Triangles, Lines, & Angles - SAT Math

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 .

Intersecting and parallel lines show up in many different geometric figures: parallelograms, trapezoids, squares, etc. Anytime you see these in a question, you have to properly leverage the essential properties of supplementary and vertical angles. On this problem, the fastest way to find y is to realize that 5x in the bottom left corner is supplementary to 2x + 5 in the bottom right (because of the intersection of two parallel lines). Therefore, 5x + 2x + 5 = 180 and x = 25. Once you have that information, you can use the fact that the sum of the interior angles of a triangle is 180 and see that x + 5x + 2y = 180 . Putting in 25 for x you see that 25+125+2y =180 and 2y =30. The correct answer is 15.

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.

If the perimeter of triangle ABC is twice as long as the perimeter of triangle DEF, and you know that the triangles are similar, that then means that each side length of ABC is twice as long as its corresponding side in triangle DEF. That also means that the heights have the same 2:1 ratio: the height of ABC is twice the length of the height of DEF.

Since the formula for area of a triangle is Base x Height, you can express the area of triangle DEF as bh and the area of ABC as . You’ll then see that the areas of ABC to DEF are and bh, for a ratio of 4:1.

NOTE: It can seem surprising that the ratio isn’t 2:1 if each length of one triangle is twice its corresponding length in the other. But keep in mind that for an area you multiply two lengths together, and go from a unit like “inches” to a unit like “square inches.” Because each length is multiplied by 2, the effect is exacerbated.

Here the SAT gives you a pair of lines with a transversal, but it does not tell you that the lines are parallel - it asks you to prove it. You are told that . Since angle and angle are vertical angles and angles and are vertical angles, you know that and . That means you can write your equation as:

, or

If that means that as well. A straight line contains 180 degrees, so you know that . And since , you can conclude that as well. From here, you can reverse engineer the same sort of equation you solved with the first set of angles. If and and are vertical angles and and are vertical angles, you can conclude that . From there you can set up the equation . Statement I is true.

In order for the horizontal lines to be parallel, you need to know that either the alternate exterior angles or the alternate interior angles are equal. Since you have already proven that , you know also that . Since you have a pair of alternate exterior angles, the two lines must be parallel. Statement II is also true.

Statement III, however, is not necessarily true. If then all angles would equal 90. However without that knowledge, you cannot come to any conclusions about the relationship between and . Statement III is not necessarily true, so the correct answer is I and II only.

Since lines x and y will add to a total of 180 degrees, you have two equations to work with:

x + y = 180

x = 3y

This means you can substitute 3y for x in order to solve for y:

3y + y = 180

4y = 180

y = 45

And since z will also sum with y to 180, then z must be 180 - 45 = 135 degrees.

If h is 121, then the angle immediately below h must be 59, as it is a supplementary angle formed by the diagonal line. Since g and k are parallel, this 59 degree angle must exactly match p as they are alternative interior angles.

Two angle rules are very important for this question:

1. The sum of the interior angles of a triangle is always 180. Here, since you have a 90-degree angle (CED) and a 35-degree angle (EDC) in the bottom triangle, you can then conclude that angle ECD must be 55.

2. Supplementary angles, angles that are adjacent to each other when two straight lines intersect, must sum to 180 degrees. If you know that ECD is 55, then ACE as a supplementary angle must form the other 125 degrees for those two angles to sum to 180. Therefore, the correct answer is 125.

An important thing to recognize in this problem is that you're dealing with two intersecting triangles that create external supplementary angles along the straight line on the bottom. To see this, consider the diagram below for which angles x and y have been added:

Angle y is an external supplementary angle to the triangle beside it so y = a + c. Why? Remember that y is supplementary to the angle beside it (x + 30) and (a + c) is supplementary to that same angle (the sum of interior angles of a triangle = 180.) Therefore y and (a + c) are identical. Anytime you have a straight line drawn off of a triangle you should recognize that the external supplementary angle equals the sum of the two opposite angles.

Using the same logic, you can see that x = b + d in the other intersecting triangle. Since the problem is asking for a + b + c + d, you should recognize that this question is really the same as what is x + y. Why? You can substitute x for b + d and y for a + c in the question stem. Since x + y = 180 - 30 on the straight line along the bottom, the correct answer is 150.

Note that another way to solve this problem involves seeing two large obtuse triangles: one with the angles a, c, and (x+30) and the other with the angles b, d, and (y+30). If you do that, you would have:

a+c+x+30=180, so a+c+x=150

b+d+y+30=180, so b+d+y=150

And you know that x+y+30=180 because x, 30, and y are all angles that make up the 180-degree straight line across the bottom of the figure. So x+y=150.

You can then sum the triangle equations:

a+c+x+b+d+y=150+150=300

And then plug in x+y = 150 and you're left with a+b+c+d=150.

This problem hinges on two important geometry rules:

1. The sum of all interior angles in a triangle is 180. Here you know that in the top triangle you have angles of 30 and 80, meaning that the angle at the point where lines intersect must be 70, since 30+80=110, and the last angle must sum to 180.

2. Vertical angles - angles opposite one another when two straight lines intersect - are congruent. Because you have identified that the angle at the bottom of the triangle at the top is 70, that also means that the top, unlabeled angle of the bottom triangle is 70. That then lets you add 70+50+ as the three angles in the bottom triangle, and since they must sum to 180 that means that .

This problem heavily leverages two rules:

1. The sum of the angles in a triangle is 180.

2. Supplementary angles - adjacent angles created when one line intersects another - must sum to 180.

Here you can first leverage the 140-degree angle to fill in that its adjacent neighbor - its supplementary partner - must then be 40. and that gives you two of the three angles in the uppermost triangle: 20 and 40. You can use that to determine that the third angle must then be 120.

From there you should see that the 120-degree angle is a vertical angle, meaning that its opposite will also be 120. And that gives you a second angle in the lower-right triangle. Knowing that you have angles of 15 and 120 means that the third angle of that triangle must be 45. And since that angle is supplementary to angle x, x must then be 135.

This problem heavily leans on two important lines-and-angles rules:

1. The sum of the three interior angles of a triangle is always 180.

2. Supplementary angles - angles next to each other formed by two lines intersecting - must also sum to 180.

Here you can then determine that the angle next to the 95-degree angle is 85, and since that angle is the lower-right hand angle of the little triangle at the top, you can close out that triangle. With angles of 40 and 85, that means that the lower left hand angle must be 55.

From there, you can use the fact that parallel lines will lead to congruent angles. Since lines and are parallel, the angle next to will be 55 degrees, meaning that will then be 125.

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.