Question

find the perimeter of rectangle whose whose whose length is 40 cm and diagonal is 41 CM

Answer 1

L = 40 cm
let the breadth be B

diagonal = 41 cm
41 = $\sqrt{L ^{2} + B^{2}}$
41 = $\sqrt{40^{2} + B^{2}}$
B^{2} = 40^{2} - 41^{2}
B^{2} = 1681-1600
B^{2} = 81.
B = 9.

P = 2(L+B)
= 2(40+9)
= 2×49
= 98 cm.

hence, the perimeter of the rectangle is 98 cm.

thanks

Answer 2

length = 40 cm
let,
breadth be = B cm
diagonal = 41cm
             41 =sqrt{ L^{2} + B^{2} }
             41=sqrt{ 40^{2}+ B^{2} }
           squaring both the sides
         41^{2} = 40^{2} + B^{2}
so, B^{2} = 41^{2} - 40^{2}
               =1681-1600
               =81
          taking sqrt on both the sides
 so,
B=9 cm
so,
breadth = 9 cm
now, perimeter of rectange = 2(length + breadth)
                                            =2(40+9)
                                            =2(49)
                                            =98 cm

HOPE IT WOULD HELP
if so, pls mark it as the brainliest