Triangles, Lines, & Angles - SAT Math

This problem tests two important rules. For one, the angle measure of a straight line is 180. Here if you follow line you can see that its angle is broken in to three segments: and the blank angle between them. Those three angles must sum to 180, so if you already know that and , then the unlabeled angle between them must equal so that .

Next, know that when lines intersect to form angles at a particular point, opposite (vertical) angles are congruent. The angle of measure is directly opposite the angle you just calculated to be degrees, so has to be as well.

The third side of a triangle is always greater than the difference of the other two sides and less than the sum of the other two sides. This applies to every side of a triangle. In other words, you can arbitrarily pick any one side to be the “third side,” and then that side must be greater than the difference of the other two and less than the sum of those two.

Here that means that the third side must greater than the difference of and . Since , that means that is not an option. It also means that the third side must be less than the sum of and . Since , that rules out as an option. You know that the third side must be greater than and less than : only , option II, fits.

One extremely helpful tool that you’ll often find in your geometry toolkit is the presence of isosceles triangles. Here, once you’ve filled in for angle , you should notice that even though there are two angles remaining to solve for within triangle ABC, those two angles each equal the same thing. So since the sum of all three has to be , and angle a already accounts for , has to equal the remaining . means that . And then you get to use the same logic all over again. Within triangle BCD, you know that and that the sum of the three angles must be . That means that , so .

An important, fundamental rule of triangles is that the sum of the interior angles equals degrees. For triangle ABC above, those three angles are expresses as , , and , meaning that the sum of the interior angles is . If , then you can divide both sides by to recognize that .

Whether you use the Pythagorean Theorem or you quickly recognize that this is a triangle, you can solve for the length of side .

You can prove that this is a triangle because the hypotenuse is twice as long as one of the legs of the triangle. This then fits the side ratio , so you can fill in the middle side as .

Or you can use Pythagorean Theorem. Since you know side YZ is the hypotenuse, you can set it up as . This means that , so and .

Once you've determined that , you can calculate the area. The area is half the base times the height, where the base and height form a -degree angle. This means that you can use and . .

There is a rule about isosceles triangles that isn't obvious the first time you see it, but that the SAT likes to test: if an isosceles triangle includes a -degree angle, then it must be an equilateral triangle.

You can prove this by testing the cases: if you know that you have an isosceles triangle with a measure of for one angle, then you can call your angles , , and . You know that must sum to (a rule of triangles), and that one of the following cases must be true:

matches . if that's true, then your three angles are , , and . Since , that means that . Here, all three angles are .

matches . The same as the above.

matches . This then means that , so . If , then so x = 60, which means that . Here, again, all angles are .

Since you know that this is an equilateral triangle, all sides have the same measure. That means that all sides are .

This problem blends two important rules related to triangles:

1. The sum of the interior angles of a triangle is .

2. In an isosceles triangle, two angles have the same measure.

Here, although you're not explicitly told which angles have the same measure, you can deduce that it must be angles XZY and YXZ - the two angles that do not measure degrees. Note that if were to be the "match," then you would already have degrees assigned to those two angles, but that would violate the rule.

Therefore, you know that your three angles are , , and (where represents the unknown, matching angle). You can then say that , so and .

One of the most convenient things about isosceles right triangles is that you can use the two shorter sides as the base and the height to find the area, since they're connected by a right angle:

So if you know that is the area in an isosceles right triangle, you can use to solve for as the length of each of the shorter sides. This means that , which you can simplify to:

And then solve for .

Because this is an isosceles right triangle, the sides will form the ratio , meaning that the hypotenuse will measure . If you sum the two shorter sides of with the hypotenuse of , you reach .

This problem forces you to visualize a right triangle from the information provided in the story. When Mirinda makes a turn from heading directly south to heading directly west, that is a -degree right angle. And the shortest distance to get from her endpoint ( km south and km west) back home is a diagonal line that connects the triangle:

So that shortest distance home is the hypotenuse of a right triangle with shorter sides and .

To solve, you can use the Pythagorean Theorem: , where and are lengths of the shorter sides and is the length of the hypotenuse. This then means that:

This simplifies to:

So and .

Note that is a relatively-common Pythagorean triplet and is one you should consider memorizing. You should absolutely memorize and as side ratios, with and as much lower priority (as they are tested much less frequently, and you can always use Pythagorean Theorem if you do not know them offhand).

The key to this problem is recognizing the two relationships here that must sum to degrees: the three angles in a triangle must sum to , and supplementary angles (those that are formed by splitting a straight line) must also sum to . Using those relationships, you should see that:

In the big triangle (JXZ), two angles are already given as and , meaning that angle KZY must equal , based on the sum of angles in that triangle needing to be .

Since that angle KZY is , that means that , as it has to sum with its supplementary angle to .

And using the small triangle to the right, KYZ, if you know that the two bottom angles (KYZ and KZY) are and , then the third angle (YKZ at the top of that triangle) must account for the remaining .

And then since a is supplementary to that -degree angle, it has to be .

So and , meaning that the sum is .

One triangle rule you must know is that the sum of a triangle's three angles must equal . Here that is important because you're given one angle and part of another , and then asked to relate two variables together. You therefore know that must equal . Using that equation:

Your job is to solve for in terms of . So first combine like terms:

Then subtract and from both sides to isolate :

And you'll see that that is your answer.

A key to solving this problem comes in recognizing that you’re dealing with similar triangles. Because lines BE, CF, and DG are all parallel, that means that the top triangle ABE is similar to two larger triangles, ACF and ADG. You know that because they all share the same angle A, and then if the horizontal lines are all parallel then the bottom two angles of each triangle will be congruent as well. You’ve established similarity through Angle-Angle-Angle.

This means that the side ratios will be the same for each triangle. And for the top triangle, ABE, you know that the ratio of the left side (AB) to right side (AE) is 6 to 9, or a ratio of 2 to 3. Note then that the remainder of the given information provides you the length of the entire right-hand side, line AG, of larger triangle ADG. If AE is 9, EF is 10, and FG is 11, then side AG is 30. And since you know that the left-hand side has a 2:3 ratio to the right, then line segment AD must be 20.

This problem tests the concept of similar triangles. First, you should recognize that triangle ACE and triangle BDE are similar. You know this because they each have the same angle measures: they share the angle created at point E and they each have a 90-degree angle, so angle CAE must match angle DBE (the top left angle in each triangle.

Because these triangles are similar, their dimensions will be proportional. Since sides, AC and BD - which are proportional sides since they are both across from the same angle, E - share a 3:2 ratio you know that each side of the smaller triangle (BDE) will be as long as its counterpart in the larger triangle (ACE). Consequently, if the bottom side CE in the larger triangle measures 30, then the proportional side for the smaller triangle (side DE) will be as long, measuring 20.

Since the question asks for the length of CD, you can take side CE (30) and subtract DE (20) to get the correct answer, 10.

This problem hinges on your ability to recognize two important themes: one, that triangle ABC is a special right triangle, a 6-8-10 side ratio, allowing you to plug in 8 for side AB. And secondly, triangles ABC and CDE are similar triangles. You know this because each triangle is marked as a right triangle and angles ACB and ECD are vertical angles, meaning that they’re congruent. Since all angles in a triangle must sum to 180, if two angles are the same then the third has to be, too, so you’ve got similar triangles here.

With that knowledge, you know that triangle ECD follows a 3-4-5 ratio (the simplified version of 6-8-10), so if the side opposite angle C in ABC is 8 and in CDE is 12, then you know you have a 9-12-15 triangle. With the knowledge that side CE measures 15, you can add that to side BC which is 10, and you have the answer of 25.

An important point of recognition on this problem is that triangles JXZ and KYZ are similar. Each has a right angle and each shares the angle at point Z, so the third angles (XJZ and YKZ, each in the upper left corner of its triangle) must be the same, too.

Given that, if you know that JX measures 16 and KY measures 8, you know that each side of the larger triangle measures twice the length of its counterpart in the smaller triangle.

You also have enough information to solve for side XZ, since you’re given the area of triangle JXZ and a line, JX, that could serve as its height (remember, to use the base x height equation for area of a triangle, you need base and height to be perpendicular; lines JX and XZ are perpendicular). Since , you can see that XZ must measure 10. And since XZ will be twice the length of YZ by the similarity ratio, YZ = 5, meaning that XY must also be 5.

Because the triangles are similar, you can tell that if the hypotenuse of the larger triangle is 15 and the hypotenuse of the smaller triangle is 10, then the sides have a ratio of 3:2 between the triangles. This allows you to fill in the sides of XYZ: side XY is 6 (which is 2/3 of its counterpart side AB which is 9) and since YZ is 8 (which is 2/3 of its counterpart side, BC, which is 12).

Since the area of a triangle is Base * Height, if you know that you have a base of 8 and a height of 6, that means that the area is .

The first important thing to note on this problem is that for each triangle, you’re given two angles: a right angle, and one other angle. Because all angles in a triangle must sum to 180 degrees, this means that you can solve for the missing angles.

In ABC, you have angles 36 and 90, meaning that to sum to 180 the missing angle ACB must be 54. And in XYZ, you have angles 90 and 54, meaning that the missing angle XZY must be 36.

Next, you can note that both triangles have the same angles: 36, 54, and 90. This means that the triangles are similar, which also means that their side ratios will be the same. You just need to make sure that you’re matching up sides based on the angles that they’re across from.

You’re given the ratio of AC to BC, which in triangle ABC is the ratio of the side opposite the right angle (AC) to the side opposite the 54-degree angle (BC). In triangle XYZ, those sides are XZ and XY, so the ratio you’re looking for is .

One important concept to recognize in this problem is that the triangles ABD and ECD are similar. Each has a right angle and they share the same angle at point D, meaning that their third angles (BAD and CED, the angles at the upper left of each triangle) must also have the same measure.

With that knowledge, you can use the given side lengths to establish a ratio between the side lengths of the triangles. If BC is 2 and CD is 8, that means that the bottom side of the triangles are 10 for the large triangle and 8 for the smaller one, or a 5:4 ratio.

You’re then told the area of the larger triangle. Knowing that the area is 25 and that area = Base x Height, you can plug in 10 as the base and determine that the height, side AB, must be 5.

Since you know that the smaller triangle’s height will be the length of 5, you can then conclude that side EC measures 4, and that is your right answer.

In beginning this problem, it is important to note that the two triangles pictured, ABC and CED, are similar. They each have a right angle and they share the vertical angle at point C, meaning that the angles at A and D must also be congruent and therefore the triangles are similar.

This means that their side lengths will be proportional, allowing you to answer this question. You’re asked to match the ratio of AB to AC, which are the side across from angle C and the hypotenuse, respectively. In triangle CED, those map to side ED and side CD, so the ratio you want is ED:CD.

As you unpack the given information, a few things should stand out:

1. Triangles ABC and ADE are similar. They each have a right angle and they each share the angle at point A, meaning that their lower-left-hand angles (at points B and D) will be the same also since all angles in a triangle must sum to 180.
2. Side BC has to measure 6, as you’re given one side (AC = 8) and the hypotenuse (AB = 10) of a right triangle. You can use Pythagorean Theorem to solve, or you can recognize the 3-4-5 side ratio (which here amounts to a 6-8-10 triangle).
3. This then allows you to use triangle similarity to determine the side lengths of the large triangle. Since the hypotenuse is 20 (segments AB and BD, each 10, combine to form a side of 20) and you know it’s a 3-4-5 just like the smaller triangle, you can fill in side DE as 12 (twice the length of BC) and segment CE as 8.

So you now know the dimensions of the parallelogram: BD is 10, BC is 6, CE is 8, and DE is 12. The sum of those four sides is 36.