Question

can anyone tell me the answer...?​

Answer 1

Answer:

Area = 4 cm²

Perimeter = 6√2 cm

Diagonal = √10 cm

Step-by-step explanation:

Area = l × b = √8 × √2 = √16 = 4cm²

Perimeter = 2(l+b) = 2 (2√2 + √2 ) = 2 × 3√2 = 6√2 cm

AC = √AB² + BC² AC = √(√8)² + (√2)² = √8 + 2 = √10 cm

Answer 2

$Given , \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ Length = \sqrt{8} \: cm \\ Breadth = \sqrt{2} \: cm$

$a)$

$Area=Length × Breadth$

$\sqrt{8} \times \sqrt{2} \\ = \sqrt{16} \: \: \: \: \\ = 4 \: \: \: \: \: \: \: \: \: \: \:$

$Area = 4 \: cm$

$Perimeter = 2 \: (Length + Breadth)$

$2( \sqrt{8} + \sqrt{2} )$

$= 2( \sqrt{(4 \times 2)} + \sqrt{2} )$

$= 2(2 \sqrt{2} + 1\sqrt{2} )$

$= 2(3 \sqrt{2} )$

$= 6 \sqrt{2}$

$Perimeter = 6 \sqrt{2} cm$

$b)$

$The \: given \: is \: a \: right \: angled \: triangle.$

$So, \: by \: Pythagoras \: Theorem ,$

${AB}^{2}+{BC}^{2}={AC}^{2}$

$AB= \sqrt{8} \\ BC= \sqrt{2}$

$({ \sqrt{8}) }^{2} + { (\sqrt{2} )}^{2} = {AC}^{2} \\ = 8 + 2 = {AC}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \:$

$10 = {AC}^{2} \\ AC = \sqrt{10}$

$Diagonal \: AC= \sqrt{10} \: cm$

$Hope \: \: this \: \: helps \: \: you \: !$