MCQ
In Figure, $\triangle\text{ABC}$ is an isosceles triangle, right-angled at $C$. Therefore.
- ✓$\mathrm{AB}^2=2 \mathrm{AC}^2$
- B$\mathrm{BC}^2=2 \mathrm{AB}^2$
- C$\mathrm{AC}^2=2 \mathrm{AB}^2$
- D$\mathrm{AB} 2=4 \mathrm{AC} 2$
✓
Answer
Correct option: A.
$\mathrm{AB}^2=2 \mathrm{AC}^2$
Since $\triangle\text{ABC}$ is a right $-$ angled triangle at $C$.
Using pythagoras theorem in $\triangle\text{ABC}$
$\mathrm{Hypotenuse^2= (Height)^2+ (Base)^2}$
$\therefore\text{AB}^2=\text{AC}^2+\text{BC}^2$
$\Rightarrow\text{AB}^2=\text{AC}^2+\text{AC}^2$
$[\therefore\text{BC = AC}]$
$\Rightarrow\text{AB}^2=2\text{AC}^2$
Hence, proved.
Using pythagoras theorem in $\triangle\text{ABC}$
$\mathrm{Hypotenuse^2= (Height)^2+ (Base)^2}$
$\therefore\text{AB}^2=\text{AC}^2+\text{BC}^2$
$\Rightarrow\text{AB}^2=\text{AC}^2+\text{AC}^2$
$[\therefore\text{BC = AC}]$
$\Rightarrow\text{AB}^2=2\text{AC}^2$
Hence, proved.
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