The area of a rectangle is given as $(24 a^{3} b + 20 a^{2} b^{2} - 4 a b)$ square units and one of its dimension is given as $2 a b$ units, find the other dimension.

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Solution

As we know that Area of rectangle = length $\times$ breadth
i.e we need two dimension

we are given one dimension i.e $2 a b$

i.e 2ab $\times 2^{n d}$ dimension $= (24 a^{3} b + 20 a^{2} b^{2} - 4 a b)$

$\Rightarrow 2^{n d}$ dimension $= \frac{(24 a^{3} b + 20 a^{2} b^{2} - 4 a b)}{2 a b}$

taking 2ab common from number

$\Rightarrow 2^{n d}$ dimension $= \frac{2 a b (12 a^{2} + 10 a b - 2)}{2 a b}$

$2^{n d}$ dimension $= 12 a^{2} + 10 a b - 2$

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State true or false:

The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. The dimensions of the rectangle are 17 units and 9 units.

The area of a rectangle gets reduced by 9 square units. if its length is reduced by 5 units and breadth is increased by 3 units. However, if the length of this rectangle increases by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Let the length of the rectangle = x units and its breadth = y units

According to first condition :
$(x - 5) (y + 3) = x y - 9$ and

According to second condition:
$(x + 3) (y + 2) = x y + 67$

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The area of a rectangle is given as (24a3b+20a2b2−4ab) square units and one of its dimension is given as 2ab units, find the other dimension.

As we know that Area of rectangle = length × breadth i.e we need two dimension we are given one dimension i.e 2abi.e 2ab×2nd dimension =(24a3b+20a2b2−4ab)⇒2nd dimension =(24a3b+20a2b2−4ab)2ab taking 2ab common from number ⇒2nd dimension =2ab(12a2+10ab−2)2ab2nd dimension =12a2+10ab−2

The area of a rectangle is given as $(24 a^{3} b + 20 a^{2} b^{2} - 4 a b)$ square units and one of its dimension is given as $2 a b$ units, find the other dimension.