# SSC EXAMS QUESTIONS ON PYRAMID

#### QUESTIONS ON PYRAMID

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on PYRAMID and PRISM of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

4. Grade-II DASS Exam conducted by Delhi Staff Subordinate Services (DSSSB)

**SHIV KISHOR**

### PYRAMID AND PRISM

** CGL SYLLABUS**

- Regular Right Pyramid with Triangular or Square base
- Right Prism

NOTE: Some of the topics here are just for extra information, means beyond the CGL syllabus. So feel free to leave them.

#### 1. PYRAMID

** Pyramid is a polyhedron with a polygonal base and triangles at sides. There are three important parts in any pyramid namely:**

- base: The base of the pyramid may be of any shape.
- face: The faces of the pyramid are mostly isosceles triangle.
- apex: All the triangular faces meet at a single point called the apex.

**The Pyramid Formula in general is given as**

Surface Area = (base area) + (1/2)*(number of sides)*(slant height)*(base length)

Volume = (1/3)*(base area)*length

**Pyramids are of different types. They are named based on the the base shape of pyramid. The types of pyramid are:**

- square pyramid
- triangular pyramid
- pentagonal pyramid
- hexagonal pyramid

i) Square Pyramid

A Square Pyramid has a square base, 4 triangular faces and an apex.

**The Square Pyramid formulas are:**

- Base Area of a Square Pyramid = b²
- Surface Area of a Square Pyramid = 2bs + h²
- Volume of a Square Pyramid = [1/3]b²h

where

b – base length of the square pyramid

s – slant height of the square pyramid

h – height of the square pyramid

ii) Rectangular Pyramid

A Rectangular Pyramid has a rectangular base, 4 triangular faces and an apex.

**The Rectangular Pyramid formulas are:**

- Base Area of a Rectangular Pyramid = ab
- Surface Area of a Rectangular Pyramid = ab + b√[h² +(a/2)²] + a√[h² +(b/2)²]
- Volume of a rectangular Pyramid = [1/3]abh

where

b – base length of the rectangular pyramid

a – base breadth of the rectangular pyramid

h – height of the rectangular pyramid.

iii) Triangular Pyramid

A Triangular Pyramid has a triangular base, 3 triangular faces and an apex.

**The Triangular Pyramid formulas are:**

- Base Area of a Triangular Pyramid = [1/2]ab
- Surface Area of a Triangular Pyramid = [b/2][a+3s]
- Volume of a Triangular Pyramid = abh/6

where

a – apothem length of the triangular pyramid.

b – base length of the triangular pyramid.

s – slant height of the triangular pyramid.

h – height of the triangular pyramid.

iv) Pentagonal Pyramid

A Pentagonal Pyramid has a pentagonal base, 5 triangular faces and an apex.

**The Pentagonal Pyramid Formulas are:**

- Base Area of a Pentagonal Pyramid = 5ab/2
- Surface Area of a Pentagonal Pyramid = [5b/2][a+s]
- Volume of a Pentagonal Pyramid = 5abh/6

where

a – apothem length of the pentagonal pyramid.

b – base length of the pentagonal pyramid.

s – slant height of the pentagonal pyramid.

h – height of the pentagonal pyramid.

v) Hexagonal Pyramid

A Hexagonal Pyramid has a hexagonal base, 6 triangular faces and an apex.

**The Hexagonal Pyramid Formulas are:**

- Base Area of a Hexagonal Pyramid = 3ab
- Surface Area of a Hexagonal Pyramid = 3b(a+s)
- Volume of a Hexagonal Pyramid = abh

where

a – apothem length of the hexagonal pyramid.

b – base length of the hexagonal pyramid.

s – slant height of the hexagonal pyramid.

#### QUERY 1

**To increase the level of a piece of land 100m*80m, soil is used from the mound with rectangular base 10m*8m. Side view of the mound is trapezium with height 2m. 8m*5m is top side dimension of the mound. Find the level increased of peace of land using this soil from the mound.**

A) 0.0148 m

B) 2 m

C) 1 m

0.4 m

**MAHA GUPTA**

First it’s important to understand what type of figure it will be. Obviously it will a three-dimensional figure in the shape of truncated rectangular pyramid, in which its upper top too will be rectangular in the shape like this

Now, volume of the truncated rectangular pyramid = (h/3)*[(area of lower surface + area of the upper surface) + √(area of lower surface* area of the upper surface)]

So the area of the given pyramid = (2/3)*[(10*80) + (8*5) + (10*8)*(8*5)

= (2/3)*(80 + 40 + √3200)

= (2/3)*(80 + 40 + 57 approx)

= (2/3)*177 = 118 m³

Therefore level increased = volume of the soil of the mound/area of the piece of land

= 118/(100*80) = 118/8000 = 0.0148 meters (approx) [option ‘A’]

#### QUERY 2

**The diagram shows a square-based pyramid with base lengths 10 cm and height 12 cm. ****What is the surface area of the pyramid?**

A) 260 cm²

B) 340 cm²

C) 360 cm²

D) 400 cm²

**MAHA GUPTA**

Surface Area of a Pyramid = 1/2 × Perimeter of Base × Side Length + Base Area

So, perimeter of the base = 4 × 10 = 40 cm

and base area = 10 cm × 10 cm = 100 cm²

and the side length =** **

If V is the vertex of the pyramid, O is the center point of the base ABCD and M is the midpoint of AB, then ΔVOM is a right triangle with base 5 in and height 12 cm.

Therefore according to the Pythagoras

VM² = 5² + 12² = 25 + 144 = 169

=> VM = √169 = 13

Now substitute these values in the above given formula

Surface Area of a Pyramid = 1/2 × Perimeter × Side Length + Base Area

= 1/2 × 40 × 13 + 100

= 360 cm² (option ‘A’)

#### QUERY 3

**The above diagram shows a pyramid with vertex V and a rectangular base ABCD. M is the midpoint of AB, N is the midpoint of BC and O is the point at the center of the base. AB = 10 ft, BC = 18 ft, VO = 12 ft, VM = 15 ft, VN = 13 ft. What is the total surface area of the pyramid?**

A) 564 ft²

B) 584 ft²

C) 664 ft²

D) 720 ft²

**MAHA GUPTA
**The total surface area of the pyramid consists of one rectangle ABCD and four triangles i.e. ΔVAB, ΔVBC, ΔVCD and ΔVDA.

Now, Area of the base i.e. rectangle ABCD = 18 × 10 = 180

Area of ΔVAB = ½ × 10 × 15 = 75

Area of ΔVBC = ½ × 18 × 13 = 117

Area of ΔVCD = ½ × 10 × 15 = 75

Area of ΔVDA = ½ × 18 × 13 = 117

Hence, total surface area of the pyramid = 180 + 75 + 117 + 75 + 117 = 564 ft² (option ‘A’)

#### QUERY 4

**A pyramid is with a triangular base ABC, where AB = 5 cm, BC = 4 cm, CA = 3 cm and CD = 8 cm. The point D is vertically above the point C. What is the volume of the pyramid?**

A) 16 cm³

B) 20 cm³

C) 24 cm³

D) 32 cm³

**MAHA GUPTA**

As the sides of the ΔABC are 3, 4 and 5, it must be right angle triangle because 3, 4 and 5 makes a Pythagoras triplet, the length 5 cm being its hypotenuse.

So, area of the base of the pyramid = ½ × 3 × 4 = 6 cm²

Now volume of the pyramid therefore = ^{1}⁄_{3}× Base Area × Height = ^{1}⁄_{3} × 6 × 8 = 16 cm³ (option ‘A’)

#### QUERY 5

**A pyramid is with a triangular base ABC, where AB = 5 cm, BC = 4 cm and CA = 3 cm. The point D is vertically above the point C. If the area of ΔABD is 20.9 cm², what is the total surface area of the pyramid?**

A) 48.9 cm²

B) 54.9 cm²

C) 58.9 cm²

D) 74.9 cm²

**MAHA GUPTA
**

As the sides of the ΔABC are 3, 4 and 5, it must be right angle triangle because 3, 4 and 5 makes a Pythagoras triplet, the length 5 cm being its hypotenuse.

Also triangles BCD and ACD are right triangles.

Now we see that the surface area of the pyramid consists of four triangles, area of those triangles:

Area of ΔABC = ½ × 3 × 4 = 6

Area of ΔBCD = ½ × 4 × 8 = 16

Area of ΔACD = ½ × 3 × 8 = 12

Area of ΔABD = 20.9 (given)

Therefore, the total surface area of the pyramid= 6 + 16 + 12 + 20.9 = 54.9 cm² (option ‘B’)