Triangles PDF

## Triangles ### Introduction #### Lines and Angles: - Sum of all angles on a straight line is 180°. - Vertically opposite angles are congruent (equal). - If any point is equidistant from the end points of a segment, then it must lie on a perpendicular bisector. - When 2 parallel lines are interse...

## Triangles ### Introduction #### Lines and Angles: - Sum of all angles on a straight line is 180°. - Vertically opposite angles are congruent (equal). - If any point is equidistant from the end points of a segment, then it must lie on a perpendicular bisector. - When 2 parallel lines are intersected by a transversal, corresponding angles are equal, alternate angles are equal and co-interior angles supplementary. (All acute angles formed are equal to each other and all obtuse angles are equal to each other) #### Concept: The ratio of intercepts formed by a transversal intersecting 3 parallel lines is equal to the ratio of corresponding intercepts formed by any other transversal. - $ab=cd=ef$ ### Triangles: - Sum of interior angles of a triangle is 180° and sum of exterior angles is 360°. - Exterior Angle= Sum of remote interior angles. - Sum of two sides is always greater than the third side and the difference between the two sides is always lesser than the third side. - Side opposite to the biggest angle is longest and the side opposite to the smallest angle is shortest. ### Area of a triangle: - $Area = \frac{1}{2} \times base\times height$ - $Area = \frac{1}{2} \times Product\ of\ sides\times Sine\ of\ included\ angle$ - $Area = \sqrt{s(s-a)\times(s-b)\times(s-c)}$, where s $ (semi-perimeter) = \frac{a+b+c}{2}$ - $Area =r\times s$ [r is the radius of the incircle) - $Area = \frac{a\times b\times c}{4R}$ [R is radius of circumcircle] ### A Median of a triangle - A median of a triangle is a line segment joining a vertex to the midpoint of the opposing side. - The three medians intersect in a single point, called the Centroid of the triangle. - Centroid divides the median in the ratio of 2:1 ### An Altitude of a triangle - An altitude of a triangle is a straight line through a vertex and perpendicular to the opposite side or an extension of the opposite side. - The three altitudes intersect in a single point, called the Orthocenter of the triangle. ### A Perpendicular Bisector - A perpendicular bisector is a line that forms a right angle with one of the triangle's sides and intersects that side at its midpoint. - The three perpendicular bisectors intersect in a single point, called the Circumcentre of the triangle. - It is the center of the circumcircle which passes through all the vertices of the triangle. ### An Angle Bisector - An angle bisector is a line that divides the angle at one of the vertices in two equal parts. - The the angle bisectors intersect in a single point, called the Incentre of the triangle. - It is the center of the incircle which touches all sides of a triangle. ### Points to note: - Centroid and incentre will always lie inside the triangle. - For an acute angled triangle, the Circumcenter and the Orthocenter will lie inside the triangle. - For an obtuse angled triangle, the Circumcenter and the Orthocenter will lie outside the triangle. - For a right-angled triangle, the Circumcenter will lie at the mid-point of the hypotenuse and the orthocenter will lie at the vertex at which the angle is 90° - The orthocenter, centroid, and circumcenter always lie on the same line known as Euler line. - The orthocenter is twice is far from the centroid as the circumcenter is. - If the triangle is isosceles, then the incentre lies on the same line. - If the triangle is equilateral, all four are the same point. ### Theorems: #### Mid-Point Theorem: The line joining the midpoint of any two sides is parallel to the third side and is half the length of the third side. #### Basic proportionality theorem: If DE||BC. Then: - $\frac{AD}{DB} = \frac{AE}{EC}$ #### Apollonius' Theorem: $AB^2 + AC^2= 2 (AD^2 + BD^2)$ #### Interior Angle Bisector Theorem: - $\frac{AE}{ED} = \frac{BA}{BD}$ ### Special Triangles: #### 1) Right angled triangle: - $\triangle{ABC} \sim \triangle{ADB} \sim \triangle{BDC}$ - $BD^2= AD \times DC\ and\ AB \times BC=BD \times AC$ #### 2) Equilateral Triangle: - All angles are equal to 60°. All sides are also equal. - $Height = \frac{\sqrt{3}}{2} \times side$ - $Area = \frac{\sqrt{3}}{4} \times side^2$ - $Inradius = \frac{1}{3} \times Height$ - $Circumradius = \frac{2}{3} \times Height$ #### 3) Isosceles Triangle: - Angles opposite to equal sides are equal. - $Area = \frac{x}{4}\sqrt{(4a^2-c^2)}$ #### 4) 30°-60°-90° Triangle: - $Area = \frac{\sqrt{3}}{2} \times (x)^2$ #### 5) 45°-45°-90° Triangle - $Area = (x)^2/2$ #### 6) 30°-30°-120° Triangle: - $Area = \frac{\sqrt{3}}{4} \times (x)^2$ ### Similarity of triangles: - Two triangles are similar if their corresponding angles are congruent and corresponding sides are in proportion. #### Tests of similarity: - AA - SSS - SAS - For similar triangles, if the sides are in the ratio a:b - Corresponding heights are in the ratio a:b - Corresponding medians are in the ratio a:b - Circumradii are in the ratio a:b - Inradii are in the ratio a:b - Perimeters are in the ratio a:b - Areas are in the ratio a²:b² ### Congruency of Triangles: - Two triangles are congruent if their corresponding sides and angles are congruent. #### Tests of congruency: - SSS - SAS - AAS - ASA All ratios mentioned in similar triangles are now 1:1. ### Solved Examples **Q.1) Find the area of a triangle whose sides are 5 cms, 6 cms and 7 cms.** **Answer: (C)** - $Area\ of\ triangle = \sqrt{s(s – a)(s – b)(s - c)}$ - Where a, b and c are the sides of the triangle and $S= \frac{a+b+c}{2}$ is the semi perimeter of the triangle. - a = 5, b = 5, c = 7 - $S= \frac{5+6+7}{2} = 9\ cm$ - $Area\ of\ triangle = \sqrt{9(9-5) (9-6)(9 – 7)} = \sqrt{9(4)(3)(2)} = 6\sqrt{6}\ cm$ - Hence, (C). **Q.2) In APQR, QS and RS are angle bisectors. If ∠QPR= 80°, find the measure of QSR. (In degrees)** **Answer: (D)** - In APQR, ∠P + ∠Q + ∠R = 180. Therefore, ∠Q + ∠R = 100. - Thus, ∠SQR + ∠SRQ = 50. - Therefore, Z∠QSR = 130° - Hence, (D). **Q.3) All possible obtuse-angled triangles with sides of integer lengths are constructed, such that two of the sides have lengths 7 and 14. How many such triangles exist?** **Answer: (C)** - Let x be the length of the third side. So, we have x < 7 + 14 and x + 7 > 14. - Therefore, 7 < x < 21. - Now for the triangle to be obtuse, - Case 1: x² + 72 < 142. This gives possible values as 8, 9, 10, 11 and 12. - Case 2: 72 + 142 < x². This gives possible values as 16, 17, 18, 19 and 20. - Therefore, total 10 such triangles are possible. - Hence, (C). **Q.4) Find the area of triangle XYZ where XY = 17.29, XZ = 18 and YK = 12, where YK is the altitude.** **Answer: (B)** - $Area= \frac{1}{2}\times 18\times 12=108$ - Hence, (B). **Q.5) Find the area of triangle PQR where, PQ = 15, PR = 15, QR=18 and Median PK = 12.** **Answer: (B)** - In an isosceles triangle, median on a non-equal side= Altitude. - Therefore, Area= $\frac{1}{2} \times 18 \times 12= 108.$ - Hence, (B). **Q.6) In triangle QPR, QS is an altitude of the side PR, which meets side PR in S such that P - S - R. Suppose side PQ = 8, side QR = 12 and angle R = 30 degrees. Calculate lengths of QS, SR and PS respectively.** **Answer: (B)** - QS= 6 (side opposite to 30° is half of side opposite to 90° in a 30°-60°-90° triangle) - SR= 6√3 (side opposite to 60° is √3 times the side opposite to 30°) - PS=√PQ2-QS2=√82-62 = 2√7, - Hence [B]. **Q.7) Calculate area of parallelogram PQRS (Double Triangle), if SR = 30 cm, PS = 18 and angle Q = 60°.** **Answer: (A)** - $Area\ of\ parallelogram = a x b x sinθ$ - Where a & b are the sides & θ is the angle between these sides. - Area of parallelogram, PQRS= 30 x 18 x sin 60° = 30 x 18 x $\frac{√3}{2} = 270√3.$ - Hence, (A). **Q.8) ABC is a right-angled triangle with ∠B = 90°, AB = 21 units and BC = 20 units, then find the circumradius.** **Answer: (B)** - By Pythagoras theorem, CA² = AB² + BC² = 21² + 20² = 441 + 400 = 841 = 29 units. - Length of Hypotenuse (CA) = 29 units. - In a right-angled triangle, circumcentre is the midpoint of the hypotenuse. - Therefore, circumradius= (1/2) *29 = 14.5 units. - Hence, (B). **Q.9) In APQR, PQ = 8 cms, PR = 6 cms and QR = 10 cms. Find the length of the median PS** **Answer: (A)** - Since PS is the median, QS = SR = 5 cm - By Apollonius theorem, 8² + 6² = 2(PS² + 5²) - 50 = PS² + 25 - PS²= 25, PS = 5 cm - Hence, (A) **Q.10) AE is the median of ∆ABC. If the area of ∆ABC is 40 sq. cm, then find the area of triangle ∆ΑΒΕ.** **Answer: (B)** - AE is the median of ∆АВС. - Area of AABE = Area of AAEC. - (1/2) * Area of ∆ABC = Area of AABE = (1/2) * 40 sq cm = 20 sq cm. - Hence, (B). **Q.11) In △ABC, DE || BC and AB = 5 cms, BC = 10 cms and AD = 3 cms. Find (DE).** **Answer: (D)** - :DE || BC, ∠DAE = ∠BAC (Common Angle) - ZADE = ZABC (Corresponding angle) - $\frac{AD}{AB} = \frac{DE}{BC}$, $\frac{3}{5}=\frac{DE}{10}$ - DE = $\frac{3}{5} \times 10$= 6 cm. - Hence, (D). **Q.12) In an equilateral triangle the circumradius is 12cm. Find out the area of the triangle.** **Answer: (B)** - From the above figure, AD = (√3/2) * 12= 18 cm. - From the property of equilateral triangles, AD = a, where 'a' is the side of the triangle. - Hence, a = (√3/2) * 18 = 12√3cm. - $Area\ of\ an-equilateral\ triangle =\frac{\sqrt{3}}{4} \times a^2 =\frac{\sqrt{3}}{4} \times (12\sqrt{3})^2 =108 \sqrt{3}\ sq\ cm.$ - Hence, (B). **Q.13) A triangle of area 60 sq. cms has two of its sides as 8 cms and 15 cms. If the inradius of the triangle is 3 cms, then find the circumradius.** **Answer: (B)** - Since, it is a right angle triangle with the perpendicular sides 8 and 15. - Thus, hypotenuse = 17. - Circumradius of a right angle triangle is half of its hypotenuse. - Thus, circumradius = 8.5 cms. - Hence, (B). **Q.14) If the area of an equilateral triangle is 25√3 sq cm, then find the inradius of the triangle.** **Answer: (A)** - Area of an equilateral triangle = (√3/4) * a², where 'a' is the side of the triangle. - (√3/4) * a²= 25√3 - Side = 10 cm - $Inradius = \frac{Area\ of\ triangle}{3 \times side} = \frac{25\sqrt{3}}{3 \times 10} = \frac{5}{\sqrt{3}} cm.$ - Hence, (A). **Q.15) In the figure, AD is the external bisector of EAC, which intersects BC produced in D. If AB = 12 units, AC = 6 units and CD = 8 units, find BC.** **Answer: (C)** - By exterior angle bisector theorem, $\frac{AB}{AC} = \frac{BD}{DC}$ - Let BC = x - $\frac{12}{6} = \frac{x+8}{8}$ - x = 8 units - Hence, (C). **Q.16) The sides of ∆ABC are 8 cms, 10 cms and 12 cms with the smallest angle 'C'. Find the length of the altitude from the vertex C.** **Answer: (B)** - Let the height of ∆ABC be h = CD. - $Area\ of\ AABC = \frac{1}{2}\times Base\times Height = \frac{1}{2} \times 8 \times h = 4h$ - $Area = \sqrt{s(s-a)(s-b)(s-c)}$. - $S= \frac{a+b+c}{2}$ is the semi perimeter of the triangle. - $S= \frac{8+10+12}{2}= 15\ cm.$ - $Area = \sqrt{15(15-8)(15 – 10)(15 – 12)} = 15\sqrt7\ sq\ cm$ - From i) and ii), 4h = $15\sqrt7$, h= $\frac{15√7}{4} cm$ - Hence, (B). **Q.17) A regular pentagon ABCDE is fitted inside a regular hexagon APQRST such that P-B-Q, Q-C-R, R-D-S and S-E-T. Find ∠SED** **Answer: (A)** - ADRC will be isosceles triangle. - ∠RDC = 180-120/2 = 30°. - m ∠SDE = 180-108-30=42°. - m ∠SED = 180-120-42= 18°. - Hence, (A). **Q.18) ΔABC is right-angled at B with AB = 6 cm and BC = 8 cm. A square PQRS is inscribed in ΔABC such that the point P and Q lie on AC, point S on AB and point R on BC. Find the approximate length of the sides of the square PQRS.** **Answer: (B)** - AB = 6 cm and BC = 8 cm, AC = 10 cm (Pythagorean triplet) - Since, SP is perpendicular to AP. - ΔAPS~ΔΑΒC (by AA test) - AP/AB = PS/BC => AP/6 = PS/8 => AP = (3/4) PS - Now, RQ is perpendicular to QC - Therefore, ACQR~ACBA (by AA test) - CQ/CB = RQ/AB => CQ = (4/3) RQ - Let the side of square PQRS be x cm. - Now, AC = AP + PQ + QC - 10 = (3/4)x + x + (4/3)x - 10 = 37/12 x - x = 3.2 cm - Hence, (B). **Q.19) Two chords of length 12 cm and 16 cm, of a circle of radius 10 cm, are perpendicular to each other. Find the distance (in cm) between the midpoints of these two chords.** **Answer: (D)** - We know that, a = 15, b =15, c = 24, R = 12.5 - $Area\ of\ triangle = \frac{a\times b\times c}{4R} = \frac{15\times 15\times 24}{4\times 12.5}= 108$ - Hence, 108. **Q.20) What is the ratio of the sides AB to AC to CB in an isosceles right-angled triangle ABC, right-angled at A?** **Answer: (A)** - In an isosceles right-angled triangle ABC, - ∠A = 90° and ∠B = ∠C = 45° - Therefore, AB = AC - Let AC = AC = 1 cm - BC² = AB² + AC² = 1 + 1 = 2. - BC = √2 - ∴AB:AC:CB = 1 : 1: √2 - Hence, (A).