Question

find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.​

Answer 1

Answer :–

• Perimeter of the rectangle = 98 cm.

Given :–

• Length of the rectangle = BC = 40cm.
• Diagonal of the rectangle = BD = 41cm.

To Find :–

• Perimeter of the rectangle.

Diagram :–

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\sf\large{A}}\put(7.7,1){ \sf\large{B}}\put(9.5,0.7){\sf{\large{40 cm}}}\put(11.5,1){ \sf\large{C}}\put(8,1){\line(1,0){3.5}}\put(8,1){\line(0,2){2}}\put(11.5,1){\line(0,3){2}}\put(8,3){\line(3,0){3.5}}\put(10.5,2){\sf{\large{41 cm}}}\qbezier(8,1)(8,1)(11.5,3)\put(11.5,3){ \sf\large{D}}\put(11.3,1){\line(0,2){0.2}}\put(11.3,1.2){\line(2,0){0.2}}\end{picture}$

Solution :–

First, we need to find the breadth of the rectangle.

In ∆BCD,

• BD = 41cm
• BC = 40cm

By the Pythagoras theorem,

$\hookrightarrow$ CD² + BC² = BD²

$\hookrightarrow$ CD² + (40)² = (41)²

$\hookrightarrow$ CD² + 1600 = 1681

$\hookrightarrow$ CD² = 1681 – 1600

$\hookrightarrow$ CD² = 81

$\hookrightarrow$ CD = √81

$\hookrightarrow$ CD = 9

So, the breadth of the rectangle is 9 cm.

Now, we have to find the perimeter of the rectangle.

• Perimeter of the rectangle = 2(l + b)

Values of length and breadth are,

• Length = 40cm.
• Breadth = 9cm.

$\hookrightarrow$ 2(40 + 9)

$\hookrightarrow$ 2 × 49

$\hookrightarrow$ 98 cm

Hence,

The perimeter of the rectangle is 98 cm.

Answer 2

$\star\:\:\:\bf\large\underline\blue{Given:—}$

⍟ Length of the rectangle,l = 40 cm

⍟ Diagonal of the rectangle, d = 41 cm

$\star\:\:\:\bf\large\underline\blue{To\:find:—}$

Perimeter of the rectangle = ?

$\star\:\:\:\bf\underline\blue{Solution:—}$

Let breadth be ' b' cm

As we know that,

$\boxed{\bf{\red{l²+b²=d²}}}$

➪ 40²+b² = 41² cm

➪ b² = 81 cm

➪ b = 9 cm

Now,

Perimeter of the rectangle is

$\boxed{\bf{\red{2(l+b)}}}$

= 2 (40+9) cm

= 98 cm

Therefore, perimeter of the rectangle is 98 cm.

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