Geometry Unit 5 Practice Test – Solutions

Problems 1 and 2 – Similar Triangles

Similar triangles are triangles that are the same shape, but with different sizes. When two triangles are

similar, their sides are proportional in length. That means if you write a ratio (a fraction) for the

corresponding sides, those ratios will be equal to one another. So, to solve for x in this problem: first,

identify the corresponding sides.

1. LK and RQ are the smallest sides of each triangle, so they are corresponding. KJ and QP are the

medium sides of each triangle, so they are corresponding. LJ and RP are the longest sides, so they

are corresponding. You are only given the lengths of two sides per triangle (KJ = x, QP = 33, LJ = 28,

RP = 42). Write the corresponding lengths as fractions with the smaller triangle over the bigger

triangle. Finally, set your two fractions equal to each other and cross multiply to solve for x.

2. The sides of each triangle on the right are corresponding with lengths: 11 and x. The sides on the

bottoms of each triangle are corresponding with lengths: 12 and 84. The sides on the top-left of

each triangle are corresponding with lengths 12 and 84. To solve for x, you need to only use two

sides from each triangle instead of all three. So, write your ratios 11/x and 12/84. Set the ratios

equal to each other and then cross-multiply to solve for x.

Problems 3 – 5

3. Which segment is an angle bisector of the triangle? Well an angle bisector is a segment/ray/line

that bisects and angle (a.k.a. cuts the angle in half). You can tell BW is an angle bisector because if

you look at the top of the triangle, you can see the congruency symbol. Angle B is cut in half.

Similarly, you can tell AT is an angle bisector because if you look at angle A, it also has the

congruency symbol for both halves of the angle. It too was cut in half.

4. Which segment is a perpendicular bisector of the triangle? A perpendicular bisector is a

segment/ray/line that bisects a line in half at a perpendicular angle. You can tell RZ is a

perpendicular bisector because it cuts the side on the right (BC) in half—this is indicated by the

congruency “hatch-marks” and it also has a right angle symbol, meaning it is perpendicular to that

side. Similarly, RS cuts the side on the left (AB) in half—again, this is indicated by the congruency

“hatch-marks” and it also has a right angle symbol, making the lines perpendicular to one another.

5. The point where perpendicular bisectors meet (called the circumcenter) is equidistant from all

vertices of the triangle. That means the length RA and RC must be congruent! So, simply set their

lengths equal to one another and solve for x.

6. This is an altitude because it is clearly the height of the triangle. Drawn

at a right angle from the opposite side, the segment extends to the

opposite vertex. That is the definition of altitude.

7. This is a median because it is a segment that reaches from one vertex of

the triangle to the midpoint of the opposite side.

8. This is a perpendicular bisector because it bisects one side of the

triangle (cuts it in half), and it has the symbol for a right angle which

makes it perpendicular.

9. This is a midsegment because it is a segment that bisects two sides of

the triangle. It falls in the exact middle of the two sides.

Problems 10-13 – Similar Triangles

There are three rules that help us decide whether we have enough information to conclude two triangles

are similar. The rules are:

Angle-Angle Similarity (AA~)

If two triangles have two sets of corresponding angles that are congruent to each other, then that

is enough information to conclude that the two triangles are similar. You do not need to write any

proportions for the sides.

Side-Angle-Side Similarity (SAS~)

If two triangles have one set of corresponding angles that are congruent. And the sides that make

those angles are proportional to one another, than that is enough information to conclude that the

two triangles are similar. To test this, you will write two ratios: one ratio will have one set of

corresponding sides (the smaller triangle over the bigger triangle), the other ration will have the

other set of corresponding sides (the smaller triangle over the bigger triangle). Set the two ratios

equal to each other and then cross multiply to ensure that they really are equal!

Side-Side-Side Similarity (SSS~)

If two triangles are similar, then all of their corresponding sides will be proportional. That means,

write three ratios for the corresponding sides. Each ratio should have the side from the smaller

triangle over the side from the bigger triangle. Then cross multiply two ratios at a time to ensure

they are all the same.

10. We are given three side lengths, so this will potentially be a SSS~. To test it, write your three ratios

(fractions). You should have one for the smallest sides of the triangles, the medium sides of the

triangles, and the longest sides of the triangles. Make sure when you write your ratios, you are

writing the length from the smaller triangle on top and the length from the bigger triangle on the

bottom. Now, cross multiply two ratios at a time to make sure their products equal one another.

On this problem, they do equal one another in both cases so we conclude the triangles are similar

by SSS~. Now, to write your similarity statement, ensure the corresponding sides match up. S

should be in the same spot as U, R with T and Q with V.

11. These triangles are overlapping, so to get a better idea of how they look, redraw them. You will

have one small triangle and one big triangle. You should mark Angle K with a congruency mark on

both triangles to show that it is the same exact angle for both of them. Now this resembles a SAS~

problem. To test it, you need to write two ratios for the corresponding sides. Remember: small

over big. Those ratios are equal so we conclude, yes, the triangles are similar by SAS~.

12. There is not enough information given to conclude these two triangles are congruent. The only

given information regards one pair of congruent vertical angles.

13. These two triangles are similar by AA~. Angle B and angle H are given to be congruent already.

Now, just mark that Angle BZA and angle GZH are congruent because they are vertical angles.

Problems 14 and 15 – Geometric Mean

The geometric mean of two numbers is a procedure that is similar to finding the regular mean. To find a

geometric mean, however, you simply multiply two numbers together and then take the square root of

the product.

16. When one angle of a triangle is bisected, the ratio of the two lengths where the opposite side is split

will equal the ratio of the lengths of the other two sides of the triangle. Just make sure corresponding

pieces are lined up correctly.

17. The midsegment of a triangle is the segment that connects two midpoints of the sides of a triangle. It

is always half the length of the third side of the triangle.

The side on the bottom of the triangle is

double the length of the midsegment.

So, write down, two times 48 is equal to the

length of that third side, 3x.

The lengths on the side opposite the bisected

angle make one ratio: 12/9

The lengths of the other two sides of the

triangle make the other ratio: x/24.

18. The altitude of a right triangle will divide the triangle into three similar triangles: a small one, a

medium one, and a big one. To identify the corresponding parts of the three triangles, you should

draw the triangles and then label each of the three sides of each triangle with the given information.

19. When three or more parallel lines are crossed by two transversals, the sides will be proportional. Just

write two ratios and make sure the corresponding segments line up correctly. Then solve for x.

20. Just like on number 18, the altitude of a right triangle divides the triangle into three similar triangles:

a small one, a medium one, and a big one. To identify the corresponding parts of the triangles, you

should draw them and then write in all of the given information.

The two x’s are on the medium triangle and

the large triangle. To solve for x, write a

proportion with the corresponding sides of

the medium and large triangles only.

The two y’s are on the small triangle and the

large triangle. To solve for x, write a

proportion with the corresponding sides of

the small and large triangles only.

The two x’s are on the medium triangle and the

large triangle. That means the small triangle will

not be relevant in this problem. To solve for x,

write a proportion with the corresponding sides

of the medium and large triangles only.

21. When the transversals of two parallel lines form a triangle, the side lengths divided up by the parallel lines are

proportional to one another. Just make sure corresponding parts are lined up correctly.

Problems 22 and 23 – Triangle Inequalities

For any triangle in the world, the sum of the lengths of two sides of the triangle must always be greater

than the length of the third side. So, to determine whether a given set of side lengths could possibly form

a triangle, you must write three inequalities. In each inequality, add two sides and make sure the sum is

bigger than the third side. If all three inequalities work, then the triangle is possible. If one or more

inequalities does not work then the triangle is not possible.

Problems 24 and 25 – Triangle Sides and Angles

In any given triangle, the largest side is always across from the largest angle. The medium side is across

from the medium angle and the smallest side is across from the smallest angle. So, draw arrows from

angles to the opposite sides (or vice versa). Note: on 25, you are missing the measure of one of the angles.

Solve for that angle measure by subtracting from 180 before you draw your arrows.

26. Draw a diagram to represent what is happening in the problem. Then identify corresponding parts in each

diagram: height with height and shadow length with shadow length. Set up two ratios for the two sets of

corresponding parts, then cross-multiply to solve for the missing dimension (height of the building).