Question

What will be the Perimeter of the isosceles right-angled $△$ LON in the given figure?
(Given: KLOJ is a square)

• A. $2\sqrt{2}$ + 1 cm
• B. $\sqrt{2}$ cm
• C. $\sqrt{2}$ + 1 cm
• D. 2 + $\sqrt{2}$ cm

Answer 1

The correct option is C

$\sqrt{2}$ + 1 cm

The figure given needs to be observed carefully.

KL is given as $\frac{1}{\sqrt{2}}$ cm

KL = LO (Sides of the same square)

LO = ON = $\frac{1}{\sqrt{2}}$ (The figure shows LON as an isosceles right-angled triangle)

$\therefore$ LN = $\sqrt{(\frac{1}{\sqrt{2}}{)}^2+(\frac{1}{\sqrt{2}}{)}^2}$

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

=$\sqrt{1}$

= 1 cm

$\therefore$ Perimeter of the $△$ LON = LO + ON + LN

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

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

= $\sqrt{2}$ + 1 cm