From the given figure, in A B C, if A D B C, C=45^ , A C=8 2, B D… - Vidyadip

Question

From the given figure, in $\triangle A B C$, if $A D \perp B C, \angle C=45^{\circ}, A C=8 \sqrt{2}, B D=5$, then for finding value of $A D$ and $BC$, complete the following activity.

Activity: In $\triangle ADC$, if $\angle ADC =90^{\circ}, \angle C =45^{\circ}$ [Given]
$\therefore \angle DAC =\square \quad$..... [Remaining angle of $\triangle ADC ]$
By theorem of $45^{\circ}-45^{\circ}-90^{\circ}$ triangle,
$ \therefore \square=\frac{1}{\sqrt{2}} AC \text { and } \square=\frac{1}{\sqrt{2}} AC$
$\therefore AD =\frac{1}{\sqrt{2}} \times \square \text { and } DC =\frac{1}{\sqrt{2}} \times 8 \sqrt{2}$
$\therefore AD =8 \text { and } DC =8$
$\therefore BC = BD + DC$
$=5+8$
$=13 $

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Answer

In $\triangle MNK , \angle MNK =90^{\circ}, \angle M =45^{\circ}$ [Given]
$\therefore \angle K=45^{\circ}$
..... [Remaining angle of $\triangle M N K$ ]
By theorem of $45^{\circ}-45^{\circ}-90^{\circ}$ triangle,
$\therefore MN =\frac{1}{\sqrt{2}} MK$ and $KN =\frac{1}{\sqrt{2}} MK$
$\therefore MN =\frac{1}{\sqrt{2}} \times 6$ and $KN =\frac{1}{\sqrt{2}} \times 6$
$\therefore MN =\frac{1 \times 6 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$ and $KN =\frac{1 \times 6 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$
... [Multiply numerator and denominator by $\sqrt{2}$ ]
$\therefore MN =\frac{6 \times \sqrt{2}}{2}$ and $KN =\frac{6 \times \sqrt{2}}{2}$
$\therefore MN =3 \sqrt{2}$ and $KN =3 \sqrt{2}$

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