Question

$ABCD$ is a square, $AC=BD=4\sqrt{2}cm,AE=DE=2.5cm$. Find the area of the adjoining figure $ABCDE$:

• A. $19c{m}^2$
• B. $22c{m}^2$
• C. $17c{m}^2$
• D. none of the above

Answer 1

The correct option is A $19c{m}^2$

Given

ABCD is square

where AC$=$BD$=4\sqrt{2}$cm (diagonals)

we know length of a diagonal of a square whose each side is 'a' cm$=\sqrt{2}a$

$\therefore 4\sqrt{2}=\sqrt{2}a$

$\therefore a=4$

That is AB$=$BC$=$CD$=$AD$=4$cm

Area of the square, ABCD$=4^2=16c{m}^2$

Next, $\Delta$ADE

$AD=4$cm, $AE=2.5$cm$=$DE

This is an isoceles triangle

using Heron's formula

Area of $\Delta$ADE$=\sqrt{s(s-a)(s-b)(s-c)}$

where s is the semi perimeter

Perimeter$=AD+AE+DE=4+2.5+2.5=9$cm

$\therefore$ semi perimeter$=\frac{9}{2}=4.5$cm

$a=AD,b=AE,c=ED$

Area of $\Delta ADE=\sqrt{4.5(4.5-4)(4.5-2.5)(4.5-2.5)}$

$=\sqrt{4.5\times 0.5\times 2\times 2}=\sqrt{9}$

$=3c{m}^2$

$\therefore$ Area of ABCDE$=$Area of ABCD$+$Area of $\Delta$ADE

$=16+3=19c{m}^2$.