Find the value of l in the frustrum above

Explanation

ΔABE and ΔACD are similar

therefore \(\frac{ x}{ x + 4} = \frac{3}{6}\)

6x = 3(x+4)
6x = 3x + 12
3x = 12
x = 4
In ΔACD; \(L^2\) = \((4+4)^2 + 6^2\)
= \(8^2 + 6^2\)
= 64 + 36
= 100
L = \(\sqrt{100}\)
   = 10cm
In ABE, \((AE)^2 = 4^2 + 3^2\)
= 16 + 9
= 25
AE = 5cm
∴L = 10 - 5
= 5cm