**Quantitative Questions on Mensuration for Campus Recruitment Tests**

**What is Campus Recruitment Process? **

Generally, a Campus Placement Process has four components:

* Pre Placement Talk

* Aptitude test

* Group Discussion

* Interview which can be various combinations of Technical and HR rounds

**Aptitude Test for Placements: **

The Aptitude test consists of basic problems in quantitative ability, logical reasoning, and verbal ability. The primary objective of this test is to check the problem-solving skills of the test-taker. Any job out there is the problem-solving skills. It is essential to understand what the problem is and how the problem should be solved. So, the application of the learning is the Aptitude test

Please read the below blogs to know the detailed campus recruitment process of respective companies

### Tips and Tricks to solve the Mensuration Questions for Campus Recruitment Test

- If you want to stand out & beat the competition in your Campus Placements, the first step is to ace the Written Aptitude Test. Quantitative Aptitude contributes a lot to increase your scores. No aptitude test is complete without questions based on Ratios & Averages
- Methods of learning in one chapter can be used in other chapters
- Whether the Question is simple or difficult you can know the options.
- Focus on all areas in quantitative will help you crack quantitative section

Watch the below video to learn Tips & Tricks to crack Questions on Mensuration in Campus Recruitment Exams

The Following questions are based on Mensuration:

**Equilateral Triangle -Perimeter to Area:**

**1. The perimeter of an equilateral triangle is 24√3 cm what is the height in(cm)?**

A. 6

B. 8

C.12

D.15

E.18

**Answer:** C

**Explanation:**

The perimeter of the equilateral triangle =3a

3a=24√3a

a=8√3

The height of the equilateral triangle=

=(√3/2)*(8√3)=24/2=12.

**Area of a triangle-Observation:**

**2. The area of a triangle whose sides are 9cm,40cm,41cm is __ sq.cm?**

A. 120

B. 150

C. 180

D. 210

E. 240

**Answer: **C

**Explanation:**

**Method-1:**

Δ=√s(s-a) (s-b) (s-c)

s=(a+b+c)/2=45 cm

Δ=√45*36*5*4

Δ=15*6*2=180sqcm.

**Method-2:**

These are the vertices of the right-angled triangle.

Area of the right-angled triangle=1/2*9*40=180.

### Linking r, Δ, and s:

**3. The perimeter of a triangle is 18cm. The radius of a circle that is inscribed in the triangle is 5 cm. what is the area of the triangle?**

A. 45cm

B. 54cm

C. 90cm

D. 180cm

E. 240cm

**Answer: **A

Explanation:

r= Inradius in a triangle

Δ= Area

s= Semiperimeter>

r=Δ/s

5=Δ/9

Δ=9*5=45cm

### Square: Finding area from Diagonal

**4. A boy walks 14 meters to cross a square field along its diagonal. What is the area of the field(approx)?**

A. 100sqm

B. 110sqm

D. cannot be determined

**Answer: **A

**Explanation:**

d=14

d=√2*a

a=14/√2=7√2

Area=a^{2}

a²=(7√2)²=49*2=98sqm

### Linking two figures: Square and circle

5. The diameter of a circle equals the perimeter of a square with the area is 3969cm². what is the circumference of the circle?

A. 396 cm

B. 792cm

C. 1188cm

D. 1584cm

Answer: B

Explanation:

The diameter of the circle=2r

The perimeter of a square= 4*a

2r=4*a

2r=4*√3969

2r=4*63

The circumference of the circle=2Πr

2ΠR=4*63*22/7

=36*22

=792cm

### Rectangle: Using the ratio of Length and Breadth

**6. The cost of leveling a rectangular piece of land came to Rs.756 when the contractor charged Rs.6 per square meter. The length and breadth of the field are in the ratio of 7:2. What is the perimeter of the rectangle?**

A. 36m

B. 45m

C. 63m

D. 54m

E. none of the above

Answer: D

Explanation:

Area of the rectangle(A)*6=756

A=756/6=126sqm

length:breadth=7x:2x

Area=7x*2x=126

x²=14*9/7*2

x=3

The Perimeter of a rectangle=2(7x+2x)

=18x

=18*3=54m

**7. The perimeter of a rectangle is 220m. Its length is 75% more than its breadth.what is the area(in sq.units)?**

A. 1400

B. 2800

C. 4200

D. 5600

**Answer: **B

**Explanation:**

Let us assume that breadth=b

The Length=1.75b

The perimeter of the rectangle=2(l+b)

2(l+b)=220m

2(b+1.75b)=220m

b(2.75)=110m

b=110/2.75=110*4/11=40m

Area=40*(1.75*40)

=40*70=2800sqm.

### Conversion: square to circle

**8. When a wire is bent in the form of a square, it encloses an area of 484sqcm. what will be the area of the figure formed when the same wire is bent to form a circle?**

A. 484sqm

B. 576sqm

C. 616sqm

D. 154sqm

E. 308sqm

**Answer: **C

**Explanation:**

When a wire is bent in the form of a square

The Length of the wire= perimeter of the square

= Circumference of the circle

a²=484

a=22cm

The perimeter of the square= Circumference of the circle

4*a=2Πr

4*22=2*22/7*r

r=14cm

The area of the circle=Πr²

=22/7*14*14=616sqm

### The Largest rectangle in a circle:

**9. The area of the largest rectangle that can be inscribed in a particular circle 784sqcm. what is the area of a circle?**

A. 1112sqcm

B. 1200sqcm

C. 1232sqcm

D. 1252sqcm

E. 1300sqcm

**Answer:** C

**Explanation:**

The Diameter of the circle= The diagonal of the rectangle

The rectangle is divided into two right-angled triangles.

If the rectangle has the maximum area then the triangle has the maximum area.

The largest rectangle is always square.

a²=784

a=28cm

d=28√2=diameter of the circle.

r=14√2

The area of the circle=Πr²=(22/7)*(14√2)²

=22/7*14*14*2

=22*28*2

=1232sqcm

### The area of compound figures:

**10. The shape of the field is such that it is bound by straight lines on three sides and a concave semi-circle on the fourth side. The three linear edges from three of the sides of a rectangle with sides measuring 30m*14m. The diameter of the semi-circular portion equals the shorter side of the rectangle.what is the area of the field (sq.mts)?**

A. 420

B. 497

C. 574

D. 723

**Answer: **B

**Explanation:**

Part1= Area of the rectangle =14*30=420sqm

Part2=Area of the semicircle=(1/2)*Πr²=(1/2)*22/7*7*7=77

=420+77=497sqm

### The Area of the quadrilateral-general form:

**11. The length of one of the diagonals of a quadrilateral is 18cm and the length of the offsets drawn to this diagonal from the other two vertices measure 5.4cm and 3.6cm. The area of the quadrilateral is___sqcm**

A. 54

B. 72

C. 81

D.105

E. 121

**Answer: **C

**Explanation:**

**
**The area of the quadrilateral=area of the two triangle

=(1/2)*18*5.4+1/2*18*3.6

=1/2*18(5.4+3.6)

= 9*9=81sqcm

### The quadrilateral with perpendicular diagonals:

**12. The two diagonals of a quadrilateral are perpendicular to each other measure 8cm and 12cm what its area(in sq cm)?**

A. 24

B. 36

C. 42

D. 48

E. 96

**Answer: **D

**Explanation:**

Area of the Quadrilateral=1/2*d*(h1+h2)

=1/2*AC*(8-y+y)

=1/2*12*8=48sqcm

where d=diagonal

h1&h2 both are perpendiculars

If the diagonals are perpendicular in Quadrilateral then

area=1/2*d1*d2

### Area of a Cyclic Quadrilateral:

**13. A quadrilateral whose sides measures 5cm, 15cm, 25cm and 30cm is inscribed in a circle. The area of the quadrilateral is ___ sq cm**

A. 20

B. 40

C. 80**√**5

D. 25**√**105

E. 35**√**75

**Answer:** D

**Explanation:**

The area cyclic Quadrilateral=A_{ca }

A_{ca}= √s (s-a) (s-b) (s-c) (s-d)

s=(a+b+c+d)/2

The cyclic Quadrilateral is a Quadrilateral whose vertices lies on the circumference of a circle.

s=(15+5+25+35)/2=40cm

A_{ca=} √35*15*25*5

=25**√**105

### The area of a parallelogram:

**14. Two sides of a parallelogram measure 15cm and 18cm while one of the diagonals measures 17cm. The area of the parallelogram is _____ sq.cm**

A. 60

B. 90

C. 120

D. 180

**Answer: **C

**Explanation:**

The diagonal (d1) should be maximum of a,b

d1>max(a,b)

The diagonal (d2) should be minimum of a,b

d2<max(a,b)

8,15,17 are the sides of a right-angled triangle

so, the area of the parallelogram= the area of the rectangle

then 8*15=120sqcm

### Using lengths of diagonal of a rhombus:

**15. what is the perimeter of a rhombus whose diagonal measure 48cm and 14cm?**

A. 50cm

B. 25cm

C. 100cm

D. 150cm

E. 200cm

**Answer:** C

**Explanation:**

In the rhombus, the diagonals bisect each other and perpendicular to each other.

The perimeter of a rhombus=4a

=4(25)=100cm

### The area of a regular hexagon:

**16. Find the perimeter of a regular hexagon whose area is 24√3 cm.**

A. 12cm

B. 24cm

C. 18cm

D. 30cm

E. 36cm

**Answer:** B

**Explanation:**

The area of the regular hexagon =24√3

The area of the regular hexagon =6Δ

6Δ=24√3

Δ=4√3

Area of the equilateral triangle=√3/4a²

√3/4a²=4√3

a²=4²

a=4

The perimeter of a regular hexagon=6a

6*4=24cm

### Area of a sector

**17. the area of the sector of a circle with radius 6cm and an arc of length 12cm is___ cm²**

A. 36

B. 48

C. 72

D. 96

E. 108

**Answer:** A

**Explanation:**

The circumference of a circle=2πr

The circumference of a total circle θ=360/π

2πr=2π*6=12π

The small sector and Arc length =12

The Length of the arc α Angle at the center α Area of the sector

The area of the sector=(θ/360)* πr²= (360/π)*(1/360)*πr²

The area of the sector=r²

where r=6

r²=36cm²

### Rolling:

**18. The wheel of a cycle has a radius of 15cm during a ride, it made 4000 revolutions along a straight road. How far did the cyclist go?**

A. 1200πm

B. 600πm

C. 300πm

D. cannot determine

**Answer:** A

**Explanation:**

Distance traveled in one revolution= Circumference=2Πr

Total distance=N*2Πr

=4000*2*Π*r

=1200*100*Π

=1200Πcm

### The path inside a rectangle:

**19. A path of width 1m is laid inside a rectangular park of dimensions 8m*20m such that it runs along the boundary. The path is laid with tiles measuring 10cm*10cm and costing Rs. 8 each what is the cost of tiles needed for the path.**

A. Rs 41,600

B. Rs 4,160

C. Rs 4,16,000

D. none of the above

**Answer: **A

**Explanation:**

The area of the path=Area of the outer rectangle-Area of the inner rectangle

=8*20-18*6=52m²

=52*100*100cm²

The cost =area *price per area

= 52*100*100*8/100

=41600Rs

### The path outside a circle

**20. A circular garden of radius 35cm is surrounded by a path of width 7m outside it. what is the area of the path(in m²)?**

A. 1038

B. 1296

C. 1352

D. 1441

E. 1694

**Answer: **E

**Explanation:**

The total radius of Outer circle= 42m

The area of the path= Area of the outer path - Area of the inner path

=π[42² -35² ]

=22/7(42+35)(42-35)

= (22/7)*77*7

=1694