One side of rectangle exceeds its other side by 4 cm. If area = 45. Find breadth.

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Solution

The correct option is A $5$
Let breadth = x
Length = x+4
Area $= l \times b = x (x + 4) = x^{2} + 4 x$
$x^{2} + 4 x = 45$
$x^{2} + 4 x - 45 = 0$
$x = \frac{- 4 \pm \sqrt{16 + 4 \times 45}}{2}$
$x = \frac{- 4 \pm \sqrt{16 + 180}}{2} = \frac{- 4 \pm \sqrt{196}}{2}$
$= \frac{- 4 \pm 14}{2} x = \frac{10}{2} = 5$

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One side of rectangle exceeds its other side by 4 cm. If area = 45. Find breadth.

The correct option is A 5Let breadth = xLength = x+4Area =l×b=x(x+4)=x2+4xx2+4x=45x2+4x−45=0x=−4±√16+4×452x=−4±√16+1802=−4±√1962=−4±142x=102=5