Question

the area of a rectangle having breadht 20 cm is 1200 cm2 what is the perimeter of the rectngle​

Answer 1

Answer:

area of rectangle =l ×b=1200cm²

l× 20=1200

l=1200/20=60cm

perimeter of rectangle =2(l+b)=2(60+20)=2×80=160cm

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Answer 2

Correct Question-:

• The area of a rectangle having breadth 20 cm is 1200 cm² . What is the perimeter of the rectangle ?

AnswEr-:

• Perimeter of Rectangle is 160 cm

Given -:

• The area of Rectangle is 1200 cm² .
• The breadth of Rectangle is 20 cm .

To Find -:

• Perimeter of Rectangle.

Solution-:

• Area of Rectangle is -: Area = length × breadth .

Here ,

• Length of Rectangle is = ??
• Breadth of Rectangle is = 20 cm .
• Area of Rectangle is = 1200 cm ²

Now ,

• 20 × length = 1200 cm²
• length = 1200 / 20
• length = 60 cm.

Therefore,

• Length of Rectangle is 60 cm .

As ,We know That ,

• Perimeter of Rectangle -: Perimeter = 2( length + breadth) .

Here ,

• Length of Rectangle is = 60 cm
• Breadth of Rectangle is = 20 cm

Now ,

• 2 (60 + 20)
• 2 × 80
• 160 cm

Therefore,

• Perimeter of Rectangle is 160 cm .

• ☆ Figure related to answer to the question -:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 60 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 20 cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

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♤ More To Know ♤

• Rectangle-: A Rectangle is a type of Quadrilateral whose opposite sides are parallel and equal to each other.

• ☆ Figure of Rectangle-:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large x cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large y cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

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