Question Bank - P2 — Maths STD 10 — Question

$\therefore \angle \mathrm{A}=\angle \mathrm{C}$ $\quad$ $\ldots$[Isosceles triangle theorem]

$\text { Let } \angle A=\angle C=x$

$\text { In } \triangle \mathrm{ABC}, \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$ $\quad$ $\ldots$[Sum of the measures of the angles of a triangle is $180^\circ]$

$\therefore \mathrm{x}+90^{\circ}+\mathrm{x}=180^{\circ}$ $\quad$ $\ldots$[From (i)]

$\therefore 2 \mathrm{x}=90^{\circ}$

$\therefore \mathrm{x}=\frac{90^{\circ}}{2}$ $\quad$ $\ldots$[From (i)]

$\therefore \mathrm{x}=45^{\circ}$

$\therefore \angle \mathrm{BAC}=\angle \mathrm{BCA}=45^{\circ}$

$\therefore \triangle A B C \text { is a } 45^{\circ}-45^{\circ}-90^{\circ} \text { triangle. }$

$\therefore A B=B C=\frac{1}{\sqrt{2}} \times A C$$\quad$ $\ldots$[Side opposite to $45^\circ$]

$=\frac{1}{\sqrt{2}} \times 2 \sqrt{2}$

$\therefore l(\mathrm{AB})=2 \text { units }$