Question

In the following figure $\angle ABC={90}^{\circ }$ and $BD\perp AC$. If $AD=8cm,CD=2cm$, then the length of $BD$ is:

• A. $4cm$
• B. $8cm$
• C. $16cm$
• D. $10cm$

Answer 1

The correct option is A $4cm$

Given that:

$\angle ABC={90}^{\circ },BD\perp AC,AD=8cm,CD=2cm$

To find:

$BD=?$

Solution:

In $△ABC,$

$AC=AD+DC=8cm+2cm=10cm$

By Pythagorean theorem,

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

or, $A{B}^2+B{C}^2=(10{)}^{2}c{m}^2$

or, $A{B}^2+B{C}^2=100c{m}^2$ $..............\left(i\right)$

In $△ADB,$

By Pythagorean theorem,

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

or, $(8cm{)}^2+B{D}^2=A{B}^2$

or, $64c{m}^2+B{D}^2=A{B}^2$ $..............\left(ii\right)$

In $△BDC,$

By Pythagorean theorem,

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

or, $(2cm{)}^2+B{D}^2=B{C}^2$

or, $4c{m}^2+B{D}^2=B{C}^2$ $..............\left(iii\right)$

From eqn.(i),(ii) and (iii) we get,

$64c{m}^2+B{D}^2+4c{m}^2+B{D}^2=100c{m}^2$

or, $68c{m}^2+2B{D}^2=100c{m}^2$

or, $2B{D}^2=100c{m}^2-68c{m}^2$

or, $2B{D}^2=32c{m}^2$

or, $B{D}^2=16c{m}^2$

or, $BD=4cm$

Therefore, $A$ is the correct option.