Question

The two sides of a rectangle are 4x and x + 2. If the length of the diagonal of the rectangle is 5, then what is the area of the rectangle?

Answer 1

Answer: 12 cm^2

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Step-by-step explanation:

Length of diagonal of a rectangle = $\sqrt{l^2 +b^2 }$ = $\sqrt{(4x)^2 + (x+2)^2}$

=$\sqrt{16x^2 + x^2 + 4x+4}$

= $\sqrt{17x^2+4x +4}$

and, $\sqrt{17x^2+4x +4}$ = 5

or, 17x^2 + 4x +4 = 25 ----> squaring both sides

or,  17x^2 + 4x -21=0

or, 17x^2 -17x +21 x -21=0

or, 17x ( x -1) +21 (x-1) =0

or, ( x -1) ( 17x +21) = 0

∴ x =1 or, -21/17 (rejected)

then x=1

Now, length = 4x = 4* 1 = 4cm

breadth = x +2 = 1 +2 = 3 cm

area of rectangle = 4*3 =12 cm^2

Answer 2

Given :-

The two sides of a rectangle are 4x and x + 2. If the length of the diagonal of the rectangle is 5

To Find :-

Area

Solution :-

We know that

Diagonal² = Length² + Breadth²

(5)² = (4x)² + (x + 2)²

Apply identity ⇒ (a + b)² = a² + 2ab + b²

25 = 16x² + x² + 2(x)(2) + (2)²

25 = 16x² + x² + 4x + 4

25 = 17x² + 4x + 4

25 - 4 = 17x² + 4x

21 = 17x² + 4x

17x² + 4x - 21 = 0

Using quadratic formula

x = -b ± √(b² - 4ac)/2a

Here

a = 17

b = 4

c = -21

x = -(4) ± √[(4)² - 4(17)(-21)]/2 × 17

x = -4 ± √[16 - (-1428)]/34

x = -4 ± √[16 + 1428]/34

x = -4 ± √[1444]/34

x = -4 ± 38/34

Either

x = -4 + 38/34

x = 34/34

x = 1

or,

x = -4 - 38/34

x = -42/34

As length can't be negative

Therefore, x = 1

Length = 4x = 4(1) = 4

Breadth = x + 2 = 1 + 2 = 3

Area = Length × Breadth

Area = 4 × 3

Area = 12 unit²