Question

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Answer 1

Let the side of First square be x cm.

And the side of Another Square be y cm.

A.T.Q

x + x = y + 4

2x = 4 +  y

y = 2x - 4          ..........(1)

Also ,

(x)^2 + (y)^2 = 400 cm

$x^{2} +(2x-4)^{2} = 400\\x^{2}  +4x^{2} + 16 - 16x = 400\\5x^{2}  - 16x = 400 - 16 \\5x^{2}  - 16x = 384\\5x^{2}  - 16x - 384 = 0$

Now , you can solve it by middle term splitting or any other relevant method and find out the dimensions .

Answer 2

${\large\bf{\mid{\overline{\underline{Correct\:Question:-}}}\mid}}$

• The side of a square exceeds the side of another square by 4cm and the sum of the areas of the two squares is 400 cm². Find the dimensions of the squares. ?

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• Area of square = side × side ..
• (a+b)² = a² + b² + 2ab

$\Large\underline{\underline{\sf{Solution}:}}$

Let sides of Square be x cm and than side of other square be (x+4)cm ..

→ Area of First Square = x²

→ Area if Second square = (x+4)² ,

Now it is given that, both total Area is 400cm² ..

so,

A/q,

${x}^{2} + (x + 4)^{2} = 400 \\ \\ \red\leadsto \: {x}^{2} + {x}^{2} + 16 + 8x = 400 \\ \\ \red\leadsto \: 2 {x}^{2} + 8x + 16 - 400 = 0 \\ \\ \red\leadsto \: 2( {x}^{2} + 4x - 192) = 0 \\ \\ \red\leadsto \: {x}^{2} + 4x - 192 = 0 \\ \\ \textbf{Splliting the Middle Term Now} \: \\ \\ \red\leadsto \: {x}^{2} + 16x - 12x - 192 = 0 \\ \\ \red\leadsto \: x(x + 16) - 12(x + 16) = 0 \\ \\ \red\leadsto \: (x + 16)(x - 12) = 0 \\ \\ x = ( - 16) \: or \: 12.$

Now, since sides of Square cant be in negative ,,,

Hence,

$\red{\large\boxed{\bold{x = 12}}}$

so, sides of Square First = 12 cm

and,

side of square second = (12+4) = 16cm ...

$\green{\large\underline\textbf{Hope it Helps You.}}$