Question

Q4. The side of a square exceeds the side of the another square by 4
cm and the sum of areas of the two squares is 400 cm2. The side of the
smaller square is.. answers me as fast as possible​

Answer 1

SOLUTION :-

Let,

Side of first square = x

Side of second square =  x + 4

Area of square = Side²

Area of first square = x²

Area of second square = ( x + 4 )²

According to the question,

$\longrightarrow\sf x^2+(x+4)^2= 400 \\\\\longrightarrow x^2+x^2+4^2+2\times x \times 4 = 400 \\\\ \longrightarrow 2x^2+16+8x = 400 \\\\ \longrightarrow 2x^2+8x = 384 \\\\\longrightarrow 2x^2+8x-384 = 0 \\\\\longrightarrow x^2+4x-192 = 0 \\\\\longrightarrow x^2+16x-12x-192 = 0 \\\\\longrightarrow x(x+16)-12(x+16)=0 \\\\\longrightarrow (x+16)(x-12) =0 \\\\\longrightarrow \bf x= -16 \ , \ x = 12$

Side of the square cannot be negative.

So, x = 12

Side of the first square = 12 cm

Side of the second square = 16 cm

Answer 2

$\underline{\underline{\green{\Large\sf Given:}}} \begin{cases} \mapsto \red{\sf sum \: of \: areas \: of \: two \: squars \: is \: 400 {cm}^{2} } \\ \\ \\ \sf \mapsto \pink{ side \: of \: a \: square \: exceeds \: the \: side \: of \: another.}\end{cases}$

$\underline{\underline{\green{\Large\sf Find:}}} \begin{cases} \\ \\ \sf \mapsto \pink{ sides \: of \: two \: squares.} \\ \\ \\ \end{cases}$

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

Let, $\sf S_1$ and $\sf S_2$ be two squares.

Let, the side of the square $\sf S_2$ be x cm

Then, side of the square $\sf S_1$ = (x + 4) cm

Now, we know that

$\sf \to Area\:of\:square= (side) \times (side)$

So,

$\sf \leadsto Area\:of\:square \: S_1= (x + 4) \times (x + 4)$

$\sf \leadsto Area\:of\:square \: S_1= {(x + 4)}^{2}$

Now,

$\sf \leadsto Area\:of\:square \: S_2= (x) \times (x)$

$\sf \leadsto Area\:of\:square \: S_2= {(x)}^{2}$

It is given that, Area of both the squares is 400cm²

So,

➟ Area of square $\sf S_1$ + Area of square $\sf S_2$ = 400cm²

$\sf \implies \: \: \: \: \: \: {(x + 4)}^{2} + {x}^{2} = 400$

➳ $\sf using \: \pink{{(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}}$

$\sf \implies \: \: \: \: \: \: ( {x}^{2} + 2(x)(4) + {b}^{2} ) + {x}^{2} = 400$

$\sf \implies \: \: \: \: \: \: ( {x}^{2} + 8x+ {4}^{2} ) + {x}^{2} = 400$

$\sf \implies \: \: \: \: \: \: {x}^{2} + 8x+ 16 + {x}^{2} = 400$

$\sf \implies \: \: \: \: \: \: 2{x}^{2} + 8x+ 16 - 400=0$

$\sf \implies \: \: \: \: \: \: 2{x}^{2} + 8x - 384=0$

$\sf \implies \: \: \: \: \: \: 2({x}^{2} + 4x - 492)=0$

$\sf \implies \: \: \: \: \: \: {x}^{2} + 4x - 492= \dfrac{0}{2}$

$\sf \implies \: \: \: \: \: \: {x}^{2} + 4x - 492= 0$

⛗ Apply middle-split term

$\sf \implies \: \: \: \: \: \: {x}^{2} + 4x - 492= 0$

$\sf \implies \: \: \: \: \: \: {x}^{2} + 16x - 12x- 492= 0$

$\sf \implies \: \: \: \: \: \: x(x+ 16) - 12(x + 16)= 0$

$\sf \implies \: \: \: \: \: \: (x - 12)(x + 16)= 0$

$\begin{gathered}\sf x - 12 = 0 \\ \sf x = 12\end{gathered} \qquad \qquad\begin{gathered}\sf x + 16 = 0 \\ \sf x = - 16\end{gathered}$

$\footnotesize{\underline{\sf \: ignoring \: - 16 \: as \:length \: of \: side \: will \: never \: be \:in \: negative}}$

$\sf \therefore x = 12cm$

Thus, the side of square $\sf S_1$ = x + 4 = 12 + 4 = 16cm

and the side of square $\sf S_2$ = x = 12cm