# Area Questions

FACTS  AND  FORMULAE  FOR  AREA  QUESTIONS

FUNDAMENTAL CONCEPTS :

I. Results on Triangles:

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

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

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

${\left(Hypotenuse\right)}^{2}={\left(Base\right)}^{2}+{\left(Height\right)}^{2}$

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.

II.Results on Quadrilaterals :

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)

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

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

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

IV.

1. Area of a triangle =$\frac{1}{2}×base×height$

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

3. Area of an equilateral triangle =$\frac{\sqrt{3}}{4}×{\left(side\right)}^{2}$

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

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

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

V.

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

2. Area of a rhombus =

3. Area of a trapezium =

VI.

1. Area of a cicle = ${\mathrm{\pi R}}^{2}$, where R is the radius.

2. Circumference of a circle = $2\mathrm{\pi R}$.

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

4. Area of a sector = $\frac{1}{2}\left(arc×R\right)=\frac{{\mathrm{\pi R}}^{2}\mathrm{\theta }}{360}$

VII.

1. Area of a semi-circle = $\frac{{\mathrm{\pi R}}^{2}}{2}$

2. Circumference of a semi - circle = $\mathrm{\pi R}$

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 = ${a}^{2}$

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

New side = 125a/100 = 5a/4

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

Increased area= $\frac{25{a}^{2}}{16}-{a}^{2}$

Increase %=$\frac{\left[9{a}^{2}/16\right]}{{a}^{2}}*100$  % = 56.25%

292 63706
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:

Let original radius = R.

New radius = $\frac{50}{100}R$$\frac{50}{100}R$

Original area =$\frac{R}{2}$  and new area = ${\mathrm{\pi R}}^{2}$

$\frac{3{\mathrm{\pi R}}^{2}}{4}*\frac{1}{{\mathrm{\pi R}}^{2}}*100$

Decrease in area = $\mathrm{\pi }{\left(\frac{\mathrm{R}}{2}\right)}^{2}=\frac{{\mathrm{\pi R}}^{2}}{4}$ = 75%

79 30775
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         =>   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

82 25846
Q:

A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be

 A) 7744 B) 8844 C) 5544 D) 4444

Explanation:

length of wire = $2\mathrm{\pi r}$= 2 x (22/7 ) x 56 = 352 cm
side of the square = 352/4 = 88cm
area of the square = 88 x 88 = 7744sq cm

47 23048
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 = $\sqrt{2x}$ = (1.41x) m.

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

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

35 22640
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

s=1/2 x (30+14+40) = 42

Area of triangle ABC = $\sqrt{\left[s\left(s-a\right)\left(s-b\right)\left(s-c\right)\right]}$

= $\sqrt{\left[42\left(12\right)\left(28\right)\left(2\right)\right]}$= 168sq m

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

44 22620
Q:

The area of the largest circle that can be drawn inside a rectangle with sides 18cm by 14cm is

 A) 49 B) 154 C) 378 D) 1078

Explanation:

The diameter is equal to the shortest side of the rectangle.

So radius= 14/2 = 7cm.

Therefore,

29 20470
Q:

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

 A) 1 B) 2 C) 3 D) 4

Explanation:

Area of the park = (60 x 40) = 2400${m}^{2}$

Area of the lawn = 2109${m}^{2}$

Area of the crossroads = (2400 - 2109) = 291${m}^{2}$

Let the width of the road be x metres. Then,

$60x+40x-{X}^{2}=291$

${x}^{2}-100x+291=0$

(x - 97)(x - 3) = 0
x = 3.