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6.) a = ½ (b + b); a = 60, b = 7, b = 3 (area of a trapezoid)

Question

6.) a = ½ (b + b); a = 60, b = 7, b = 3 (area of a trapezoid)

Explanation:

Step1: Identify the formula and known values

The formula for the area of a trapezoid is \( A=\frac{1}{2}(B + b)h \) (wait, the given formula here is \( A=\frac{1}{2}(B + b) \), maybe it's a typo and should be \( A=\frac{1}{2}(B + b)h \), but with the given values \( A = 60\), \( B=7\), \( b = 3\), we can assume we need to find \( h \) (height) if it's the area of trapezoid formula. Let's correct the formula to the standard area of trapezoid: \( A=\frac{1}{2}(B + b)h \). So we have \( A = 60\), \( B=7\), \( b = 3\), and we need to solve for \( h \).

Step2: Substitute the known values into the formula

Substitute \( A = 60\), \( B = 7\), \( b=3\) into \( A=\frac{1}{2}(B + b)h \):
\( 60=\frac{1}{2}(7 + 3)h \)

Step3: Simplify the right - hand side

First, calculate \( 7+3=10 \), then \( \frac{1}{2}\times10 = 5 \). So the equation becomes \( 60 = 5h \)

Step4: Solve for \( h \)

Divide both sides of the equation \( 60 = 5h \) by 5: \( h=\frac{60}{5}=12 \)

Answer:

If we assume the formula is for the area of a trapezoid (\( A=\frac{1}{2}(B + b)h \)) and we are solving for \( h \), the value of \( h \) is 12. If the given formula is correct as \( A=\frac{1}{2}(B + b) \), then substituting the values gives \( A=\frac{1}{2}(7 + 3)=\frac{1}{2}\times10 = 5\), but this contradicts \( A = 60\), so the more probable case is that the formula is missing the height \( h \) and we solve for \( h = 12\).