Question

In triangle ABC , AD is an altitude and angle A is right angle. If AB = root 20,BD =4 then find CD

Answer 1

The value of $CD$ is $1$ units

Step-by-step explanation:

Given,

$\triangle ABC$, $AD$ is an altitude and $\angle A$ is right angle. If $AB =\sqrt{20}$,$BD =4$

From figure,

$\triangle ABD$ also right triangle,

So,

$AD^{2}=AB^{2}-BD^{2}$

⇒$AD=\sqrt{20-4^{2} }$

⇒$AD=\sqrt{20-16}$

⇒$AD=\sqrt{4} =2$

Take $BC$ as base.

Area of $\triangle ABC=\frac{1}{2}\times BC\times AD$

                         $=\frac{1}{2}\times BC\times 2=BC$

Also, take $AB$ as base.

Area of $\triangle ABC=\frac{1}{2}\times AB\times AC$

                         $=\frac{1}{2}\times \sqrt{20} \times AC$

                         $=\sqrt{5}AC$

∴ Area of same triangle is equal.

∴ $BC=\sqrt{5}AC$

From $\triangle ABC$,

$BC^{2}=AC^{2}+AB^{2}$

⇒$5AC^{2} =AC^{2}+AB^{2}$

⇒$4AC^{2}=20$

⇒$AC^{2}=5$

∴ $AC=\sqrt{5}$

Then, $BC=\sqrt{5}\times \sqrt{5}$

⇒$BC=5$

∴ $CD=BC-BD$

⇒$CD=5-4$

∴ $CD=1$

So, The value of $CD$ is $1$ units

Answer 2

Answer:

Step-by-step explanation: