Question

the area of rectangle gets reduced by 9 square units,if it's 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.find the dimensions of the rectangle ​

Answer 1

Answer:

Let length of rectangle = x units

And width of rectangle = y units

Area of rectangle = length * width = x*y

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.

So

Decrease the length by 5 unit so new length = x - 5

Increase the width by 3 unit so new width = y + 3

New area is reduced by 9 units

So new area  = xy – 9

Plug the value in formula length * width = area we get

(x  - 5)(y + 3) = xy  - 9

Xy  + 3x – 5y – 15  = xy – 9

Subtract xy both side we get

3x - 5y  = 6                 …(1)

If we increase the length by 3units and the breadth by 2 units, the area increases by 67 square units.

Increase the length by 3 unit so new length = x +3

Increase the width by 2 unit so new width = y + 2

New area is increased  by 67 units

So new area  = xy + 67

Plug the value in formula length * width = area we get

(x  +3)(y + 2) = xy  +  67

Xy  + 2x  +  3y  +  6  = xy + 67

Subtract xy both side we get

2x  +  3y = 61                         …(2)*3

3x - 5y  = 6                 …(1)*2

Cross multiply the coefficient of x we get

6x + 9 y = 183

6x -10y  =12

Subtract now we get

19 y = 171

Y = 171/19 = 9

Plug this value of y in equation first we get

2x + 3* 9 = 61

2x  = 61 – 27

2x = 34

X = 34/2 = 17

So length is 17 units and width is 9 units

Answer 2

$\large\underline{Given:-}$

• Base ⇢ 24cm
• Area of Isosceles triangle ⇢ 192cm²

$\large\underline{To find:-}$

• find the perimeter ?

$\large\underline{Solutions:-}$

$\: \: \: \: \: Area \: \: of \: \: triangle \: \: = \: \: \frac{1}{4} \: b \: \sqrt{{4a}^{2} \: - \: {b}^{2}}$

»★ Given Area of Isosceles triangle ⇢ 192cm², and base 24cm.

$\: \: \: \: \: \leadsto {192} \: \: = \: \: \frac{1}{4} \: b \: \sqrt{{4a}^{2} \: - \: {b}^{2}}$

$\: \: \: \: \: \leadsto {192} \: \: = \: \: \frac{1}{4} \: \times \: {24} \: \times \: \sqrt{{4a}^{2} \: - \: {576}}$

$\: \: \: \: \: \leadsto {192} \: \: = \: \: {6} \: \times \: \sqrt{{4a}^{2} \: - \: {576}}$

$\: \: \: \: \: \leadsto \frac{192}{6} \: \: = \: \: \sqrt{{4a}^{2} \: - \: {576}}$

$\: \: \: \: \: \leadsto {32} \: \: = \: \: \sqrt{{4a}^{2} \: - \: {576}}$

$\: \: \: \: \: \leadsto {(32)}^{2} \: \: = \: \: {4a}^{2} \: - \: {576}$

$\: \: \: \: \: \leadsto {1024} \: \: = \: \: {4a}^{2} \: - \: {576}$

$\: \: \: \: \: \leadsto {1024} \: + \: {576} \: \: = \: \: {4a}^{2}$

$\: \: \: \: \: \leadsto {1600} \: \: = \: \: {4a}^{2}$

$\: \: \: \: \: \leadsto \frac{1600}{4} \: \: = \: \: {a}^{2}$

$\: \: \: \: \: \leadsto {400} \: \: = \: \: {a}^{2}$

$\: \: \: \: \: \leadsto {a} \: \: = \: \: \sqrt{400}$

$\: \: \: \: \: \leadsto {a} \: \: = \: \: {20}$

$\: \: \: \: \: Perimeter \: \: of \: \: the \: \: Isosceles \: \: triangle \: \: \leadsto \: \: {2a} \: + \: {b}$

$\: \: \: \: \: \leadsto \: \: {2} \: \times \: {20} \: + \: {24}$

$\: \: \: \: \: \leadsto \: \: {40} \: + \: {24}$

$\: \: \: \: \: \leadsto \: \: {64cm}^{2}$

»★ Hence,

$\: \: \: \: \: Perimeter \: \: of \: \: the \: \: Isosceles \: \: triangle \: \: \leadsto \: \: {64cm}^{2}.$

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