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sierravick123owr441
07.07.2020 •
Mathematics
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Ответ:
Area of a trapezium = 1/2(a+b)×h
where a and b are parallel sides of the trapezium
h is the height
First question
We must first find the height of the trapezium using Pythagoras theorem
That's
h² = 8² -3²
h =√ 64 - 9
h = √ 55m
a = 7m
b = 10+3 = 13m
Area of the trapezoid = 1/2(7+13)×√55
= 1/2×20×√55
= 74.16
= 74m² to the nearest tenth
Second question
We use sine to find the height
sin30° = h/12
h = 12 sin 30°
h = 6 in
Let the other half of the parallel side be x
To find the other half of the parallel side we use Pythagoras theorem
That's
x² = 12²- 6²
x = √144-36
x = √108
x = 6√3 in
So for this trapezoid
a = 9 in
b = (9 + 6√3) in
h = 6 in
Area of the trapezoid = 1/2(9 + 9+6√3) × 6
= 1/2(18+6√3)×6
= 85.176 in²
= 85 in² to the nearest tenth
Hope this helps you
Ответ:
The first (left) trapezoid's area is
m or ≈ 74.2m²
The second (right) trapezoid's area is
or ≈ 85.2 in²
Step-by-step explanation:
First trapezoid (left):
Because the first trapezoid is a normal trapezoid, we can use the equation
Where a is equal to one base length and b is equal to the other base length and h is the height of the trapezoid.
a = 7
b = 13
h =
(
)
Plug into the equation:
A =
or ≈74.2m²
Second trapezoid (right):
Not a normal trapezoid (split into a triangle and a square)
Let's solve for the triangle first:
using
to find the right-hand side of the triangle we get x = 6
because this is a 30 60 triangle, the last side has to be![6\sqrt{3}](/tpl/images/0702/3456/981e8.png)
Now we can calculate the area of the figure:
Triangle is![\frac{1}{2} * 6 * 6\sqrt{3} = 18\sqrt{3}](/tpl/images/0702/3456/b4fa0.png)
Rectangle is 6 * 9 = 54
Area =
or ≈ 85.2 in²
Ответ:
5.25 ounces
step-by-step explanation:
divided 3/4 and 6