The diagonal of a rectangle is displaystyle sqrt-41- cm and its area is 20 sq cm The perimeter of the rectangle must be-9 cm18 cm41 cm20 cm

Question

Toppr Admin · Accepted Answer

From the figure we have-l2-b2-x221A-41-2-41-xA0-Also lb-20We know that-l-b-2-l2-b2-2lb-41-2-xD7-20-41-40-81-x21D2-l-b-x221A-81-9-x2234- Perimeter-xA0- -2-l-b-2-xD7-9-18-xA0-cm

The diagonal of a rectangle is $\sqrt{41}$ cm and its area is 20 sq cm The perimeter of the rectangle must be:
$9 c m$
$18 c m$
$41 c m$
$20 c m$

From the figure we have,
$l^{2} + b^{2} = {(\sqrt{41})}^{2} = 41.$
Also $l b = 20$
We know that,
$(l + b)^{2} = l^{2} + b^{2} + 2 l b$
$= 41 + 2 \times 20 = 41 + 40 = 81$
$\Rightarrow l + b = \sqrt{81} = 9$
$∴$ Perimeter $= 2 (l + b) = 2 \times 9 = 18 c m$

The diagonal of a rectangle is √41 cm and its area is 20 sq cm The perimeter of the rectangle must be:9 cm18 cm41 cm20 cm

From the figure we have,l2+b2=(√41)2=41. Also lb=20We know that,(l+b)2=l2+b2+2lb=41+2×20=41+40=81⇒l+b=√81=9∴ Perimeter =2(l+b)=2×9=18 cm

The diagonal of a rectangle is $\sqrt{41}$ cm and its area is 20 sq cm The perimeter of the rectangle must be:
$9 c m$
$18 c m$
$41 c m$
$20 c m$

From the figure we have,
$l^{2} + b^{2} = {(\sqrt{41})}^{2} = 41.$
Also $l b = 20$
We know that,
$(l + b)^{2} = l^{2} + b^{2} + 2 l b$
$= 41 + 2 \times 20 = 41 + 40 = 81$
$\Rightarrow l + b = \sqrt{81} = 9$
$∴$ Perimeter $= 2 (l + b) = 2 \times 9 = 18 c m$