Question

The diagonal of the rectangle is square root of 73. If the width is equals to 3cm, what is the semi-perimeter of the rectangle?

Answer 1

GiveN:-

The diagonal of the rectangle is square root of 73. If the width is equals to 3cm,

To FinD:-

What is the semi-perimeter of the rectangle?

SolutioN:-

We know that,

$\large{\green{\underline{\boxed{\bf{Diagonal=\sqrt{(Length)^2+(Breadth)^2}}}}}}$

where,

• Breadth = 3 cm.
• Diagonal = √73 cm.

Putting the values,

$\large\implies{\sf{\sqrt{73}=\sqrt{(Length)^2+(3)^2}}}$

Squaring both the sides,

$\large\implies{\sf{73=(Length)^2+(3)^2}}$

$\large\implies{\sf{73=(Length)^2+9}}$

$\large\implies{\sf{73-9=(Length)^2}}$

$\large\implies{\sf{64=(Length)^2}}$

Square rooting both the sides,

$\large\implies{\sf{\sqrt{64}=Length}}$

$\large\implies{\sf{8=Length}}$

$\large\therefore\boxed{\bf{Length=8.}}$

VerificatioN:-

$\large\implies{\sf{Diagonal=\sqrt{(Length)^2+(Breadth)^2}}}$

$\large\implies{\sf{\sqrt{73}=\sqrt{(8)^2+(3)^2}}}$

Squaring both the sides,

$\large\implies{\sf{73=(8)^2+(3)^2}}$

$\large\implies{\sf{73=64+9}}$

$\large\implies{\sf{73=73}}$

$\large\therefore\boxed{\bf{LHS=RHS.}}$

• Hence verified.

Now the Semiperimeter:-

We know that,

$\large{\green{\underline{\boxed{\bf{Perimeter=2(Length+Breadth)}}}}}$

where,

• Length = 8 cm
• Breadth = 3 cm

Putting the values,

$\large\implies{\sf{Perimeter=2(8+3)}}$

$\large\implies{\sf{Perimeter=2\times11}}$

$\large\therefore\boxed{\bf{Perimeter=22\:cm.}}$

●Semiperimeter:-

$\large\boxed{\sf{Semiperimeter=\dfrac{Perimeter}{2}}}$

$\large\implies{\sf{Semiperimeter=\dfrac{22}{2}}}$

$\large\implies{\sf{Semiperimeter=\dfrac{\cancel{22}}{\cancel{2}}}}$

$\large\therefore\boxed{\bf{Semiperimeter=11\:cm.}}$

Semiperimeter of the rectangle is 11 cm.