# 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:

The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter

 A) Rs.16500 B) Rs.15500 C) Rs.17500 D) Rs.18500

Explanation:

l=5.5m w=3.75m
area of the floor = 5.5 x 3.75 = 20.625 sq m
cost of paving = 800 x 20.625 = Rs. 16500

Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

17 27119
Q:

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

 A) 20 B) 30 C) 40 D) 50

Explanation:

Let x and y be the length and breadth of the rectangle respectively.
Then, x - 4 = y + 3 or x - y = 7 ----(i)
Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)
(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)
Solving (i) and (ii), we get x = 16 and y = 9.
Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

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Job Role: Bank PO

21 26978
Q:

The area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field?

 A) 20 m B) 30 m C) 40 m D) 50 m

Explanation:

Let breadth=x meters.   Then, Length =$115x100$ meters

Given that, $x×115x100=460$

=> x = 20

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Exam Prep: Bank Exams
Job Role: Bank PO

34 26883
Q:

The inner circumference of a circular race track, 14 m wide, is 440 m. Find radius of the outer circle

 A) 44 B) 22 C) 33 D) 84

Explanation:

Let inner radius be r metres. Then, 2$π$r = 440 ; r = $440×744$= 70 m.

Radius of outer circle = (70 + 14) m = 84 m.

Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

35 26421
Q:

The diagonal of the floor of a rectangular closet is 7$12$ feet. The shorter side of the closet is 4$12$ feet. What is the area of the closet in square feet?

 A) 9 B) 18 C) 27 D) 36

Explanation:

Other side = $d2-s2$$2254-814$ = $1444$ = 6 ft

Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.

Filed Under: Area
Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

13 26353
Q:

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

 A) 2520 B) 2420 C) 2320 D) 2620

Explanation:

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) = 2520 sq.m

Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

29 26188
Q:

A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart. How many poles will be needed?

 A) 65m B) 45m C) 55m D) 56m

Explanation:

perimeter of the plot = 2(90+50) = 280m
no of poles =280/5 =56m

Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

19 25568
Q:

The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per metre.

 A) 2808 B) 3808 C) 4808 D) 5808

Explanation:

Area = (13.86 x 10000) sq.m = 138600 sq.m

$πR2=138600⇒R2=138600×722⇒R=210m$

Circumference = $2πR2=2×227×210=1320m$

Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.