Question

Question 122

In the given figure, $\Delta AEC$ is right angled at E, B is a point on EC, BD is the altitude of $\Delta ABC$, AC = 25 cm BC = 7 cm and AE = 15 cm. Find the area of $\Delta ABC$ and the length of DB.

Answer 1

Given, AC = 25 cm, BC = 7 cm, and AE = 15 cm
In $\Delta AEC,$ using Pythagoras theorem,
$A{C}^2=A{E}^2+E{C}^2$
$\Rightarrow E{C}^2=A{C}^2-A{E}^2$
$\Rightarrow E{C}^2=(25{)}^2-(15{)}^2=625-225$
$=400$
$EC=\sqrt{400}=20cm$
and $EB=EC-BC=20-7$
$=13cm$
Area of $\Delta AEC=\frac{1}{2}\times AE\times EC$
$=\frac{1}{2}\times 15\times 20=150c{m}^2$
and Area of $\Delta AEB=\frac{1}{2}\times AE\times EB=\frac{1}{2}\times 15\times 13=97.5c{m}^2$
$\therefore$ Area of $\Delta ABC=$ Area of $\Delta AEC-$ Area of $\Delta AEB$
$=150-97.5$
$=52.5c{m}^2$
Again, Area of $\Delta ABC=\frac{1}{2}\times BD\times AC$
$52.5=\frac{1}{2}\times BD\times 25$
$\Rightarrow BD=\frac{52.5\times 2}{25}=4.2cm$
Hence, the area of $\Delta ABC$ is $52.5c{m}^2$ and the length of DB is 4.2 cm.