Question

There is a rectangle whose length is more than breadth by 7m and the diagonal is more than the length by 1m find the lengtg and breadth​

Answer 1

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

Answer 2

Answer:

Let the breadth (BC) of a rectangle ABCD = B m

Length (AB) of a rectangle ABCD = (B + 7) m

Diagonal, the hypotenuse (AC) of the rectangle = (B + 7 + 1) m

AC^2 = AB^2 + BC^2

OR (B + 8)^2 = (B + 7)^2 + B^2

OR B^2 + 16B + 64 = B^2 + 14B + 49 + B^2

OR B^2 - 2 B - 15 = 0

OR (B - 5) (B + 3) = 0

Therefore, Breadth = 5 m, Length = (5 + 7) = 12 m

Step-by-step explanation: