Question

HELP PLEASE!!! I need your guys help on this question.

Answer 1

Answer:

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

Answer 2

Answer:

The first (left) trapezoid's area is $10\sqrt{55}$m or ≈ 74.2m²

The second (right) trapezoid's area is $54 +18\sqrt{3}$ or ≈ 85.2 in²

Step-by-step explanation:

First trapezoid (left):

Because the first trapezoid is a normal trapezoid, we can use the equation $A = \frac{a+ b}{2} * h$ 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 = $\sqrt{55}$   ($3^{2}+h^{2} = 8^{2}$)

Plug into the equation:

$A = \frac{7+13}{2} *\sqrt{55}$

A = $10\sqrt{55}$ 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 $sin(30) = \frac{x}{12}$ 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 $6\sqrt{3}$

Now we can calculate the area of the figure:

Triangle is $\frac{1}{2} * 6 * 6\sqrt{3} = 18\sqrt{3}$

Rectangle is 6 * 9 = 54

Area = $54 +18\sqrt{3}$ or ≈ 85.2 in²