Question

192

1. find the perimeter of this trapezoid.

dimensions are in meters.

1. a base of the right solid 7 inches high is shown. find the volume of the right solid.

dimensions are in inches.

Explanation:

Problem 29

Step1: Convert to improper fractions

$3\frac{5}{6}=\frac{23}{6}$, $4\frac{1}{4}=\frac{17}{4}$, $9\frac{1}{2}=\frac{19}{2}$, $12\frac{5}{12}=\frac{149}{12}$

Step2: Find common denominator (12)

$\frac{23}{6}=\frac{46}{12}$, $\frac{17}{4}=\frac{51}{12}$, $\frac{19}{2}=\frac{114}{12}$

Step3: Sum all side lengths

$\frac{46}{12}+\frac{51}{12}+\frac{114}{12}+\frac{149}{12}=\frac{46+51+114+149}{12}$

Step4: Calculate total and simplify

$\frac{360}{12}=30$

Step1: Calculate base area (rect + semicircle)

Area of rectangle: $10\times(4\times2)=80$; Area of semicircle: $\frac{1}{2}\pi(4)^2=8\pi$; Total base area: $80+8\pi$

Step2: Compute volume (base × height)

$V=(80+8\pi)\times7$

Step3: Expand the expression

$V=560+56\pi$

Answer:

30 meters

---

Problem 30