Let the radius of the sphere be r.
Surface area of sphere = 4\(\pi\)r2 = 154 cm2
Volume of a sphere = \(\frac{4}{3}\pi\)r3
\(⇒\) Surface area of the sphere = 4\(\pi\)r2 = 154cm²
r2 = \(\frac{154\ cm^2 }{ 4\pi}\)
r2 = (154 cm2) \(÷\) (4 × \(\frac{22}{7}\))
r = \(\sqrt{\frac{49}{4}}\)cm²
r = \(\frac{7}{2}\) cm
Now, radius of the sphere = \(\frac{7}{2}\) cm
So, volume of the sphere = \(\frac{4}{3}\pi\)r3
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) cm × \(\frac{7}{2}\) cm × \(\frac{7}{2}\) cm
= \(\frac{539}{3}\) cm3
Therefore, volume of the sphere is \(\frac{539}{3}\) cm3.
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)