Question

find the area and breath of rectangle its length 8 m and perimeter is25 m​

Answer 1

Area of rectangle = L*B

.°. L*B = 25

=) 8*B = 25

=) B = 25/8

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Answer 2

SOLUTION

Length of the rectangle = 8m.

Perimeter of the rectangle = 25m.

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 8m}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large b}\put(-0.5,-0.4){\bf }\put(-0.5,3.2){\bf }\put(5.3,-0.4){\bf }\put(5.3,3.2){\bf }\end{picture}$

Perimeter of a rectangle = 2(l + b) , Where l is the length and b is the breadth of the rectangle.

$\implies$ 25 = 8 × b

$\implies$ 25 = 8b

$\implies \sf b = \dfrac{25}{8}$

$\implies$ b = 3.125 m.

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 8m}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 3.125}\put(-0.5,-0.4){\bf }\put(-0.5,3.2){\bf }\put(5.3,-0.4){\bf }\put(5.3,3.2){\bf }\end{picture}$

Area of a rectangle = l × b , Where l is the length and b is the breadth of the rectangle.

Here,

l = 8m.

b = 3.125 m.

$\implies$ Area of the rectangle = 8 × 3.125

$\implies$ Area of the rectangle = 25m.