Work out the area of this trapezium 6cm 10cm 8cm 17cm
1 answer:
Answer:
(81√15)/4 ≈ 78.43 cm²
Step-by-step explanation:
It can help to draw a diagram.
The base angle at the side of length 6 can be found using the Law of Cosines. Its opposite side is of length 8, and the two adjacent sides are of length 6 and 7. So, we have ...
8² = 6² +7² -2·6·7·cos(α)
cos(α) = (36+49-64)/(84) = 1/4
α = arccos(1/4) ≈ 75.522°
Then the height of the trapezium is ...
6·sin(75.522°) ≈ 5.80948 . . . . cm
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The area of the trapezium is given by the formula ...
A = 1/2(b1 +b2)h
A = (1/2)(17 +10)(5.80948) . . . . cm²
A ≈ 78.43 cm²
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Of course, sin(arccos(1/4)) = (√15)/4, so the exact formulation of the area is ...
A = (81√15)/4 cm²
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Answer:
12 ft
Step-by-step explanation:
Hi there!
<u>1) Find the length of the other leg in the right triangle</u>
There are two ways we can go about this. The first way is to use trigonomic ratios:
We're given one of the legs and we're trying to solve for the other, so we can use the tangent ratio:
Let x be equal to the length of the other leg. Plug in the known values:
The other way to find this is to observe the angles. Because we're given that two of the angles are 45 and 90 degrees, we know that the third angle is 45 degrees. This makes this an isosceles triangle, meaning the other leg is 7 ft long.
<u>2) Determine the height of the hoop</u>
Add 7 ft and 5 ft (the person's height):
7+5=12 ft
I hope this helps!