Question

two square have side x km and x+4 km the sum of their area is 656cm find the side of the squares

Answer 1

Hey Friend.....☺

HERE IS YOUR'S ANSWER

Sum of their areas = 656cm

Their sides, side. Of square 1 = x
Side of square 2 = x+4

Area of square = a^2
Sum of their areas = x^2 + (x+4)^2
By identity

=> x^2 + x^2 + 4^2 + 2(x)(4) =656
=> 2x^2 + 16+ 8x = 656
=> 2x^2 + 8x = 640
=> x^2 + 8x = 320
=> x^2 + 8x - 320 = 0. ......(1)
We got our equation then,

We have to solve it , you can solve it with any method.

Here I am using splitting the middle term.

x^2 +8x -320 =0 {. Using (1). }

x^2 + 4x -320 =0

x^2 +20x -16x -320 =0

x (x +20) -16 (x + 20)

(x+20) (x-16)
Hence x = 16 since side of a square cannot be negative.

x = 16 and x+ 4 ,i.e., 16 +4 = 20

Are the 2 sides

Hope it helps you

Thanks. :-)

Sorry for my past answer , it was incorrect.

Answer 2

$\Large{\underline{\underline{\bf{Holamate:-}}}}$

$\Large{\underline{\underline{\bf{Given:</p><p>-}}}}$

➣Side of Ist square is (x) cm

➣Side of 2nd square is (x+4)cm

➣Sum of areas of both sqauare is 656 cm²

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$\Large{\underline{\underline{\bf{To \: Find:-}}}}$

Length of sides of the square =?

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$\Large{\underline{\underline{\bf{Solution:-}}}}$

${\bold{\underline{\underline{Area\: of \: first \: square:-}}}}$

Area of the square = a²

⟹Area = x²

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${\bold{\underline{\underline{Area \: of \: second\: square:-}}}}$

Area of the square = (x+4)²

By using identity (a+b)²= a²+b²+2ab

$area \: = {(x + 4)}^{2} \\ \\= {x}^{2} + {4}^{2} + 2 \times x \times 4 \\ \\ = {x}^{2} + 16 + 8x$

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Sum of areas of square = 656cm²

${\bold{\underline{\underline{According\: to \: the \: question:-}}}}$

$⟹{x}^{2} + (x + 4) = 656 \: \: {cm}^{2} \\ \\⟹ {x}^{2} + {x}^{2} + 16 + 8x = 656 \: {cm}^{2} \\ \\⟹ {2x}^{2} + 16 + 8x = 656 \\ \\ ⟹{2x}^{2} + 8x + 16 - 656 = 0 \\ \\⟹ {2x}^{2} + 8x - 640 = 0$

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${\bold{\underline{\underline{By \:middle\: term splitting:-}}}}$

$⟹{2x}^{2} + 8x - 640 = 0\\ \\⟹ 2( {x}^{2} + 4x - 320) = 0 \\ \\⟹ {x }^{2} + 4x - 320 = \frac{0}{2} \\ \\⟹ {x}^{2} + 4x - 320 = 0 \\ \\⟹ {x}^{2} + 20x - 16x - 320 = 0 \\ \\ ⟹x(x + 20) - 16(x + 20) = 0 \\ \\ ⟹(x + 20)(x - 16) = 0$

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${\bold{\underline{\underline{For \: zeroes \: Case-1:-}}}}$

$⟹x + 20 = 0 \\ \\⟹ x = - 20$

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${\bold{\underline{\underline{For \: zeroes\: case-2:-}}}}$

$⟹x - 16 = 0 \\ \\⟹ x = 16$

➪We cannot take negative number as side of a square cannot be negative

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${\bold{\underline{\underline{So, Length \: of \: sides \: of \: square:-}}}}$

➟Side of first square = x = 16 cm

➟Side of second square = x+4 = 16+4 = 20 cm

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$\Large{\underline{\underline{\bf{Thanks}}}}$