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I really can't do Parts 3 and 4 until I take the time to create the necessary illustrations. Sorry. So here's Part 2. If you missed Part 1, it's here.

### Part 2

**29. ** *The coordinates of the endpoints of BC are B(5, 1) and C(-3, -2). Under the transformation R _{90}, the image of BC is B'C'. State the coordinates of points B' and C'.*

Remember, if they don't state which direction you are rotating, the default is *counterclockwise*. That might seem counter-intuitive, but you are moving from quadrant I to II, from II to III, etc.

Under R90, a point P(x, y) becomes P'(-y, x). So these points will become B'(-1, 5) and C'(2, -3). Interestingly enough, according to the key, you didn't need to say which was which, but you did need to include the parentheses.

**30. ** * As shown in the diagram below [COMING SOON], AS is a diagonal of trapezoid STAR, RA || ST, m<ATS = 48, m<RSA = 47, and m<ARS = 68.
Determine and state the longest side of triangle SAT.*

You needed to calculate that m<AST = 65 degrees. You can get this using triangle RAS or knowing that the same-side interior angles are supplementary.

Once you know that, you can use 65 + 48 + x = 180 to find m<SAT, which is 67.

Side ST is the longest side because it is opposite the largest angle. You needed to give the side AND show some kind of work that indicates that you didn't go eeny-meeny... Seriously.

ST without any work -- even just angle sizes after you did the subtraction in your head or on your calculator -- is worth 0 points.

**31. ** *In right triangle ABC shown below [I'LL GET TO IT!], altitude BD is drawn to hypotenuse AC. If AD = 8 and DC = 10, determine and state the length of AB.*

There's a short way and a long way to do this. Both are okay for full credit.

First, remember that there are three right triangles in this diagram. Both of the smaller ones are similar (and therefore proportional) to the larger one and to each other.

You could write a proportional comparing ( base / hypotenuse) = ( base / hypotenuse) of triangle BAD and triangle ABC.

You would have ( 8 / x ) = ( x / 18 ). (It's 18 because it is 10+8.)

Cross-multiply: x^{2} = 144.

x = 12, which is the length of AB.

The longer method: If you forgot about that proportionality, but remembered the Altitude Rule, you would have found that altitude (BD)^{2} = (AD)(DC), so y^{2} = (8)(10) or 80. So the altitude is the square root of 80. Okay, that's good for partial credit, but we aren't done.

Now you can use Pythagorean Theorem: 8^{2} + 80 = x^{2}. Remember: (the square root of 80) squared is just 80.

64 + 80 = x^{2}

144 = x^{2}

x = 12, again.

Either method is valid and will get full credit. I saw quite a few of each type. The only problem with the longer method was that some students started estimating an answer in the middle and that introduced an error into the rest of the calculations.

**32. ** *Two prisms with equal altitudes have equal volumes. The base of one prism is a square with a side length of 5 inches. The base of the second prism is a rectangle with a side length of 10 inches. Determine and state, in inches, the measure of the width of the rectangle.*

I graded a lot of these. Two things: First, I can't believe how many Geometry students can do so well and still not know the Area of a rectangle or the Volume of a rectangular prism.

Second: This is a good case of **Less is More**. What do I mean by that? A lot of students talked or worked themselves out of a 1 or 2 points. Seriously.

Despite the fact that the word **prism** is used repeatedly, students draw pyramids, which by themselves, might not have been work and could've been overlooked. But then they used formulas for **Volume** that included **1/3** in them. That's a conceptual error. What makes this worse is that the (1/3)s cancel out and don't affect the outcome -- just the final score. I had to call over an arbitrator on this one because I didn't want to make that call.

Another error that was made repeatedly was that students assumed that the first prism was a **cube** even though there is nothing supporting this. They substituted 5 for the height! There is nothing about 5 being the height. This led to a discussion about whether that could be a case of "guess and check" and not an actual "conceptual error". Again, the mistake factors out.

But the worst part of it all is that because the heights were the same, the formula for **Volume** wasn't even needed! You only needed to compare the **Area** of the two bases.

This became confusing as well. People did perimeter of one and Area of the other. Or they assume that both bases were squares or other crazy stuff.

If was a very simple problem. How simple?

5^{2} = 10x.

That simple: **x = 2.5**.

*Funniest answer:* Someone wrote 2.5in = w, with a little bit of cursive and I thought it said "2 Sin w". Okay, it was funny to me.

**33. ** *As shown in the diagram below, BO and tangents BA and BC are drawn from external point B to circle O. Radii OA and OC are drawn. If OA = 7 and DB = 18, determine and state the length of AB.*

A very similar question was asked in August 2011, but that was multiple choice.

Again, there are TWO methods to solve this: Pythagorean Theorem, and Tangent-Secant.

First, realize that OD is also a radius, with length 7. That means that OB is 25. OA is perpendicular to the tangent line BA. Therefore, OAB is a right triangle with OB as its hypotenuse and OA, a radius, as one of its legs. So 7^{2} + (AB)^{2} = 25^{2}.

Solving the equation, we find that **AB has a length of 24**.

The other way: Draw **diameter** DOE (make point E on the circle opposite from D). BE is now a secant line with a length of 18 + 7 + 7 = 32.

Tangent-secant rule: (BA)^{2} = (BD)(BE) = 18 X 32 = 576. **AB = 24**.

Both methods are valid.

**34. ** *Triangle RST is similar to triangle XYZ with RS = 3 inches and XY = 2 inches. If the area of triangle RST is 27 inches, determine and state the area of triangle XYZ, in square inches.*

Short way: The ratio of the areas of two similar triangles is the *square* of the ratio of their sides. The scale factor of the sides is 3:2, so the scale factor for the areas is 9:4. If the bigger one has area 27, then ( 9 / 4 ) = ( 27 / x ).

Cross-multiply and 9x = 108, and **x = 12**.

LONG WAY: If you forgot about that, you could have done the following: if the large triangle has Area 27 and base 3, and A = 1/2 b h, then 27 = (1/2)(3)(h), and h = 18.

Then to find the height of the smaller triangle, ( 3 / 2 ) = ( 18 / h ). Then 3h = 36 and h = 12 inches. But that isn't the answer -- that's just the height.

A = (1/2) b h = (1/2) (2) (12) = 12 sq inches. That's the answer. Even though both numbers are 12, if you didn't complete it, you lost a point.

**35. **

## 6 comments:

For the similar triangles how many points do you lose for not squaring 3/2

Conceptual error. You would lose 1 point on a 2-point problem.

Not about pt 2 but in pt 3 with the equaterial triangle did you have to label the 30 degree angle

For number 32 if I originally put Bh=Bh but wrote to the side that height is irrelevant to finding the answer because it would just cancel out would I lose a point?

Also thank you for taking the time to type this and answer questions in the comments

You don't need to label the 30 degree angle in the construction.

If you made the equilateral triangle and then bisected the correct angle, you're good.

Question 32: The height is irrelevant. All you needed to work out as that area of the bases because the heights were the same.

Equations using the height are not incorrect, but they aren't necessary either.

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