find the perimeter ofrectangle whose length is 40 cm and a diagonal is 41 cm
Answers
Answer:
98 cm
Step-by-step explanation:
let ABCD be rectangle.
Then AC will be diagonal. And angle B = 90.
Length(BC) = 40 cm
diagonal(AB) = 41 cm
By using pythagoras theorem, we have:-
AB^{2} + BC^{2} = AC^{2}AB
2
+BC
2
=AC
2
x^{2} + 40^{2} = 41^{2}x
2
+40
2
=41
2
AB^{2} = 41^{2} - 40^{2}AB
2
=41
2
−40
2
AB^{2} = 1681 - 1600 = 81AB
2
=1681−1600=81
AB = \sqrt{81} = 9 cmAB=
81
=9cm
Now, Perimeter of rectangle = 2(L+B) = 2(BC+ AB)
= 2(9 + 40) = 2(49)
= 98 cm
Answer:
98 cm
Step-by-step explanation:
As we can see in the rectangle,
Now as given in the question,
Diagonal = 41 cm.
Length = 40 cm.
So, Putting these value we get,
Hence the width of the rectangle is 9 cm.
So
The perimeter of the rectangle = 2 ( Length + Width )
= 2 ( 40 cm + 9 cm )
= 2 x 49 cm
= 98 cm
Hence the perimeter of the rectangle is 98 cm.
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