In the given figure, AB and DE are perpendicular to BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm, calculate AD.
$[4 M a r k s]$

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Solution

Given: $A B ⊥ B C, D E ⊥ B C, A B = 9 c m, D E = 3 c m$ and $A C = 24 c m$

In $Δ A B C$ and $Δ D E C$ ,
$∠ A B C = ∠ D E C$ [Each of $90^{\circ}$ ]
$∠ A C B = ∠ D C E$ [Common]
$∴ Δ A B C \sim Δ D E C$ [By AA similarity] $[1.5 M a r k s]$
Now $\frac{A B}{D E} = \frac{A C}{D C}$
$\Rightarrow \frac{9}{3} = \frac{24}{D C}$
$\Rightarrow D C = \frac{24 \times 3}{9} c m$
$= 8 c m$ $[1.5 M a r k s]$
Hence, AD = (24-8)cm = 16 cm $[1 M a r k]$

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In the given figure, AB and DE are perpendicular to BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm, calculate AD. [4 Marks]

Given: AB⊥BC,DE⊥BC,AB=9 cm,DE=3 cm and AC=24 cm In ΔABC and ΔDEC, ∠ABC=∠DEC [Each of 90∘] ∠ACB=∠DCE [Common] ∴ΔABC∼ΔDEC [By AA similarity] [1.5 Marks] Now ABDE=ACDC ⇒93=24DC ⇒DC=24×39 cm =8 cm [1.5 Marks] Hence, AD = (24-8)cm = 16 cm [1 Mark]

In the given figure, AB and DE are perpendicular to BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm, calculate AD.
$[4 M a r k s]$