**Draw Your Own Figure**

Question 2 with Solution

Question 2 with Solution

Last week, I went over a strategy for solving certain math problems on standardized tests by drawing a figure, and I gave you four problems to try on your own. You can see that post here: Math Strategy: Draw Your Own Figure

Today I would like to give a solution to the second of those four problems. You can see a solution for the first problem here: Draw Your Own Figure: Question 1.

**Example: **If line *m *is perpendicular to segment *PQ* at point *R,* and *PR* = *RQ*, how many points on line m are equidistant from point *P *and point *Q *?

A) One

B) Two

C) Three

D) More than three

Try to solve the problem yourself before checking the solution below.

**Solution: **Recall first that **equidistant** means at the same distance. So we’re looking for points on line *m* that are at the same distance from *P* as they are from *Q*. Let’s begin by drawing a picture:

Notice that *R* is equidistant from *P* and *Q*, so there is at least one. Let’s draw some more.

Since there are 5 shown in the above picture, the answer is choice D.

**Note: **Every point on line *m* is actually equidistant from *P* and *Q*. *m* is the **perpendicular bisector** of line segment *PQ*.

The **perpendicular bisector **of a line segment is a line perpendicular to the segment that passes through the midpoint of the segment.

**More Problems with Explanations**

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**Draw Your Own Figure**

Question 1 with Solution

Question 1 with Solution

Last week, I went over a strategy for solving certain math problems on standardized tests by drawing a figure, and I gave you four problems to try on your own. You can see that post here: Math Strategy: Draw Your Own Figure

Today I would like to give a solution to the first of those four problems. I will give solutions to the other three problems throughout this week.

**Example: **What is the area of a right triangle whose sides have lengths 14, 48, and 50?

Try to solve the problem yourself before checking the solution below.

**Solution: **Let’s begin by drawing a picture

**Remember: **The** hypotenuse** of a right triangle (the side opposite the right angle) is always longer than both **legs**.

In a right triangle we can always take the two legs to be the base and the height (in either order). So *b* = 14, *h* = 48, and

*A* = 1/2 *bh* = 1/2(14)(48) = **336**.

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**Draw Your Own Figure**

Today I would like to remind you of a very simple but effective strategy for solving math problems on standardized tests such as the ACT, SAT, and GRE.

If a math problem does not have a figure above it, then do not hesitate to draw your own. Sometimes drawing a quick picture of a situation makes a problem very easy, or at least easier. This is especially helpful with geometry problems.

**Example: **Segment *PQ* has midpoint M. If the length of *PM* is *t*, what is the length of *PQ* in terms of *t *?

Try to solve the problem yourself before checking the solution below.

**Solution: **Let’s begin by drawing a picture

From the picture, we see that *PQ* has twice the length of *PM*. Thus, the length of *PQ* is 2*t*.

Here are a few more problems for you to try. Try to draw a picture. I will provide solutions to these over the next few days.

1. What is the area of a right triangle whose sides have lengths 14, 48, and 50?

2. If line *m *is perpendicular to segment *PQ* at point *R,* and *PR* = *RQ*, how many points on line m are equidistant from point *P *and point *Q *?

A) One

B) Two

C) Three

D) More than three

3. Point *A* is a vertex of a 6-sided polygon. The polygon has 6 sides of equal length and 6 angles of equal measure. When all possible diagonals are drawn from point *A* in the polygon, how many triangles are formed?

4. In rectangle *PQRS*, point *T *is the midpoint of side *PQ*. If the area of quadrilateral* QRST* is 0.8, what is the area of rectangle *PQRS* ?

**More Problems with Explanations**

If you are preparing for the SAT, ACT, or an SAT math subject test, you may want to take a look at the Get 800 collection of test prep books.

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Geometry Question with Solution

Geometry Question with Solution

Yesterday I went over a method for solving certain geometry problems by moving the sides of a figure around. You can see that post here: Moving the Sides of a Figure Around

Today I would like to solve yesterday’s problem.

**Problem: **In the figure below, *AB* = 2, *BC* = 8, and *AD* = 10. What is the length of line segment *CD *?

**Solution: **The problem becomes much simpler if we “move” *BC* to the left and *AB* to the bottom as shown below.

We now have a single right triangle and we can either use the Pythagorean Theorem, or better yet notice that 10 = 5 *⋅ *2 and 8 = 4 *⋅ *2. Thus, the other leg of the triangle is 3 *⋅ *2 = 6. So we see that *CD* must have length 6 – 2 = **4**.

**Remark: **If we didn’t notice that this was a multiple of a 3-4-5 triangle, then we would use the Pythagorean Theorem as follows.

(*x* + 2)^{2} + 8^{2} = 10^{2
}(*x* + 2)^{2} + 64 = 100

(*x* + 2)^{2} = 36

*x* + 2 = 6

*x* = **4**

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**Geometry Strategy – Moving the Sides of a Figure Around**

Today I would like to teach you a more advanced geometry strategy that can sometimes be used on standardized tests such as the ACT, SAT, and GRE.

A seemingly difficult geometry problem can sometimes be made much easier by moving the sides of the figure around.

This procedure is best understood with an example:

**Example: **What is the perimeter, in meters, of the figure below?

Try to solve the problem yourself before checking the solution below.

**Solution: **Recall that to compute the perimeter of the figure we need to add up the lengths of all 8 line segments in the figure. We “move” the two smaller horizontal segments up and the two smaller vertical segments to the right as shown below.

Note that the “bold” length is equal to the “dashed” length. Thus, the perimeter is

(2)(7) + (2)(5) = 14 + 10 = **24**.

**Warning: **Although lengths remain unchanged by moving line segments around, areas will be changed. This method should **not** be used in problems involving areas.

Here is one more problem for you to try. See if you can find the best way to move the sides of the figure around to make the problem easy to solve. I will provide a solution to this problem tomorrow.

**Example: **In the figure below, *AB* = 2, *BC* = 8, and *AD* = 10. What is the length of line segment *CD *?

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**Ratio Question 3 with Solutions**

Earlier this week I went over my method for setting up a ratio, and I gave you three ratio problems to try on your own. You can see that post here: Setting Up a Ratio

Today I would like to give a solution to the third of those three ratio problems. You can see solutions for the first and second problems here: Ratio Q1 Ratio Q2

**Problem 3: **The height of a solid cone is 22 centimeters and the radius of the base is 15 centimeters. A cut parallel to the circular base is made completely through the cone so that one of the two resulting solids is a smaller cone. If the radius of the base of the small cone is 5 centimeters, what is the height of the small cone, in centimeters?

For those students that are getting the hang of this, here’s the quick computation right away:

**Quick solution: ** **(**22/15) *⋅ *5 = **22/3**.

And here are the details for the rest of us:

**Detailed solution**. A picture of the problem looks like this:

In the above picture we have the original cone together with a cut forming a smaller cone. We have also drawn two triangles that represent the 2 dimensional cross sections of the 2 cones. Let’s isolate the triangles:

The two triangles formed are **similar**, and so the ratios of their sides are equal. We identify the 2 key words “height” and “radius.”

height 22 *h*

radius 15 5

We now find *h* by cross multiplying and dividing.

22/15 = *h*/5

110 = 15*h*

*h *= 110/15 = **22/3**

**Definition:** Two triangles are **similar** if they have the same angles.

**Notes:** (1) To show that two triangles are similar we need only show that two angles of one of the triangles are equal to two angles of the other triangle (the third is free because all triangles have 180 degrees).

(2) Similar triangles do not have to be the same size.

(3) Similar triangles have sides that are all in the same proportion.

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**Ratio Question 2 with Solutions**

Earlier in the week I went over my method for setting up a ratio, and I gave you three ratio problems to try on your own. You can see that post here: Setting Up a Ratio

Today I would like to give a solution to the second of those three ratio problems. You can see solutions for the first problem here: Ratio Question 1 with Solutions.

**Problem 2: **A copy machine makes 4800 copies per hour. At this rate, in how many __minutes__ can the copy machine produce 920 copies?

**Solution: **We identify 2 key words. Let’s choose “copies” and “minutes.”

copies 4800 920

minutes 60 *x*

At first glance it might seem to make more sense to choose “hours” as our second key word, but choosing the word “minutes” is more efficient because

- The answer that we’re looking for must be in minutes.
- It’s extremely simple to convert 1 hour into 60 minutes.

We now find *x* by cross multiplying and dividing.

4800/60 = 920/*x
*4800

*x*= 55200

*x*= 11.5

So, the answer is **11.5**. ⋅

*** Quick computation: **(920/4800) ⋅ 60 = **11.5**.

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