Question

Aun cuadrado se le aumenta 7 cm de largo y 3 de ancho con lo que se forma un rectángulo cuya área es...


¿Cuáles en las dimensiones del rectángulo construido?


¿Si las medidas de las dimensiones del rectángulo cambian aunmentando en 6 de largo del cuadrado y en 3 el ancho del cuadrado cuál será el área del nuevo rectángulo?


¿El área que encontraste del rectángulo anterior se iguala a cuarenta, encuentra la medida de los lados?

Answer 1

The question in English

A) A square is increased by 7 cm in length and 3 cm in width, forming a rectangle whose area is...

B) What are the dimensions of the constructed rectangle?

C) If the measures of the dimensions of the rectangle change joining in 6 of length of the square and in 3 the width of the square what will be the area of the new rectangle?

D) The area you found of the previous rectangle equals forty, does it find the measure of the sides?

Answer:

Part A) the area of the rectangle is equal to

$A=(x+7)(x+3)$ or $A=x^{2} +10x+21$

Part B) The dimensions are $Length=x+7$ and $Width=x+3$

Part C) the new area of the rectangle is equal to

$A=(x+6)(x+3)$ or $A=x^{2} +9x+18$

Part D) $Length=8\ units$ and $Width=5\ units$

Step-by-step explanation:

Let

x------> the length of the original square

y--------> the width of the original square

we know that

the area of a square is equal to

$A=xy$

but

$x=y$

so

$A=x^{2}$

Part A) A square is increased by $7$ cm in length and $3$ cm in width

so

$x=x+7$

$y=x+3$

The area of the rectangle is equal to

$A=(x+7)(x+3)$

$A=x^{2} +3x+7x+21$

$A=x^{2} +10x+21$

Part B) What are the dimensions of the constructed rectangle?

The dimensions are

$Length=x+7$

$Width=x+3$

Part C) If the measures of the dimensions of the rectangle change joining in $6$ of length of the square and in $3$ the width of the square what will be the area of the new rectangle?

The new dimensions will be

$Length=x+6$

$Width=x+3$

$A=(x+6)(x+3)$

$A=x^{2} +3x+6x+18$ ------> $A=x^{2} +9x+18$

Part D) The area you found of the previous rectangle equals forty, does it find the measure of the sides?

$A=x^{2} +9x+18$

$A=40\ units^{2}$

so

$x^{2} +9x+18=40$

$x^{2} +9x-22=0$

Solve the quadratic equation

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2} +9x-22=0$

so

$a=1\\b=9\\c=-22$

substitute in the formula

$x=\frac{-9(+/-)\sqrt{9^{2}-4(1)(-22)}} {2(1)}$

$x=\frac{-9(+/-)\sqrt{169}} {2}$

$x=\frac{-9(+/-)13} {2}$

$x=\frac{-9+13} {2}=2$

$x=\frac{-9-13} {2}=-11$

The value of x is equal to $x=2\ units$

the dimensions are

$Length=x+6=2+6=8\ units$

$Width=x+3=2+3=5\ units$

Answer 2

(a). The area of rectangle is $\boxed{{x^2} + 10x + 21}.$

(b). The dimension of the rectangle is $\boxed{{\text{Length}} = x + 7{\text{ and Breadth}} = x + 3}.$

(c). The new area of rectangle is $\boxed{{x^2} + 9x + 18}.$

(d). The length and the width of the rectangle is $\boxed{8{\text{ units and 5 units}}}$

Further Explanation:

The formula of area of rectangle can be expressed as follows,

$\boxed{Area = \left( l \right) \times \left( w \right)}$

The formula of perimeter of the rectangle can be expressed as follows,

$\boxed{{\text{Perimeter}} = 2\left( {l + w} \right)}$

Here, “$l$” represents the length of the rectangle and “$w$” represents the width of the rectangle.

Explanation:

Part (a)

Side of the square is increased by 7 cm in length.

Side of the square is increased by 3 cm in width.

${\text{length}} = x + 7$

${\text{width}} = x + 3$

The area of the rectangle so obtained can be obtained as follows,

$\begin{aligned}{\text{Area}}&=\left( {x + 7} \right)\left( {x + 3} \right)\\&= {x^2} + 10x + 21\\\end{aligned}$

Part (b)

The dimensions of the rectangle obtained are as follows,

${\text{length}} = x + 7$

${\text{width}} = x + 3$

Part (c)

Length of the rectangle is 6 more than the side of square.

Width of the rectangle is 3 more than the side of square.

The area of the rectangle so obtained can be obtained as follows,

$\begin{aligned}{\text{Area}} &= \left( {x + 6} \right)\left( {x + 3} \right)\\&= {x^2} + 3x + 6x + 18\\&= {x^2} + 9x + 18 \\\end{aligned}$

Part (d)

The area of the rectangle is $40{\text{ Unit}}{{\text{s}}^2}$

The dimension of the rectangle can be obtained as follows,

$\begin{aligned}{\text{Area}} 7= 40\\{x^2} + 9x + 18 &= 40\\{x^2} + 9x - 22&= 0\\\end{aligned}$

Solve for the value of $x$.

$\begin{aligned}x&= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\ x&= \frac{{ - 9 \pm \sqrt {{9^2} - 4 \times 1 \times \left( { - 22} \right)} }}{{2 \times 1}}\\ x&= \frac{{ - 9 \pm \sqrt {169} }}{2}\\x&= \frac{{ - 9 + 13}}{2}\,\,\,\,{\text{or}}\,\,\,\,x = \frac{{ - 9 - 13}}{2}\\ x&= 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{or}}\,\,\,\,x = - 11\\\end{aligned}$

The length of the rectangle is $\boxed{8{\text{ units}}}.$

The width of the rectangle is $\boxed{5{\text{ units}}}.$

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Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Mensuration

Keywords: rectangle, length, width, breadth, area of rectangle, dimension obtained, 3 wide, 7 cm long, constructed rectangle, combining 6 in length, new rectangle, area is 40, forty, measure of sides, dimensions of rectangle, expression, square.