In the figure,if angle ACB = angle CDA,AC = 8 | vedclass

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(A) Given: $AC = 8 \, cm$,$AD = 3 \, cm$ and $\angle ACB = \angle CDA = 90^{\circ}$ (from the figure).In right-angled $	riangle ADC$,by Pythagoras th

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$\frac{55}{3}$

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(A) Given: $AC = 8 \, cm$,$AD = 3 \, cm$ and $\angle ACB = \angle CDA = 90^{\circ}$ (from the figure).In right-angled $	riangle ADC$,by Pythagoras theorem:$AC^2 = AD^2 + CD^2$$8^2 = 3^2 + CD^2$$64 = 9 + CD^2$$CD^2 = 55$$CD = \sqrt{55} \, cm$.Now,consider $	riangle CDB$ and $	riangle ADC$:$1$. $\angle BDC = \angle ADC = 90^{\circ}$ (given).$2$. $\angle DBC = \angle DCA$ (since both are equal to $90^{\circ} - \angle A$).By $AA$ similarity criterion,$	riangle CDB \sim 	riangle ADC$.Therefore,the ratio of corresponding sides is equal:$\frac{CD}{BD} = \frac{AD}{CD}$$CD^2 = AD 	imes BD$$55 = 3 	imes BD$$BD = \frac{55}{3} \, cm$.

In the figure,if $\angle ACB = \angle CDA$,$AC = 8 \, cm$ and $AD = 3 \, cm$,find $BD$ (in $cm$).

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