Question

Find the peri meter of rectangle whose length is 40cm and a diagonal is 41 cm

Answer 1

by Pythagoreas theorem
(h)^2 = (p) ^2+(b) ^2. (in which
h= 41 CM and
b= 40cm)
(41)^2 =(p)^2 +(40)^2
1681 = (p)^2 + 1600


(p) ^2 =1681 - 1600
(p) ^2 = 81
p = √81
p = 9
so breadth = 9cm
perimeter of rectangle = 2(l + b)
= 2(40 +9)
= 2*49
= 98 cm

Answer 2

Given,
Length=40cm
Diagonal=41cm


So, Draw a figure according to question as shown in figure.
In triangle abc
AB=40cm
CD=41cm
AC^2=CB^2-AB^2 (Phytagorean fomula)
AC=rootCB^2-rootAB^2
AC=ROOT 41^2-40^2
AC=Root 1681 cm^2-root 1600 cm^2
AC=root 81 cm^2
AC=9cm



So, now breadth= 9cm
Hence Perimeter of rectangle=2(length+breadth)
=2(40cm +9cm)
=2*49cm
=98cm.

I hope you can understand this question.