# CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry

**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

A cuboid has its length, breadth and height as 5 cm, 4 cm and 3 cm respectively. Find its lateral surface area (in sq cm).

Answer: 54

**Explanation** :

Lateral surface area (LSA) of a cuboid = 2(l + b)h

∴ LSA = 2 × (5 + 4) × 3 = 54 cm^{2}.

Hence, 54.

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**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

A prism has a square base whose side is 5 cm. Its height is 8 cm. Find its total surface area (in sq cm).

- A.
240

- B.
312

- C.
210

- D.
440

Answer: Option C

**Explanation** :

Total surface area of a prism = (Base perimeter) × height + 2 × (Base area)

∴ TSA = (4 × 5) × 8 + 2 × 5^{2}.

⇒ TSA = 160 + 50 = 210 cm^{2}.

Hence, option (c).

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**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

What is the volume of a cylinder which has a lateral surface area 40π cm^{2} and height 5 cm?

- A.
40π cm

^{3} - B.
80π cm

^{3} - C.
120π cm

^{3} - D.
200π cm

^{3}

Answer: Option B

**Explanation** :

Lateral surface area of a cylinder = 2πrh, where r is the radius of the base cylinder and h is the height of the cylinder.

∴ 2 × π × r × 5 = 40π

∴ r = 4 cm

∴ Volume of the cylinder = π × r^{2} × h= π × 4^{2} × 5 = 80π cm^{3}

Hence, option (b).

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**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

A pyramid has a slant height of 5 cm and a square base of side 5 cm. Find its lateral surface area (in sq cm).

Answer: 50

**Explanation** :

Lateral Surface Area of a pyramid = ½ × (Base perimeter) × (Slat height)

∴ LSA = ½ × 20 × 5 = 50 cm^{2}.

Hence, 50.

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**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

Find the total surface area of a hemisphere with radius 14 cm.

- A.
294π

- B.
1176π

- C.
588π

- D.
None of these

Answer: Option C

**Explanation** :

For a hemisphere,

Total Surface Area (TSA) = 2πr^{2} + πr^{2} = 3πr^{2}

∴ Total surface area of a hemisphere with radius 14 cm = 3π × (14)^{2} = 588π cm^{2}.

Hence, option (c).

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**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

Eight solid metal spheres with same radii are melted and recast into a sphere without losing any metal. What is the percentage change in the total surface area of the solid?

- A.
100%

- B.
20%

- C.
25%

- D.
50%

Answer: Option D

**Explanation** :

Let the radii of the eight spheres and the new hemisphere be r and R respectively.

Volume of the eight original spheres = Volume of the new sphere

∴ 8 × (4/3) × πr^{3} = (4/3) × πR^{3}

∴ R = 2r

Original total surface area of the eight spheres = 8 × (4πr2) = 32πr^{2}

New total surface area of the new sphere = 4πR^{2} = 4π(2r)^{2} = 16πr_{1}^{2}

∴ Reduction in total surface area = 32πr^{2} – 16πr^{2} = 16πr^{2}

∴ Percentage reduction = (16πr^{2}/32πr^{2}) × 100 = 50%

Hence, option (d).

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**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

A cone is made of a sector with a radius of 21 cm and an angle of 600. What is total surface area of the cone?

- A.
269.5 cm

^{2} - B.
291.87 cm

^{2} - C.
256.5 cm

^{2} - D.
None of these

Answer: Option A

**Explanation** :

Let the radius of cone be R and radius of sector = r, then the slant height of cone ( l) = r

Circumference of the sector = Perimeter of base circle of the cone.

∴ 2πR = 2πr × 60/360

⇒ R = r/6 = 21/6 = 7/2 cm

Total surface area of cone = πr (l + r) = $\frac{22}{7}\times \frac{7}{2}\times \left(21+\frac{7}{2}\right)$ = 269.5 cm^{2}

Hence, option (a).

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**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

A right circular cone is cut parallel to its base at half its height. Find the ratio of the volume of the frustum formed and that of the original cone.

- A.
7 : 8

- B.
1 : 2

- C.
1 : 8

- D.
1 : 3

Answer: Option A

**Explanation** :

For frustum of a cone we know, $\frac{\mathrm{Volume}\mathrm{of}\mathrm{smaller}\mathrm{cone}}{\mathrm{Volume}\mathrm{of}\mathrm{bigger}\mathrm{cone}}$ = ${\left(\frac{r}{R}\right)}^{3}$

Since the original cone is cut at half its height from the base, hence ratio of height of smaller cone and original cone = 1 : 2

Ratio of heights of these cones will be same as ratio of their radii as well.

∴ (Volume of smaller cone)/(Volume of bigger cone) = (1/2)^{3} = 1/8

⇒ If the volume of bigger cone is 8V, volume of smaller cone will be V.

∴ Volume of the frustum = 8V – V = 7V.

⇒ $\frac{\mathrm{Volume}\mathrm{of}\mathrm{frustum}}{\mathrm{Volume}\mathrm{of}\mathrm{bigger}\mathrm{cone}}$ = $\frac{7V}{8V}$ = $\frac{7}{8}$.

Hence, option (a).

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**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 12 feet, and the sum of the lengths of the painted sides is 32 feet, find out the area of the parking space in square feet?

- A.
160 sq. ft.

- B.
120 sq. ft.

- C.
100 sq. ft.

- D.
144 sq. ft.

Answer: Option B

**Explanation** :

Length of the parking space (l) = 12 ft.

Given, painted length = l + 2b = 32

⇒ b = 10

Area = l × b = 12 × 10 = 120 sq. ft.

Hence, option (b).

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**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

A regular hexagonal prism has its perimeter of base as 240 cm and height 100 cm. Find weight of petrol (in kg) it can hold if density is 0.9 gm/cc.

- A.
800

- B.
687.2

- C.
486

- D.
486√3

Answer: Option D

**Explanation** :

Side of the base hexagon = 240/6 = 60 cm

Area of the regular hexagon = 6 × √3/4 × (60)^{2} = 5400√3

Volume = Base × Area = 5400√3 × 100 = 540000√3 cm^{3}

⇒ Weight = 540000√3 × 0.9 = 486000√3 gms = 486√3 kg.

Hence, option (d).

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**CRE 4 - Mensuration | Geometry, Mensuration & Trigonometry**

An iron sheet of 221 cm × 218 cm × 1 cm dimension is molten. A sphere and a cube are formed from the molten material. If the side of cube and radius of the sphere formed are equal, then what is the radius of the sphere (in cm)?

Answer: 21

**Explanation** :

Volume of iron sheet = 882 × 109 × 0.5 cm^{3}

Let r be the radius of sphere which is equal to side of cube

Volume of sphere = $\frac{4}{3}{\mathrm{\pi r}}^{3}$ = $\frac{4}{3}\times \frac{22}{7}\times {\mathrm{r}}^{3}$ = $\frac{88}{21}{r}^{3}$

Volume of cube = (side)^{3} = r^{3}

Total volume of iron after reformation = $\frac{88}{21}{r}^{3}$^{ }+ r^{3} = $\frac{109}{21}{r}^{3}$

Volume of iron before reformation and after reformation should be same

⇒ 221 × 218 × 1 = $\frac{109}{21}{r}^{3}$

⇒ r^{3} = 21 × 221 × 2

⇒ r^{3} = 21 × 441

⇒ r = 21

Hence, 21.

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