MCQ: Regular Right Pyramid with Triangular or Square Base - SSC CGL MCQ

Given:

The curved surface area of the right square base pyramid is 1920 cm 2.

The diagonal length of the square base is 12√2 cm.

Formula used:

Where, 

A, is the curved surface area of the right square base pyramid.
a, is the side length of the square
L, is the slant height of the square pyramid

(2) d = √2a

Where,
d, is the diagonal length of the square base
a, is the side length of the square base
(3) The slant height (L):

Where, 

r = a/2
a, is the side length of the square base 
h, is the height of the pyramid
L, is the slant height

Calculation:

According to the question, the required figure is:

The diagonal length of the square base = 12√2 cm

The side length of the square base, 
⇒ √2a = 12√2
⇒ a = 12 cm
The perimeter of the square base = 4a = 48 cm
The curved surface area of the right square base pyramid, A = 1920 cm²
Let L be the slant height of the pyramid.
According to the question,

∴ The (approximate) height of the pyramid is 80 cm.

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 = 7

Let point A be the apex of the pyramid. C and D are the opposite vertices of the square base.

Draw AB ⊥ CD.

Volume of pyramid = 1/3 x area of base x height

Let the side of the square = x cm.

CB = CD/2 (where B is the midpoint of CD)

∴  The length of the slant edges is 14 cm.

Given: 

A triangular pyramid (tetrahedron).

Concept used: 

A triangular pyramid has 4 vertices.

Calculation:

These vertices are the points where the four triangular faces of the pyramid meet.

Each vertex is connected to three edges and forms the corners of the triangular pyramid.

⇒ The number of vertices in a triangular pyramid is 4.

∴ The triangular pyramid has 4 vertices.

Number of faces = x = 6

Number of edges = y = 10

Number of vertices = z = 6

∴ The correct answer is 26.

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) x Perimeter of the base x Slant height

Calculations:

Slant height = EF 
Height = EO = 10 cm
OF = 10/2 = 5 cm

In ΔEOF,
EF² = OE² + OF²
⇒ 10² + 5² 
⇒ 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) x 40 x 5√5 
⇒ 100√5 cm
∴ The lateral surface are of the pyramid is 100√5 cm²

Given:

The base of right pyramid is an equilateral triangle, each side, (a) = 20 cm

Each slant edge, (E) = 30 cm

Concept used:

Circumradius of the equilateral triangle 

Calculation:

According to the question

Each side of an equilateral triangle = 20 cm

We know that, Circumradius of the equilateral triangle = a/√3

∴ The vertical height of the pyramid is cm.

Base area of the pyramid = Area of square = (10)² = 100 cm²

∵ Volume of pyramid = 1/3 x Base area x Height

⇒ 400 = 1/3 x 100 x Height

⇒ Height of pyramid (h)= 12 cm

Now,

⇒ Slant height of pyramid = √[(height)² + (side/2)²] = √(144 + 25) = √169 = 13 cm
⇒ Area of a triangular face = 1/2 x side x slant height = 1/2 x 10 x 13 = 65 cm²
∵ A square pyramid has four triangular faces & a square base,
∴ Total surface area of pyramid = 4(65) + 100 = 360 cm²

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

Now, Total surface area = Lateral surface area + Area of the base
⇒ 1/2 x Perimeter of base × slant height + Area of the base
⇒ 1/2 x (4 x 32) x 20 + (32)²
⇒ 1280 + 1024
⇒ 2304 sq. cm

Given: 

Perimeter of base is 56 cm and height of pyramid is 12 cm.

Concept: 

Volume of pyramid = 1/3 × area of base × height 

Calculation: 

Side of square = 56/4
⇒ 14 cm
Area of base = 14 × 14
⇒ 196 cm²
Now, Volume of pyramid = 1/3 x 196 x 12
⇒ 196 x 4
⇒ 784 cm³ 
∴ 784 cm³

Given:

The base of a triangular pyramid is an isosceles triangle of sides length are 5 cm, 5 cm and 6 cm
The height of pyramid is 10 cm

Calculation:

Let equal sides be a and unequal side be b

⇒ 12 cm²
So, the volume of pyramid = 1/3 x 12 x 10
∴ 40 cm³

Height of cylinder = 28 cm
Radius of cylinder = 10 cm
Volume of cylinder = πr2h
⇒ 22/7 x 28 x 10 x 10
⇒ 8800 cm³
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 x height)/3
Length of the base of pyramid = 20 cm
Area of base of pyramid = 20 x 20 = 400 cm²
⇒ 1/3 x H x 400 = 2200
⇒ H = 16.5 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
17² = 15² + (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