Question

For finding AB and BC wit the help of information given in the figure, complete following activity .
$AB=BC............$
$\therefore \angle BAC=...........$
$\therefore AB=BC=........\times AC$
$=........\times \sqrt{8}$
$=........\times 2\times \sqrt{2}$
$=.......$

Answer 1

In $△ABC$

$\angle B={90}^{\circ },AC=\sqrt{8}AB=BC,\therefore \angle A=\angle C={45}^{\circ }$

By ${45}^{\circ }-{45}^{\circ }-{90}^{\circ }$ theorem, $AB=BC=\frac{1}{\sqrt{2}}\times AC$

$=\frac{1}{\sqrt{2}}\times \sqrt{8}$

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

$=2$

Hence, AB = 2 and BC = 2

Hence, the completed activity is

$AB=BC$ .... Given

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

$\therefore AB=BC=\frac{1}{\sqrt{2}}\times AC$

$=\frac{1}{\sqrt{2}}\times \sqrt{8}$

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

$=2$