Find the area of a rectangle whose length and breadth are - units and - units respectively-

We know area = length × breadth Area = (y + 4) × (y 3) = gt; Area= y^2 + y(4 3) 12 recall (x + a)(x + b) = x^2 + (a + b)x + ab = gt; Area= y^2 + y 1 ...

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Quantitative Aptitude

Coordinate Geometry

Find the area...

Question

Find the area of a rectangle whose length and breadth are $(y + 4)$ units and $(y - 3)$ units respectively.

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Solution

We know area = length × breadth
$Area = (y + 4) \times (y - 3)$
$=> Area = y^{2} + y (4 - 3) - 12$
$recall (x + a) (x + b) = x^{2} + (a + b) x + a b$
$=> Area = y^{2} + y - 12$

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Find the area of a rectangle whose length and breadth are (y+4) units and (y−3) units respectively.

We know area = length × breadth Area=(y+4)×(y−3) =>Area=y2+y(4−3)−12 recall(x+a)(x+b)=x2+(a+b)x+ab =>Area=y2+y−12

Find the area of a rectangle whose length and breadth are $(y + 4)$ units and $(y - 3)$ units respectively.

We know area = length × breadth
$Area = (y + 4) \times (y - 3)$
$=> Area = y^{2} + y (4 - 3) - 12$
$recall (x + a) (x + b) = x^{2} + (a + b) x + a b$
$=> Area = y^{2} + y - 12$