Question

HOPE is a square of perimeter 24 cm and T is the midpoint of side EP. If PN = EN = 5 cm, then the area of the triangle PEN is .

• A. $24c{m}^2$
• B. $36c{m}^2$
• C. $12c{m}^2$

Answer 1

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


Given: Perimeter of the square
HOPE = 24 cm.
Length of each side
$=\frac{24}{4}=6cm$

Given: T is the mid point of EP.
$PT=\frac{6}{2}=3cm$

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

By Pythagoras’ theorem, we have
$TN=\sqrt{PN²-PT²}=\sqrt{5²-3²}=\sqrt{25-9}=\sqrt{16}=4cm$

$\text{Area of}△PEN=\frac{1}{2}\text{(Base}\times \text{Height)}\text{Base EP = 6 cm, Height TN = 4 cm.}$

$\text{Area of}△PEN=\frac{1}{2}(6cm\times 4cm)=12c{m}^2$