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Q:

If a parallelogram with area p, a triangle with area R and a triangle with area T are all constructed on the same base and all have the same altitude, then

A) P=R	B) P=A
C) P=A/2	D) P=2R

Answer & Explanation Answer: A) P=R

Explanation:

let each have base = b and height = h
then p = b*h, R = b*h and T = (1/2) * b*h
so P = R, P = 2T and T = (1/2)*R are all correct statements

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Q:

The ratio of bases of two triangles is x:y and that of their areas is a:b. Then the ratio of their corresponding altitudes wll be

A) ay:bx	B) a:b
C) x:y	D) bx:ay

Answer & Explanation Answer: A) ay:bx

Explanation:

a : b =(1/2 * x * H) : (1/2 * y * h)

bxH = ayh => H/h =ay/bx

Hence H:h = ay:bx

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Q:

An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that of the square?

A) 1:2	B) 1:3
C) 2:3	D) 3^(1/2) :2

Answer & Explanation Answer: D) 3^(1/2) :2

Explanation:

area of a square = a² sq cm

length of the diagonal = $\sqrt{2 a}$ cm

area of equilateral triangle with side $\sqrt{2 a}$ = $\frac{\sqrt{3}}{4} * {(\sqrt{2 a})}^{2}$
required ratio = $\frac{\sqrt{3}}{4} * {(\sqrt{2 a})}^{2} : a^{2} = \sqrt{3} : 2$

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Q:

If the area of a square with side a s equal to the area of a triangle with base a, then the altitude of the triangle is

A) a	B) 2a
C) 5a	D) 3a

Answer & Explanation Answer: B) 2a

Explanation:

Area of a square with side a = a² sq unts

Area of a triangle with base a = (1/2) * a * h sq.unts

a² =1/2 *a *h

=> h = 2a

altitude of the triangle is 2a

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Q:

If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle.

A) 110	B) 120
C) 130	D) 140

Answer & Explanation Answer: B) 120

Explanation:

let length = x and breadth = y then

2(x+y) = 46 => x+y = 23

x²+y² = 17² = 289

now (x+y)² = 23²

=> x²+y²+2xy= 529

=> 289+ 2xy = 529

=> xy = 120

area = xy = 120 sq.cm

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Q:

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

A) 50	B) 60
C) 70	D) 80

Answer & Explanation Answer: B) 60

Explanation:

let ABC be the isosceles triangle, the AD be the altitude

Let AB = AC = x then BC= 32-2x [because parameter = 2 (side) + Base]

since in an isoceles triange the altitude bisects the base so

BD = DC = 16-x

In a triangle ADC, ${(A C)}^{2} = {(A D)}^{2} + {(D C)}^{2}$

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

BC = 32-2x = 32-20 = 12 cm

Hence, required area = $\frac{1}{2} * B C * A D$ = $\frac{1}{2} * 12 * 10$ = 60 sq cm

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Q:

In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error percent in the area, calculate from the those measurements.

A) .7%	B) 0.8%
C) 0.9%	D) 0.3%

Answer & Explanation Answer: B) 0.8%

Explanation:

let x and y be the sides of the rectangle then

correct area = $\frac{105}{100} x \times \frac{96}{100} y$

Error% = $\frac{504}{500} x y - x y = \frac{4}{500} x y$ %

$\frac{4}{500} x y \times \frac{1}{x y} \times 100 = \frac{4}{5} = 0.8$

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Q:

The length of a rectangle is twice its breadth if its length is decreased by 5cm and breadth is increased by 5cm, the area of the rectangle is increased by 75 sq cm. Find the length of the rectangle.

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

Answer & Explanation Answer: B) 40

Explanation:

let length = 2x and breadth = x then
(2x-5) (x+5) = (2x * x)+75
5x-25 = 75 => x=20
length of the rectangle = 40 cm

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