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Q:

In two triangles, the ratio of the areas is 4 : 3 and the ratio of their heights is 3 : 4. Find the ratio of their bases.

A) 13:9	B) 16:9
C) 15:9	D) 14:9

Answer & Explanation Answer: B) 16:9

Explanation:

Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively.
Then, $\frac{\frac{1}{2} \times x \times 3 h}{\frac{1}{2} \times y \times 4 h} = \frac{4}{3} \Rightarrow \frac{x}{y} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$

Required ratio = 16 : 9.

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Q:

Find the length of the altitude of an equilateral triangle of side $3 \sqrt{3}$ cm.

A) 4.5	B) 3.5
C) 2.5	D) 6.5

Answer & Explanation Answer: A) 4.5

Explanation:

Let the length of the altitude be h.
Then, h = $a \frac{\sqrt{3}}{2}$ = $3 \sqrt{3} \times \frac{\sqrt{3}}{2}$ = 4.5

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Q:

The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.

A) 24	B) 48
C) 60	D) 72

Answer & Explanation Answer: B) 48

Explanation:

Let ABC be the isosceles triangle and AD be the altitude

Let AB = AC = x. Then, BC = (32 - 2x).

Since, in an isosceles triangle, the altitude bisects the base,
so BD = DC = (16 - x).
In triangle ADC, ${(A C)}^{2} = {(A D)}^{2} + {(D C)}^{2}$

$\Rightarrow x^{2} = {(8)}^{2} + {(16 - x)}^{2} \Rightarrow x = 10$

BC = (32- 2x) = (32 - 20) cm = 12 cm.

Hence, required area = $\frac{1}{2} * B C * A D = \frac{1}{2} * 12 * 8 = 48 c m^{2}$

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Q:

Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13cm.

A) 30	B) 40
C) 50	D) 60

Answer & Explanation Answer: A) 30

Explanation:

Height of the triangle = $\sqrt{(13^{2} - 12^{2})} = \sqrt{25}$ = 5 cm.

Its area = $\frac{1}{2} * b a s e * h e i g h t$ = $\frac{1}{2} * 12 * 5$ = $30 c m^{2}$ .

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Q:

Find the area of a triangle whose sides measure 13 cm, 14 cm and 15 cm.

A) 84	B) 64
C) 44	D) 22

Answer & Explanation Answer: A) 84

Explanation:

Let a = 13, b = 14 and c = 15. Then, $s = \frac{1}{2} (a + b + c)$ =21

(s- a) = 8, (s - b) = 7 and (s - c) = 6.

Area = $\sqrt{[s (s - a) (s - b) (s - c)]}$ = $\sqrt{(21 \times 8 \times 7 \times 6)}$ = 84 sq.cm

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Q:

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

A) 20	B) 30
C) 40	D) 50

Answer & Explanation Answer: D) 50

Explanation:

Let x and y be the length and breadth of the rectangle respectively.
Then, x - 4 = y + 3 or x - y = 7 ----(i)
Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)
(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)
Solving (i) and (ii), we get x = 16 and y = 9.
Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

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Q:

The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

A) 20%	B) 30%
C) 40%	D) 50%

Answer & Explanation Answer: D) 50%

Explanation:

Let original length = x and original breadth = y.

Original area = xy.

New length = x/2 and New breadth=3y

New area = $\frac{3}{2} x y$

Therefore, Increase in area = New area-original area = $\frac{3}{2} x y - x y = \frac{1}{2} x y$

Therefore, Increase % = $\frac{i n c r e a s e i n a r e a}{o r i g i n a l a r e a} * 100 = (\frac{\frac{1}{2} x y}{x y}) * 100 = 50$ %

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Q:

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

A) 2520	B) 2420
C) 2320	D) 2620

Answer & Explanation Answer: A) 2520

Explanation:

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) = 2520 sq.m

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