Area Questions

Q:

If each side of a square is increased by 25%, find the percentage change in its area?

 A) 65.25 B) 56.25 C) 65 D) 56

Explanation:

let each side of the square be a , then area = a x a

New side = 125a / 100 = 5a / 4

New area =(5a x 5a) / (4 x 4) = (25a²/16)

increased area== (25a²/16) - a²

Increase %= [(9a²/16 ) x (1/a² ) x 100]% = 56.25%

107 20037
Q:

If the radius of a circle is decreased by 50%, find the percentage decrease in its area.

 A) 55% B) 65% C) 75% D) 85%

Explanation:

New radius = $\inline \fn_cm \frac{50}{100}R$$\inline \fn_cm \frac{R}{2}$

Original area =$\inline \fn_cm \tiny \prod R^{2}$  and new area = $\inline \fn_cm \tiny \prod(\frac{R}{2})^{2}=\frac{\prod R^{2}}{4}$

Decrease in area = $\inline \fn_cm \tiny \frac{3\prod R^{2}}{4}\times \frac{1}{\prod R^{2}}\times 100$ = 75%

29 10037
Q:

A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

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

Explanation:

Let the side of the square(ABCD) be x meters.

Then, AB + BC = 2x metres.

AC = $\inline \fn_jvn \sqrt{2}x$ = (1.41x) m.

Saving on 2x metres = (0.59x) m.

Saving % =$\inline \fn_jvn \frac{0.59x}{2x}\times 100$ = 30% (approx)

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Q:

The diagonal of a rectangle is $\inline \sqrt{41}$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

 A) 18 B) 28 C) 38 D) 48

Explanation:

$\inline \sqrt{l^{2}+b^{2}}=\sqrt{41}$ (or)   ${\color{Black}&space;l^{2}+b^{2}=41}$

Also, $\inline lb=20$

${\color{Black}&space;(l+b)^{2}=l^{2}+b^{2}+2lb}$

= 41 + 40 = 81

(l + b) = 9.

Perimeter = 2(l + b) = 18 cm.

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Q:

One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field?

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