**Mensuration Exercise 15.3 – Questions**

**1.Flower vase is in the form of a frustum of a cone. The perimeter of the ends are 44 cm and 8.4π cm. If the depth is 14 cm, fond how much water it can hold?**

**A bucket is in the shape of a frustum with the top and bottom circles of radii 15 cm and 10 cm. Its depth is 12 cm. Find its curved surface area and total surface area.(express your answer in terms of π)****From the top of a cone of base radius 24 cm and height 45 cm, a cone of slant height 17 cm is cut off What is the volume of the remaining frustum of the cone?****A vessel is in the form of a frustum of a cone. Its radius at one end is 8 cm and the height is 14 cm. If its volume is**^{5676}/_{3}cm^{3}, find the radius of the other hand.

**Mensuration Exercise 15.3 – Solutions**

**1.Flower vase is in the form of a frustum of a cone. The perimeter of the ends are 44 cm and 8.4π cm. If the depth is 14 cm, fond how much water it can hold?**

Solution:

Given, Perimeter, P_{1} = 2πr_{1} = 44 cm

Then, r_{1} = ^{44}/_{2π} = ^{22×7}/_{22} = 7 cm

Perimeter, P_{2} = 2πr_{2} = 8.4π cm

Then, r_{2} = ^{8.4π}/_{2π} = 4.2 cm

Given: h = 14 cm

Volume of the flower vase = ^{1}/_{3 }πh(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})

= ^{1}/_{3} x π x 14 [(7)^{2} + (4.2)^{2} + (7 x 4.2)]

= ^{1}/_{3} x ^{22}/_{7} x 14(49 + 17.64 + 29.4)

= 1408.58 cm^{3}

**A bucket is in the shape of a frustum with the top and bottom circles of radii 15 cm and 10 cm. Its depth is 12 cm. Find its curved surface area and total surface area.(express your answer in terms of π)**

Solution:

r_{1} = 15 cm

r_{2} = 10 cm

h = 12 cm

l^{2} = h^{2} + (r_{2} – r_{1})^{2}

l^{2} = 12^{2} + (15 – 10)^{2}

= 12^{2} + 5^{2}

= 144 + 25

= 169

l = 13 cm

Curved surface area of the frustum = π(r_{1 }+ r_{2})l

= ^{22}/_{7} x (15 + 10) x 13

= 1042.43 cm^{2}

**From the top of a cone of base radius 24 cm and height 45 cm, a cone of slant height 17 cm is cut off. What is the volume of the remaining frustum of the cone?**

Solution:

L^{2} = R^{2} + H^{2}

= 24^{2} + 45^{2}

= 576 + 2025

= 2601

l = √2601

l = 51 cm

We know,

^{r}/_{R} = ^{l}/_{L} and ^{h}/_{H }= ^{l}/_{L}

^{r}/_{24} = ^{17}/_{51} and ^{h}/_{45} = ^{17}/_{51}

r = ^{17}/_{51} x 24 and h = ^{17}/_{51} x 45

r = 8 cm and h = 15 cm

Height of the frustum = 45 – 15 = 30 cm

Volume of the frustum = ^{1}/_{3} πh(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})

= ^{1}/_{3} x ^{22}/_{7} x 30 x [8^{2} + 24^{2} + 8 x 24]

= 26148.57 cm^{3}

**A vessel is in the form of a frustum of a cone. Its radius at one end is 8 cm and the height is 14 cm. If its volume is**^{5676}/_{3}cm^{3}, find the radius of the other hand.

Solution:

r_{1} = 8 cm

h = 14 cm

Volume of the frustum = ^{5676}/_{3} cm^{3}

Volume of the frustum = ^{1}/_{3} πh(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})

^{5676}/_{3} = ^{1}/_{3}x ^{22}/_{7} x 14(8^{2} + r_{2}^{2} + 8xr_{2})

^{5676}/_{3} x ^{3×7}/_{22×14} = (64 + r_{2}^{2} + 8r_{2})

^{5676}/_{3} x ^{3×7}/_{22×14} = 64 + r_{2}^{2} + 8r_{2}

129 – 64 = r_{2}^{2} + 8r_{2}

r_{2}^{2} + 8r_{2} = 65

r_{2}^{2} + 8r_{2} – 65 = 0

r_{2}^{2} + 13r_{2} – 5r_{2} – 65 = 0

r_{2}(r_{2} + 13) – 5 (r_{2} + 13) = 0

(r_{2} + 13)(r_{2} – 5) = 0

r_{2} = -13 and r_{2} = 5

Therefore, radius of the other hand, r_{2} = 5 cm.