Question

In the given figure, $\angle ABC={90}^{\circ }$ and $BD\perp AC$. If $AB=5\cdot 7cm,BD=3\cdot 8cm$ and $CD=5\cdot 4cm,$ find $10BC$ in cm.

Answer 1

$BD\perp AC,$ $AC=5.7cm,BD=3.8cm$ and $CD=5.4cm$ and $\angle ABC={90}^o$ [ Given ]

In $△ABC$ and $△BDC$

$\Rightarrow$ $\angle ABC=\angle BDC$ [ Both ${90}^o$ ]

$\Rightarrow$ $\angle ACB=\angle BCD$ [ Common angle ]

$\therefore$ $△ABC\sim △BDC$ [ By $AA$ similarity test ]

We know, triangles are similar, hence corresponding sides will be proportional.

$\Rightarrow$ $\frac{AB}{BD}=\frac{BC}{CD}$

$\Rightarrow$ $\frac{5.7}{3.8}=\frac{BC}{5.4}$

$\Rightarrow$ $BC=\frac{5.7\times 5.4}{3.8}$

$\Rightarrow$ $BC=8.1cm$

$\Rightarrow$ $10BC=10\times 8.1=81cm$