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In the given figure, $\triangle \text{ABC}$ is right angled triangle with $\angle A = 90^\circ$. $\text{AD}$ is perpendicular to $\text{BC}$.
Prove that:
(i) $\triangle \text{DBA} \sim \triangle \text{DAC}$
(ii) $\text{DA}^2 = \text{DB \times DC}$
(iii) Find the area of $\triangle \text{ABC}$ when $\text{DB = 9 cm}$ and $\text{DC = 16 cm}$.
Prove that:
(i) $\triangle \text{DBA} \sim \triangle \text{DAC}$
(ii) $\text{DA}^2 = \text{DB \times DC}$
(iii) Find the area of $\triangle \text{ABC}$ when $\text{DB = 9 cm}$ and $\text{DC = 16 cm}$.
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Solution:
(a) (i) In $\triangle \text{DBA}$ and $\triangle \text{DAC}$
$\angle \text{BDA} = \angle \text{CDA} = 90^\circ$ (1 Mark)
$\angle \text{BAD} = 90^\circ - \angle \text{B} = \angle \text{ACB}$ (1 Mark)
$\triangle \text{DBA} \sim \triangle \text{DAC}$ (by AA similarity)
(ii) As $\triangle \text{DBA} \sim \triangle \text{DAC} : \frac{\text{DA}}{\text{DC}} = \frac{\text{DB}}{\text{DA}}$ (1 Mark)
$\Rightarrow \text{DA}^2 = \text{DC \times DB}$
(iii) taking $\text{DC = 16 cm}$ and $\text{DB = 9 cm}$, $\text{DA}^2 = 9 \times 16 = 144$ (1 Mark)
$\Rightarrow \text{DA = 12 cm}$
Area $\triangle \text{ABC} = \frac{1}{2} \times \text{BC} \times \text{AD} = \frac{1}{2} \times (9+16) \times 12 = \frac{1}{2} \times 25 \times 12 = 150 \text{ cm}^2$ (1 Mark)
(a) (i) In $\triangle \text{DBA}$ and $\triangle \text{DAC}$
$\angle \text{BDA} = \angle \text{CDA} = 90^\circ$ (1 Mark)
$\angle \text{BAD} = 90^\circ - \angle \text{B} = \angle \text{ACB}$ (1 Mark)
$\triangle \text{DBA} \sim \triangle \text{DAC}$ (by AA similarity)
(ii) As $\triangle \text{DBA} \sim \triangle \text{DAC} : \frac{\text{DA}}{\text{DC}} = \frac{\text{DB}}{\text{DA}}$ (1 Mark)
$\Rightarrow \text{DA}^2 = \text{DC \times DB}$
(iii) taking $\text{DC = 16 cm}$ and $\text{DB = 9 cm}$, $\text{DA}^2 = 9 \times 16 = 144$ (1 Mark)
$\Rightarrow \text{DA = 12 cm}$
Area $\triangle \text{ABC} = \frac{1}{2} \times \text{BC} \times \text{AD} = \frac{1}{2} \times (9+16) \times 12 = \frac{1}{2} \times 25 \times 12 = 150 \text{ cm}^2$ (1 Mark)