Question

A rectangle has perimeter 28cm. Its area is 42sq.cm. Determine the dimensions of the rectangle. Include a diagram in your solution. Round your answer to the nearest hundredth of a centimeter.

Answer 1

The dimensions of the rectangle are: b=9.65 cm and h=4.35 cm or b=4.35 cm and h= 9.65 cm.

Quadrilaterals

There are different types of quadrilaterals, for example, square, rectangle, rhombus, trapezoid, and parallelogram.  Each type is defined accordingly to its length of sides and angles. For example, in a rectangle,  the opposite sides are equal and parallel and their interior angles are equal to 90°.

The area of a rectangle is equal to base x height (A=bh). The perimeter of a geometric figure is the sum of its sides. Thus, for a rectangle, the perimeter is 2b+2h.

System of Linear Equations

System of linear equations is the given term math for two or more equations with the same variables. The solution of these equations represents the point at which the lines intersect.

The question gives:

Perimeter=28cm

Area= 42 cm²

If the perimeter is the sum of all sides of the rectangle, you have:

Perimeter= 2b+2h=28

If the area of the rectangle is 42cm², you have:

Area=bh=42 cm²

You can write a system of linear equations.

2b+2h=28 (1)

bh=42 (2)

From equation 2, you have  $b=\frac{42}{h}$. Then, you can replace it in equation 1.

$2*\frac{42}{h}+2h=28 \\ \\ 84+2h^2=28h\\ \\ 2h^2-28h+84=0 \; dividing\; by\; 2\\ \\ h^2-14h+42=0$

Now, you should solve the quadratic equation.

$h_{1,\:2}=\frac{-\left(-14\right)\pm \sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:42}}{2\cdot \:1}$

$h_{1,\:2}=\frac{-\left(-14\right)\pm \:2\sqrt{7}}{2\cdot \:1}$

$h_1=\frac{-\left(-14\right)+2\sqrt{7}}{2\cdot \:1}=7+\sqrt{7}=9.65 cm$

$h_2=\frac{-\left(-14\right)-2\sqrt{7}}{2\cdot \:1}=7-\sqrt{7}=4.35cm$

If h1=9.65 cm, then $b_1=\frac{42}{9.65} =4.35 cm$, and if  h2=4.35 cm, then $b_1=\frac{42}{4.35} =9.65 cm$.

Read more about the quadratic function here:

brainly.com/question/1497716

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