Question

the length of rectangle is 4 times of its breadth the area of rectangle is 1600 m square find the diagonal of rectangle​

Answer 1

Answer:

$\large{\underline{\boxed{\mathfrak\blue{Diagonal \: of \: rectangle = 82.46 \: m }}}}$

Reference of diagram is given below:-

$\setlength{\unitlength}{0.78 cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ A }\put(5.5,5.8){ B }\put(11.1,5.8){ C }\put(11.05,9.1){ D }\put(4.5,7.5){ 20\:m }\put(8.1,5.3){ 80 \:m }\put(11.5,7.5){ 20 \:m }\put(8.1,9.5){ 80\:m }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(6,6){\line(5,3){5}}\end{picture}$

Given:-

• Length(AB/CD) = 4 × breadth
• Area of rectangle = 1600 sq.m

To find:-

• Diagonal of rectangle (AC/BD) =?

Now, let breadth be x m and length be 4x m.

$\rm We \:know, \\\\\dag \: {\boxed{\rm{Area_{rectangle} = Length \times Breadth }}}$

A/Q,

$\dashrightarrow\sf 4x \times x = 1600 \\\\\dashrightarrow\sf 4x^2 = 1600 \\\\\dashrightarrow\sf x^2 = 1600/4 \\\\\dashrightarrow\sf x^2 = 400 \\\\\dashrightarrow\sf x = \sqrt{400} \\\\\dashrightarrow\sf x ={\boxed{\sf\green{20 \: m }}}$

$\rule{100}{2}$

$\dashrightarrow\sf Breadth = x = {\boxed{\sf\green{20\: m }}}$

$\dashrightarrow\sf Length = 4x = 4(20) ={\boxed{\sf\blue{80\: m }}}$

$\rule{200}{1}$

$\star{\underline{\sf{According \:to \: Pythagoras \: theorem :- }}}$

$\dashrightarrow\sf BD^2 = CD^2 + BC^2 \\\\\dashrightarrow\sf BD^2 = (80)^2 + (20)^2 \\\\\dashrightarrow\sf BD^2 = 6400 + 400 \\\\\dashrightarrow\sf BD^2 = 6800 \\\\\dashrightarrow\sf BD = \sqrt{6800} \\\\\dashrightarrow{\underline{\boxed{\sf\blue{BD = 82.46 \: m }}}}$

$\therefore{\underline{\sf\red{Diagonal \: of \: rectangle = 82.46 \: m }}}$

Answer 2

$\huge{\underline{\underline{\bf{Solution}}}}$

$\rule{200}{2}$

$\tt Given\begin{cases} \sf{Area \: of \: rectangle \: 1600 \: m^2. } \\ \sf{Length \: of \: rectangle \: is \: 4 \: times \: of \: its \: breadth} \end{cases}$

$\rule{200}{2}$

$\Large{\underline{\underline{\bf{To \: Find :}}}}$

We have to find the Diagonal of the rectangle.

$\rule{200}{2}$

$\Large{\underline{\underline{\bf{Explanation :}}}}$

Let the breadth of rectangle be x.

So, length of rectangle be 4x.

We know that,

$\Large{\star{\boxed{\rm{Area = Length \times Breadth}}}}$

________________[Put Values]

$\tt{→ 4x \times x = 1600} \\ \\ \tt{→4x^2 = 1600} \\ \\ \tt{→x^2 = \frac{\cancel{1600}}{\cancel{4}}} \\ \\ \tt{→x^2 = 400} \\ \\ \tt{→x = \sqrt{400}} \\ \\ \tt{→x = (\sqrt{20})^2} \\ \\ \tt{→x = \pm 20}$

As, breadth and length if rectangle can't be negative.

So, Breadth = + 20 m

Length = 4(+20) = 80 m

$\rule{200}{2}$

As, we have to find diagonal. We can apply Pythagoras theorem

$\Large{\boxed{\star{\rm{(Diagonal)^2 = (Length)^2 + (Breadth)^2}}}}$

$\tt{→Diagonal = \sqrt{(80)^2 + (20)^2}} \\ \\ \tt{→Diagonal = \sqrt{6400 + 400}} \\ \\ \tt{→Diagonal = \sqrt{6800}} \\ \\ \tt{→diagonal = 82.46}$

$\Large{\implies{\boxed{\boxed{\sf{Diagonal = 82.46 \: m}}}}}$