NCERT class 9 ex 13 . 4
Answers
Some NCERT class 9 ex 13 . 4
Question 1 :Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Given:
(i) Radius of the sphere (r) = 10.5 cm
Surface area of a sphere= 4πr²
= (4 × 22/7 × 10.5 × 10.5)
= 1386 cm²
ii)Given: Radius of the sphere (r) = 5.6 cm
Surface area of a sphere= 4πr²
= (4 × 22/7 × 5.6 × 5.6)
= 394.24cm²
(iii)Given: Radius of the sphere (r) = 14 cm
Surface area of a sphere= 4πr²
= (4 × 22/7 × 14 × 14)
= 2464 cm²
Question 3 : Find the total surface area of a hemisphere of radius 10 cm. [Use π = 3.14]
Given: radius(r) = 10 cm
Total surface area of hemisphere = 3πr²
= (3 × 3.14 × 10 ×10)
= 9.42×100
= 942 cm²
Hence, the total surface area of a hemisphere is 942 cm²
Question 5: A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm^2 .
Given:
Diameter of the bowl = 10.5 cm
Inner Radius(r)=10.5/2= 5.25 cm
Curved surface area of the hemispherical bowl = 2πr²
= (2×22/7× 5.25 × 5.25)
= 173.25 cm²
Rate of tin plating is = ₹16 per 100cm²
cost of 1cm²= ₹16/100
Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100
= ₹27.72
Hence, cost of tin plating on the the hemispherical bowl = ₹27.72
Learn more :
Find the surface area of a sphere of diameter :
(i) 14 cm
(ii) 21 cm
(iii) 3.5 cm
https://brainly.in/question/15912348
Find the total surface area of a hemisphere of radius 10 cm.
https://brainly.in/question/15912593
Answer:
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