Question

HOPE is a square of perimeter 24 cm and T is the midpoint of side EP. PEN is an isosceles triangle where PN = EN = 5 cm. Find the area of triangle PEN.

• A. 8 $c{m}^2$
• B. 36 $c{m}^2$
• C. 12 $c{m}^2$
• D. 10 $c{m}^2$

Answer 1

The correct option is C 12 $c{m}^2$


Given:
Perimeter of HOPE = 24 cm.
Length of each side = $\frac{24}{4}$
= 6 cm
(T is the mid point of EP)
PT = $\frac{6}{2}=3cm$


Thus, in right-angled ∆PNT, we have
PN = 5 cm, PT = 3 cm.

By Pythagoras’ theorem, we have

TN = $\sqrt{PN²-PT²}$

=$\sqrt{5²-3²}$

=$\sqrt{25-9}$

= $\sqrt{16}$

= 4
Area of $△PEN$ = $\frac{1}{2}(base\times height)$

base EP = 6 cm, height TN = 4 cm.

Area of $△PEN$ = $\frac{1}{2}(6cm\times 4cm)$

= 12 $c{m}^2$