Question

Area of rectangle whose diagnol is 41 cm and width is 9

Answer 1

Answer: 261.70 $cm^{2}$

Step-by-step explanation:

Diagonal = 41

⇒ l√2 = 41

⇒ l$=\frac{41}{\sqrt{2} }$

b = 9 cm

area = l . b

        =$\frac{41}{\sqrt{2}}*9$

        =$\frac{369}{\sqrt{2}}$

        = 261.70 $cm^{2}$

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Answer 2

Answer:

360 cm²

Step-by-step explanation:

l²=41²-9²

(by pythagoras triplet)

l=40 cm

area=l×b

       =40×9

       =360 cm²

HOPE YOU FOUND IT USEFUL.