I would like to give a question which
i could not solve.
Rd sharma class 10 maths exercise-16.3 mensuration.
In this question we have to find the area
of its whole surface and volume given that
a frustum of a right circular cone has a diameter
of base 20 cm, of top 12 cm and height 3 cm.
Given,
Base diameter of cone (d1) = 20 cm
So the radius (r1) = 20/2 cm = 10 cm
Top diameter of Cone (d2) = 12 cm
So, the radius (r2) = 12/2 cm = 6 cm
Height of the cone (h) = 3 cm
Volume of the frustum of a right circular cone = 1/3 π(r22 + r12 + r1 r2 )h
= π/3(102 + 62 + 10 × 6)3
= 616 cm3
Let ‘L’ be the slant height of cone, then we know that
L = √(r1 – r21)2 + h2
L = √(10 – 6)2 + 32
L = √(16 + 9)
L = 5cm
So, the slant height of cone (L) = 5 cm
Thus,
Total surface area of the frustum = π(r1 + r2) x L + π r12 + π r22
= π(10 + 6) × 5 + π × 102 + π × 62
= π(80 + 100 + 36)
= π(216)
= 678.85 cm2