Question

C) Solve the following Questions
4m
1. If diagonal of a rectangle is 26 cm and one side is 24 cm, find the
other side.​

Answer 1

Answer:

Other side of the rectangle or AD = 10 cm

Step-by-step explanation:

Given :

Diagonal of a rectangle = 26 cm

One side of the rectangle = 24 cm

To find :

The other side of the rectangle

Taken :

To find the other side of the rectangle use the Pythagoras theorem .

According to the attachment -:

Let the AB = 24 cm

AC = 26 cm

And AD = other side of the rectangle

So , the formula will like this -:

$\boxed{\bf{{AB}^{2}+{AD}^{2}={AC}^{2}}}$

Solution :

$:\Rightarrow\bf{{24cm}^{2}+{AD}^{2}={26cm}^{2}}$

$:\Rightarrow\bf{{AD}^{2}={26cm}^{2}-{24cm}^{2}}$

$:\Rightarrow\bf{AD=\sqrt{{26cm}^{2}-{24cm}^{2}}}$

$:\Rightarrow\bf{AD=\sqrt{(26\times26)cm-(24\times24)cm}}$

$:\Rightarrow\bf{AD=\sqrt{676cm-576cm}}$

$:\Rightarrow\bf{AD=\sqrt{100cm}}$

$:\Rightarrow\bf{AD=10cm}$

So , the other side of the rectangle or AD is 10 cm .

Answer 2

Question :-

• If diagonal of a rectangle is 26 cm and one side is 24 cm, find the
• other side.

Diagram :-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,2.74){\framebox(0.25,0.25)}\put(4.74,0.01){\framebox(0.25,0.25)}\put(2,-0.7){\sf\large 24 cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\qbezier(0,0)(0,0)(5,3)\put(1.65,1.8){\sf\large 26 cm}\end{picture}$

Given :-

• Diagonal ( Hypotenuse ) = 26cm
• One side ( Base ) = 24cm

To Find :-

• Second Side ( Perpendicular ) ?

Solution :-

★ Using Pythagoras Theorem ★

⇒ ( AC )² = ( AB )² + ( AD )²

⇒ ( 26 )² = ( 24 )² + ( AD )²

⇒ 676 = 576 + ( AD )²

⇒ 676 - 576 = ( AD )²

⇒ 100 = ( AD )²

⇒ √100 = AD

⇒ 10 = AD

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Verification :-

⇒ ( AC )² = ( AB )² + ( AD )²

⇒ ( 26 )² = ( 24 )² + ( 10 )²

⇒ 676 = 576 + 100

⇒ 676 = 676

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∴ The Second Side is 10cm

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More For Knowledge :-

Formulas Related to Rectangle

• Perimeter = 2 ( l + b )
• Area = l × b
• Length = Area/Breadth
• Breadth = Area/Length
• Diagonal = √ l² + b²

Where l and b stands for Length and Breadth respectively.

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