Question

$ABCD$ is a quadrilateral $AC=19cm$. The lengths of perpendiculars from $B$ and $D$ on $AC$ are $5cm$ and $7cm$ respectively. Then, the area of $ABCD$ (in $c{m}^2)$ is:

• A. $162$
• B. $144$
• C. $228$
• D. $114$

Answer 1

Answer: (B)

$144$

Given: $AC=19$cm

Perpendicular from $B=5$cm

Perpendicular from $D=7$cm

is $BP=5$cm and $DQ=7$cm

Therefore area of ABCD$=$Area of $\Delta$ABC$+$Area of $\Delta$ADC

Area of $\Delta$ABC$=\frac{1}{2}\times BP\times AC$

$=\frac{1}{2}\times 5\times 19=47.5c{m}^2$

Area of $\Delta ADC=\frac{1}{2}\times DQ\times AC=$

$=\frac{1}{2}\times 7\times 19=66.5c{m}^2$

Therefore, area of ABCD$=47.5+66.5$

$=114c{m}^2$

Area of ABCD$=114c{m}^2$.