(i) Slant height (l) of cone = 14 cm
Let the radius of the circular end of the cone be r.
We know, CSA (Curved surface area) of cone = \(\pi rl\)
\(\pi rl\) = 308
\(r = \frac{308 cm²}{\pi l}\)
\(r =\frac{ 308 cm²}{14 cm} \times \frac{7}{22}\)
= 7 cm
Therefore, the radius of the circular end of the cone is 7 cm.
(ii) Total surface area of cone = CSA of cone + Area of base
\(=\pi r (l + r)\)
= \(\frac{22}{7} \)× 7 cm × (7 cm + 14 cm)
= 22 cm × 21 cm
= 462 cm²
Therefore, the total surface area of the cone is 462 cm2 .
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ACP = ∠ QCD

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∠∆BAD
(iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)