[1000] Find (a) the area and (b) the length of the perimeter of this triangle correct to 2 decimal places.
First, let's find the area.
The base of this triangle goes from x = -2 to x = 4. So, the length of the base is 4 - (-2) = 6.
The height of this triangle goes from y = 0 to y = 4. So, the height is 4 - 0 = 4.
The area of this triangle is (6)(4)/2 = 12.
To find the perimeter we need the lengths of the left side and the
right side using
the Pythagorean theorem.
You'll notice that I drew a line for the height. If you're smart you
noticed that it divides the triangle into 2 right angled triangles.
Let's call the left side "a" and the right side "b" (for those of us who
are Three Stooges fans, we could call the 3 sides Moe, Larry and
Curley).
For the triangle on the left, the base is 2 and the height is 4. So we
have:
a^{2} = 2^{2} + 4^{2}
a^{2} = 4 + 16
a^{2} = 20
Taking the square root of both sides we get
a = 4.47
rounding the answer to 2 decimal places.
Doing the same exercise for the right hand triangle, with the base now 4 and the height still 4, we get b = 5.66.
Adding everything together, the length of the perimeter is 4.47 + 5.66 + 6 = 16.13.
[1001] Find (a) the circumference and (b) the area of this circle correct to 2 decimal places.
Since we're rounding the answer to 2 decimal places, let Pi = 3.14.
Since the diamater is 4, that means the radius is 2.
So, the circumference
is (2)(3.14)(2) = 12.56. The area is
(3.14)(2^{2}) = 12.56.
This is the one situation where the area of a circle equals its circumference.
[1002] Find (a) the area and (b) the length of the perimeter of this rhombus correct to 2 decimal places.
The area is easier to find, so let's do that one first.
The base goes from x = 2 to x = 4. So the length of the base is 2.
Choosing either of the vertical lines, the height goes from y = 1 to y = 4. So the height is 3.
Ergo (there's some Latin for you), the area of this rhombus is (2)(3) = 6.
To find the length of the perimeter, the geniuses among you will notice that the top and bottom are the same length and that the left and right sides are the same length. All we have to do is find the length of the left side and we're home.
To do that, notice the right angle triangle formed by the vertical line at x = 3. The length of the base of this triangle is 1 and the height is 3. Let's call the left side "a". Using the Pythagorean theorem, we get:
a^{2} = 1^{2} + 3^{2}
a^{2} = 1 + 9
a^{2} = 10
Taking the square root of both sides we get a = 3.16 rounding the answer to 2 decimal places.
Adding everything together, the length of the perimeter is 2 + 2 + 3.16 + 3.16 = 10.32.
[3000] You have a rectangle where the width is 1/3 the height. If the perimeter of this rectangle is 48 cm, what are its dimensions?
Solution: Let W = width and H = height. Since the width is 1/3 the height, we can write H = 3W which is another way of saying the height is 3 times the width.
We are given the perimeter to be 48 cm. We can write 2W + 2H = 48 or W + H = 24, by dividing both sides of the equation by 2. But, since H = 3W we can rewrite W + H = 24 as W + 3W = 24 or 4W = 24. Solving for W, we get W = 6 and by plugging this solution into H = 3W, we get H = 18. So our rectangle is 6 cm wide and 18 cm high.
[3001] You have an isosceles triangle where the two equal sides are each twice as long as the base which is 6 cm long. What is the height of this triangle correct to 3 decimal places?
Let's call either equal side L. Since the length of L is twice the base, L = 12. You will notice in the picture that the triangle can be split into 2 right angle triangles. Let's call the height H. Since we have a right-angle triangle, we can use the Pythagorean theorem. The thing to keep in mind is that the base of each right angle triangle is 3 cm in length, not 6. Solving for H, we have
12^{2} = 3^{2} + H^{2}
144 = 9 + H^{2}
135 = H^{2}
Taking the square root of both sides, we have H = 11.619 round to 3 decimals places. So the height of our triangle is 11.619 cm.
[3002] You have a square where the distance from 1 corner to its opposite corner is 2 cm. What are the dimensions of this square correct to 3 decimal places?
Solution: Call the distance from one corner to the other D. Call one side X. Using the Pythagorean theorem we have D^{2} = X^{2} + X^{2} or D^{2} = 2X^{2}
Since D = 2, we have
2^{2} =
2X^{2}
4 = 2X^{2}
2 = X^{2}
by dividing both sides by 2. Taking the square root of both sides, we have X = 1.414 rounding to 3 decimal places.
[2000] 1) What are the coordinates of the point at the top of this
triangle?
2) What is angle b in degrees correct to 4 decimal places?
3) How long is side C correct to 4 decimal places?
Let's call the triangle on the left A and the one on the right B.
We'll also call the height of the triangle h.
Question 1): To solve this problem, we'll work with triangle A. It's obvious the x-coordinate is 3 because the length of the base is 3. To find the y-coordinate (a.k.a. the height of triangle A), we can use a little trig. If you recall, if you have a right-angle triangle, the tangent of an angle is opposite over adjacent. In our case, we have:
Since tan 60^{o} = 1.7321 correct to 4 decimal places, we have h = (3)(1.7321) = 5.1962. So, the coordinates at the top of the triangle are (3,5.1962)
Question 2): This time, we'll use triangle B. This question is a snap now that we have the height of the triangle. The length of the base is 2. Have we been here already? Since we have the lengths of both the adjacent and opposite sides, once again we can use the tangent of angle b to solve for it:
or that tan b = 2.5981. To solve for b, we take the tan inverse of both sides: b = tan^{-1}(2.5981). Using our trusty calculator, we get b = 68.9483^{o}
Question 3): There are a couple of ways to slay this dragon. The easiest is to use the Pythagorean theorem. We have c^{2} = 5.1962^{2} + 2^{2} or c^{2} = 31. Solving for c, we get c = 5.5678.
[2001] 1) Find the area of this polygon.
2) What are the angles of the 4 corners correct to 4 decimal places?
The first thing we do is split the polygon into 3 triangles and 1 rectangle.
Question 1) To find the area of the polygon, we get the areas of the triangles and the rectangle. We summarize it here:
Figure | Width | Height | Area |
---|---|---|---|
Triangle A | 2 | 4 | 4 |
Triangle B | 3 | 3 | 4.5 |
Triangle C | 1 | 7 | 3.5 |
Rectangle D | 3 | 4 | 12 |
Adding up the areas of the 4 figures, we get the total area: 4 + 4.5 + 3.5 + 12 = 24. So, the total area of the polygon is 24 sq. units.
Question 2) Contrary to what some of you might be thinking, we don't have to find the lengths of the hypotenuses of the 3 triangles in order to find the angles of the triangles. Since we're given the lengths of the adjacent/opposite sides of all 3, we can use the tan inverse function instead. We can also take advantage of the fact that the angles in a triangle add up to 180^{o}. This table shows how the answers are derived:
Angle location | Step 1 | Step 2 | Final answer |
---|---|---|---|
Lower left | tan^{-1}(4/2) | 63.4349 | 63.4349 |
Upper left | (90 - 63.4349) + 90 + tan^{-1}(3/3) | 26.5651 + 90 + 45 | 161.5651 |
Lower right | tan^{-1}(7/1) | 81.8699 | 81.8699 |
Upper right | (90 - 45) + (90 - 81.8699) | 45 + 8.1301 | 53.1301 |
If you add up the 4 angles, you will see that they total 360^{o} which they are supposed to.
[2002] If the height of this triangle is 5 units, what are the lengths of sides A, B, and C correct to 4 decimal places?
The easiest approach is to find the lengths of sides B and C first, then use the Pythagorean theorem to find the length of side A since we have a right-angle triangle.
Solving for B, we have sin(30^{o}) = 5/B or 0.5 = 5/B. Cross-multiplying to solve for B we get B = 10.
We can do the same thing for side C: sin(60^{o}) = 5/C or 0.8660 = 5/C. Doing the same thing as we did for side B, we get C = 5.7735
Now we can use the Pythagorean theorem to solve for side A: A^{2} = 5.7735^{2} + 10^{2} or A^{2} = 133.3333. Taking the square root of both sides, we get A = 11.5471
[2003] 1) What are the angles of B and C?
2) What are the co-ordinates of (a,b) at the top of the triangle?
Solution to 1): We'll solve this problem using the Law of Cosines. If we call the side opposite the 60^{o} angle x, we have
x^{2} = 3^{2} + 6^{2} - (2)(3)(6)(cos
60^{o})
x^{2} = 9 + 36 - (36)(0.5)
x^{2} = 27
Taking the square root of both sides, we get x = 5.1962
Now that we have the lengths of all 3 sides, we can solve for one of the angles, again using the Law of Cosines. Let's tackle angle B:
6^{2} = 3^{2} + 5.1962^{2} - (2)(3)(5.1962)(cos B)
36 = 9 + 27 - (31.1769)(cos B)
36 = 36 - (31.1769)(cos B)
Solving for cos B we get cos B = 0. Since cos 90^{o} = 0, we have angle B = 90^{o}. Ergo, angle C = 30^{o} because the 3 angles in the triangle have to add up to 180^{o}.
Solution to 2): The easiest approach is to split the triangle
into 2 right angle triangles:
This is almost too easy. Since cos 60^{o}
= sin 30^{o} = 0.5 we have
a/3 = 0.5 and
b/5.1962 = 0.5.
Solving for a and b, we get
a = 1.5 and
b = 2.5981. So, the co-ordinates of the
point at the top of the triangle are
(1.5,2.5981)
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