Answer

Verified

455.4k+ views

**Hint:**Use perimeter to find the length and breadth of the given rectangle. With the length and breadth of the rectangle, diagonal can be easily found by using Pythagoras theorem on the triangle of sides consisting of length, breadth, and diagonal.

We know that the opposite sides of the rectangle are equal. The angle between two adjacent sides is ${90^ \circ }$.

The angle between two adjacent sides of a rectangle is $90^\circ $.

Remember that: Pythagoras theorem can be applied to every right-angled triangle.

Pythagoras theorem: square of the hypotenuse is equal to the sum of the square of base and square of perpendicular.

In a right-angled triangle, base and perpendicular are at the angle of $90^\circ $ to each other and hypotenuse is the longest side.

The comparison of two quantities in terms of ‘how many times’ is known as a ratio.

For example: There are 24 girls and 16 boys in a class. Ratio of the numbers of girls to the numbers of boys $ = \dfrac{{24}}{{16}} = \dfrac{3}{2}= 3:2$

Therefore, we can get equivalent ratios by multiplying or dividing the numerator and denominator by the same number.

A ratio can be written as a fraction, thus the ratio $3:17$ can be written as $\dfrac{3}{{17}}$.

**Complete step by step solution:**

Step 1: Draw the rectangle ABCD

Properties of rectangle:

LENGTH: Side $AB$ = side $CD$

BREADTH: Side $BC$ = side $AD$

Side $AB \ne $side $BC$

All the interior angles of the rectangle ABCD $ = {90^ \circ }$

Step 2: Given that

Perimeter of rectangle $ABCD = 24cm$

Ratio of two unequal sides of rectangle $ABCD = 1:2$

$ \Rightarrow \dfrac{{{\text{length}}}}{{{\text{breadth}}}} = \dfrac{1}{2}$

Step 3: multiply numerator and denominator by $x$

$ \Rightarrow \dfrac{{{\text{length}}}}{{{\text{breadth}}}} = \dfrac{x}{{2x}}$

$ \Rightarrow \dfrac{{{\text{AB}}}}{{{\text{BC}}}} = \dfrac{x}{{2x}}$ (from step 1)

$

\Rightarrow {\text{AB}} = x{\text{ cm}} \\

{\text{ BC}} = 2x{\text{ }}cm \\

$

Step 4: find length and breadth of rectangle using parameter of rectangle

Given Parameter of rectangle = 24 cm

$2 \times \left( {{\text{length}} + {\text{breadth}}} \right) = 24$

$ \Rightarrow 2 \times \left( {{\text{AB}} + {\text{BC}}} \right) = 24$

On substituting the corresponding values,

$ \Rightarrow 2 \times \left( {x + 2x} \right) = 24 $

On simplifying expressions,

$ \Rightarrow {\text{ }}2\left( {3x} \right) = 24$

$ \Rightarrow {\text{ }}6x = 24 $

On further simplification, we get

$ \Rightarrow {\text{ }}x = \dfrac{{24}}{6} $

$ \Rightarrow x = 4cm$

Hence, length $AB = x = 4 cm$

And breadth $BC = 2x = 2(4) = 8 cm$

Step 5: Calculate diagonal BD using Pythagoras theorem

Join the Vertex BD, i.e. diagonal of rectangle ABCD.

We know, $\angle BCD = {90^ \circ }$,

$\vartriangle BCD$ is a right angled triangle, right angle at vertex C.

Apply Pythagoras theorem on $\vartriangle BCD$.

i.e. Pythagoras theorem: square of the hypotenuse is equal to the sum of the square of base and square of perpendicular.

$\mathop {{\text{Hypotenuse}}}\nolimits^{\text{2}} {\text{ = }}\mathop {{\text{ Base}}}\nolimits^{{\text{2 }}} {\text{ + }}\mathop {{\text{ Perpendicular}}}\nolimits^{\text{2}} $

$\mathop {{\text{BD}}}\nolimits^2 = \mathop {{\text{BC}}}\nolimits^2 + \mathop {{\text{CD}}}\nolimits^2 $

On substituting the corresponding values,

\[\mathop {{\text{BD}}}\nolimits^2 = \mathop 8\nolimits^2 + \mathop 4\nolimits^2 \] ( $\because $AB = CD = 4 cm)

$ \mathop { \Rightarrow {\text{BD}}}\nolimits^2 = 64 + 16 $

On simplification of the above values,

$\Rightarrow \mathop {{\text{BD}}}\nolimits^2 = 80 $

$\Rightarrow {\text{BD = }}\sqrt {80} $

On further simplification, we get

$ \Rightarrow {\text{BD = 4}}\sqrt 5 $

**$\therefore$ The length of the diagonal of a given rectangle is $4\sqrt 5 cm$. The correct option is (C).**

**Note:**

Area of rectangle = length $ \times $ breadth sq. units

The diagonal of the rectangle bisects its area.

Students may get confused in labeling hypotenuse, perpendicular, and base sides of the right-angled triangle. Remember that the hypotenuse is the longest side in the right-angled triangle. It is the side opposite to the right angle (or \[{90^ \circ }\])

Pythagoras theorem only applies to the right-angled triangle, not to every triangle.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

What percentage of the solar systems mass is found class 8 physics CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you graph the function fx 4x class 9 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Why is there a time difference of about 5 hours between class 10 social science CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE