Question

find lengths using area and perimeter mastery check a rectangle has a perimeter of 38 inches and an area of 48 square inches. find its length and width. length: inches width: inches

Explanation:

Step1: Set up equations

Let the length be $l$ and the width be $w$. The perimeter formula is $P = 2(l + w)$, so $2(l + w)=38$, which simplifies to $l + w=19$ (dividing both sides by 2), so $l = 19 - w$. The area formula is $A=lw$, and $A = 48$, so $(19 - w)w=48$.

Step2: Rearrange the area - equation

Expand $(19 - w)w=48$ to get $19w-w^{2}=48$, then rearrange it to the quadratic - form $w^{2}-19w + 48=0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=-19$, $c = 48$), the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-19)^{2}-4\times1\times48=361 - 192 = 169$. Then $w=\frac{19\pm\sqrt{169}}{2}=\frac{19\pm13}{2}$.

Step4: Find the values of $w$

$w_1=\frac{19 + 13}{2}=\frac{32}{2}=16$ and $w_2=\frac{19 - 13}{2}=\frac{6}{2}=3$.

Step5: Find the corresponding values of $l$

If $w = 3$, then $l=19 - 3=16$. If $w = 16$, then $l=19 - 16=3$. Since length is usually longer than width, $l = 16$ inches and $w = 3$ inches.

Answer:

length: 16 inches
width: 3 inches