Question 2: 1. The parallel sides of an isosceles trapezium are 20cm and 10cm. Its non-parallel sides are both equal, each being 13cm, find the area of trapezium. 2. Two years ago,... Question 2: 1. The parallel sides of an isosceles trapezium are 20cm and 10cm. Its non-parallel sides are both equal, each being 13cm, find the area of trapezium. 2. Two years ago, Dilip was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their present ages 3. Person A has invested ₹2500 at the rate of 6% p.a for 2 years and 6 months. Person B has invested ₹2500 at the rate of 5% p.a for 3 years. Who got more simple interest and how much? Question 3: 1. Using compass and ruler only, construct a quadrilateral where, AB = 4cm, BC = 3.5cm, CD = 4cm, AD = 3cm and ∠BAD = 60° 2. Factorise: 16(x + y)²-81(x-2y)² Read more

Steps to Solve

1. Area of the trapezium

First, we need to find the height of the trapezium. We can do this by dropping perpendiculars from the shorter parallel side to the longer one. This creates two right-angled triangles and a rectangle. The length of the base of each triangle is $(20-10)/2 = 5$ cm. Using the Pythagorean theorem, we can find the height $h$ of the trapezium:

$h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm

Now we can calculate the area of the trapezium using the formula: $Area = \frac{1}{2} \times (sum \ of \ parallel \ sides) \times height$ $Area = \frac{1}{2} \times (20 + 10) \times 12 = \frac{1}{2} \times 30 \times 12 = 15 \times 12 = 180 \ cm^2$

2. Ages of Dilip and his son

Let Dilip's present age be $D$ and his son's present age be $S$. Two years ago, Dilip's age was $D-2$ and his son's age was $S-2$. According to the problem, $D-2 = 3(S-2)$. Two years hence, Dilip's age will be $D+2$ and his son's age will be $S+2$. According to the problem, $2(D+2) = 5(S+2)$. We now have a system of two equations:

$D-2 = 3(S-2)$ $2(D+2) = 5(S+2)$

Expanding these equations, we get: $D - 2 = 3S - 6 \implies D = 3S - 4$ $2D + 4 = 5S + 10 \implies 2D = 5S + 6$

Substituting the first equation into the second: $2(3S - 4) = 5S + 6$ $6S - 8 = 5S + 6$ $S = 14$

Now, substitute $S$ back into the first equation: $D = 3(14) - 4 = 42 - 4 = 38$ So, Dilip's present age is 38 and his son's present age is 14.

3. Simple interest calculation

For person A: Principal ($P$) = ₹2500 Rate ($R$) = 6% per annum Time ($T$) = 2 years and 6 months = 2.5 years Simple Interest ($SI_A$) = $\frac{P \times R \times T}{100} = \frac{2500 \times 6 \times 2.5}{100} = 25 \times 6 \times 2.5 = 150 \times 2.5 = ₹375$

For person B: Principal ($P$) = ₹2500 Rate ($R$) = 5% per annum Time ($T$) = 3 years Simple Interest ($SI_B$) = $\frac{P \times R \times T}{100} = \frac{2500 \times 5 \times 3}{100} = 25 \times 5 \times 3 = 125 \times 3 = ₹375$

Both got the same simple interest. Therefore, the difference is ₹0.

4. Geometric construction

This question requires a geometric construction using a compass and ruler. You would need to:

1. Draw line segment AB of length 4cm.

2. Construct an angle of 60 degrees at point A.

3. Mark point D on the 60-degree line such that AD = 3cm.

4. Using A as the center and a radius of 3 cm, draw an arc.

5. Using B as the center and a radius of 3.5 cm, draw another arc to intersect the previous arc at point C.

6. Join BC and CD.

7. ABCD is the required quadrilateral. Since I cannot perform the construction here, I will move to the next question and provide the solution for it below.

8. Factorization

We are given the expression $16(x + y)^2 - 81(x - 2y)^2$.

We can rewrite this as $[4(x+y)]^2 - [9(x-2y)]^2$.

Using the difference of squares formula $a^2 - b^2 = (a+b)(a-b)$, we have:

$[4(x+y) + 9(x-2y)][4(x+y) - 9(x-2y)]$ $[4x + 4y + 9x - 18y][4x + 4y - 9x + 18y]$ $[13x - 14y][-5x + 22y]$ $(-5x + 22y)(13x - 14y)$ or $(22y - 5x)(13x - 14y)$