Math Problem Statement

Find the area of the trapezoid and the missing base measurement of the triangle, given specific dimensions.

Solution

Let's tackle each question one by one.


Problem 5: Area of the Figure

The figure shown in Problem 5 is a trapezoid. The formula for the area AA of a trapezoid is:

A=12×(b1+b2)×hA = \frac{1}{2} \times (b_1 + b_2) \times h

Where:

  • b1b_1 and b2b_2 are the lengths of the two parallel sides.
  • hh is the height of the trapezoid.

From the figure:

  • b1=38b_1 = \frac{3}{8} yd
  • b2=78b_2 = \frac{7}{8} yd
  • h=13h = \frac{1}{3} yd

Step-by-Step Calculation:

  1. Add the bases:
    b1+b2=38+78=108=54b_1 + b_2 = \frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}

  2. Plug into the formula:
    A=12×(54)×(13)A = \frac{1}{2} \times \left(\frac{5}{4}\right) \times \left(\frac{1}{3}\right)

  3. Multiply the fractions:
    A=12×54×13=524A = \frac{1}{2} \times \frac{5}{4} \times \frac{1}{3} = \frac{5}{24}

  4. Multiply by simplifying.

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Math Problem Analysis

Mathematical Concepts

Geometry
Area of Trapezoid
Area of Triangle
Fractions
Solving for Variables

Formulas

Area of Trapezoid = 1/2 × (b1 + b2) × h
Area of Triangle = 1/2 × base × height

Theorems

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Suitable Grade Level

Grades 6-8