Question

In the given figure, $\Delta ABC$ is a right angle triangle, right-angled at B and DE is perpendicular to AC. The approximate length of DE is equal to

• A. 3.8 cm
• B. 5.6 cm
• C. 4.2 cm
• D. 6.2 cm

Answer 1

The correct option is C 4.2 cm

In ΔABC, by Pythagoras theorem,

$(AC{)}^2=(AB{)}^2+(BC{)}^2$
$=(8{)}^2+(15{)}^2$
$=64+225$
$=289$
$\Rightarrow AC=17cm$
Now, Area of $\Delta ADC$ = Area of $\Delta ABC$ - Area of $\Delta ABD$
$=\frac{1}{2}\times BC\times AB-\frac{1}{2}\times BD\times AB$
$=\frac{1}{2}\times AB\times (BC-BD)$
$=\frac{1}{2}\times 8\times \left(9\right)$
$=36c{m}^2$

Also, area of $\Delta ADC=\frac{1}{2}\times AC\times DE=36c{m}^2$
$\therefore \frac{1}{2}\times 17\times DE=36$
$\Rightarrow DE=\frac{36\times 2}{17}\approx 4.2cm$
Thus, the approximate length of DE is equal to 4.2 cm.
Hence, the correct answer is option (c).