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Trava [24]
3 years ago
11

What is the area of a a trapezoid with the following dimensions h=5cm b1=9 cm b2= 15cm ?

Mathematics
1 answer:
e-lub [12.9K]3 years ago
7 0

Answer:

60cm^2

Step-by-step explanation:

The formula for the area of a trapezium is : \frac{a+b}{2}*height

To work out the area of the trapezium you would first have to add the lengths at the top and bottom (a and b) which is 9+15, this gives you 24cm.

The next step is to divide the value of 24 by 2, this gives you 12cm.

The final step is to multiply the value of 12cm by the height of 5, this gives you  60cm^2. This is because if you were to substitute the values into the formula you would get the same answer.

1) Add 9 and 15.

15+9=24

2) Divide 24 by 2.

24/2=12

3) Multiply 12 and 5.

12*5=60cm^2

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<h3>62.5 cm^2</h3>

The figure is a trapezoid, albeit not possessing the same lengths for the left and right sides. If I'm not mistaken, it has another name or term. But I forgot.

Anyway, to solve the problem, let us chop-chop or break the figure into two parts or two polygons that we are familiar with - a rectangle and a triangle. See the attached image for a clearer presentation.

After breaking the original polygon into two parts as portrayed by the red broken line, we are also going to divide the equivalent value of each side.

Remember, a rectangle has two pairs of common sides. Since the upper side of the polygon has a length of 10 cm, then, the length of the lower side (the side at the left side of the broken line) is 10, too. The same goes for the length of the left side which is 5 cm, and therefore the length of the broken line is 5 cm. Please refer to the attached image, particularly the figures in green.

From the standpoint of the triangle, based on thee figure, it has a height of 5 cm and a base of 5 cm (15 cm- 10 cm). Did you get it? If yes, let's continue.

Solving Time! :D

1. Solve for the area of the rectangle.

\begin{gathered}A_{rectangle}=L\times W\\=5\ cm \times 10\ cm\\=\bold{50\ cm^2}\end{gathered}

Arectangle=L×W=5 cm×10 cm=50 cm²

2. Solve for the area of the triangle.

\begin{gathered}A_{triangle}=\dfrac{b\times h}{2}\\=\dfrac{5\ cm\times5\ cm}{2}\\=\dfrac{25\ cm^2}{2}\\=\bold{12.5\ cm^2}\end{gathered}

Atriangle=2b×h=25 cm×5 cm=225 cm²=12.5 cm²

3. Combine (add) the two areas to find the area of the original figure.

\begin{gathered}A_{composite\ figure}=A_{rectangle}+A_{triangle}\\=50\ cm^2 +12.5\ cm^2\\=\bold{62.5\ cm^2}\end{gathered}

Acomposite figure=Arectangle+Atriangle=50 cm²+12.5 cm2=62.5 cm

If there's a Latex Error, Please click the second image attached below :)

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