170
In the given figure, $\triangle ABC$ is right angled triangle with $\angle A = 90\degree$. $AD$ is perpendicular to $BC$.
Prove that:
(i) $\triangle DBA \sim \triangle DAC$
(ii) $DA^2 = DB \times DC$
(iii) Find the area of $\triangle ABC$ when $DB = 9$ cm and $DC = 16$ cm.
Prove that:
(i) $\triangle DBA \sim \triangle DAC$
(ii) $DA^2 = DB \times DC$
(iii) Find the area of $\triangle ABC$ when $DB = 9$ cm and $DC = 16$ cm.
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Solution:
(a) (i) In $\triangle DBA$ and $\triangle DAC$
$\angle BDA = \angle CDA = 90\degree$
$\angle BAD = 90\degree - \angle B = \angle ACB$
triangle DBA
sim
triangle DAC$$\begin{aligned}& (by AA similarity) (1 Mark) \\ & (ii) As\end{aligned}$$
triangle DBA
sim
triangle DAC :
frac{DA}{DC} =
frac{DB}{DA}$\\$
Rightarrow DA^2 = DC
times DB$$\begin{aligned}& (1 Mark) \\ & (iii) taking\end{aligned}$$DC = 16$ cm and $DB = 9$ cm, $DA^2 = 9
times 16$\\$
Rightarrow DA = 12$$\begin{aligned}& cm (1 Mark) \\ & Area\end{aligned}$$
triangle ABC =
frac{1}{2}
times 25
times 12 = 150$ cm$^2$ (1 Mark)
(a) (i) In $\triangle DBA$ and $\triangle DAC$
$\angle BDA = \angle CDA = 90\degree$
$\angle BAD = 90\degree - \angle B = \angle ACB$
triangle DBA
sim
triangle DAC$$\begin{aligned}& (by AA similarity) (1 Mark) \\ & (ii) As\end{aligned}$$
triangle DBA
sim
triangle DAC :
frac{DA}{DC} =
frac{DB}{DA}$\\$
Rightarrow DA^2 = DC
times DB$$\begin{aligned}& (1 Mark) \\ & (iii) taking\end{aligned}$$DC = 16$ cm and $DB = 9$ cm, $DA^2 = 9
times 16$\\$
Rightarrow DA = 12$$\begin{aligned}& cm (1 Mark) \\ & Area\end{aligned}$$
triangle ABC =
frac{1}{2}
times 25
times 12 = 150$ cm$^2$ (1 Mark)