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

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

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Regular Right Pyramid with Triangular or Square Base

MCQ: Regular Right Pyramid with Triangular or Square Base for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Regular Right Pyramid with Triangular or Square Base questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Regular Right Pyramid with Triangular or Square Base MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Regular Right Pyramid with Triangular or Square Base below.
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MCQ: Regular Right Pyramid with Triangular or Square Base - Question 1

### The curved surface area of the right square base pyramid is 1920 cm2. If the diagonal length of the square base is 12√2 cm, then find the (approximate) height of the pyramid. (in cm)

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 1

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.

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 2

### The whole surface area of a pyramid whose base is a regular polygon is 260 cm2 and the area of its base is 120 cm2. The area of each lateral face is 20 cm2. Then the number of lateral faces is:

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 2

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

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MCQ: Regular Right Pyramid with Triangular or Square Base - Question 3

### The volume of a pyramid with a square base is 200 cm3. The height of the pyramid is 13 cm. What will be the length of the slant edges (i.e., the distance between the apex and any other vertex), rounded to the nearest integer?

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 3

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.

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 4

Number of vertices of a triangular pyramid are -

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 4

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.

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 5

If x, y, and z are the number of faces, edges, and vertices, respectively, of a pentagonal pyramid, then what is the value of  ?

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 5

Number of faces = x = 6

Number of edges = y = 10

Number of vertices = z = 6

∴ The correct answer is 26.

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 6

The base of a pyramid is a rectangle whose length and breadth are 16 cm and 12 cm, respectively. If the length of all the lateral edges passing through the vertex of the right rectangular pyramid is 26 cm, then find the volume of the pyramid in cubic centimeter.

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 7

The total surface area of a right pyramid, with base as a square of side 8 cm, is 208 cm². What is the slant height (in cm) of the pyramid?

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 8

The base of a pyramid is a square of side 10 cm. If its height is 10 cm, then the area (in cm2) of its lateral surface is:

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 8

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

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 9

The base of right pyramid is an equilateral triangle, each side of which is 20 cm. Each slant edge is 30 cm. The vertical height (in cm) of the pyramid is:

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 9

Given:

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

Each slant edge, (E) = 30 cm

Concept used:

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.

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 10

A square pyramid with side 10 cm has a volume of 400 cm3. Find the total surface area of the pyramid.

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 10

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

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 11

The base of a right pyramid is a square and the length of the side of square is 32 cm and height of the pyramid is 12 cm, then what is the total surface area of the square pyramid?

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 11

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

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 12

Find the volume of square pyramid, If perimeter of base is 56 cm, and height of pyramid is 12 cm.

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 12

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

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 13

The base of a triangular pyramid is an isosceles triangle of sides length are 5 cm, 5 cm and 6 cm. If the height of pyramid is 10 cm then find the volume of the pyramid.

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 13

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

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 14

A cylinder of height 28 cm and radius 10 cm is melted and formed into 4 square base pyramids each has the same height H and the base length of each pyramid is equal. Find the height of the pyramid, if the length of its base is 20 cm.

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 14

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

MCQ: Regular Right Pyramid with Triangular or Square Base - Question 15

A pyramid has a square base and its slant height is 17 cm and its vertical height is 15 cm. What is the volume of the pyramid?

Detailed Solution for MCQ: Regular Right Pyramid with Triangular or Square Base - Question 15

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

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## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests