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.

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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