Height (h) of cone = 15 cm
Let the radius of the cone be r.
Volume of cone = 1570 cm3
\(\frac{1}{3}\pi\)r²h = 1570 cm³
r² =\(\frac{\text{ (1570 cm³ × 3) }}{ \pi h}\)
r² = \(\frac{\text{(1570 cm³ × 3) }}{\text{ (3.14 × 15 cm) }}\)= 100 cm²
r = \(\sqrt{100}\) cm²
r = 10 cm
Therefore, the radius of the base of cone is 10 cm.
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)