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.

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 the length of the diagonal of a square is 20cm,then its perimeter must be

a. $402$cm     b. $302$cm    c. 10cm    d. 15$2$cm

 A) a B) b C) c D) d

Explanation:

We know that d=√2s

Given diagonal = 20 cm

=> s = 20/$2$ cm

Therefore, perimeter of the square is 4s = 4 x 20/$2$ = 40$2$   cm.

109 195809
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 = $a2$

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

New side = 125a/100 = 5a/4

New area = $5a42$

Increased area= $25a216-a2$

Increase %=$9a2/16a2*100$  % = 56.25%

989 147448
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$m2$

Area of the lawn = 2109$m2$

Area of the crossroads = (2400 - 2109) = 291$m2$

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

$60x+40x-X2=291$

$x2-100x+291=0$

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

103 109322
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

299 86129
Q:

A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq m is

 A) Rs. 258 B) Rs. 358 C) Rs. 458 D) Rs. 558

Explanation:

Area to be plastered = $2l+b×h+l×b$

=$225+12×6+25×12=744sq.m$

Cost of plastering = $744×75100=Rs.558$

106 74153
Q:

The percentage increase in the area of a rectangle, if each of its sides is increased by 20%

 A) 22% B) 33% C) 44% D) 55%

Explanation:

Let original length = x metres and original breadth = y metres.

Original area = xy sq.m

Increased length  = $120100$ and Increased breadth = $120100$

New area =

The difference between the Original area  and New area  is:

$3625xy-xy$

$1125xy$ Increase % =$1125xyxy*100$= 44%

119 73590
Q:

The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

 A) 16 B) 18 C) 20 D) 22

Explanation:

$2l+bb=51$      => 2l + 2b = 5b     => 3b = 2l

b=(2/3)l

Then, Area = 216 cm2

=> l x b = 216     => l x (2/3)l =216

l = 18 cm.

58 65433
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 = $2x$ = (1.41x) m.

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

Saving % =$0.59x2x*100$ = 30% (approx)