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CBSE Class 9 Mathematics Question 8 of 44

Surface Areas and Volumes — Question 8

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Question 8

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take 1.04\sqrt{1.04} = 1.02)

Answer

Given,

Diameter of cone (d) = 40 cm = 40100\dfrac{40}{100} m = 0.4 m

Radius of cone (r) = Diameter2=0.42\dfrac{\text{Diameter}}{2} = \dfrac{0.4}{2} = 0.2 m

Height of cone (h) = 1 m

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m^2, what will be the cost of painting all these cones? (Use π = 3.14 and take 1.04 = 1.02). NCERT Class 9 Mathematics CBSE Solutions.

Slant height of cone (l) = r2+h2\sqrt{r^2 + h^2}

Substituting values we get :

l=(0.2)2+(1)2=0.04+1=1.04=1.02 m.l = \sqrt{(0.2)^2 + (1)^2} \\[1em] = \sqrt{0.04 + 1} \\[1em] = \sqrt{1.04} = 1.02 \text{ m}.

By formula,

Curved surface area of each cone = πrl

= 3.14 × 0.2 m × 1.02 m

= 0.64056 m2

Curved surface area of 50 cones = 50 × 0.64056 m2 = 32.028 m2.

Given,

Cost of painting = ₹ 12 per m2.

Cost of painting of 32.028 m2 area = ₹ (32.028 × 12)

= ₹ 384.34 (approx.)

Hence, the cost of painting 50 such hollow cones = ₹ 384.34 (approx.)