Posts tagged with: triangles

Thomas is making a sign in the shape of a regular hexagon with 4-inch sides, which he will cut out from a rectangular sheet of metal. What is the sum of the areas of the four triangles that will be removed from the rectangle?

So it’ll look like this:

It’s helpful just to know that a regular hexagon’s interior angles all measure 120°, but you can also calculate that using (n-2)\times 180^\circ:

\dfrac{(6-2)\times 180^\circ}{6}=120^\circ

That means that the four triangles you’re cutting off the rectangle are each 30°-60°-90° triangles with 4-inch hypotenuses.

Those will have legs of 2 and 2\sqrt{3}, and therefore areas of \dfrac{1}{2}(2)(2\sqrt{3})=2\sqrt{3}. Since there are four such rectangles, the total area you’re cutting off is 8\sqrt{3}

A triangle’s base was increased by 15%. If its area is increased by 38%, what percent was the height of the triangle increased by?

The easiest way to get this question is to plug in! Say the base and height of the original triangle are each 10. The formula for finding the area of a triangle is A=\frac{1}{2}bh, where b and h are the base and height, so the area of our original triangle is A=\frac{1}{2}(10)(10)=50.

Increasing the base by 15% brings it from 10 to 10\times 1.15=11.5. Increasing the area by 38% brings it from 50 to 50\times 1.38=69. Plug those back into the formula to solve for the new height:

    \begin{align*}69&=\frac{1}{2}11.5h\\138&=11.5h\\12&=h\end{align*}

If the original height was 10 and the new height is 12, then the height increased by 20%.

I’ll draw this as best I can:

image

 

Look OK? Now let me draw a few more segments in blue…

image

See what’s going on there? All of the small triangles in the figure are the same! (You can prove this with triangle similarity/congruence rules easily enough—I won’t spend the time doing so here, though.)

image

We know that the area of the big triangle is (½)(10)(10) = 50. Further, we know that of the 8 small congruent triangles in the figure, 4 of them are inside the square and 4 aren’t. Because those 8 triangles together have an area of 50, the square has an area of half that: 25.

from Tumblr https://ift.tt/2vZoYXb

Question from March SAT: Section 4 #33

In triangle ABC, angle A measures 48 degrees, angle B measures 88 degrees, and angle C measures 44 degrees. Triangle ABC is similar to triangle LMN, such that LM/AB = MN/BC = LN/AC = 3. What is the measure, in degrees, of angle L ?

When triangles are similar, their corresponding angles are congruent. Notationally, when we say triangle ABC is similar to triangle LMN, we list the corresponding angles in order, so A corresponds with L, B corresponds with M, and C corresponds with N. The question also reassures us of this by telling us which sides correspond–angle A will be across from BC, and angle L will be across from MN.

Long story short: because the question tells you that angle A measures 48°, and that triangles ABC and LMN are similar, you know that angle L also measures 48°.

Hello Mr. Mike. Can you please explain how to solve Similar triangle problems? I especially find confusing identifying equal sides and making proportions. I have an example problem, please see the link below.
http://ibb.co/fLJ1Hc

First, you need to know the similarity rules. Triangles are similar if you can establish that two sets of their angles are congruent (Angle-Angle), if two sets of sides are proportional and the angles between them are congruent (Side-Angle-Side), or if all three sets of sides are proportional (Side-Side-Side). In the case of the screenshot you’ve provided, we have Angle-Angle: both triangles have 90° angles and both triangles contain angle L.

Once you’ve established that you’ve got similar triangles, you kinda have to follow where the question leads—there are lots of combinations of angles and sides that could be given and/or asked for.

Continuing with the example you’ve provided, we can solve for LY using the Pythagorean theorem:

    \begin{align*}AL^2+AY^2&=LY^2\\3.5^2+7^2&=LY^2\\61.25&=LY^2\\\sqrt{61.25}&=LY\end{align*}

Now this is important: LY is the hypotenuse of the small triangle, but it’s the short leg of the big triangle! At this point in these questions I find it helpful to draw both triangles in the same orientation. For example, you might draw these ones like so:

Doing that makes it much easier to see that the long leg of triangle WLY will be twice the length of the short leg, just like triangle YLA.

From there, all we need to do is the Pythagorean theorem again to find LW.

    \begin{align*}LY^2+WY^2&=LW^2\\\sqrt{61.25}^2+\left(2\sqrt{61.25}\right)^2&=LW^2\\306.25&=LW^2\\17.5&=LW\end{align*}

How do you do Test 2 Section 3 Number 18?

The key here is to recognize that you’re dealing with similar triangles (pro tip: similar triangle questions often take this “hourglass” form). The two angles at point B are vertical, so they must be congruent. And because segments AE and CD are parallel, you’ve got alternate interior angles for the rest. When all the angles in two triangles are congruent, those triangles are similar.

(It might be easier to see the alternate interior angles if you extend the lines…expand the image on the right.)

Anyway, now that you’ve established that these are similar triangles, you just need to use ratios to solve. If AB=10 and [latexBD=5[/latex], then each set of corresponding sides will be in the same ratio. So we can solve for BC thusly:

    \begin{align*}\dfrac{AB}{BD}&=\dfrac{BE}{BC}\\\\\dfrac{10}{5}&=\dfrac{8}{BC}\\\\2&=\dfrac{8}{BC}\\BC&=4\end{align*}

But wait—you’re not quite done! The question asks for CE, not BC!

CEBCBE = 4 + 8 = 12

There. Now you’re done.

Test 7 Section 3 #17

The first thing to do on any find-the-measure-of-a-certain-angle problem is complete any 180°s you can. In this case, triangle PQR can be completed (the measure of angle PRQ must be 50°) and \overline{MQ} can be completed (the measure of angle MPR must be 120°).

From there, you’re almost home. The question tells you that MPPR, so you know that triangle MPR is isosceles. That means that angles PMR and PRM must be congruent! Because that triangle already has a 120° angle in it, the two unknowns must add up to 60°. Because they must be congruent, they must each be 30°.

Therefore, the measure of angle QMR is 30°.

Could you help me with question 9 on page 38 please? I don’t understand how to solve it even using back-solving

Yeah, this one is pretty tough! 🙂

To backsolve it successfully, it’s helpful to also plug in. Say that the base and height of triangle A are both 10. Then if you try, say, answer choice C, which says that p = 23, you know that r = 18, and therefore that the base of triangle B is 12.3 (23% more than 10) and the height is 8.2 (18% less than 10). Does that give you the same area for both triangles?

\text{Area}_A=\dfrac{1}{2}(10)(10)=50

\text{Area}_B=\dfrac{1}{2}(12.3)(8.2)=50.43

Not quite, but you’re close! For that reason, probably makes sense to try the next closest choice, D. Sure enough, that one works.

I don’t know how to navigate your site yet, so please forgive me if you have already answered this question. Regarding question number 8 in your book, will you further explain why y=180-x is the same angle degree as the unmarked angle? I know y=2x because of geometry, but I do not understand how y can also equal x?

You’re talking about this question from the triangles chapter, right?

You’ve actually hit on a totally legit way to solve the question: If y=2x due to the exterior angle theorem, and y=180-x, then you can substitute and solve for x:

    \begin{align*}2x&=180-x\\3x&=180\\x&=60\end{align*}

Once you know that x=60, you can determine that the triangle is equilateral, so you need a perimeter that’s a multiple of 3.

To answer your question more directly, look at just the top part of the figure:

See how you’ve got a straight line there? That means the angle that’s NOT marked y^\circ (i.e., the angle inside the triangle) has a measure of (180-y)^{\circ}. Since the question tells you that y=180-x, you can substitute:

    \begin{align*}180-(180-x)=180-180+x=x\end{align*}

 

How do you do #18 in Test 6 Section 3 without a calculator?

The key to this one is recognizing that you’ve got similar triangles. First, both triangle CDB and triangle CEA contain angle C. The question also tells you that \overline{BD}\parallel\overline{AE}, which means that angle CDB is also a right angle. Therefore, you have angle-angle similarity.

It’s also good to recognize a 6-8-10 triangle quickly. Know your Pythagorean triples! 🙂

From there, all you need to do is recognize that the length of \overline{AE} is 3 times the length of \overline{BD}, which means all the sides of triangle CEA are 3 times the length of their corresponding sides in triangle CDB. Therefore, because the length of \overline{CD} is 10, the length of \overline{CE} must be 30.

Dear Mike,

Can you please do #16 in the Official Practice Test #4 NO CALCULATOR section
(aka Section 3)?

You bet!

This is a question about similar triangles. Here’s a simplified figure we can use to work through it.

Note that I’ve taken the step of drawing a vertical line from the top vertex to the base. Because the shelves are parallel, we know the angles they form with the vertical segment I drew and with the sides of the triangle will all be the same. That tells us that we have angle-angle similarity, so we know that the 2:3:1 ratio we have on the left side will also apply down the middle, like so:

Since we know that the height of the unit is 18 inches, we can solve for y:

y + 3y + 2y = 18
6y = 18
y = 3

From there, we know that the shampoo needs to fit on the middle shelf, so it needs to fit into a space that’s 3y = 3(3) = 9 inches tall.

Can you please do #20 from the no calculator section in test 5?

Sure. Here’s (basically) the image you’re given:

To find the value of x, you need to remember that the angles that make up straight lines and the angles that make up triangles will both sum to 180°. You’ve got a straight line (marked in red below) that helps you calculate that the upper left angle in the triangle is 180° – 106° = 74°.

From there, you can calculate the measure of the third angle in the triangle: 180° – 23° – 74° = 83°. Just one more 180° straight line to find the value of x:

= 180º – 83°
= 97°

A note since I’m sure someone would bring it up in the comments if I didn’t: another way to go here is the Exterior Angle Theorem, which states that the measure of an exterior angle of a triangle (the 106° angle in this case) equals the sum of the measures of the nonadjacent interior angles (the 23° and the 83°). You can use that theorem as a “shortcut” to find the 83° angle in one fewer step if you like. 🙂

Can you please explain College Board Test 5 Section 4 #36?

I understand that angle A is half of x- the only part of the college board explanation I’m not understanding is why we do (360-x) while setting up the equation. Thanks!

Sure. Without looking at CB’s solution, here’s how I do it.

First, as you pointed out, the measure of angle A is half of . That’ll be useful. The other thing I think is useful is to draw segment BC and mark the angles you create (I’ll say their measures are  and ). Now you’ve got two triangles, ABC and PBC.

Each triangle, of course, has angles that add up to 180°. You can write two equations that show this:

Triangle ABC: \dfrac{1}{2}x+20+20+b+c=180

Triangle PBC: x+b+c=180

All you need to do to solve for x is set those left sides equal to each other since they both equal 180 (b and c will cancel right out!).

    \begin{align*}\dfrac{1}{2}x+20+20+b+c&=x+b+c\\\dfrac{1}{2}x+40&=x\\40&=\dfrac{1}{2}x\\80&=x\end{align*}

Test 3 Section 3 #18

First, here’s a mockup of the figure (not to scale):

You’re told that 180 – z = 2y, and then you’re told that y = 75. (This is typical of the SAT: give you the information in the reverse order in which you should use it.) So let’s solve for z, and then mark up the figure a bit.

180 – z = 2(75)
180 – z = 150
z = –30
z = 30

To move forward from here, you need to remember the base angle theorem: in isosceles triangles, the angles across from the congruent sides are also congruent. So the blue angles and the red angles below are congruent.

We want x, so let’s figure out what those red angles are (I’ll use a for them):

30 + 2a = 180
2a = 150
a = 75

If the red angles each measure 75°, then remembering that a straight line measures 180°, we can solve for x:

75 + x = 180
x = 105

Triangles ABC and ABD share side AB.
Triangle ABC has area Q and triangle ABD has area R. If AD is longer than AC and BD is longer than BC, which of the following could be true?
I R> Q
II R=Q
III R < Q
I chose “I” only but the answer was E (all of them could be.) How can the second and third condition be true?
Thanks in advance!

Untitled_1

Above is an example of R < Q. If you imagine dragging D downward, eventually the area of triangle ABD will equal the area of triangle ABC, and then after that it will be bigger.