Mensuration Practice Test 6

Area of the circle = $\mathit{\pi }{\mathit{r}}^2$
= $\mathit{\pi }{\left(10\right)}^2$ = 100π
If 40% of area is cut, then 60% of area is left to make cone.
60% area of circle = $\frac{60}{100}\times 100\mathit{\pi }$
= 60π
So, lateral surface area of cone = 60π
π Rl = 60π
R × l = 60 ......(1)
Circumference of a circle = 2π r = 2π (10) = 20π
If 40% part is cut and 60% is used to make cone, then this 60% of circumference will be the circumference of the base of the cone.
60% circumference of the circle = $\frac{60}{100}\times 20\mathit{\pi }$
= 12 π
12π = circumference of the base of the cone
12π = 2πR
So, R (radius of the cone) = 6 cm
Put value of R in ______(1) we get
l = 10 cm
Height = $\sqrt{{\mathit{l}}^2-{\mathit{R}}^2}$
= $\sqrt{100-36}$
Height = 8 cm
Required ratio = Radius : height
6 : 8
3 : 4