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.
(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that
(i) ∆ ABD ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.
