Question

OA = 4 cm dan SB = 2 cm. jarak OS = 10 cm. panjang garis singgung persekutuan luar AB adalah..

Answer 1

AB^2= OS^2-(OA+SB)^2
= 10^2-(4+2)^2
= 100-(6)^2
= 100-36
=64
AB = akar 64 = 8cm

Answer 2

Diket: OA=4 cm, OB=2 cm, OS=10 cm.
misal: Persekutuan luar=L, maka.......,:
L²= OS² - (OA-OB)²
L²=10²-(4-2)²
L²=100 - 4
L²=96
L= 9,79 cm