Simple and Compound Pyramid MCQ Quizquestion for your practice provided with shortcuts and tricks to help get to the solution faster. These Pyramid Question Answers will help the candidates practice for several competitive, entrance and interviews such as Bank PO, IBPS PO, SBI PO, RRB PO, RBI Assistant, LIC,SSC, MBA - MAT, XAT, CAT, NMAT, UPSC, NET exams. Real time examples will help the candidates get concepts of the Pyramid Objective Questions easily and faster.

Option 1 : 1536

Length of the rectangle l = 16 cm

Breadth of the rectangle b = 12 cm

Diagonal of the rectangle = √(l^{2} + b^{2}) = √(12^{2} + 16^{2}) = 20

Length of lateral edge = 26 cm

Let the height of the pyramid be H.

As we know,

(Length of lateral edge)^{2} = (Height of the pyramid)^{2} + [(Length of diagonal of the base)/2]^{2}

26^{2} = H^{2} + (20/2)^{2}

⇒ H^{2} = 26^{2} - 10^{2}

⇒ H = √576 = 24

As we know,

Volume of the pyramid = (1/3) × Area of Base × height

Volume of the pyramid = (1/3) × 16 × 12 × 24 = 1536 cm^{3}

Option 1 : 1280 cu.cm

**Given:**

Height = 15 cm

Slant height = 17 cm

**Concept:**

For a pyramid with a square base of side a

(slant height)^{2} = (height)^{2} + (a/2)^{2}

**Formula used:**

The volume of pyramid = 1/3 × area of base × height

**Calculation:**

Let the side of the square be a cm

17^{2} = 15^{2} + (a/2)^{2}

⇒ a/2 = √289 - 225 = 8

⇒ a = 16 cm

∴ Volume = 1/3 × 16 × 16 × 15

⇒ V = 1280 cu.cm

**∴ The volume is 1280 cu.cm**

Option 2 : 16.5 cm

Height of cylinder = 28 cm

Radius of cylinder = 10 cm

Volume of cylinder = πr^{2}h

⇒ 22/7 × 28 × 10 × 10

⇒ 8800 cm^{3}

Now the cylinder is melted and cast into 4 pyramids of equal volume.

Volume of each pyramid = 8800/4 = 2200 cm3

Volume of pyramid = (area of base × height)/3

Length of the base of pyramid = 20 cm

Area of base of pyramid = 20 × 20 = 400 cm^{2}

⇒ 1/3 × H × 400 = 2200

⇒ H = 16.5 cm

Option 2 : 960 cm^{2}

Side of the square a = 24 cm

Height of the square h = 16 cm

As we know,

(Slant height)^{2} = (24/2)^{2} + 16^{2}

⇒ (Slant height)^{2} = 144 + 256 = 400

⇒ Slant height = √400 = 20 cm

As we know,

Lateral surface area of the pyramid = 1/2 × Perimeter of base × slant height

⇒ 1/2 × 4 × 24 × 20

⇒ 960 cm^{2}

Option 2 : 6√3 cm

**Given:**

Side of Base triangle = 8 cm

Volume of pyramid = 96 cm^{3}

**Concept:**

Using the formula of volume of the pyramid, calculate the height.

**Formula used:**

Volume of pyramid = (1/3) × (Area of base) × height

Area of equilateral triangle = (√3/4) × (side)^{2}

**Calculation:**

Let, the Height of the pyramid = ‘x’

Area of equilateral triangular base = (√3/4) × 8 × 8

= 16√3 cm^{2}

∴ Volume of pyramid = (1/3) × (Area of base) × height

⇒ 96 = (1/3) × 16√3 × x

⇒ x = (96 × 3)/16√3

⇒ x = 18/√3 cm

⇒ x = 6√3 cm

Option 3 : 1296 cubic m

Given,

⇒ Base of pyramid = 12m

⇒ Height of pyramid = 27m

⇒ Area of base = 12 × 12 = 144 sq m

Now,

⇒ Volume of pyramid = (Area of base × height)/3 = (144 × 27)/3 = 1296 cubic mOption 3 : 160 cm^{2}

Volume of pyramid = (1/3) × Base area × Height

400 = (1/3) × 10^{2} × h

Height = h = 12 cm

Slant height of pyramid = √[(10/2)^{2} + 12^{2}] = √[5^{2} + 12^{2}] = 13 cm

Lateral surface area of pyramid = (1/2) × Perimeter of Base × Lateral height

= (1/2) × (4 × 10) × 13 = 260 cm^{2}

Base area of pyramid = 10^{2} = 100 cm^{2}

Option 2 : 100√5

**Given:**

The base of the pyramid is a square

Side of square = 10 cm

Height of pyramid = 10 cm

**Concept used:**

Lateral surface = (1/2) × Perimeter of the base × Slant height

**Calculations:**

Slant height = EF

Height = EO = 10 cm

OF = 10/2 = 5 cm

In ΔEOF,

EF^{2} = OE2 + OF2

⇒ 10^{2} + 5^{2}

⇒ EF = √125 = 5√5

Slant height = EF = 5√5 cm

Perimeter of the base = 4 × 10 = 40 cm

Lateral surface = (1/2) × Perimeter of the base × Slant height

⇒ (1/2) × 40 × 5√5

⇒ 100√5 cm

**∴ The lateral surface are of the pyramid is 100√5 cm ^{2}**

Option 1 : 360 cm^{2}

**Given:**

Length of side of square base = 10 cm

Height of pyramid = 12 cm

**Concept:**

First, calculate the slant height of the pyramid. The surface area of the pyramid will be the area of each slant face and the area of the square base. Each slant face is a triangle. So, slant height means the height of the triangle.

**Formula used:**

Slant height = √[(Height of pyramid)^{2} + (Half of side of base)^{2}] ------(In case of square base)

Surface area of pyramid = [(1/2) × (Perimeter of base) × slant height] + Area of base

Area of square = side × side

**Calculation:**

∵ Slant height = √[(Height of pyramid)^{2} + (Half of side of base)^{2}]

⇒ Slant height = √[(12)^{2} + (5)^{2}]

⇒ Slant height = √(169)

⇒ Slant height = 13 cm

∴ Surface area of pyramid = [(1/2) × (Perimeter of base) × slant height] + Area of square base

= [(1/2) × (4 × 10) × 13] + (10 × 10)

= 260 + 100

= 360 cm^{2}

Option 2 : 96√7 cm^{3}

The perimeter of the base of a square pyramid is 48 cm

The length of slant edge is 10 cm

**Formula used:**

(Slant height)^{2 }= (slant edge)^{2 }– {(side of base)/2}^{2}

(Height)^{2 }= (slant height)^{2 }– {(side of base)/2}^{2}

**Calculation:**

Let the side of base of the pyramid be a cm

So, according to question

Perimeter of base = 48 cm

⇒ 4a = 48

So, a = 12 cm

Since, (Slant height)^{2 }= (slant edge)^{2 }– {(side of base)/2}^{2}

So, (Slant height)^{2 }= (10)^{2 }– {(12)/2}^{2}

⇒ (Slant height)^{2 }= 8^{2}

⇒ Slant height = 8 cm

Now, (Height)^{2 }= 8^{2} - 6^{2}

So, height = 2√7 cm

So, the volume of pyramid = 1/3 × (12)^{2 }× 2√7

Option 2 : 2304 sq. cm

**Given: **

Base of a right pyramid is a square and the length of the side of square is 32 cm

Height of the pyramid is 12 cm

**Formula used: **

Total surface area = Lateral surface area + Area of the base

Lateral surface area = 1/2 × Perimeter of base × slant height

**Calculation:**

In ΔOAB

The slant height, l

⇒ l^{2} = h^{2} + AB^{2}

⇒ l^{2} = (12)^{2} + (16)^{2}

⇒ l^{2} = 400

⇒ l = 20 cm

Now, Total surface area = Lateral surface area + Area of the base

⇒ 1/2 × Perimeter of base × slant height + Area of the base

⇒ 1/2 × (4 × 32) × 20 + (32)^{2}

⇒ 1280 + 1024

⇒ 2304 sq. cm

Option 4 : 3200 cm^{3}

Radius of incircle = side/2

⇒ 10 = side/2

⇒ Side = 20 cm

Using Pythagoras theorem,

⇒ (Height)^{2} = (slant height)^{2} – (side/2)^{2}

⇒ (Height)^{2} = (26)^{2} – (20/2)^{2}

⇒ Height = √(676 – 100) = √576 = 24 cm

Volume of the pyramid = (1/3) × Base area × height

⇒ Volume = (1/3) × (side)^{2} × height

⇒ Volume = (1/3) × (20)^{2} × 24 = 400 × 8

⇒ Volume = 3200 cm^{3}

∴ Required volume of pyramid = 3200 cm^{2}

Option 4 : \(\sqrt {\frac{{611}}{{12}}}\)

Area of equilateral triangle = 16√3 cm^{2}

(√3/4) × a^{2} = 16 √3

a^{2} = 16 × 4

⇒ a = √(16 × 4) = 4 × 2

⇒ a = 8 cm

In radius of the equilateral triangle (r) = side/2√3 = 8/2√3 = 4/√3

Area of one of its lateral faces = 30 cm^{2}

As we know, lateral face is in triangular face so,

(1/2) × Base (side of equilateral triangle) × h = 30

(1/2) × 8 × l = 30

l = 30/4 = 15/2 cm

As we know,

Slant height of the pyramid (l) = 15/2 cm

l^{2} = r^{2} + h^{2}

⇒ (15/2)^{2} = (4/√3)^{2} + h^{2}

⇒ h^{2} = (225/4) – (16/3)

⇒ h^{2} = (675 – 64)/12 = 611/12

Option 3 : 2352 cm^{3}

Given:

Area of square = side^{2}

Top area = 12^{2} = 144 sq.cm

Base area = 20^{2} = 400 sq.cm

Volume of frustum = (h/3) × (A_{1} + A_{2} + √A_{1}√A_{2})

Volume of frustum = (9/3) × (400 + 144 + 12 × 20)

= 3 × 784

Option 4 : 3a^{3}/2 cm^{3}

⇒ Area of base = 6 × (√3/4) × a^{2} = 3√3a^{2}/2 cm^{2}

⇒ Height of the pyramid = √(slant edge^{2} – side^{2}) = √(4a^{2} – a^{2}) = √3a

Option 1 : 7

Total surface area = Lateral surface area + Area of Base

260 = Lateral surface area + 120

Lateral surface area = 260 – 120 = 140 cm2

Area of each lateral face = 20 cm2

Number of lateral faces = 140/20 = 7Option 1 : 384

**Given:**

The base of a right pyramid is an equilateral triangle with side 8 cm, and the height of the pyramid is 24√3 cm.

**Formula used:**

Volume of pyramid = (1/3) × Area of base × height

**Calculation:**

As we know,

The volume of any pyramid = (1/3) × Area of base × height

Volume of the pyramid whose base is an equilateral triangle = (1/3) × (√3/4) × 8^{2} × 24√3

⇒ (1/3) × (√3/4) × 8 × 8 × 24√3

⇒ 384 cm^{3}

**∴ The volume (in cm3) of the pyramid is 384.**

Option 3 : 360 cm^{2}

**Given:**

A square pyramid with side 10 cm has a volume of 400 cm3

**Formula used:**

Volume of pyramid = 1/3 × Base area × Height** **

**Calculation:**

Base area of the pyramid = Area of square = (10)^{2} = 100 cm^{2}

Volume of pyramid = 1/3 × Base area × Height

⇒ 400 = 1/3 × 100 × Height

⇒ Height of pyramid = 12 cm

Now,

⇒ Slant height of pyramid = √[(height)^{2} + (side/2)^{2}] = √(144 + 25) = √169 = 13 cm

⇒ Area of a triangular face = 1/2 × side × slant height = 1/2 × 10 × 13 = 65 cm^{2}

∵ A square pyramid have four triangular faces & a square base,

Total surface area of pyramid = 4(65) + 100 = 360 cm^{2}

**∴ the correct answer is 360 cm ^{2}**.

Option 3 : 480 cm3

**Given:**

Side of equilateral triangle = 8 cm

Height = 30 \(√{3}\) cm

**Formula used:**

The volume of the pyramid (V) = (1/3) × Base Area × Height

Area of equilateral triangle = (√3/4) × a^{2} (where a is the side of the equilateral triangle)

**Calculations:**

⇒ Area of triangle = (√3/4) × 8 × 8

⇒ Area of equilateral triangle = 16√3 cm^{2}

⇒ The volume of pyramid = (1/3) × 16√3 × 30√3

⇒ V = 480 cm^{3}

**∴ The volume of the pyramid is 480 cm3**

Option 3 : 15.1875 cm

**Given:**

Side of the base of pyramid = 12 cm

Side of cube = 9 cm

**Formula used:**

Volume of cube = (side)^{3}

The volume of the pyramid = 1/3 × Area of base × height

**Calculation:**

The volume of square base pyramid = Volume of cube

⇒ Volume of cube = 9 × 9 × 9 = 729 cubic cm

Area of the base of pyramid = 12 × 12 = 144 sq. cm

Let the height of the pyramid is h cm.

Therefore,

Volume of the pyramid = 1/3 × Area of base × height

⇒ 729 = 1/3 × 144 × h

⇒ 48h = 729

⇒ h = 729/48

⇒ h = 243/16 cm

⇒ h = 15.1875 cm