Question

The length and width of the rectangle are in the ratio \[{\text{3:4}}\]. If the rectangle has an area of \[{\text{300c}}{{\text{m}}^{\text{2}}}\], what is the length of the diagonal ?
A.\[{\text{15cm}}\]
B.\[{\text{12cm}}\]
C.\[{\text{25m}}\]
D.\[{\text{25cm}}\]

Answer 1

Hint:Let the length and width of the rectangle be x and y . As we know the formula to calculate the area of the rectangle which is \[{\text{A = x}}{\text{.y}}\] and ratio of x and y is also given. So , finally we just have to calculate the length of the diagonal after finding out the value of x and y.

Complete step-by-step answer:

Here we can see that as \[\vartriangle {\text{ABD is }} \bot \](right angle triangle) , so we can calculate length of triangle directly by applying Pythagoras theorem in \[\vartriangle {\text{ABD}}\]. So, length of diagonal is \[\sqrt {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}} \].
As area of rectangle is \[{\text{300c}}{{\text{m}}^{\text{2}}}\]and the ratio of sides is \[{\text{3:4}}\],
Hence, \[\dfrac{{\text{x}}}{{\text{y}}}{\text{ = }}\dfrac{{\text{3}}}{{\text{4}}}\] and \[{\text{x}}{\text{.y = 300}}\]
While solving both the equation we can simplify them as
\[
  {\text{4x = 3y}} \\
  {\text{x}}{\text{.y = 100}} \\
  \]
Substitute the value of x from one equation and put it into the another equation,
\[ \Rightarrow {\text{x = }}\dfrac{{{\text{3y}}}}{{\text{4}}}\]
On substituting in \[{\text{x}}{\text{.y = 300,}}\]
\[
   \Rightarrow \dfrac{{{\text{3y}}}}{{\text{4}}}{\text{.y = 300}} \\
   \Rightarrow {{\text{y}}^{\text{2}}}{\text{ = 400}} \\
  \]
On taking root we get,
\[ \Rightarrow y = 20\]
Now , again placing the value of y in the substituted equation in order to obtain value of x.
\[
   \Rightarrow {\text{x = }}\dfrac{{{\text{3(20)}}}}{{\text{4}}} \\
   \Rightarrow {\text{x = 15}} \\
  \]
Hence , on solving both the above equations we can reach up to the conclusion as
\[
  {\text{x = 15cm}} \\
  {\text{y = 20cm}} \\
  \]
Now, length of diagonal is \[\sqrt {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}} \]
\[
  {\text{ = }}\sqrt {{\text{1}}{{\text{5}}^{\text{2}}}{\text{ + 2}}{{\text{0}}^{\text{2}}}} \\
  {\text{ = }}\sqrt {{\text{ 225 + 400}}} \\
  {\text{ = }}\sqrt {{\text{625}}} \\
  {\text{ = }}\sqrt {{\text{2}}{{\text{5}}^{\text{2}}}} \\
  {\text{ = 25cm}} \\
  \]
Hence , option (d) is our required correct answer.

Note: A rectangle has two diagonals. Each one is a line segment drawn between the opposite vertices (corners) of the rectangle. The diagonals have the following properties:
The two diagonals are of the same length. Each diagonal divides the rectangle into two congruent right triangles. Because the triangles are congruent, they have the same area, and each triangle has half the area of the rectangle