Question

In Fig. 7.242, ABC is a triangle right angled at B and BD ⊥ AC. If AD = 4 cm and CD = 5 cm, then BD = _______ and AB = __________.

Answer 1

ABC is a right triangle right angled at B and BD ⊥ AC.

Now,

∠A + ∠ABD = 90º .....(1)

∠ABD + ∠CBD = 90º .....(2)

From (1) and (2), we have

∠A + ∠ABD = ∠ABD + ∠CBD

⇒ ∠A = ∠CBD

In ∆ABD and ∆BCD,

∠A = ∠CBD (Proved)

∠BDA = ∠BDC (90º each)

∴ ∆ABD $~$ ∆BCD (AA similarity)

$\Rightarrow \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{AD}}{\mathrm{BD}}$ (If two triangles are similar, then the ratio of their corresponding sides is proportional)

$$\begin{gathered} \Rightarrow \frac{\mathrm{BD}}{5\mathrm{cm}}=\frac{4\mathrm{cm}}{\mathrm{BD}} \\ \Rightarrow {\mathrm{BD}}^2=20{\mathrm{cm}}^2 \\ \Rightarrow \mathrm{BD}=2\sqrt{5}\mathrm{cm} \end{gathered}$$

In right ∆ABD,

$$\begin{gathered} {\mathrm{AB}}^2={\mathrm{BD}}^2+{\mathrm{AD}}^2\left(\mathrm{Pythagoras}\mathrm{theorem}\right) \\ \Rightarrow {\mathrm{AB}}^2={\left(2\sqrt{5}\right)}^2+{\left(4\right)}^2 \\ \Rightarrow {\mathrm{AB}}^2=20+16=36 \\ \Rightarrow \mathrm{AB}=\sqrt{36}=6\mathrm{cm} \end{gathered}$$

In Fig. 7.242, ABC is a triangle right angled at B and BD ⊥ AC. If AD = 4 cm and CD = 5 cm, then BD = $\underline{2\sqrt{5}\mathrm{cm}}$ and AB = __6 cm__.