How to find area of rectangle when perimeter 14 and diagonal 5 is given?
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See bro....
Area of rectangle=length × breadth
Now the real game begins....,
Perimeter of rectangle=2×(l+b)
14cm=2(AB+BC) (Because AB=l and BC=b)
So,
AB + BC=7cm........ (1)
Now, As ABCD is a rrectangle....
Then ∆ABC will be a right angled triangle.....
So,
By pythogoras theorem...
AC²=BA²+BC²
5²=AB²+BC²
25=AB²+BC².... (2)
Now,
we know that,
(a+b)²=a²+b²+2ab
So,
(AB+BC)²=AB²+BC²+2AB×BC
From (1) and (2)...
7²=25+2AB×BC
49=25+2AB×BC
2AB×BC=49-25=24
AB×BC=12
And thats the area brother
Area of rectangle=length × breadth
Now the real game begins....,
Perimeter of rectangle=2×(l+b)
14cm=2(AB+BC) (Because AB=l and BC=b)
So,
AB + BC=7cm........ (1)
Now, As ABCD is a rrectangle....
Then ∆ABC will be a right angled triangle.....
So,
By pythogoras theorem...
AC²=BA²+BC²
5²=AB²+BC²
25=AB²+BC².... (2)
Now,
we know that,
(a+b)²=a²+b²+2ab
So,
(AB+BC)²=AB²+BC²+2AB×BC
From (1) and (2)...
7²=25+2AB×BC
49=25+2AB×BC
2AB×BC=49-25=24
AB×BC=12
And thats the area brother
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