Question

Find the perimeter of a rectangle with the diagonal 13 cm and length of one side is 12cm
a .33
b.34
c.35
d.36
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Answer 1

Given :-

• Diagonal of rectangle = 13 cm
• Length of rectangle = 12cm

To find :-

• Perimeter of rectangle

Solution :-

As we know that

★ Diagonal = √(breadth)² + (length)²

➟ D = √b² + l²

➟ 13 = √b² + (12)²

➟ 13 = √b² + 144

➟ (13)² = b² + 144

➟ 169 = b² + 144

➟ b² = 169 - 144

➟ b² = 25

➟ b = √25

➟ b = ± 5

★ Breadth never in negative ★

Hence, breadth of rectangle is 5 cm

Now, perimeter of rectangle

★ 2(length + breadth)

➟ 2(12 + 5)

➟ 2 × 17

➟ 34 cm

Hence,

• Perimeter of rectangle is 34 cm

Additional Information

• Area of rectangle = length × breadth
• Area of rhombus = ½ × product of diagonals
• Area of trapezium = ½ × sum parallel sides × height
• Area of Parallelogram = base × height
• Perimeter of square = 4 × side
• Area of square = side × side

Answer 2

QUESTION:-

✯Find the perimeter of a rectangle with the diagonal 13 cm and length of one side is 12cm

ANSWER:-

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow diagonal=13cm$

$\sf\dashrightarrow length\:of\:rectangle=12cm$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow perimeter\:of\:rectangle.$

USING FORMULA,

$\sf\large\:therefore diagonal\:of\:the\:rectangle = \sqrt{(breadth)^2+(length)^2}$

$\sf\therefore 13= \sqrt{(b)^2+(12)^2}$

$\sf\therefore 13= \sqrt{(b)^2+144}$

$\sf\therefore (13)^2= (b)^2+144$

$\sf\therefore 169=(b)^2+144$

$\sf\therefore 169-144=b^2$

$\sf\therefore b^2= 25$

$\sf\therefore b= \sqrt{25}$

$\sf\therefore b= \pm 5$

$\sf\therefore note:- \:breadth\:is\:never\:negative,$

therefore,

$\sf\dashrightarrow breadth\:of\:rectangle\:is\:5\:cm$

$\large{\boxed{\bf{\therefore breadth=5cm}}}$

NOW,

WE KNOW,

BREADTH =5CM

LENGTH=12CM

$\sf\therefore finding\:the\:perimeter\:of\:rectangle,$

USING FORMULA,

$\sf\large\therefore perimeter=2 \times (length+breadth)$

$\sf\implies 2 \times (12+5)$

$\sf\implies 2 \times (17)$

$\sf\implies 34cm$

$\large{\boxed{\bf{\therefore perimeter\:of\:rectangle\:is\:34cm}}}$

__________________

✯ADDITIONAL INFORMATION,

✯DIAGRAM OF RECTANGLE,

$\setlength{\unitlength}{1.6mm}\begin{picture}(5,6)\put(0,25){\line(1,0){35}}\put(0,0){\line(1,0){35}}\put(0,0){\line(0,1){25}}\put(35,0){\line(0,1){25}}\put(17,-2){5cm}\put(18,-2){}\put(35,15){34cm}\put(18,0){ }\end{picture}$

FEW FORMULAS FOR LEARNING,

• AREA OF SQUARE=(SIDE)²
• PERIMETER OF SQUARE = 4 ×(SIDE)
• AREA OF TRIANGLE=1/2 × base ×height
• AREA OF TRAPEZIUM= 1/2 × (A+B)
• VOLUME OF CUBE =(L)³
• VOLUME OF CUBOID= L×B×H
• VOLUME OF CONE = 1/3 πr²h
• VOLUME OF CYLINDER=πr²h

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