Find the area of a rectangle of length 4cm and whose diagonal is 8cm, (Leave your answer in surd form)
- 8√3cm2
- 12√3cm2
- 16√2cm2
-
16√3cm2
The correct answer is: D
Explanation
|AB| = 4cm
|BC| = 8cm
In right-angled BAC;
8\(^2\) = 4\(^2\) + |AC|\(^2\)
|AC|\(^2\) = 8\(^2\) - 4\(^2\)
|AC\\(^2\) = 64 - 16 → 48
|AC| = \(\sqrt{48}\)cm → \(4\sqrt{3}cm\)
The area of rectangle = L x B
= |AB| x |AC|
= \((4 \times 4\sqrt{3}cm^2\)
=\(16\sqrt{3}cm^2\)