Triangles

The answer is A. Among other formulas, the CLT provides you with $a^2+b^2=c^2$ for right triangles. (We recommend memorizing the reference formulas so you don’t have to spend time consulting them on test day!) In this formula, $a$ and $b$ are interchangeable, representing the two legs of the triangle. The most common mistake is to confuse one of the legs with the hypotenuse, plugging in the wrong value for $c$. For this reason, it’s helpful to think of the formula in the other $c^2=a^2+b^2$ and make sure to identify c first. In this case, we know the value of $c$: it’s 13. So we can write the formula either as $13^2=12^2+b^2$ or $13^2=a^2+12^2$. Either way, solving for our variable will yield $5$ ($169-144=25$, which is $5^2$). We shouldn’t need to use answer elimination on this question, but if you used UnCLES method step of looking at the answers, you may have noticed that answer choice D is impossible because it makes one of the legs larger than the hypotenuse.

Note: $5:12:13$ is a Pythagorean triple, so if you have that memorized, you don’t need to use the Pythagorean theorem at all!

The answer is B. Similar triangles have congruent corresponding angles, so when the question tells us that angle $A$ corresponds to angle $F$, we know that those two angles are congruent. In addition, we may assume that the alphabetical order of the angles shows us the other corresponding pairs: $B$ and $G$ are congruent, as are $C$ and $H$. That means that to discover the measure of angle $F$, we could also calculate the measure of angle $A$. The information given pertains to two different triangles, so we need to replace one of the angles with its corresponding partner from the other triangle. For example, if angle $B$ has a measure of $48°$, then so does angle $G$. Now that we have the measure of angles $G$ and $H$, we can subtract their measure from $180°$: $180-72-48=60$.

The answer is B. If you’ve completed the course module on logic, you have encountered counterexample questions, which ask you to disprove a statement by finding an exception to the rule proposed. In this case, we need to find the hypotenuse of each triangle presented in the answer choices; don’t forget about the helpful Pythagorean triples! The classic $3:4:5$ triangle eliminates choice $A$ right away; the hypotenuse is certainly an integer. If you remember that Pythagorean triples are ratios and that their multiples are also triples, you might notice that $9$ and $12$ (choice $D$) are multiples of $3$ and $4$, respectively, and therefore that this another trio of integers.

From here, you have a clear path to the answer if you remember that $8:15:17$ is also a triple. That means answer $C$ is eliminated as containing three integers. If you didn’t remember that triple, it would be better to test answer $B$ since it has smaller numbers. Trying the Pythagorean theorem, we get $7^2+8^2=c^2$. If, therefore, $c=\sqrt{113}$, we need to determine whether $113$ is a perfect square. Since $10^2=100$ and $11^2=121$, it becomes evident that $\sqrt{113}$ is not an integer, confirming choice $B$ as the answer.

The answer is A. As is typical, drawing is the first step! If the base of the figure is $24$ and the other two sides are congruent, we know from our Approach question to draw a vertical line down the middle of the triangle–especially since that’s exactly the length we’re trying to find in this case. That vertical line, or altitude, divides the isosceles triangle into two right triangles, and it divides the base evenly into $12$ and $12$. That means each of our triangles has one leg of length $12$ and a hypotenuse of $13$. You can use $a^2+b^2=c^2$, but do you see the Pythagorean triple? It’s $5:12:13$. $5$ is the missing height.

The answer is B. Examining the answers, we can eliminate choices $C$ and $D$ right away because an isosceles triangle has two equal sides and this triangle does not. This triangle is scalene, meaning that all of its side lengths are different. Is it a right triangle? We answer that question by plugging in the side lengths into the Pythagorean theorem to see if they satisfy the equation. $5^2+12^2=25+144=169$. But $169$ is not the square of $14$ (the square of $14$ is $196$), so this is not a right triangle.

One more note: once again, knowing Pythagorean triples helps us answer more quickly. If you remember that $5:12:13$ is one of the Pythagorean triples, then those numbers constitute the lengths of a right triangle. Changing the largest number to 14 changes the hypotenuse, so we no longer have a right triangle.

The answer is C. This problem provides partial information that tells us everything we need to know to complete the triangle. First off, we know that the angle not mentioned must be $45°$, because $180-90-45-45$. So we have an isosceles triangle, also known as a $45-45-90$. Consulting the CLT reference formula list, we see that the ratio of sides in a $45-45-90$ is $1:1:\sqrt{2}$.

What does this mean for the third side, whose length is not mentioned here? The clue is that we already know the length of two equal sides, so those must be the legs; the hypotenuse is missing. The tricky part is to apply the ratio to the known length of the legs in order to find the length of the hypotenuse. If we see that the two lengths of $5\sqrt{2}$ must correspond to the “1’s” in the ratio, then the hypotenuse must correspond to the $\sqrt{2}$. That means we need to multiply the leg length by $\sqrt{2}$ to find the hypotenuse. What is $\sqrt{2}$ times itself? Simply $2$, because we are essentially squaring a square root, thereby canceling it. This means that $5\sqrt{2}\times \sqrt{2}=5\times 2=10$. The hypotenuse is $10$. We can now add that hypotenuse to the two legs and discover our perimeter of $10\sqrt{2}+10$.

In case it helps you to see that ratio calculation done a different way, consider setting up a proportion as follows:

$1/5\sqrt{2}=\sqrt{2}/x$

In this proportion, the ratio parts ($1$ and $\sqrt{2}$) are in the numerator and the side lengths ($5\sqrt{2}$ and x for the unknown hypotenuse) are in the denominator. Once we set up a proportion we should always cross-multiply; doing so gives us $x=(5\sqrt{2})\sqrt{2}=10$.

The answer is D. Why this question in a lesson on triangles? Consider it a hint! If you get a non-standard figure on the CLT, such as this trapezoid, you should feel free to draw extra lines on the figure to divide it into more accessible parts. If we drop a vertical line from the top of the figure down to point $D$, we’ll create a right triangle, as shown below.

The answer is A. In geometry, you probably learned about theorems side as $SSS$ (“side-side-side”) and $SAS$ (“side-angle-side”) that can be used to prove two triangles are congruent. Here are some reminders about that process:

1. Three congruent side pairs ($SSS$) mean the triangles are congruent; three congruent angle pairs ($AAA$) only prove that the triangles are similar.
2. If you have two congruent side pairs and one congruent angle pair, the angle pair must be the included angle between the two congruent sides in order to prove congruence. “Side-side-angle,” in that order, does not prove congruence.
3. Two congruent angle pairs and one congruent side pair always proves congruence; it doesn’t matter where the side is relative to the angles.

Given these principles, what can we conclude in this case? In the first place, this problem can’t be solved without noticing that we already have two congruent side pairs, not just one, because $AC$ is equal to itself and forms a side of both triangles. This means that we can prove congruence with either 1) a third side pair or 2) the included angle pair. Choice $C$ gives us the third side pair and choice $B$ shows us that the included angles (that is, the angles between the two sides we already know are part of congruent pairs) are indeed congruent. Both of those choices can be eliminated since we’re looking for the choice that is not sufficient.

Where from here? Think about what we learn from choice $D$. If the triangles are similar, then all their corresponding angle pairs are congruent. So we have the angles we needed to prove congruence. Choice $A$ is the answer that does not prove congruence because it only supplies that insufficient “side-side-angle” relationship mentioned in principle #2 above.

The answer is C. Why is a hexagon question in this lesson? Because a regular hexagon divides nicely into six equilateral triangles, as shown in the figure below.

The answer is D. Jumping to Triangle $B$, we can use the area of an equilateral triangle formula to find the area as $4\sqrt{3}$. The perimeter is all the sides added together: $4+4+4=12$. As mentioned in Flashcard Fodder, $\sqrt{2}$ is approximately $1.4$ and $\sqrt{3}$ is approximately $1.7$. So the area of this triangle is approximately $5$, which is much smaller than its perimeter. Using smart test strategy at this stage, we can eliminate any answer suggesting that the perimeter of Triangle $B$ is less than its area. That takes care of choice $C$.

Triangle $A$ is not as straightforward. As with just about any isosceles triangle problem, it helps to draw the triangle with the equal sides diagonally on the right and left and draw a vertical line down the middle. In this case, the vertical line divides the triangles into two $30-60-90$ right triangles. Using those side ratios and discerning that the side of length $3$ would be the hypotenuse of each of these triangles, we can infer that the smallest side would $1.5$ and the middle side would be $1.5\sqrt{3}$ or $3/2\sqrt{3}$. So the height of the isosceles triangle, being across from the $30°$ angle, is $1.5$, and the base is both right triangle bases added together, or $3\sqrt{3}$. That means its area is $2.25\sqrt{3}$ and its perimeter is $6+3\sqrt{3}$. How do these numbers compare with the ones for Triangle B? $2.25\sqrt{3}$ is the smallest number we’ve found yet, as it’s clearly smaller than $4\sqrt{3}$. This means that the area of triangle $A$ is the smallest value, so even just from that information, we can see that $D$ is the only possible answer. To confirm, we can note that the perimeter of triangle $A$ is about $11$, just a little bit less than the perimeter of triangle $B$. This also matches the list in answer choice $D$.

This is a difficult problem, and if you have to make a guess, you might note that the side lengths for both triangles are fairly small, reasoning that, since the area formula for a triangle includes a $½$, the areas of these triangles are likely going to smaller than the perimeters. (By contrast, if we imagine an equilateral triangle with sides of length $100$, its area will be way larger than its perimeter.) That reasoning would lead you to the answer listing the areas as the two smallest quantities and the perimeter as the two largest quantities.