BC ⇒ = =4 Area of triangle CDE DE Therefore, we divide the area of the trapezium BCED in the ratio 1: 4 to find the area of triangle CDE. 1 The required area = = × 150 = 30. 5 In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1, the ratio of CD: PQ is (CAT 2003- Leaked) 1. 1: 0.69 2. 1: 0.75 3. 1: 0.72 4. None of these Answer: Let BQ = a and DQ = b, as shown in the figure below. PQ b Triangle ABD and triangle PQD are similar. Therefore, =. Also triangle CBD and triangle PBQ AB a + b PQ a are similar. Therefore, = CD a+b AB a CD a+b 4 Dividing the second equality by the first, we get. = = 3. Therefore, = = = 1 : 0.75 CD b PQ a 3 13 http://www.totalgadha.com In triangle ABC, lines DG, EH, and FI are parallel to the base BC. Then it can be proved that AD AG DE GH EF HI = , = , = DE GH EF HI FB IC In the figure (not drawn to scale) given below, P is a point on AB such that AP: PB = 4: 3. PQ is parallel to AC and QD is parallel to CP. In ∆ARC, ∠ARC = 90°, and in ∆PQS, D PSQ = 90°. The length of QS is 6 cm. What is ratio AP: PD? (CAT 2003) 1. 10: 3 2. 2: 1 3. 7: 3 4. 8: 3 AP CQ 4 Answer: PQ is parallel to AC ⇒ = =. PB QB 3 Let AP = 4x and PB = 3x. PD CQ 4 4PB 12x QD is parallel CP ⇒ = = ⇒ PD = = DB QB 3 7 7 12x ⇒ AP: PD = 4x: = 7: 3 7 In the adjoining figure, I and II are circles with centres P and Q respectively. The two circles touch each other and have a common tangent that touches them at points Rand S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in the ratio 4: 3. It is also known that the length of PO is 28 cm. (CAT 2004) What is the ratio of the length of PQ to that of QO? 1. 1 : 4 2. 1 : 3 3. 3 : 8 4. 3 : 4 What is the radius of the circle II? 1. 2 cm 2. 3 cm 3. 4 cm 4. 5cm The length of SO is 1. 8 3 cm 2. 10 3 cm 3. 12 3 cm 4. 14 3 cm Answer: 14 http://www.totalgadha.com Join R and P, and S and Q. ∠PRO = ∠QSO = 90°. Therefore, ∆PRO and ∆QSO are similar. Therefore, PR PO 3 1 = ⇒ QO = × PO = 21 ⇒ PQ = × PO = 7 ⇒ PQ : QO = 1 : 3. Also, PQ = 7 and the radii are in SQ QO 4 4 the ratio 4: 3. Therefore, radius of circle II = 3. Now SO = QO2 − SQ2 = 441 − 9 = 432 = 12 3 NOTE: In similar triangles, sides opposite to equal angles are in proportion. Consider the triangle ABC shown in the following figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and ∠BCD = ∠BAC. What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC? (CAT 2005) 7 8 6 5 1. 2. 3. 4. 9 9 9 9 Answer: In ∆BAC and ∆BCD, ∠BCD = ∠BAC, ∠B is common ⇒ ∠BDC = ∠BCA. Therefore, the two triangles are similar. AB AC BC BC2 BC × CD = = ⇒ AB = = 16 ⇒ AD = 7, Similarly, AC = =8 BC CD BD BD BD 7 Perimeter ∆ADC = 7 + 6 + 8 = 21, perimeter ∆BDC = 27. Therefore, ratio = 9 15 http://www.totalgadha.com  REGULAR POLYGONS A regular polygon is a polygon with all its sides equal and all its interior angles equal. All vertices of a regular lie on a circle whose center is the center of the polygon. 360 Each side of a regular polygon subtends an angle θ = at the centre, as shown in the figure. n 360 180 − n = 180(n − 2) 180(n − 2) Also X = Y =. Therefore, interior angle of a regular polygon = X + Y =. 2 2n n 180(n − 2) Sum of all the angles of a regular polygon = n × = 180(n − 2). n What is the interior angle of a regular octagon? 180(8 − 2) Answer: The interior angle of a regular octagon = = 135° 8 The formula for sum of all the angles of a regular polygon, i.e. 180(n − 2) , is true for all n-sided convex simple polygons. Let’s look at some polygons, especially quadrilaterals: Quadrilateral: A quadrilateral is any closed shape that has four sides. The sum of the measures of the angles is 360o. Some of the known quadrilaterals are square, rectangle, trapezium, parallelogram and rhombus. Square: A square is regular quadrilateral that has four right angles and parallel sides. The sides of a square meet at right angles. The diagonals also bisect each other perpendicularly. If the side of the square is a, then its perimeter = 4a, area = a2 and the length of the diagonal = 2a 16 http://www.totalgadha.com Rectangle: A rectangle is a parallelogram with all its angles equal to right angles. Properties of a rectangle: Sides of rectangle are its heights simultaneously. Diagonals of a rectangle are equal: AC = BD. A square of a diagonal length is equal to a sum of squares of its sides’ lengths, i.e. AC² = AD² + DC². Area of a rectangle = length × breadth Parallelogram: A parallelogram is a quadrangle in which opposite sides are equal and parallel. Any two opposite sides of a parallelogram are called bases, a distance between them is called a height. Area of a parallelogram = base × height Perimeter = 2(sum of two consecutive sides) Properties of a parallelogram: 1. Opposite sides of a parallelogram are equa l(AB = CD, AD = BC ). 2. Opposite angles of a parallelogram are equal ( A = C, B = D ). 3. Diagonals of a parallelogram are divided in their intersection point into two ( AO = OC, BO = OD ). 4. A sum of squares of diagonals is equal to a sum of squares of four sides: AC² + BD² = AB² + BC² + CD² + AD². Rhombus: If all sides of parallelogram are equal, then this parallelogram is called a rhombus. Diagonals of a rhombus are mutually perpendicular ( AC BD ) and divide its angles into two ( DCA = BCA, ABD = CBD etc. ). 1 1 Area of a rhombus = × product of diagonals = × AC × BD 2 2 17 http://www.totalgadha.com Trapezoid: Trapezoid is a quadrangle two opposite sides of which are parallel. Here AD || BC. Parallel sides are called bases of a trapezoid, the two others (AB and CD ) are called lateral sides. A distance between bases (BM) is a height. The segment EF, joining midpoints E and F of the lateral sides, is called a midline of a trapezoid. A midline of a trapezoid is equal to a half-sum of bases: and parallel to them: EF || AD and EF || BC. A trapezoid with equal lateral sides ( AB = CD ) is called an isosceles trapezoid. In an isosceles trapezoid angles by each base, are equal ( A = D, B = C).. Sum of parallel sides AD + BC Area of a trapezoid = × height = × BM 2 2 In a trapezium ABCD with bases AB and CD , the sum of the squares of the lengths of the diagonals is equal to the sum of the squares of the lengths of the non-parallel sides and twice the product of the lengths of the parallel sides: AC + BD = AD + BC + 2 ⋅ AB ⋅ CD 2 2 2 2 Here is one more polygon, a regular hexagon: Regular Hexagon: A regular hexagon is a closed figure with six equal sides. If we join each vertex to the centre of the hexagon, we get 6 equilateral triangles. Therefore, if the side of the hexagon is a, each equilateral triangle has a side a. Hence, area of the regular hexagon = 3 2 3 3 2 6× a = a. 4 2 18 http://www.totalgadha.com  CIRCLE A circle is a set of all points in a plane that lie at a constant distance from a fixed point. The fixed point is called the center of the circle and the constant distance is known as the radius of the circle. Arc: An arc is a curved line that is part of the circumference of a circle. A minor arc is an arc less than the semicircle and a major arc is an arc greater than the semicircle. Chord: A chord is a line segment within a circle that touches 2 points on the circle. Diameter: The longest distance from one end of a circle to the other is known as the diameter. It is equal to twice the radius. Circumference: The perimeter of the circle is called the circumference. The value of the circumference = 2πr, where r is the radius of the circle. Area of a circle: Area = π x (radius)2 = πr2. Sector: A sector is like a slice of pie (a circular wedge). θ Area of Circle Sector: (with central angle θ) Area = × π × r2 360 θ Length of a Circular Arc: (with central angle θ) The length of the arc = × 2π × r 360 Tangent of circle: A line perpendicular to the radius that touches ONLY one point on the circle If 45° arc of circle A has the same length as 60° arc of circle B, find the ratio of the areas of circle A and circle B. Answer: Let the radius of circle A be r1 and that of circle B be r2. 45 60 r 4 πr12 16 ⇒ × 2π × r1 = × 2π × r2 ⇒ 1 = ⇒ Ratio of areas = =. 360 360 r2 3 πr22 9 The perpendicular from the center of a circle to a chord of the circle bisects the chord. In the figure below, O is the center of the circle and OM ⊥ AB. Then, AM = MB. 19 http://www.totalgadha.com Conversely, the line joining the center of the circle and the midpoint of a chord is perpendicular to the chord. In a circle, a chord of length 8 cm is twice as far from the center as a chord of length 10 cm. Find the circumference of the circle. Answer: Let AB and CD be two chords of the circle such that AB = 10 and CD = 8. Let O be the center of the circle and M and N be the midpoints of AB and CD. Therefore OM ⊥ AB, ON ⊥ CD, and if ON = 2x then OM = x. BM2 + OM2 = OB2 and DN2 + ON2 = OD2. OB = OD = r (2x)2 + 42 = r2 and x2 + 52 = r2. Equating both the equations we get, 4x2 + 16 = x2 + 25 Or x = 3 r = 2 7. Therefore circumference = 2πr = 4π 7. What is the distance in cm between two parallel chords of length 32 cm and 24 cm in a circle of radius 20 cm? (CAT 2005) 1. 1 or 7 2. 2 or 14 3. 3 or 21 4. 4 or 28 Answer: The figures are shown below: The parallel chords can be on the opposite side or the same side of the centre O. The perpendicular (s) dropped on the chords from the centre bisect (s) the chord into segments of 16 cm and 12 cm, as shown in the figure. From the Pythagoras theorem, the distances of the chords from the centre are 202 − 162 = 12 and 202 − 122 = 16 , respectively. Therefore, the distances between the chords can be 16 + 12 = 28 cm or 16 − 12 = 4 cm. 20 http://www.totalgadha.com In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE:EB = 1:2, and DF is perpendicular to MN such that NL:LM = 1:2. The length of DH in cm is (CAT 2005) Answer: In the above figure, AB = MN = 3 cm and AE: EB = NL: LM = 1: 2 ⇒ AE = NL = 1 cm. Now AO = NO = 1.5 cm ⇒ OE = HL = OL = 0.5 cm. Join O and D 1 2 2 −1 ⇒ OD2 = OL2 + DL2 ⇒ DL2 = OD2 − OL2 = 1.52 − 0.52 = 2 ⇒ DH = DL − HL = 2 − = 2 2 Equal chords are equidistant from the center. Conversely, if two chords are equidistant from the center of a circle, they are equal. In the following figure, two chords of a circle, AB and CD, intersect at point P. Then, AP × PB = CP × PD. In the following figure, length of chord AB = 12. O-P-C is a perpendicular drawn to AB from center O and intersecting AB and the circle at P and C respectively. If PC = 2, find the length of OB. Answer: Let us extend OC till it intersects the circle at some point D. 21 http://www.totalgadha.com CD is the diameter of the circle. Since OP is perpendicular to AB, P is the midpoint of AB. Hence, AP = PB = 6. Now DP × PC = AP × PB DP = 18. Therefore, CD = 20 OC = 10. OB = OC = radius of the circle = 10. In a circle, equal chords subtend equal angles at the center. The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circumference. In the figure shown above, a = 2b. Angles inscribed in the same arc are equal. In the figure angle ACB = angle ADB. An angle inscribed in a semi-circle is a right angle. Let angle ACB be inscribed in the semi-circle ACB; that is, let AB be a diameter and let the vertex C lie on the circumference; then angle ACB is a right angle. In the figure AB and CD are two diameters of the circle intersecting at an angle of 48°. E is any point on arc CB. Find angle CEB. 22 http://www.totalgadha.com Answer: Join E and D. Since arc BD subtends an angle of 48° at the center, it will subtend half as many degrees on the remaining part of circumference as it subtends at the center. Hence, angle DEB = 24°. Since angle CED is made in a semicircle, it is equal to 90°. Hence, angle CEB = angle CED + angle DEB = 90° + 24° = 114°. In the above figure, AB is a diameter of the circle and C and D are such points that CD = BD. AB and CD intersect at O. If angle AOD = 45°, find angle ADC. Answer: Draw AC and CB. CD = BD ⇒ ∠DCB = ∠DBC = θ (say). ∠ACB = 90° ⇒ ∠ACD = 90°− θ. ∠ABD = ∠ACD = 90°− θ ⇒ ∠ABC = θ − (90°− θ) = 2θ − 90. In ∆OBC, 45° + 2θ − 90 + θ = 180° ⇒ 3θ = 225° ⇒ θ = 75°. ∠ADC = ∠ABC = 2θ − 90 = 60°. In the adjoining figure, chord ED is parallel to the diameter AC of the circle. If angle CBE = 65°, then what is the value of angle DEC? (CAT 2004) 1. 35˚ 2. 55˚ 3. 45˚ 4. 25˚ Answer: ∠ABC = 90° ⇒ ∠ABE = 90 − ∠EBC = 25°. ∠ABE = ∠ACE = 25°. ∠ACE = ∠CED = 25° (alternate angles) The straight line drawn at right angles to a diameter of a circle from its extremity is tangent to the circle. Conversely, If a straight line is tangent to a circle, then the radius drawn to the point of contact will be perpendicular to the tangent. Let AB be a diameter of a circle, and let the straight line CD be drawn at right angles to AB from its extremity B; then the straight line CD is tangent to the circle. 23 http://www.totalgadha.com If two tangents are drawn to a circle from an exterior point, the length of two tangent segments are equal. Also, the line joining the exterior point to the centre of the circle bisects the angle between the tangents. In the above figure, two tangents are drawn to a circle from point P and touching the circle at A and B. Then, PA = PB. Also, ∠APO = ∠BPO. Also, the chord AB is perpendicular to OP. In the following figure, lines AP, AQ and BC are tangent to the circle. The length of AP = 11. Find the perimeter of triangle ABC. Answer: let AB = x and BP = y. Then, BD = BP because they are tangents drawn from a same point B. Similarly CD = CQ and AP = AQ. Now perimeter of triangle ABC = AB + BC + CA = AB + BD + DC + AC = AB + BP + CQ + AC = AP + AQ = 2AP = 22. From an external point P, a secant P-A-B, intersecting the circle at A and B, and a tangent PC are drawn. Then, PA × PB = PC2. In the following figure, if PC = 6, CD = 9, PA = 5 and AB = x, find the value of x Answer: Let a tangent PQ be drawn from P on the circle. Hence, PC × PD = PQ2 = PA × PB 6 × 15 = 5 × (5 + x) x = 13 In the following figure, PC = 9, PB = 12, PA = 18, and PF = 8. Then, find the length of DE. 24 http://www.totalgadha.com  9 27 Answer: In the smaller circle PC × PB = PF × PE PE = 12 × =. In the larger circle, PB × PA = PE × 8 2 2 PD PD = 12 × 18 × = 16. Therefore, DE = PD – PE = 16 – 13.5 = 2.5 27 The angle that a tangent to a circle makes with a chord drawn from the point of contact is equal to the angle subtended by that chord in the alternate segment of the circle. In the figure above, PA is the tangent at point A of the circle and AB is the chord at point A. Hence, angle BAP = angle ACB. In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If ∠ATC = 30° and ∠ACT = 50°, then the angle ∠BOA is (CAT 2003) 1. 100° 2. 150° 3. 80° 4. not possible to determine Answer: Tangent TC makes an angle of 50° with chord AC. Therefore, ∠TBC = 50°. In triangle TBC, ∠BCT = 180° − (30° + 50°) = 100°. Therefore, ∠BCA = ∠BCT − ∠ACT = 100° − 50° = 50°. ∠BOA = 2∠BCA = 100°. Two circles touch internally at P. The common chord AD of the larger circle intersects the smaller circle in B and C, as shown in the figure. Show that , ∠APB = ∠CPD. Answer: Draw the common tangent XPY at point P. 25 http://www.totalgadha.com Now, for chord DP, ∠DPX = ∠DAP, and for chord PC, ∠CPX = ∠CBP ⇒ ∠CPD = ∠CPX − ∠DPX = ∠CBP − ∠DAP. In triangle APB, ∠CBP is the exterior angle ⇒ ∠CBP = ∠CAP + ∠APB ⇒ ∠CBP − ∠CAP = ∠APB ⇒ ∠CPD = ∠CPX − ∠DPX = ∠CBP − ∠DAP = ∠APB When two circles intersect each other, the line joining the centers bisects the common chord and is perpendicular to the common chord. In the figure given above, the line joining the centers divides the common chord in two equal parts and is also perpendicular to it. Two circles, with diameters 68 cm and 40 cm, intersect each other and the length of their common chord is 32 cm. Find the distance between their centers. Answer: In the figure given above, the radii of the circles are 34 cm and 20 cm, respectively. The line joining the centers bisects the common chord. Hence, we get two right triangles: one with hypotenuse equal to 34 cm and height equal to 16 cm, and the other with hypotenuse equal to 20 cm and height equal to 16 cm. Using Pythagoras theorem, we get the bases of the two right triangles equal to 30 cm and 12 cm. Hence, the distance between the centers = 30 + 12 = 42 cm. 26 http://www.totalgadha.com  SOLIDS A solid figure, or solid, is any portion of space bounded by one or more surfaces, plane or curved. These surfaces are called the faces of the solid, and the intersections of adjacent faces are called edges. Let’s have a look at some common solids and their properties. Parallelepiped: A parallelepiped is a solid bounded by three pairs of parallel plane faces. Each of the six faces of a parallelepiped is a parallelogram. Opposite faces are congruent. The four diagonals of a parallelepiped are concurrent and bisect one another. Cuboid: A parallelepiped whose faces are rectangular is called a cuboid. The three dimensions associated with a cuboid are its length, breadth and height (denoted as l, b and h here.) The length of the three pairs of face diagonals are BF = b2 + h2 , AC = l2 + h2 , and DF = l2 + b2. The length of the four equal body diagonals AF = l2 + b2 + h2. The total surface area of the cuboid = 2(lb + bh + hl) Volume of a cuboid = lbh Diagonal l2 + b2 + h2 The radius of the sphere circumscribing the cuboid = =. 2 2 Note that if the dimensions of the cuboid are not equal, there cannot be a sphere which can be inscribed in it, i.e. a sphere which touches all the faces from inside. Euler’s Formula: the number of faces (F), vertices (V), and edges (E) of a solid bound by plane faces are related by the formula F + V = E + 2 gives here 6 + 8 = 12 + 2. Cube: A cube is a parallelepiped all of whose faces are squares. Total surface area of the cube = 6a2 Volume of the cube = a3 27 http://www.totalgadha.com  Length of the face diagonal b = 2a Length of the body diagonal c = 3a 3a Radius of the circumscribed sphere = 2 a Radius of the inscribed sphere = 2 a Radius of the sphere tangent to edges = 2 Prism: A prism is a solid bounded by plane faces, of which two, called the ends, are congruent figures in parallel planes and the others, called side-faces are parallelograms. The ends of a prism may be triangles, quadrilaterals, or polygons of any number of sides. The side- edges of every prism are all parallel and equal. A prism is said to be right, if the side-edges are perpendicular to the ends: In this case the side faces are rectangles. Cuboids and cubes are examples. Curved surface area of a right prism = perimeter of the base × height Total surface area of a right prism = perimeter of the base × height + 2 × area of the base Volume of a right prism = area of the base × height Right Circular Cylinder: A right circular cylinder is a right prism whose base is a circle. In the figure given below, the cylinder has a base of radius r and a height of length h. Curved surface area of the cylinder = 2πrh Total surface area of the cylinder = 2πrh + πr2 Volume of the cylinder = πr2h Pyramid: A pyramid is a solid bounded by plane faces, of which one, called the base, is any rectilinear figure, and the rest are triangles having a common vertex at some point not in the plane of the base. The slant height of a pyramid is the height of its triangular faces. The height of a pyramid is the length of the perpendicular dropped from the vertex to the base. 28 http://www.totalgadha.com In a pyramid with n sided regular polygon as its base, Total number of vertices = n + 1 Perimeter Curved surface area of the pyramid = × slant height. 2 Perimeter Total surface area of the pyramid = × slant height + area of the base. 2 Base area Volume of the pyramid = × height 3 Tetrahedron: A tetrahedron is a pyramid which has four congruent equilateral triangles as it four faces. The figure below shows a tetrahedron with each face equal to an equilateral triangle of side a. Total number of vertices = 4 The four lines which join the vertices of a tetrahedron to the centroids of the opposite faces meet at a point which divides them in the ratio 3: 1. In the figure. AH: HF = 3: 1. 3 3a2 Curved surface area of the tetrahedron =. 4 Total surface area of the tetrahedron = 3a2. 6a Height of the tetrahedron = 3 2a3 Volume of the tetrahedron = 12 Right Circular Cone: a right circular cone is a pyramid whose base is a circle. In the figure given below, the right circular cone has a base of radius r and a height of length h. 29 http://www.totalgadha.com  Slant height l = h2 + r2 Curved surface area of the cone = πrl Total surface area of the cone = πrl + πr2 πr2h Volume of the cone = 3 Frustum of a Cone: When a right circular cone is cut by a plane parallel to the base, the remaining portion is known as the frustum. Slant height l = h2 + (R − r)2. Curved surface area of the frustum = π(r + R)l Total surface area = π(r + R)l + π(r2 + R2) πh(r2 + R2 + R r) Volume of the frustum = 3 Sphere: A sphere is a set of all points in space which are at a fixed distance from a given point. The fixed point is called the centre of the sphere, and the fixed distance is the radius of the sphere. Surface area of a sphere = 4πr2 4 3 Volume of a sphere = πr 3 Spherical Shell: A hollow shell with inner and outer radii of r and R, respectively. 30 http://www.totalgadha.com  4 Volume of the shell = π(R3 − r3 ) 3 31 http://www.totalgadha.com  180 1. A regular polygon with n sides has interior angles measuring 178o. What is the value of ? n 2. A regular hexagon is inscribed in a circle of radius 6. What is the area of the hexagon? 3. A rhombus has a perimeter of 52 cm and a diagonal measuring 24 cm. What is the length, in centimeters, of the other diagonal? 4. A rhombus has diagonals measuring 6 cm and 10 cm. What is its area in square centimeters? (a) 30 (b) 32 (c) 60 (d) 64 (e) None of these 5. Two gears are circular, and the circles are tangent as shown. If the centers are fixed and the radii are 30 cm and 40 cm, how many revolutions will the larger circle have made when the smaller circle has made 4 revolutions? 6. A regular hexagon has a perimeter of 12 cm. What is its area ? (a) 6 3 (b) 72 3 (c) 144 3 (d) 216 3 (e) None of these 7. If the expressions shown are the degree measures of the angles of the pentagon, find the value of x + y. 8. One angle of a regular polygon measures 177°. How many sides does it have? (a) 89 (b) 120 (c) 177 (d) 183 (e) None of these 9. Octagon ABCDEFGH is similar to octagon JKLMNOPQ. If AB = 10 , JK = 8 , and m∠A = 120° , what is m∠J in degrees? (a) 96° (b) 120° (c) 135° (d) 150° (e) None of these 10. Find the sum of the measures of one interior and one exterior angle of a regular 40-gon. (a) 168° (b) 174° (c) 180° (d) 186° (e) None of these 11. One angle of a parallelogram measures ( 2x + y ) °. Another angle of the same quadrilateral (but not the opposite angle) measures ( x + 2 y)°. What is ( x + y ) ? (a) 30 (b) 60 (c) 90 (d) 120 (e) None of these 32 http://www.totalgadha.com 12. An isosceles trapezoid has a mid segment measuring 13 cm and an area of 52 cm2. If one base has length 10 cm, find the perimeter of the trapezoid in centimeters. 13. A right triangle with legs measuring 12 cm and 16 cm is inscribed in a circle. What is the circumference of the circle in centimeters? (a) 14π (b) 16π (c) 20π (d) 28π (e) None of these 14. A square has a diagonal measuring 8 cm. When its area is expressed as 2 K square centimeters, what is K? (a) 4 (b) 5 (c) 6 (d) 7 (e) None of these 15. One-fourth of the area of a square with each side measuring 2x cm is sectioned off and removed. (“Before And After” pictures of the procedure appear to the right.) The area removed is itself square-shaped. What is the perimeter of the resultant figure in centimeters? (a) 6x (b) 7x (c) 8x (d) 9x (e) None of these 16. A central angle measuring M ° intercepts an arc in a circle of radius r cm. The length of the subtended arc is 8π cm. The area of the sector formed by (and including) the angle is 48π cm2. ⎛M ⎞ Evaluate ⎜ ⎟. ⎝ r ⎠ (a) 5 (b) 10 (c) 20 (d) 40 (e) None of these 17. A regular hexagon with a perimeter of 12 2 units lies in the Cartesian plane in such a way that its center is on the origin, two of the vertices lie on the x-axis, and the midpoints of two of its sides lie on the y-axis. If the portion of the hexagon that lies in Quadrant I is completely revolved around the x-axis, a solid whose volume is X cubic units results. If the same portion is completely 2 ⎛X⎞ revolved around the y-axis, a solid with a volume of Y cubic units results. Evaluate ⎜ ⎟. ⎝Y ⎠ 48 4 16 (a) (b) 1 (c) (d) (e) None of these 49 3 9 33 http://www.totalgadha.com  18. A circle is inscribed inside a square. The square is inscribed inside another circle. If the area of the small circle is π cm2, what is the area of the large circle, in square centimeters? (a) π 2 (b) 2π (c) 2π 2 (d) 4π (e) None of these 19. A circle is inscribed in a triangle with sides measuring 4 cm, 6 cm, and 8 cm. What is the area of the circle in square centimeters? 7π 3π 5π 7π (a) (b) (c) (d) (e) None of these 6 2 3 4 20.An equilateral triangle T1 has area 100 3 sq. cm. A second triangle, T2, is drawn with vertices on the midpoints of the sides of T1. The midpoints of the sides of T2 are the vertices of triangle T3, and so on. What is the sum of the perimeters, in centimeters, of all the triangles, T1, T2, T3… etc.? 21.In a 30°- 60°- 90° triangle, the longest side and the shortest side differ in length by 2002 units. What is the length of the longest side? 22.What is the area of a triangle with sides of lengths 7, 8 and 9? 23.The base of an isosceles triangle is 80 cm long. If the area of the triangle cannot exceed 1680 square centimeters, what is the maximum number of centimeters in the perimeter of the triangle? 24.A triangle has sides measuring 41 cm, 41 cm and 18 cm. A second triangle has sides measuring 41 cm, 41 cm and x cm, where x is a whole number not equal to 18. If the two triangles have equal areas, what is the value of x? 25.In a triangle ABC, AB = 16 units, ∠CAB = 30°, and ∠ACB = 45°. What is the length of BC? 26.You have 6 sticks of lengths 10, 20, 30, 40, 50 and 60 cm. The number of non-congruent triangles that can be formed by choosing three of the sticks to make the sides is (a) 3 (b) 6 (c) 7 (d) 10 (e) 12 27.A triangle has sides of lengths 10, 24 and n, where n is a positive integer. The number of values of n for which this triangle has three acute angles is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 34 http://www.totalgadha.com  ABC forms an equilateral triangle in which B is 2 km from A. A person starts walking from B in a direction parallel to AC and stops when he reaches a point D directly east of C. He, then, reverses direction and walks till he reaches a point E directly south of C. 28.Then D is (a) 3 km east and 1 km north of A (b) 3 km east and 3 km north of A (c) 3 km east and 1 km south of A (d) 3 km west and 3 km north of A 29.The total distance walked by the person is (a) 3 km (b) 4 km (c) 2 3 km (d) 6 km 30.How many non-congruent triangles with perimeter 7 have integer side lengths? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 31.When the base of a triangle is increased by 10% and he altitude to this base is decreased by 10%, the area is (a) increased by 10% (b) decreased by 10% (c) increased by 1% (d) decreased by 1% (e) unchanged 32.In the figure given below, triangle ABC is right-angled. What is the area of triangle ABD? (a) 6 (b) 7 (c) 8 (d) 9 (e) 10 33.Let ABC be an equilateral triangle with sides x. Let P be the point of intersection of the three angle bisectors. What is the length of AP? x 3 x 3 5 3 2x 3 4x 3 (a) (b) (c) (d) (e) 3 6 6 6 6 34.In the figure below, ∠ABC and ∠BDA are both right angles. If v + w = 35 and x + y = 37, then what is the value of y? 35 http://www.totalgadha.com  (a) 11 (b) 12 (c) 13 (d) 14 (e) 15 35.The area of ∆ABC is 60 square units. If BD = 8 units and DC = 12 units, what is the area of ∆ABD? (a) 24 (b) 40 (c) 48 (d) 36 (e) 6 36.In right triangle ABC, AX = AD and CY = CD, as shown in the figure below. What is the measure of ∠XDY? (a) 35° (b) 40° (c) 45° (d) 50° (e) cannot be determined 37.In triangle ABC, ∠A is equal to 120°. A point D is inside the triangle such that ∠DBC = 2∠ABD and ∠DCB = 2∠ACD. What is the measure of ∠BDC? (a) 135 (b) 140 (c) 145 (d) 150 (e) 155 In the figure shown, PQR is an isosceles triangle with PQ = PR. S is a point on QR such that PS = PT. Also, ∠QPS = 30°. 38.What is the measure of ∠RST? (a) 7.5° (b) 15° (c) 20° (d) 45° 39.Two sides of a triangle are of length 15 and 7 centimeters. If the length of the third side is an integer value, what is the sum of all the possible lengths of the third side? (a) 253 (b) 231 36 http://www.totalgadha.com  (c) 210 (d) 195 An agriculturist is conducting an experiment in a rectangular field ABCD. He sows seeds of a crop evenly across the field, but he uses a new variety of manure in the area CEF (E and F are midpoints of BD and BE, respectively) whereas he uses the old variety in the rest of the field. The yield per unit area of the crop in ∆CEF is 3 times the yield in rest of the field. 40.What is the ratio of the amount of crop produced in ∆CEF to that produced in the rest of the field? (a) 3: 4 (b) 1: 2 (c) 3: 7 (d) 3: 8 41.In triangle ABC, the longest side BC is of length 20 and the altitude from A to BC is of length 12. A rectangle DEFG is inscribed in ABC with D on AB, E on AC, and both F and G on BC. The maximum possible area of rectangle DEFG is (a) 60 (b) 100 (c) 120 (d) 150 (e) 200 42.The degree measure of an angle whose complement is eighty percent of half the angle’s supplement is (a) 60 (b) 45 (c) 30 (d) 15 43.Two congruent 90°-60°-30° triangles are placed, as shown, so that they overlap partly and their hypotenuses coincide. If the hypotenuse is 12 cm, find the area common to both triangles (a) 6 3 cm2 (b) 8 3 cm2 (c) 9 3 cm2 (d) 12 3 cm2 (e) 24 cm2 In triangle ABC, side AC and the perpendicular bisector of side BC meet in point D, and BD bisects ∠ABC. If AD = 9, and DC = 7. 44.What is the area of triangle ABD? (a) 14 5 (b) 21 37 http://www.totalgadha.com  (c) 28 (d) 21 5 Between two parallel lines PQ and RS, two transversals are drawn, as shown in the figure. The angles which the two transversals make with each other and with their respective lines are also shown. 45.What is the value of x? (a) 20° (b) 25° (c) 30° (d) 35° 46.The perimeter of an isosceles right triangle is 2a. Then its area is (a) 4a2 2 (b) 3a2/2 (c) a2(3 − 2 2 ) (d) 4a2(1 + 3) Triangle ABC (shown in the figure) is equilateral with side length of 16. Also, AD ⊥ BC and AE ≅ ED. 47.What is the value of BE? (a) 4 3 8 (b) 3 16 (c) 3 (d) 4 7 In the square ABCD, AB is extended and E is a point on AB such that CE intersects AD at F and BD at G. The length of FG and GC are 3 and 5 units, respectively. 48.What is the length of EF? (a) 4 (b) 5 13 (c) 3 16 (d) 3 38 http://www.totalgadha.com 49.Two telegraph poles of height a and b meters are on opposite sides of a road. Wires are drawn from top of one pole to the bottom of the other. If the wires are completely taut, then how many feet above the ground will the wires cross each other? 2ab (a) a+b ab (a) 2(a + b) ab (b) a+b a+b (c) 2 Triangle ABC is right angled at B. AB = 7, AC = 25 and D is a point on BC such that AD is the bisector of angle A, as shown in the figure. 50.What is the length of AD? (a) 9.00 (a) 8.75 (b) 12.5 (c) 13.0 Sides of triangle ABC are AB = 12, BC = 18, and AC = 10. There is a point D, on BC, such that both incircles of triangles ABD and ACD touch AD at a common point E, as shown in the adjacent figure. 51.The length of CD (a) is 8 (a) is 12 (b) is 16 (c) cannot be determined ABC is an isosceles triangle right angled at B. A square is inscribed inside the triangle with three vertices of the square on three sides of the triangle as shown in the adjacent figure. It is known that the ratio x to y is equal to 2 to 1. 52.The ratio of the area of the square to the area of the triangle is equal to (a) 2: 5 (a) 1: 10 (b) 1: 3 (c) 2: 3 39 http://www.totalgadha.com  In triangle ABC, sides AB, AC, and BC are extended till Q, P and R such that AC = AP, BC = CR, and AB = BQ, as shown in the adjacent figure. It is known that the area of triangle ABC is 10 square centimetres. 53.What is the area of triangle PQR? (a) 40 cm2 (a) 70 cm2 (b) 80 cm2 (c) 90 cm2 54.A water lily with a rigid straight stem extends one meter above the surface of the water. When it bends at the bottom of its stem, it disappears under the water at a distance three meter from where the stem originally came out of the water. How deep is the lake? (a) 6 (a) 3 (b) 4 (c) 5 In the adjacent figure, ABCD is a square and ABE is an equilateral triangle. 55.Angle DEC is equal to (a) 15° (a) 30° (b) 45° (c) 20° In triangle ABC, D and E are any points on AB and AC such AD = AE. The bisector of angle C meets DE at F. It is known that angle B = 60º. 56.What is the degree measure of angle DFC? (a) 25° (a) 30° (b) 45° (c) 60° In the adjacent figure, triangle ABC is equilateral and D, E, and F AB are points on AB, BC, and AC such AD = BE = CF =. BF, CD, 3 and AE intersect to form triangle PQR inside ABC. 57.What is the ratio of the area of triangle PQR to that of triangle ABC? (a) 1: 9 (a) 1: 7 40 http://www.totalgadha.com  (b) 1: 8 (c) 1: 12 58.Four lines parallel to the base of a triangle divide each of the other sides into five equal segments and the area into five distinct parts. If the area of the largest of these parts is 27, then what is the area of the original triangle? (a) 135 (a) 75 (b) 225 (c) 175 In the diagram AB = 35, AE = CD = x, EC = 8, ED = 7. Also, angle DEC = angle ABC. 59.What is the value of x? (a) 1 (a) 2 (b) 3 (c) 4 In triangle PQR, points X, Y and Z are on PQ, PR and QR, RY a QZ respectively, such that PX = XQ, = , and = 3. Also YP b ZR (area ∆PXY)2 = (area ∆QXZ) × (area ∆RYZ) 60.The ratio a: b is 3 + 105 (a) 6 2 + 35 (a) 6 3 + 31 (b) 3 105 − 3 (c) 6 Triangle ABC is equilateral. D, the midpoint of BC, is the centre of the semicircle whose radius is R which touches AB and AC, as well as a smaller circle with radius r which also touches AB and AC. R 61.What is the value of ? r 41 http://www.totalgadha.com 62.In the figure, AB = AC = AD and ∠BCA = 15°. Find ∠BDA. (a) 30° (b) 40° (c) 60° (d) 70° (e) 75° 63.In the adjacent figure (not drawn to scale) ∠ PQR = 30°. Find the other two angels of ∆PQR if PQ and PR are the angle bisector of ∠APB and ∠APC, respectively. (a) 30° and 120° (b) 60° and 90° (c) 70° and 80° (d) 50° and 100° (e) 75° and 75° 64.In the following figure, PS bisects ∠ QPR. The area of ∆ PQS = 40 sq. cm. and PR is 2.5 times of PQ. Find the area of ∆ PQR. (a) 35 sq. cm. (b) 70 sq. cm. (c) 105 sq. cm. (d) 140 sq. cm. (e) 175 sq. cm. 65.In ∆ MNO, MP is a median. NQ bisects MP and meets MO in R. Find the length of MR if MO = 30 cm. (a) 5 cm. (b) 6 cm. (c) 10 cm. (d) 15 cm. (e) 20 cm. 66.S is the point on the side QR of a triangle PQR such that ∠ PSR = ∠ QPR. The length of the side PR is 8cm. Find the maximum possible length of QS if it is known that both QR and SR take integral values greater than one. (a) 16 cm. (b) 18 cm. (c) 30 cm. (d) 32 cm. (e) none of the above 42 http://www.totalgadha.com 67.In a right angled triangle with sides p, q, r (where p < q < r), 2p + 7r = 9q. If p =12cm, find the value of r (a) 25 cm (b) 25.5 cm (c) 26 cm (d) 26.5 cm (e) none of the above 68.In triangle ABC, AB = 10, AC = 7, and BC = 8. How do we need to slide it along AB so that the area of the overlapping region (the shaded triangle A’BD) is one-half the area of the triangle ABC? 69.In the figure, PS and QR are parallel lines. If PO: OR = 1:4 and the length of QO = 12cm, find out the length of SO. (a) 2 cm. (b) 3 cm. (c) 4 cm. (d) 4.5 cm. (e) 5.5 70.In the figure, PD is the median of ∆PQR. The bisectors of ∠PDQ and ∠PDR meet the sides PQ and PR at E and F, respectively. If EQ = 4EP then find the length of PR, given that PF is 2cm. (a) 8 cm (b) 9 cm (c) 10 cm (d) 11 cm (e) None of the above 71.D is the mid point of the side QR of a triangle PQR. O is a point on PD such that PO is 4 times OD. QO and RO produced meet PR and PQ in E and F, respectively. Find the length of the side PQ if FQ = 3cm. (a) 3 cm (b) 6 cm (c) 8 cm (d) 9 cm (e) 12 cm 72.In ∆ MNO, the bisector of ∠ NMO intersects NO at P. MN = 9cm, MO = 12cm. PO = PN + 1. Find the length of PO. (a) 2 cm (b) 3 cm (c) 4 cm 43 http://www.totalgadha.com  (d) 5 cm (e) 6 cm 73.In ∆ PQR, the line segment MN intersects PQ in M and PR in N such that MN is parallel to QR. Find the ratio of QM: QP if it is known that the area of the ∆ PMN is half of the ∆ PQR. 2 (a) 2 −1 2 −1 (b) 2 2 +1 (c) 2 2 +2 (d) 2 (e) None of the above 74.ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED = FD. Find the area of the shaded figure. 5 (a) 9 1 (b) 3 1 (c) 2 13 (d) 16 1 (e) 5 75.Find the sum of squares of the medians MP and OQ drawn from the two acute angled vertices of a right angled triangle MNO. The longest side of ∆ MNO is 20cm. (a) 200 sq. cm (b) 300 sq. cm (c) 400 sq. cm (d) 500 sq. cm (e) cannot be determined 76.Right triangle ABC, with AB = 48, and BC = 20, is kept on a horizontal plane. Another right-triangle ADC (right-angled at C) is kept on the triangle with DC = 39 cm. A vertical line is drawn through point D, intersecting AB at E. Then the length of BE is equal to (a) 15 (b) 21 (c) 12 (d) 18 44 http://www.totalgadha.com  (e) cannot be determined 77.A concave dodecagon (the cross shown to the right) is inside of, and shares four nonconsecutive sides with, a regular octagon. Each reflex angle of the dodecagon measures 270° , and every other interior angle of the dodecagon is a right angle. If the octagon has a perimeter of 16 cm, what is the area of the dodecagon in square centimeters? (a) 4 + 8 2 (b) 16 (c) 4 + 16 2 (d) 20 (e) None of these 78.Two concentric circles are drawn so that the tangent segment (shown) to the smaller circle is a chord of the larger circle. If the area of the annulus (region outside the smaller circle and inside the larger one) is 100 π , find the length of the chord shown ( AB ). 79.Two circles are externally tangent, and have a common external tangent line, with points of tangency P and Q. PQ = 10 cm , and the radii of the circles are R cm and R − 1 cm. The center of the smaller circle (the one which contains point Q) is S, and a line through S is drawn tangent to the larger circle. The point of tangency of this line is T, as shown. If ST= 2 19 cm, find the length of the larger radius R. 80.Two circles are externally tangent with a common external tangent. If the radii of the circles are 9 and 16, what is the distance (x) between points of tangency? 81.In the rectangle ABCD, the perpendicular bisector of AC divides the longer side AB in a ratio 2:1. Then the angle between AC and BD is (a) 30° (b) 45° (c) 60° (d) 90° 82.An isosceles right triangle is inscribed in a square. Its hypotenuse is a midsegment of the square. What is the ratio of the triangle’s area to the square’s area? 45 http://www.totalgadha.com  1 2 2 1 (a) (b) (c) (d) (e) None of these 4 5 4 2 83.When each side of a square is increased by 4 centimeters, the area is increased by 120 square centimeters. By how many centimeters should each side of the original square be decreased in order to decrease the area of the original square by 120 square centimeters? (a) 5 (b) 6 (c) 7 (d) 8 Three circles are drawn touching each other, their centers lying on a straight line. The line PT is 16 units long and is tangent to the two smaller circles, with points P and T lying on the larger circle. 84.The area inside the largest circle but outside the smaller two circles is equal to (a) 4π (b) 8π (c) 16π (d) 32π The sum of the number of sides of two regular polygons S1 and S2 is 12 and the sum of the number of diagonals of S1 and S2 is 19. 85.Then S1 and S2 are (a) square and octagon (b) heptagon and pentagon (c) both hexagons (d) triangle and nonagon 86.If an arc of 45o on circle A has the same length as an arc of 30o on circle B, then the ratio of the area of circle A to the area of circle B is (a) 2: 3 (b) 3: 2 (c) 4: 9 (d) 9: 4 Monty is playing with geometrical shapes made of paper. He cuts four equilateral triangles of side length 2 and joins them together to form a parallelogram ABCD, as shown in the figure. 87.What is the length of the diagonal AC? (a) √8 (b) √14 (c) 2√7 (d) 6.5 88.In a circle, chords AB and CD intersect perpendicularly at P. if AP = 20, PB = 36 and CP = 24, then the perimeter of the circle is 46 http://www.totalgadha.com  (a) 2π√119 (b) 2π√793 (c) 2π√65 (d) 2π√484 A square is inscribed inside another square, as shown in the figure. Each vertex of the inner square divides the side of the outer square in the ratio x: y. The area of the inside square is 4/5th of the area of the bigger square. 89.The value of x/y is equal to (a) 1 + 5 5 (b) 2 (c) 4 − 15 (d) 4 + 15 From petal arrangements of roses to shape of our galaxy, the number phi, or the ‘golden ratio’, is present in many natural phenomena, even in the structure of human body. To find the value of golden ratio, a square of side unity is drawn and midpoint of a side is joined to an opposite vertex, as shown in the figure. Then, an arc of radius r, meeting AB at E is drawn, and the rectangle BEFC is constructed. The value of a + b gives the golden ratio. 90.The value of the golden ratio is 1+ 3 (a) 2 3 −1 (b) 2 5 −1 (c) 2 5 +1 (d) 2 ABCD is an isosceles trapezoid with AB = 10 and CD = 6. The length of the altitude EF = 8. 91.Then the perimeter of ABCD is (a) 16 + 2√11 (b) 16 + 8√15 (c) 16 + 4√17 (d) 16 + 4√13 47 http://www.totalgadha.com  ABCD is a square with side length 10. A circle is drawn through A and D so that it is tangent to BC. 92.What is the radius of circle? (a) 5 (b) 6 (c) 6.25 (d) 6.75 93.On a circle with center O, ten points A1, A2, A3…A10 are equally spaced. The value of ∠ A1A5O is (a) 180° (b) 72° (c) 36° (d) 18° Use the information given below to answer the question that follows. Three squares of side lengths 3, 5, and 8 are kept side by side. A corner of the smallest square is joined to a corner of the biggest square, as shown in the figure. 94.What is the area of the shaded figure? (a) 10 (b) 12.5 (c) 13.75 (d) 15 Square ABCD is inscribed inside a circle. Another square is inscribed between square ABCD and the circle such that its two vertices are on the circle and one side lies along AB, as shown in the figure. 95.The ratio of the length of the sides of the smaller square and bigger square is (a) 1/5 (b) 2/7 (c) 3/8 (d) 4/9 48 http://www.totalgadha.com  ABCD is a square with side length 2 cm. It is divided into five rectangles of equal areas, as shown in the figure. 96.The perimeter of the rectangle BEFG is (a) 51/16 (b) 36/11 (c) 58/15 (d) 47/13 Three circles are in a row touching each other such that all three of them have two common tangents. The radii of the largest and the smallest circle are 9 and 4 respectively. Line segment AB passes through the centres of the circles and lies on the two outer circles. 97.What is the length of AB? (a) 38 (b) 26 + 3√32 (c) 26 + 2√38 (d) 19 In triangle ABC the length of the sides AB, BC, and AC are 12, 18 and 20 units, respectively. D is a point on AC such that AB = DB. 98.The value of the ratio AD: DC is (a) 3: 2 (b) 11: 9 (c) 7: 3 (d) 3: 1 A star is formed from five points A, B, C, D and E lying on a circle, as shown in the figure. 99.What is the sum of the angles at these five points? (a) 180° (b) 270° (c) 360° (d) 540° 49 http://www.totalgadha.com  A circle is inscribed inside an isosceles trapezoid with lengths of its parallel sides as 75 and 108 units, as shown in the figure. 100. The diameter of the inscribed circle is (a) 87.5 (b) 90 (c) 91.5 (d) 100 101. A square and an equilateral triangle have the same perimeter. What is the ratio of the area of the circle circumscribing the square to the area of the circle inscribed in the triangle? (a) 16: 9 (b) 18: 5 (c) 24: 7 (d) 27: 8 A cubic container of edge 16 cm is 5/8 full of liquid. It is tilted along an edge. The diagram shows the cross section of the container and the liquid in it. The ratio of length of line segment LC to length of line segment BK is 3: 2 exactly. 102. The length of line segment LC is (a) 6 (a) 9 (b) 12 (c) 15 103. Hexagon ABCDEF is inscribed in a circle. The sides AB, CD, and EF are each x units in length whereas the sides BC, DE, and FA are each y units in length. Then, the radius of the circle is (a) [(x2 + y2 + xy)/3]1/2 (b) [(x2 + y2 + xy)/2]1/2 (c) [(x2 + y2 – xy)/3]1/2 (d) [(x2 + y2 – xy)/2]1/2 Two circles touch each other externally and also touch a bigger circle of diameter 10 cm internally, as shown in the figure. A triangle is formed by joining the centers of the three circles. 104. The perimeter of the triangle, in cm, is (a) 5 (b) 10 (c) 15 (d) 20 50 http://www.totalgadha.com  Three circles, each of diameter 4 cm, are kept touching each other. The smallest square circumscribing these circles is drawn, as shown in the figure. 105. What is the length of the side of the square? (a)7.14 (b)7.58 (c)7.86 (d)7.92 In a right-angled isosceles triangle ABC, a circle and two separate squares are inscribed, as shown in the figure. I II III 106. The increasing order of the areas of the inscribed figures in the three cases is (a) I > III > II (b) II > I > III (c) I > II > III (d) II > III > I 107. In the above figure, arcs AB, BC and CD are equal. Then the value of ∠ECA is equal to 108. A jogging park has two identical circular tracks touching each other, and a rectangular track enclosing the two circles. The edges of the rectangles are tangential to the circles. Two friends, A and B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track. A jogs along the rectangular track, while B jogs along the two circular tracks in a figure of eight. Approximately, how much faster than A does B have to run (in percentage), so that they take the same time to return to their starting point? 109. The length of the side of the square is 2. Find the radius of the smaller circle. 51 http://www.totalgadha.com 110. A cone of volume V is cut into three pieces by planes parallel to the base. If the planes are at h 2h heights and above the base, the volume of the piece of the cone between the two planes is 3 3 111. In triangle ABC, D and E are points on AC and AB such DE // BC and length of DE is one-third of BC. If the area of triangle ABC is 216 square units, find the area of the shaded triangle. 112. What is the value of the shaded area, common to both the squares? 113. The length of the chord AB is 114. 9 squares are arranged as shown in the figure above. If the area of square A is 1cm2 and that of square B is 81 cm2, find the area of square I. 52 http://www.totalgadha.com 115. Four points A, B, C, and D lie on a straight line in the X-Y plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle. 116. In the figure, ABCD is a square with side length 17 cm. Triangles AGB, BFC, CED and DHA are congruent right triangles. If EC = 8, find the area of the shaded figure. 117. A cow is tied with a 50 m rope to a corner of a 20 m by 30 m rectangular field. The field is completely fenced and the cow can graze on the outside only. What area of the land can the cow graze? 118. Find the length of the side AB. 119. Rectangular tiles each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 110 cm by 130 cm, such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is 120. At the top of the 100th floor, how far is Carmen from the bottom of the staircase? 53 http://www.totalgadha.com  121. P is a point inside rectangle ABCD such that AP = 4 units, BP = 3 units and PD = 5 units. Find the length of PC. 122. All three sides of a triangle have integer side lengths of 11, 60 and n cm. For how many values of n is the triangle acute-angled? 123. What is the total height of the bottle? 124. In the figure, the square and the circle are intersecting each other such that AB = BC. If the 1 radius of the circle, with the centre as O, is 1 unit and OB = , then find the length of AB. 2 125. In the figure, CD is perpendicular to chord AB. Find the radius of the circle. 54 http://www.totalgadha.com 126. Find the length of DE. 127. In triangle ABC, AB = 5 cm, BC = 6 cm, and CA = 7 cm. There is a point P inside the triangle such that P is at a distance of 2 cm from AB and 3 cm from BC. How far is P from CA? 128. In a trapezium PQRS, PQ is parallel to RS and 2PQ = 3RS. UV is drawn parallel to PQ and cuts SP in U & RO in V such that SU: UP = 1: 2. Find the ratio UV : SR 129. In the given figure BCDE is a parallelogram and F is the midpoint of the side DE. Find the length of AG, If CG = 3cm. 130. P, Q, S, and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR? 131. In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE:EB = 1:2, and DF is perpendicular to MN such that NL:LM = 1:2. The length of DH in cm is 55 http://www.totalgadha.com 132. In the given figure, ABCD is a square of side 3cm. If BEMN is another square of side 5cm and BCE is a triangle right-angled at C, then the length of CN will be (a) 56cm (b) 57cm (c) 58cm (d) 59cm 133. In the given figure, if PQ || RS, ∠QPM = 30°, ∠SRT = 85° and RM = RP, then ∠RMN is equal to (a) 90° (b) 110° (c) 115° (d) 125° 134. If O is the centre of the given circle and BC = AO, then (a) 2x = y (b) x = 3y (c) 3x = y (d) x = 2y 135. A circle passes through the vertex C of a rectangle ABCD and touches its sides AB and AD at M and N respectively. If the distance from C to the line segment MN is equal to 5 units find the area of rectangle ABCD. 56 http://www.totalgadha.com  2 2 2 2 (a) 25 units (b) 30 units (c) 35 units (d) 40 units 136. In the given figure, ABCD is a square of side 1 unit. E and F are midpoints of AD and AB respectively. I is the point of intersection of BE and CF. Then the area of quadrilateral IEDC is 11 11 11 7 (a) unit 2 (b) unit 2 (c) unit 2 (d) unit 2 16 18 20 24 137. In the adjoining figure, AP is tangent to the circle at P, ABC is a secant and PD is the bisector of ∠BPC. Also, ∠BPD = 25o and ratio of angle ∠ABP and ∠APB is 5:3. Find ∠APB. (a) 75° (b) 125° (c)) 65° (d) None of these 138. In a triangle ABC, the internal bisector of the angle A meets BC at D. If AB=4, AC=3 and ∠A = 60° , then the length of AD is 12 3 15 3 6 3 (a) 2 3 (b) (c) (d) 7 8 7 139.The length of the common chord of two circles of radii 15 cm and 20 cm, whose centres are 25 cm apart, is (in cm) (a) 24 (b) 25 (c) 15 (d) 20 2 140. In the figure given above, ABCD is a rectangle. The area of the isosceles right triangle ABE= 7cm ; EC=3(BE). The area of ABCD in cm ( 2 ) is (a) 21 (b) 28 (c) 42 (d) 56 57 http://www.totalgadha.com 141. In the given figure, AB is a diameter of the circle and points C and D are on the circumference such that ∠CAD = 30° and ∠CBA = 70°. What is the measure of ∠ACD ? (a) 40° (b) 50° (c) 30° (d) 90° 142. The line AB is 6metres in length and tangent to the inner one of the two concentric circles at point C. It is known that the radii of the two circles are integers. The radius of the outer circle is (a) 5 m (b) 10 m (c) 6 m (d) 4 m 143. In triangle DEF shown below, points A, B, and C are taken on DE, DF and EF respectively such that EC=AC and CF = BC. IF angle D=40 degrees then what is angle ACB in degrees? (a) 140 (b) 70 (c) 100 (d) None of these 144. If a, b and c are the sides of a triangle, and a + b + c = bc + ca + ab , then the triangle is 2 2 2 (a) equilateral (b) isosceles (c) right-angle (d) obtuse-angled 145. Shown above are three circles, each of radius 20 and centres at P, Q and R; further AB=5, CD=10 and EF=12. What is the perimeter of the triangle PQR? 58 http://www.totalgadha.com  (a) 120 (b) 66 (c) 93 (d) 87 146. In the given figure, points A, B, C and D lie on the circle. AD=24 and BC=12. What is the ratio of the area of ∆CBE to that of ∆ADE? (a) 1:4 (b) 1:2 (c) 1:3 (d) Data insufficient 147. In the given figure, EADF is a rectangle and ABC is a triangle whose vertices lie on the sides of EADF. AE = 22, BE = 6, CF = 16 and BF = 2. Find the length of the line joining the midpoints of the sides AB and BC. (a) 4 2 (b) 5 (c) 3.5 (d) None of these 148. The figure shows a circle of diameter AB and radius 6.5 cm. If chord CA is 5cm long, find the area of ∆ABC. 149. AB ⊥ BC, BD ⊥ AC and CE bisects ∠C , ∠A = 30°. Then what is ∠CED ? o o o o (a) 30 (b) 60 (c) 45 (d) 65 59 http://www.totalgadha.com 150. In “AABC, ∠B is a right angle, AC=6cm, and D is the midpoint of AC. The length of BD is (a) 4cm (b) 6cm (c) 3cm (d) 3.5cm 151. In ∆ABC, a point D is on AC such that AB = AD and ∠ABC − ∠ACB = 40°. Then the value of ∠CBD is equal to 152. In ∆ABC, median AM is such that m ∠BAC is divided in the ratio 1:2, and AM is extended through to D so that ∠DBA is a right angle, then the ratio AC: AD is equal to (a) 1: 2 (b) 1: 3 (c) 1: 1 (d) 2: 3 (e) 3: 4 60 http://www.totalgadha.com 153. In square ABCD, E is the midpoint of AB. A line perpendicular to CE at E meets AD at F. What fraction of the area of square ABCD is the area of triangle CEF? 154. In ∆ABC, DE BC , FE DC , AF = 8, and FD = 12. Find DB. 155. The measure of the longer base of a trapezoid is 97. The measure of the line segment joining the midpoints of the diagonals is 3. Find the measure of the shorter base. 156. In ABCD, E is on BC. AE cuts diagonal BD at G and DC at F. If AG = 6 and GE = 4, find EF. 157. In ∆ABC, median AD is perpendicular to median BE. Find AB if BC = 6 and AC = 8. 61 http://www.totalgadha.com 158. On sides AB and DC of rectangle ABCD, points F and E are chosen so that AFCE is a rhombus. If AB = 16 and BC = 12, find EF. 159. The radius of a cylinder is increased by 16.67%. By what percent should the height of the cylinder be reduced to maintain the volume of the cylinder? (a) 10% (b) 14.28% (c) 16.67% (c) 20% (e) 25% 160. The ratio of the radius and height of a cylinder is 2:3. Further the ratio of the numerical of value of its curved surface area to its volume is 1:2. Find the total surface area of the cylinder. (a) 3π units (b) 6π units (c) 12π units (d) 24π units (e) 48π units 161. An ant starts from a point on the bottom edge of a circular cylinder and moves in a spiral manner along the curved surface area such that it reaches the top edge exactly as it completes two circles. Find 12 the distance covered by the ant if the radius of the cylinder is and height is 20 units? π (a) 2 61 units (b) 4 61 units (c) 24 units (d) 52 units (e) 60 units 162. A sphere is carved out of a cone with height 15cm and radius of base circle 12cm. What is the maximum volume of the cylinder? 163. A right circular cone of volume ‘P’, a right circular cylinder of volume ‘Q’ and sphere of Volume ‘R’ all have the same radius, and the common height of the cone and the cylinder is equal to the diameter of the sphere. Then : (a) P – Q + R = 0 (b) P + Q = R (c) 2P = Q + R (d) P2 – Q2 + R2 = 0 (e) 2P + 2Q = 3R 164. The volume of the solid generated by the revolution of an isosceles right angled triangle about its hypotenuse of length 3x is: 8πx3 (a) 3 (b) 8πx3 9 3 (c) πx 4 27πx3 (d) 3 62 http://www.totalgadha.com  32πx3 (e) 3 165. A sphere of 10.5cm radius is melted and cast into a cuboid of maximum volume. The total surface area of such cuboid is approximately : (a) 1720 (b) 1650 (c) 1810 (d) 1932 (e) 1680 166. One day sanjeev planned to make lemon tea and used a portion of the spherical lemon as shown in the figure. Find out the volume of the remaining lemon. Radius of the lemon is 6cm. 167. A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to make ‘n’ solid cones of height 1 cm and base radius 1 mm. Find the value of ‘n’. 168. Three cubes of volumes, 1 cm3, 216 cm3 and 512 cm3 are melted to form a new cube. What is the diagonal of the new cube? A cylinder with height and radius in a ratio of 2: 1 is full of soft drink