In an isosceles triangle ABC if AC = BC and AB2 = 2AC2, then ∠C =

Question:

In an isosceles triangle $\mathrm{ABC}$ if $\mathrm{AC}=\mathrm{BC}$ and $\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$, then $\angle \mathrm{C}=$

(a) $30^{\circ}$

(b) $45^{\circ}$

(C) $90^{\circ}$

(d) $60^{\circ}$

Solution:

Given: In Isosceles $\triangle \mathrm{ABC}, \mathrm{AC}=\mathrm{BC}$ and $\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$.

To find: Measure of angle C

In Isosceles ΔABC,

AC = BC

∠B=∠A    (Equal sides have equal angles opposite to them)

$\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$

$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{AC}^{2}$

$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}(\mathrm{AC}=\mathrm{BC})$

$\Rightarrow \triangle \mathrm{ABC}$ is right angle triangle, with $\angle \mathrm{C}=90^{\circ}$

Hence the correct answer is $(c)$.

 

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