Question

Find the perimeter of the rectangle whose length is 40cm and a diagonal is 41cm.
Explain it too.​

Answer 1

Step-by-step explanation:

Length = 40 cm(Given)

Diagonal = 41 cm

Diagonal = $\sqrt{(length)^{2}+(breadth)^{2} }$

41=$\sqrt{(40)^{2}+b^{2} }$

$1600+b^{2} =41^{2}$

$1600+b^{2} =1681$

$b^{2} =1681-1600$

$b^{2} =81$

$b=\sqrt{81}$

$b=\sqrt{9*9}$

∴b=9 cm

Perimeter of a rectangle = 2(l+b)

                                         = 2(40+9)

                                         = 2*49

                                         = 98 cm

Hence, the perimeter of the rectangle is 98 cm.

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Answer 2

Answer:

98cm

Step-by-step explanation:

Let us consider a rectangle ABCD    TIP: Draw diagram for better understanding                                            

Therefore given,

AB = CD = 40cm;

BD(diagonal) = 41cm

∠A = ∠B = ∠C = ∠D = 90°

To find,

AD and perimeter

Solution:

therefore, in triangle ABD of the rectangle, AB = 40cm, BD = 41cm and ∠BAD = 90°

therefore, using pythagoras theorm, i.e., P² + B² = H²

∴ AB² + AD² = BD²

  40² + AD² = 41²

 1600 + AD² = 1681

            AD²  = 1681 - 1600 = 81

Therefore,

AD = √81 = 9cm

therefore, perimeter of rectangle = 2( L + B )

                                                        =  2 × ( 40 + 9 )

                                                        = 2 × 49

                                                        = 98cm