Question

$\angle B$ is a right angle in $△ABC$ and $\overline{BD}$ is an altitude to hypotenuse. $AB=8$ cm and $BC=6$ cm. Find the area of $△BDC$

Answer 1

Given: $AB=8;BC=6cm$

ABC is a right-angled triangle, right-angled at B.

By Pythagoras theorem, we have

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=8^2+6^2$

$AC=\sqrt{100}=10cm$

Let $AD=x$, then $CD=AC-AD=10-x$

In right-angled triangle ABD, we have

$B{D}^2=A{B}^2-A{D}^2$

$h^2=8^2-x^2$ [Suppose $BD=h$]

$h^2+x^2=64\dots \dots \left(i\right)$

In right-angled triangle BDC, we have

$B{D}^2=B{C}^2-C{D}^2$

$\Rightarrow h^2=6^2-(10-x{)}^2$

$\Rightarrow h^2+(100+x^2-20x)=36$

$\Rightarrow h^2+100+x^2-20x=36$

$\Rightarrow (h^2+x^2)+100-36=20x$

$\Rightarrow 64+100-36=20x$ [Using eq.(i)]

$\Rightarrow 128=20x$

$\Rightarrow x=\frac{128}{20}=6.4$

So, $CD=10-x=10-6.4=3.6cm$

$\therefore h^2+{6.4}^2=64$

$\Rightarrow h^2=64-40.96$

$\Rightarrow BD=h=\sqrt{23.04}=4.8cm$

Area of triangle $BDC=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times CD\times BD$

$=\frac{1}{2}\times 4.8\times 3.6$

$=8.64c{m}^2$