Question

the area of a rectangular carpet is 120 m sq and its perimeter is 46m . find the length of its diagonal ?​

Answer 1

Given:-

Area of rectangular carpet=$120 {m}^{2}$

And perimeter= 46 m

To find:-

Length of it's diagonal

Answer:-

Length of it's diagonal= 17 cm

Solution:-

$area \: of \: rectangle = 120 \: {m}^{2}$

$l \times b = 120 \: {m}^{2}$

Perimeter= 46 m

2(l+b)= 46 m

$(l + b) = \frac{46}{2}$

$(l + b) = 23$

Squaring on both sides

${(l +b)}^{2} = {(23)}^{2}$

${l}^{2} + 2lb +{b}^{2} = 529$

${l}^{2} + 2(120) +{b}^{2} = 529 \: \: \: {(as \: lb = 120 {m}^{2} )}$

${l}^{2} + 240 + {b}^{2} = 529$

${l}^{2} + {b}^{2} = 529 - 240$

${l}^{2} + {b}^{2} = 289$

Now;

We know that

$diagonal \: of \: rectangle = \sqrt{ {l}^{2} + {b}^{2} }$

$= \sqrt{289}$

$= 17 \: m$

___________________

Length of it's diagonal= 17 m

Answer 2

Answer is 17..

Solutions:-

According to the question,

Area of rectangular carpet = 120 m sq

Perimeter = 46 m

Length of its diagonal =?

We know that,

Area of rectangle = l × b

120 = l × b -----------(1)

Perimeter = 2(l+b)

46 = 2(l + b)

23 = l + b ---------(2)

Squaring both sides of (2)

$( {l + b})^{2} = ( {23})^{2}$

${l}^{2} + {b}^{2} + 2lb = 529$

Putting the value of lb from (1)

${l}^{2} + {b}^{2} + 2(120) = 529$

${l}^{2} + {b}^{2} + 240 = 529$

${l}^{2} + {b}^{2} = 529 - 240$

${l}^{2} + {b}^{2} = 289$

Now,

Diagonal of rectangle = $\sqrt{ {l }^{2} + {b}^{2} }$

$= \sqrt{289} = 17$