Question

5. The length of a rectangle is three times of its width. If the length of the diagonal is 8 10 m,
then the perimeter of the rectangle is
(a) 15/10 m
(b) 16/10 m
(c)
o
(d) 64 m​

Answer 1

Answer:

let be the width of rectangle x m and is length is 3x m.

so,x²+(3x)²=810

⇒x²+9x²=810

⇒10x²=810

⇒x²=81

⇒x=9

hence,its length is 3x=27m and width is x=9m.

now,the perimeter of given rectangle is 2(l+b)=2(27+9)=2×36=72m .

HERE IS YOUR RIGHT ANSWER.

Step-by-step explanation:

Answer 2

Answer:

• The perimeter of the rectangle is 64 m.

Step-by-step explanation:

Given that:

• The length of a rectangle is three times of its breadth.
• The length of the diagonal of the rectangle is 8√10 m.

To Find:

• Perimeter of the rectangle.

Let us assume:

• Breadth of the rectangle is x.

Then,

• Length of the rectangle will be 3x.

Now,

• The diagonal, length and breadth of the rectangle together forms a right angled triangle, with diagonal as hypotenuse, length as base and breadth as height.

As we know that:

$\sf{\circ\;(Hypotenuse)^2 = (Base)^2 + (Height)^2}$

Substituting the values,

$\sf{\longmapsto(8\sqrt{10})^2 = (3x)^2 + (x)^2}$

$\sf{\longmapsto(\sqrt{8\times8\times10})^2 = (3x)^2 + (x)^2}$

$\sf{\longmapsto(\sqrt{640})^2 = (3x)^2 + (x)^2}$

Opening the brackets,

$\sf{\longmapsto640 = 9x^2 + x^2}$

Adding 9x² and x²,

$\sf{\longmapsto640 = 10x^2}$

Transposing 10 from RHS to LHS and changing its sign,

$\sf{\longmapsto\dfrac{640}{10} = x^2}$

Dividing 640 by 10,

$\sf{\longmapsto64 = x^2}$

Solving further,

$\sf{\longmapsto\sqrt{64} = x}$

$\sf{\longmapsto8 = x}$

$\sf{\longmapsto x=8}$

Hence,

• x = 8

Therefore,

• Length of the rectangle = 3x = 3 × 8 = 24 m
• Breadth of the rectangle = x = 8 m

Finding perimeter of the rectangle:

As we know that,

• Perimeter of rectangle = {2(length + breadth)} sq. units

Substituting the values,

Perimeter = {2(24 + 8)} m²

= {2 × 32} m²

= 64 m²

Hence, perimeter of the rectangle is 64 m.