Answer:
3. 79.9 mm²
4. 6.4 in
5. 177.5 mi²
6. 60.3°
Explanation:
Given various quadrilaterals and their dimensions, you want to find missing dimensions.
Trig relations
In all cases, one or more area formulas and trig relations are involved. The trig relations are summarized by the mnemonic SOH CAH TOA. The relevant relation for these problems is ...
Tan = Opposite/Adjacent
It is also useful to know that 1/tan(x) = tan(90°-x).
Area formulas
The formula for the area of a trapezoid is ...
A = 1/2(b1 +b2)h
The relevant formula here for the area of a parallelogram is ...
A = bs·sin(α) . . . . . where α is the angle between sides of length b and s
3. Parallelogram area
Using the area formula above, we find the area to be ...
A = (21 mm)(9 mm)·sin(155°) ≈ 79.9 mm²
The area is about 79.9 square mm.
4. Trapezoid base 2
The given figure shows two unknowns. We can write equations for these using the area formula and using a trig relation.
If we draw a vertical line through the vertex of the marked angle, the base of the triangle to the right of it is (b2-4). The acute angle at the top of that right triangle is (121°-90°) = 31°. The tangent relation tells us ...
tan(31°) = (b2 -4)/h ⇒ h = (b2 -4)/tan(31°)
Using the area formula we have ...
A = 1/2(b2 +4)h
and substituting for A and h, we get ...
20.8 = 1/2(b2 +4)(b2 -4)/tan(31°)
2·tan(31°)·20.8 = (b2 +4)(b2 -4) = (b2)² -16 . . . . . multiply by 2tan(31°)
(b2)² = 2·tan(31°)·20.8 +16 . . . . . . . add 16
b2 = √(2·tan(31°)·20.8 +16) ≈ 6.4 . . . . . . take the square root
Base 2 of the trapezoid is about 6.4 inches.
5. Trapezoid area
To find the area, we need to know the height of the trapezoid. To find the height we can solve a triangle problem.
If we draw a diagonal line parallel to the right side through the left end of the top base, we divide the figure into a triangle on the left and a parallelogram on the right. The triangle has a base width of 10 mi, and base angles of 60° and 75°.
Drawing a vertical line through the top vertex of this triangle divides it into two right triangles of height h. The top angle is divided into two angles, one being 90°-60° = 30°, and the other being 90°-75° = 15°. The bases of these right triangles are now ...
and their sum is 10 mi.
The height h can now be found to be ...
h·tan(30°) +h·tan(15°) = 10
h = 10/(tan(30°) +tan(15°))
Back to our formula for the area of the trapezoid, we find it to be ...
A = 1/2(b1 +b2)h = 1/2(20 +10)(10/(tan(30°) +tan(15°)) ≈ 177.5
The area of the trapezoid is about 177.5 square miles.
6. Base angle
The final formula we used for problem 5 can be used for problem 6 by changing the dimensions appropriately.
A = 1/2(b1 +b2)(b1 -b2)/(tan(90-x) +tan(90-x))
112 = 1/2(20+12)(20-12)/(2·tan(90-x)) = (20² -12²)·tan(x)/4
tan(x) = 4·112/(20² -12²)
x = arctan(4·112/(20² -12²)) = arctan(7/4) ≈ 60.3°
Angle x° in the trapezoid is about 60.3°.
__
Additional comment
There is no set "step by step" for solving problems like these. In general, you work from what you know toward what you don't know. You make use of area and trig relations as required to create equations you can solve for the missing values. There are generally a number of ways you can go at these.
A nice scientific calculator has been used in the attachment for showing the calculations. A graphing calculator can be useful for solving any system of equations you might write.
The second attachment shows a graphing calculator solution to problem 5, where we let y = area, and x = the portion of the bottom base that is to the left of the top base. Area/15 represents the height of the trapezoid. This solution also gives an area of 177.5 square miles.