Question

Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm. [2 marks]

Answer 1

Given diagonal (PR) = 41 cm, length (PQ) = 40 cm
Let breadth (QR) be x cm.
Now, in right angled triangle PQR,
$(PR{)}^2=(RQ{)}^2+(PQ{)}^2$
[0.5 marks]

[By Pythagoras theorem]

$\Rightarrow (41{)}^2=x^2+(40{)}^2\Rightarrow 1681=x^2+1600\Rightarrow x^2=1681-1600\Rightarrow x^2=81\Rightarrow x=\sqrt{81}=9cm$
[0.5 marks]

Therefore the breadth of the rectangle is 9 cm.
Perimeter of rectangle = 2(length +breadth)
= 2 (9 +40)
= 2 $\times$ 49 = 98 cm
[1 mark]

Hence the perimeter of the rectangle is 98 cm.