# How to Solve Mensuration Questions?

By SATISH KUMAR GUPTA|Updated : July 26th, 2019

## How to Solve Mensuration Questions?

We are providing you Important Short Tricks on Mensuration Questions which are usually asked in CSAT Exams. Use these below given short cuts to solve questions within minimum time. These shortcuts will be very helpful for your upcoming Aptitude Exam in 2019.

To make the chapter easy for you all, we are providing you all some Important Short Tricks to Mensuration  Questions which will surely make the chapter easy for you all.

Mensuration questions are an important part of the Quant section in the CSAT exam. Mensuration question asked in the CSAT exam related to Perimeter and Area.

To put simply, Area measures the area of shape i.e. the space that shape takes up. Perimeter is the measurement of the boundary of the figure.

1. Square:- A square four-sided polygon characterized by right angles and sides of equal length.

Area = side2,

Perimeter = 4side
2. Rectangle:- A four-sided flat shape with straight sides where all interior angles are right angles (90°). Also, the opposite sides are parallel and of equal length.

Perimeter = 2(L+B)

3. Circle;- Circle is the locus of points equidistant from a given point, the centre of the circle. The common distance from the centre of the circle to its points is called radius

Let’s look at some questions asked:-
Que 1. The length of the rectangular plot is 20 m more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per meter is Rs. 5300, what is the length of the plot?
Sol. Cost = Rate × Perimeter
⇒ Perimeter = 5300/26.50 = 200
⇒ 2 (L + B) = 200
⇒ 2 (L + L – 20) = 200
⇒ L – 10 = 50
⇒ L = 60 m

Que 2. If the length is to be increased by 10% and breadth of a rectangle plot to be decreased by 12%, then find % change in area?
Sol:- In such questions, use formula: Increment in area = [L%+B%+{(L%*B%)/100)}]
Increment in area = 10 – 12(because breadth is decreased) {+ 10+(-12) (-120/100)} = -3.2%

Que 3. The length of rectangular plot is increased by 60%. By what percentage should the width be decreased to maintain the same area?
Sol:- In such questions, use the formula:

Required % decrease in breadth = [%change in L{100/(100+%change in L)}]
Required % decrease in breadth = 60 (100/160) = 37.5%

Que 4. If the radius of the circle is increased by 5% find the percentage change in its area.
In such questions, use formula:

Sol:-
Change in area = (2x + x2/100) %
Change in area = 2×5 + 52/100 = 10 + ¼ = 10.25% increment.

Note: In such questions, the negative sign implies decrements while positive sign shows increment.

Que 5. The circumference of a circle is 100 cm. Find the side of the square inscribed in the circle.
Sol. Always use this formula, Side of a square inscribed in a circle of radius r = Root 2
Circumference = 100 ⇒ 2πr = 100
r = 50/π
Side of square = Root 2 (50/π)

Note: Similar formula:
1) Area of the largest triangle inscribed in a semi-circle of radius r =  r2
2) Area of the largest circle that can be drawn in a square of side x = π(x/2)2

Que 6. The length and breadth of the floor of the room are 20 by 10 feet respect. Square tiles of 2 feet are to be laid. Black tiles are laid in the first row on all sides, white tiles on 1/3rd of remaining sides and blue tiles on the rest. How many blue tiles are required?
Sol:- Side of a tile = 2 feet ⇒ Area of 1 tile = 22 = 4 sq ft. ----- (1)

Length left after lying black tile on 4 sides = 20 – 4

⇒ Area left after black tiles = (20 – 4)×(10-4) = 96 sq. ft.
Area left after white tiles = 2/3×96 =  64 sq. ft.
⇒ Area for blue tiles = 64 sq. ft.
Number of blue tiles = 64/4 = 16   (using 1)

Que 7. A cow is tethered in the middle of the field with a 14 ft long rope. If the cow grazes 100 sq. ft. per day, then the approximate time is taken to graze the whole field?
Sol: Here, the rope of cow is like radius

Area = π (14)2

No. of days = (Area of field)/Rate of cow = π (14)2/100 = 6 days(approx)

Que 8. A circle and a rectangle have the same perimeter. Sides of the rectangle are 18 by 26 cm. What is the area of the circle?
Sol:-  2πr = 2 (18 + 26) ⇒ r = 14 cm
Area = π r2 = 616 cm2

Que 9. What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as a base and a vertex on the opposite side of the rectangle?
Sol:-

Area of triangle = ½ × L × B

Area of rectangle = L × B
Area of a rectangle: area of a triangle = L × B: ½ × L × B = 2: 1

Que 10. In a rectangular plot, a cow is tied down at a corner with a rope of 14m long. Find the area that cow can graze?
Sol. The area that cow can graze can be illustrated as the shaded area:

Here, the shaded area is a quarter of a circle with radius 14m,

Area of grazed field = ¼ × π (14)2 = 154 m2

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