PE 1 - Mensuration | Geometry - Mensuration

Answer: Option A

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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 cm²

Hence, option (a).

Answer: Option B

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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 × r² × 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).

Answer: Option C

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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).

Answer: Option D

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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 r₁ and h₁, then

8 × π × r₁²  × h₁ = π × r² × h

∴ 8 × (r/2)² × h₁ = r² × 8

∴ h₁ = 4

Hence, option (d).

Answer: Option C

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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.

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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).

Answer: Option A

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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.

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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 = 5 × 5 = 25 meters = 2500 cm.

∴ Time taken by the insect = $\frac{2500}{3}$ seconds.

Hence, option (a).

Answer: Option C

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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 cm².

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

Total surface area of a smaller cube = 6 × (1 × 1) = 6 cm².
Total surface area of 125 smaller cube = 125 × 6 = 750 cm².

Now, out of 750 cm², 150 cm² is painted, hence the unpainted surface area = 750 - 150 = 600 cm².

∴ 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).

Answer: Option C

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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).

Answer: Option B

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Explanation :

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

Now,

Volume of top cylinder = π × r² × h = 22/7 × 3² × 7 = 198 cm².

Volume of bottom cylinder = π × r² × h = 22/7 × 6² × 7 = 792 cm².

Now, for the frustum of the cone, BCED,

∆ADE ~ ∆ABC
∴ AX : AY = DE : BC = 3 : 6 = 1 : 2
⇒ AX = 7 and AY = 14

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Volume of frustum = Volume of bigger cone (ABC) - Volume of smaller cone (ADE)
                              = 1/3 × π × R² × h - 1/3 × π × r² × h
                              = 1/3 × 22/7 × 6² × 14 - 1/3 × 22/7 × 3² × 7
                              = 528 - 66
                              = 462

∴ Total volume of the bottle = 198 + 792 + 462 = 1452 cm².

Hence, option (b).

Answer: Option A

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Explanation :

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

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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.

∴ OQ² = PO² + PQ²
⇒ 12 = h² + 4
⇒ h² = 8
⇒ h = 2√2

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

Hence, option (a).