Question

For finding AB and BC with the help of information given in the figure, complete following activity.

AB = BC ..........$\boxed{}$

$$\begin{gathered} \therefore \angle \mathrm{BAC}=\boxed{} \\ \therefore \mathrm{AB}=\mathrm{BC}=\boxed{}\times \mathrm{AC} \\ =\boxed{}\times \sqrt{8} \\ =\boxed{}\times 2\sqrt{2} \\ =\boxed{} \end{gathered}$$

Answer 1

In ∆ABC,
∠B = 90^∘, AC = $\sqrt{8}$, AB = BC, ∴ ∠A = ∠C = 45^∘

By 45^∘ − 45^∘ − 90^∘ theorem,

$$\begin{gathered} \mathrm{AB}=\mathrm{BC}=\frac{1}{\sqrt{2}}\times \mathrm{AC} \\ =\frac{1}{\sqrt{2}}\times \sqrt{8} \\ =\frac{1}{\sqrt{2}}\times 2\sqrt{2} \\ =2 \end{gathered}$$

Hence, AB = 2 and BC = 2.

Hence, the completed activity is

AB = BC ..........$\boxed{\mathrm{given}}$

$$\begin{gathered} \therefore \angle \mathrm{BAC}=\boxed{{45}^{°}} \\ \therefore \mathrm{AB}=\mathrm{BC}=\boxed{\frac{1}{\sqrt{2}}}\times \mathrm{AC} \\ =\boxed{\frac{1}{\sqrt{2}}}\times \sqrt{8} \\ =\boxed{\frac{1}{\sqrt{2}}}\times 2\sqrt{2} \\ =\boxed{2} \end{gathered}$$