# Area Questions

FACTS  AND  FORMULAE  FOR  AREA  QUESTIONS

FUNDAMENTAL CONCEPTS :

I. Results on Triangles:

1. Sum of the angles of  a triangle is $\inline \fn_cm 180^{o}$

2. The sum of any two sides of a triangle is greater than the third side.

3. Pythagoras Theorem : In a right - angled triangle,

$\inline \fn_cm (Hypotenuse)^{2}=(Base)^2+(Height)^2$

4. The line joining the mid-point of a side of a triangle to the opposite vertex is called the median.

5. The point where the three medians of a triangle meet, is called Centroid. The centroid divides each of the medians in the ratio 2 : 1.

6. In an Isosceles triangle, the altitude from the vertex bisects the base.

7. The median of a triangle divides it into two triangles of the same area.

8. The area of the triangle formed by joining the mid-points of the sides of a given triangle is one-fourth of the area of the given triangle.

1. The diagonals of a parallelogram bisect each other

2. Each diagonal of a parallelogram divides it into two triangles of the same area.

3. The diagonals of a rectangle are equal and bisect each other.

4. The diagonals of a square are equal and bisect each other at right angles

5. The diagonals of a rhombus are unequal and bisect each other at right angles

6. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.

7. Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.

IMPORTANT FORMULAE

I.

1. Area of a rectangle = (length x Breadth)

$\inline \fn_cm \therefore Length =\left [ \frac{Area}{Breadth} \right ] and \; Breadth =\left [ \frac{Area}{Length} \right ]$

2. Perimeter of a rectangle = 2( length + Breadth)

II. Area of square = $\inline \fn_cm (side)^2=\frac{1}{2}(diagonal)^2$

III. Area of 4 walls of a room = 2(Length + Breadth) x Height

IV.

1. Area of a triangle =$\inline \fn_cm \frac{1}{2}\times Base\times Height$

2. Area of a triangle = $\inline \fn_cm \sqrt{s(s-a)(s-b)(s-c)}$, where a, b, c are the sides of the triangle and $\inline \fn_cm s=\frac{1}{2}(a+b+c)$

3. Area of an equilateral triangle =$\inline \fn_cm \frac{\sqrt{3}}{4}\times (side)^2$

4. Radius of incircle of an equilateral triangle of side $\inline \fn_cm a=\frac{a}{2\sqrt{3}}$

5. Radius of circumcircle of an equilateral triangle of side $\inline \fn_cm a=\frac{a}{\sqrt{3}}$

6. Radius of incircle of a triangle of area $\inline \fn_cm \Delta$ and semi-perimeter $\inline \fn_cm s=\frac{\Delta }{s}$

V.

1. Area of a parallelogram = (Base x Height)

2. Area of a rhombus = $\inline \fn_cm \frac{1}{2}\times (product\; of\; diagonals)$

3. Area of a trapezium = $\inline \fn_cm \frac{1}{2}\times (sum \; of\; parallel\; sides)\times distance\; between\; them$

VI.

1. Area of a cicle = $\inline \fn_cm \prod R^{2}$, where R is the radius.

2. Circumference of a circle = $\inline \fn_cm 2\prod R$.

3. Length of an arc = $\inline \fn_cm \frac{2\prod R\theta }{360}$, where $\inline \fn_cm \theta$ is the central angle.

4. Area of a sector = $\inline \fn_cm \frac{1}{2}(arc\times R)$$\inline \fn_cm \frac{\prod R^{2}\theta }{360}$

VII.

1. Area of a semi-circle = $\inline \fn_cm \frac{\prod R ^{2}}{2}$

2. Circumference of a semi - circle = $\inline \fn_cm \prod R$

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

As given that The side is increased by 25%, then

New side = $\inline \fn_jvn \small \frac{125a}{100}$ = $\inline \fn_jvn \small \frac{5a}{4}$

New area = $\inline \fn_jvn \small \left ( \frac{5a}{4} \right )^{2}$

increased area= $\inline \fn_jvn \small \frac{25a^{2}}{16}-a^{2}$

Increase %= $\inline \fn_jvn \small \frac{\left [ \frac{9a^{2}}{16} \right ]}{a^{2}}x100$ % = 56.25%

214 51102
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%

60 23549
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)

33 18054
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

Explanation:

let length = x and breadth = y then

2(x+y) = 46  $\inline \fn_cm \Rightarrow$  x+y = 23

x²+y² = 17² = 289

now (x+y)² = 23²

$\inline \fn_cm \Rightarrow$x²+y²+2xy= 529

289+ 2xy = 529

$\inline \fn_cm \Rightarrow$ xy = 120

area = xy = 120 sq.cm

60 17225
Q:

A parallelogram has sides 30m and 14m and one of its diagonals is 40m long. Then its area is

 A) 136 B) 236 C) 336 D) 436

Explanation:

let ABCD be the given parallelogram

area of parallelogram ABCD = 2 x (area of triangle ABC)

now a = 30m, b = 14m and c = 40m

$\inline \fn_jvn s=\frac{1}{2}\times \left ( 30+14+40 \right ) = 42$

Area of triangle ABC = $\inline&space;{\color{Black}\sqrt{s(s-a)(s-b)(s-c)}}$

= $\inline&space;{\color{Black}\sqrt{42(12)(28)(2)}}$= 168sq m

area of parallelogram ABCD = 2 x 168 = 336 sq m