Questions & Answers

Question

Answers

A) 15 m

B) 16 m

C) 17 m

D) 20 m

Answer
Verified

The area of the rectangle is given by:-

\[area = length \times breadth\]

The perimeter of the rectangle is given by:

\[perimeter = 2(length + breadth)\]

The Pythagoras theorem states that the sum of squares of the base and height of a triangle is equal to the square of its hypotenuse.

\[{\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {height} \right)^2}\]

It is given that area of the rectangle is \[120{m^2}\]

The perimeter of the rectangle is $46m$.

Let ABCD be the rectangle with length=\[l\] and breadth= \[b\]

Then, the area of the rectangle is given by:-

\[area = length \times breadth\]

Therefore putting in the values we get:-

\[120 = l \times b.....................\left( 1 \right)\]

Also, the perimeter of the rectangle is given by:

\[perimeter = 2(length + breadth)\]

Putting in the values we get:-

\[ 46 = 2\left( {l + b} \right) \]

$ l + b = \dfrac{{46}}{2} $

On simplification,

$ l + b = 23.............\left( 2 \right) $

Now solving equations 1 and 2 we get:-

From equation 2 we get:-

\[b = 23 - l.....................\left( 3 \right)\]

Putting this value in equation 1 we get:-

\[

l\left( {23 - l} \right) = 120 \\

23l - {l^2} = 120 \\

{l^2} - 23l + 120 = 0 \\

\]

Solving this quadratic equation by middle term split we get:-

\[ {l^2} - 15l - 8l + 120 = 0 \]

\[ l\left( {l - 15} \right) - 8\left( {l - 15} \right) = 0 \]

on simplification,

\[ \left( {l - 15} \right)\left( {l - 8} \right) = 0 \]

\[ l = 15;\, l = 8 \]

Putting the value of $l$ back in equation 3 we get:-

When $l=15m$

Then,

\[b = 23 - 15 \]

$\Rightarrow b = 8 $

When $l=8m$

\[\Rightarrow b = 23 - 8 \]

$\Rightarrow b = 15$

Now applying Pythagoras theorem in \[\Delta ABC\] we get:-

Since the formula of Pythagoras is given by:-

\[{\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {height} \right)^2}\]

Therefore,

\[{\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2}\]

Which implies:

\[{\left( {diagonal} \right)^2} = {\left( {length} \right)^2} + {\left( {breadth} \right)^2}\]

Now putting in the values we get:-

\[\Rightarrow {\left( {diagonal} \right)^2} = {\left( 8 \right)^2} + {\left( {15} \right)^2} \]

$\Rightarrow {\left( {diagonal} \right)^2} = 64 + 225$

$\Rightarrow {\left( {diagonal} \right)^2} = 289 $

$\Rightarrow diagonal = \sqrt {289} $

$\Rightarrow diagonal = 17m $

The student can also apply quadratic formula to solve the quadratic equation and find the value of \[l\]

For any quadratic equation of the form \[a{x^2} + bx + c = 0\]

The roots of the equation are given by:-

\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Hence applying this formula for the equation

\[{l^2} - 23l + 120 = 0\]

We get:-

\[

l = \dfrac{{ - \left( { - 23} \right) \pm \sqrt {{{\left( { - 23} \right)}^2} - 4\left( 1 \right)\left( {120} \right)} }}{{2\left( 1 \right)}} \\

l = \dfrac{{23 \pm \sqrt {529 - 480} }}{2} \\

l = \dfrac{{23 \pm \sqrt {49} }}{2} \\

l = \dfrac{{23 \pm 7}}{2} \\

l = \dfrac{{23 + 7}}{2}or{\text{ }}l = \dfrac{{23 - 7}}{2} \\

l = \dfrac{{30}}{2}or{\text{ }}l = \dfrac{{16}}{2} \\

l = 15cm{\text{ }}or{\text{ }}l = 8cm \\

\].