There is a rectangle whose length is more than its breadth by
7m and the diagonal is more than the length by 1 m. Find
the length and breadth of the rectangle?
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Answers
Step-by-step explanation:
There is a rectangular onion storehouse in the farm of Mr. Ratnakar at Delhi. The length of rectangular base is more than its breadth by 7 metre and the diagonal is more than length by 1 metre. Find the length and breadth of this storehouse.
Step by step explanation :
Let the breadth of the storehouse be 'x' metres.
Thus, As per your question,
Length = (x + 7) m
Diagonal = (x + 7 + 1) m = (x + 8) m
Thus, We can solve this question by Pythagoras theorem,
\tt{x{}^{2} + (x + 7){}^{2} = (x + 8){}}^{2}x
2
+(x+7)
2
=(x+8)
2
\tt \small{x{}^{2} + x{}^{2} + 14x + 49 = x{}^{2} + 16x + 64}x
2
+x
2
+14x+49=x
2
+16x+64
Cancelling x^2 on both sides,
\tt \small{x{}^{2} + 14x - 16x + 49 - 64 = 0}x
2
+14x−16x+49−64=0
\tt{x{}^{2} -2x - 15 = 0}x
2
−2x−15=0
Thus, Now solving this equation by factorization method.
\tt{x{}^{2}- 5x +3x -15 = 0}x
2
−5x+3x−15=0
\tt{x(x - 5)+3(x - 5)= 0}x(x−5)+3(x−5)=0
\tt{(x+3) =0\:or\:(x-5) =0}(x+3)=0or(x−5)=0
\tt{x=-3 \:or\:x = 5}x=−3orx=5
But,
We know that,
Length is never negative.
\tt{\therefore x≠ - 3}∴x
=−3
So, x = 5.
Breadth of storehouse = 5m.
Length = x + 7 = 5 + 7 = 12m.
Thus,
Length of the base of storehouse is 12 m whereas breadth is 5 m.
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Answer:
given ,l - b = 7
√l^2 + b^2 - l = 1
squaring both sides after transposing l we get,
l^2 + b^2 = 1 + l^2 + 2l
so b^2 = 1 + 2l
b^2 -2(7+b) -1 = 0
b^2 - 2b - 15 = 0
hence b = 5 or -3 but side can't take negative value so b = 5
hence l = 12