Aptitude - Area - Discussion
Discussion Forum : Area - General Questions (Q.No. 9)
9.
The diagonal of a rectangle is 41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:
Answer: Option
Explanation:
l2 + b2 = 41.
Also, lb = 20.
(l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81
(l + b) = 9.
Perimeter = 2(l + b) = 18 cm.
Discussion:
15 comments Page 2 of 2.
Silmn said:
8 years ago
Why (a + b)^2 formula is used here?
(1)
Prabir said:
7 years ago
How +40 came?
It should be -40 isn't it?
It should be -40 isn't it?
(1)
EZHIL said:
7 years ago
By PYTHAGORAS THEOREM;
L^2+B^2=D^2,
=5^2+4^2 =41^2,
=5X4=20.
Then;
The perimeter of Rectangle =2(5+4)=40.
L^2+B^2=D^2,
=5^2+4^2 =41^2,
=5X4=20.
Then;
The perimeter of Rectangle =2(5+4)=40.
(3)
Yazhini said:
4 years ago
How +40 came?
l^2 + b^2 + 2(lb) = 41.
l^2 + b^2 + 2(20) =41.
Then it should be like
l^2 + b^2 = 41 - 40,
l^2 + b^2 = 1.
How 81 will come?
I am not able to understand this. Please, anyone, help me to get it.
l^2 + b^2 + 2(lb) = 41.
l^2 + b^2 + 2(20) =41.
Then it should be like
l^2 + b^2 = 41 - 40,
l^2 + b^2 = 1.
How 81 will come?
I am not able to understand this. Please, anyone, help me to get it.
(1)
Hii said:
3 months ago
Solution.
As we know the perimeter of a rectangle is = 2 (l+b).
So, As we have. Area = 20m and by applying Pythagoras (l^2+b^2=41).
So, the parameters of rectangle= 2 (l+b) ^2 =4 (l^2+b^2+2lb)
= 4 (41+40)
= (81×4)
={324}.
P^2= 324.
Then P= 18m.
As we know the perimeter of a rectangle is = 2 (l+b).
So, As we have. Area = 20m and by applying Pythagoras (l^2+b^2=41).
So, the parameters of rectangle= 2 (l+b) ^2 =4 (l^2+b^2+2lb)
= 4 (41+40)
= (81×4)
={324}.
P^2= 324.
Then P= 18m.
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