====== Question 5, Exercise 5.3 ======
Solutions of Question 5 of Exercise 5.3 of Unit 05: Polynomials. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 5=====
{{ :math-11-nbf:sol:unit05:math-11-nbf-ex5-3-q5.png?nolink&400|Picture}}
The area of rectangle ACED is represented by $6 x^{2}+38 x+56$. Its width is represented by $2 x+8$. Point B is the midpoint of AC. ABFG is a square. Find the length of rectangle $ACED$ and the area of square $ABFG$.


** Solution. **
Given:

Area of $ACED$ = $6 x^{2}+38 x+56$

Width = $2 x+8$

We have \begin{align*}
& 6 x^{2}+38 x+56 \\
= & 2(3x^2+19x+28) \\
= & 2(3x^2+12x+7x+28) \\
= & 2(3x(x+4)+7(x+4)) \\
=& 2(x+4)(3x+7) \\
=& (2x+8)(3x+7)
\end{align*}

Now \begin{align*}
& Length \times Width = Area\\
\implies & Length \times (2x+8) = 6 x^{2}+38 x+56 \\
\implies & Length \times (2x+8) = (2x+8)(3x+7) \\
\implies & Length = 3x+7 \\
\end{align*}

Hence length of rectangle $ACED$ = $3x+4$

Now Length of side of square $ABFG$ = $\dfrac{1}{2}(2 x+8)$ = $x+4$

Area of square $ABFG$ = (x+4)^2. GOOD

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