# PE 1 - Mensuration | Geometry - Mensuration

**PE 1 - Mensuration | Geometry - Mensuration**

What is the area of the copper sheet required to prepare a cone of base radius 30 cm with the height 40 cm?

- A.
7543 cm

^{2} - B.
5146 cm

^{2} - C.
5432 cm

^{2} - D.
7246 cm

^{2}

Answer: Option A

**Explanation** :

The area of the copper sheet required to prepare a cone of base radius 30 cm and height 40 cm will be equal to the total surface area of the cone.

Total Surface area of cone = π × r(r + l)

Where l is the slant height of the cone and r is the radius of its base.

Also, *l* =$\sqrt{{\mathrm{r}}^{2}+{\mathrm{h}}^{2}}$

∴ Using r = 30 and h = 40, we get l = 50 cm

∴ Total Surface area of cone = 22/7 × 30 × (30 + 50) ≈ 7543 cm^{2}

Hence, option (a).

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**PE 1 - Mensuration | Geometry - Mensuration**

If the heights of two cones are in the ratio 7 : 3 and their diameters are in the ratio 6 : 7, what is the ratio of their volumes?

- A.
6 : 14

- B.
12 : 7

- C.
3 : 7

- D.
5 : 7

Answer: Option B

**Explanation** :

The height of the two cones is in the ratio 7: 3.

Also, the ratio of the radii will be the same as the ratio of the diameters. Hence, the ratio of the radii will be 6 : 7.

For a cone, the volume V is given by,

V = 1/3 × π × r^{2} × h

Hence, the ratio of the volume of the cones will be the product of the ratio of the squares of the radius and the ratio of the heights of the two cones.

i.e., $\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}$ = $\frac{{{\mathrm{r}}_{1}}^{2}}{{{\mathrm{r}}_{2}}^{2}}$ × $\frac{{\mathrm{h}}_{1}}{{\mathrm{h}}_{2}}$ = $\frac{36}{49}$ × $\frac{7}{3}$ = $\frac{12}{7}$

∴ The ratio is 12 : 7.

Hence, option (b).

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**PE 1 - Mensuration | Geometry - Mensuration**

The number of circular pipes with an inside diameter of 1 inch which will carry the same amount of water as a pipe with an inside diameter of 6 inches is:

- A.
6π

- B.
12

- C.
36

- D.
36π

Answer: Option C

**Explanation** :

Water carried by pipe ∝ cross sectional area.

∴ Water carried by pipe with 6 inches = k${\left(\frac{6}{2}\right)}^{2}$ = 9k and water carried by pipe with 1 inch diameter = k${\left(\frac{1}{2}\right)}^{2}$ = $\frac{\mathrm{k}}{4}$

∴Number of 1 inch pipes = $\frac{9\mathrm{k}}{\mathrm{k}/4}$ = 36

Hence, option (c).

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**PE 1 - Mensuration | Geometry - Mensuration**

A child consumed an ice cream of inverted right-circular conical shape from the top and left only 12.5% of the cone for her mother. If the height of the ice cream-cone was 8 cm, what was the height of the remaining ice cream-cone?

- A.
2.5 cm

- B.
3 cm

- C.
3.5 cm

- D.
4 cm

Answer: Option D

**Explanation** :

Cone left for the mother after the child consumes = 12.5% = 1/8

Now, the change in the volume of the coin is proportional to the cube of change in radius.

Thus the radius of the cone that is left = r(1/8)^{1/3} = r/2

If the radius of the base and height of the cone left be r_{1} and h_{1}, then

8 × π × r_{1}^{2} × h_{1} = π × r^{2 }× h

∴ 8 × (r/2)^{2} × h_{1} = r^{2} × 8

∴ h_{1} = 4

Hence, option (d).

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**PE 1 - Mensuration | Geometry - Mensuration**

A right circular hollow cylinder, kept vertically on its circular base has a height of 20 cm and radius of 10 cm. A sugar grain is kept inside this cylinder on its circular base at the periphery. If an ant is at the top rim of the same cylinder and diagonally opposite the sugar grain, the minimum distance the ant should travel to reach the sugar grain is approximately:

- A.
82.86 cm

- B.
51.43 cm

- C.
37.25 cm

- D.
65.96 cm

Answer: Option C

**Explanation** :

If we think of the cylinder as a folded sheet of paper, then on opening the cylinder the ant is at position A and the sugar grain is kept at C as shown in the following figure.

AB = height of the cylinder = 20 cm

BC = 0.5 × the circumference of the base of the cylinder = 10π cm

∴ AC is the shortest distance that the ant has to travel.

∴ AC = $\sqrt{{20}^{2}+{\left(10\mathrm{\pi}\right)}^{2}}$ ≈ 37.25 cm

Hence, option (c).

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**PE 1 - Mensuration | Geometry - Mensuration**

An insect is sitting at the bottom of tree that is 20 meters tall. The insect starts to crawl up the trunk of the tree in a regular spiral. It makes 5 rounds of the tree and reaches the top of the tree at a point directly above the point from which it started. If the speed of the insect is 3 cm/s, what is the time (in seconds) it takes to reach the top of the tree from the moment it starts walking up in a spiral? Assume that the trunk of the tree is a right circular cylinder of radius and it takes the insect 1 minute and 40 seconds to crawl around the base of the truck once.

- A.
2000/3

- B.
1000

- C.
1500

- D.
660

Answer: Option A

**Explanation** :

The insect takes 1 minute and 40 seconds = 100 second to crawl around the base of trunk at 3 cm/s.

∴ Circumference of the base = 100 × 3 = 300 cm = 3 meters.

Figure (1) shows the spiral when the insect crawls up the tree.

Now, if we open up the truck, it will make a rectangle (Figure (2)) of height 20 meters and circumference 3 meters.

⇒ From figure AB = $\sqrt{{4}^{2}+{3}^{2}}$ = 5 meters

⇒ From figure the total distance travelled by the insect = 4 × 5 = 20 meters = 2000 cm.

∴ Time taken by the insect = $\frac{2000}{3}$ seconds.

Hence, option (a).

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**PE 1 - Mensuration | Geometry - Mensuration**

A cube of dimension 5 cm × 5 cm × 5 cm is painted red on all six faces. Now this cube is cut to form 1 cm × 1 cm × 1 cm identical cubes. What is the ratio of total area of painted surfaces to the total area of unpainted surfaces?

- A.
2 : 7

- B.
1 : 3

- C.
1 : 4

- D.
1 : 5

- E.
None of these

Answer: Option C

**Explanation** :

A 5 cm × 5 cm × 5 cm cube is cut into 1 cm × 1 cm × 1 cm identical cubes.

Total surface area of the original cube = 6 × (5 × 5) = 150 cm^{2}.

Number of smaller cubes obtained = 5 × 5 × 5 = 125

Total surface area of a smaller cube = 6 × (1 × 1) = 6 cm^{2}.

Total surface area of 125 smaller cube = 125 × 6 = 750 cm^{2}.

Now, out of 750 cm^{2}, 150 cm^{2} is painted, hence the unpainted surface area = 750 - 150 = 600 cm^{2}.

∴ Ratio of painted surface to unpainted surface = 150/600 = 1 : 4

**Alternatively**,

To cut the given cube in to 125 (5 × 5 × 5) smaller cubes, we have to make 4 cuts in all the 3 dimensions, i.e. a total of 12 cuts.

Every cut will expose unpainted area equal to 2 faces of the original cube.

So in total area of unpainted surfaces = 24 faces of original cube

Area of painted surface = 6 surfaces of original cube

∴ Ratio of painted to unpainted surfaces = 6/24 = 1/4

Hence, option (c).

Workspace:

**PE 1 - Mensuration | Geometry - Mensuration**

25 deciliters of paint is required to paint a cardboard box on all sides. How much paint (in liters) will be required to paint all sides of another cardboard box whose sides are 20% more than the sides of the orginal carboard box.

- A.
36

- B.
25

- C.
3.6

- D.
2.5

Answer: Option C

**Explanation** :

Since the box is painted on all the sides, the amount of paint required will be directly proportional to the surface area of the box.

If sides increase by 20% (i.e., 1/5^{th}), the surface area will become, 6/5 × 6/5 = 36/25 times.

∴ The amount of paint required will also become 36/25 times.

∴ The amount of paint required = 36/25 × 25 = 36 deciliters = 3.6 liters.

Hence, option (c).

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**PE 1 - Mensuration | Geometry - Mensuration**

A bottle has the following cross-section. The radius of the bigger cylinder is 8 cm, and that of the smaller one is 4 cm. The height of the frustum of the cone is 7 cm, which is also equal to the height of each of the two cylinders. The volume of the bottle is? [π = 22/7]

- A.
1525

- B.
1452

- C.
1624

- D.
1820

Answer: Option B

**Explanation** :

Total volume of the bottle = Volume of top cylinder + Volume of frustum + Volume of bottom cylinder

Now,

Volume of top cylinder = π × r^{2} × h = 22/7 × 3^{2} × 7 = 198 cm^{2}.

Volume of bottom cylinder = π × r^{2} × h = 22/7 × 6^{2} × 7 = 792 cm^{2}.

Now, for the frustum of the cone, BCED,

∆ADE ~ ∆ABC

∴ AX : AY = DE : BC = 3 : 6 = 1 : 2

⇒ AX = 7 and AY = 14

Volume of frustum = Volume of bigger cone (ABC) - Volume of smaller cone (ADE)

= 1/3 × π × R^{2} × h - 1/3 × π × r^{2} × h

= 1/3 × 22/7 × 6^{2} × 14 - 1/3 × 22/7 × 3^{2} × 7

= 528 - 66

= 462

∴ Total volume of the bottle = 198 + 792 + 462 = 1452 cm^{2}.

Hence, option (b).

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**PE 1 - Mensuration | Geometry - Mensuration**

In a right pyramid with square base of side 4 cm. The vertical edge makes an angle of 60° with the side of the base. Find the volume of the pyramid.

- A.
32√2/3

- B.
32/3

- C.
32√2

- D.
None of these

Answer: Option A

**Explanation** :

The vertical edge makes an angle of 60° with the side of the base, Let us consider the vertical triangle OAB.

In triangle OAB, the two base angles are 60°, hence the vertical angle will also be 60°.

∴ OAB is an equilateral triangle,

⇒ The slant height of the pyramid = height of triangle OAB = √3/2 × 4 = 2√3

Now, in right triangle OPQ,

OP = height of the pyramid,

PQ = 4/2 = 2

OQ = 2√3.

∴ OQ^{2} = PO^{2} + PQ^{2}

⇒ 12 = h^{2} + 4

⇒ h^{2} = 8

⇒ h = 2√2

Now, Volume of pyramid = 1/3 × (base area) × (height) = 1/3 × 16 × 2√2 = 32√2/3.

Hence, option (a).

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