Right Triangles - PSAT Math

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Question

Acute angles x and y are inside a right triangle. If x is four less than one third of 21, what is y?

Answer

We know that the sum of all the angles must be 180 and we already know one angle is 90, leaving the sum of x and y to be 90.

Solve for x to find y.

One third of 21 is 7. Four less than 7 is 3. So if angle x is 3 then that leaves 87 for angle y.

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Question

If a right triangle has one leg with a length of 4 and a hypotenuse with a length of 8, what is the measure of the angle between the hypotenuse and its other leg?

Answer

The first thing to notice is that this is a 30o:60o:90o triangle. If you draw a diagram, it is easier to see that the angle that is asked for corresponds to the side with a length of 4. This will be the smallest angle. The correct answer is 30.

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Question

In the figure above, what is the positive difference, in degrees, between the measures of angle ACB and angle CBD?

Answer

In the figure above, angle ADB is a right angle. Because side AC is a straight line, angle CDB must also be a right angle.

Let’s examine triangle ADB. The sum of the measures of the three angles must be 180 degrees, and we know that angle ADB must be 90 degrees, since it is a right angle. We can now set up the following equation.

x + y + 90 = 180

Subtract 90 from both sides.

x + y = 90

Next, we will look at triangle CDB. We know that angle CDB is also 90 degrees, so we will write the following equation:

y – 10 + 2_x_ – 20 + 90 = 180

y + 2_x_ + 60 = 180

Subtract 60 from both sides.

y + 2_x_ = 120

We have a system of equations consisting of x + y = 90 and y + 2_x_ = 120. We can solve this system by solving one equation in terms of x and then substituting this value into the second equation. Let’s solve for y in the equation x + y = 90.

x + y = 90

Subtract x from both sides.

y = 90 – x

Next, we can substitute 90 – x into the equation y + 2_x_ = 120.

(90 – x) + 2_x_ = 120

90 + x = 120

x = 120 – 90 = 30

x = 30

Since y = 90 – x, y = 90 – 30 = 60.

The question ultimately asks us to find the positive difference between the measures of ACB and CBD. The measure of ACB = 2_x_ – 20 = 2(30) – 20 = 40 degrees. The measure of CBD = y – 10 = 60 – 10 = 50 degrees. The positive difference between 50 degrees and 40 degrees is 10.

The answer is 10.

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Question

Which of the following sets of line-segment lengths can form a triangle?

Answer

In any given triangle, the sum of any two sides is greater than the third. The incorrect answers have the sum of two sides equal to the third.

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Question

In right $\Delta ABC$ , $\angle ABC$ $=$ $2x$ and $\angle BCA= \frac{x}{2}$ .

What is the value of $x$ ?

Answer

There are 180 degrees in every triangle. Since this triangle is a right triangle, one of the angles measures 90 degrees.

Therefore, $90 + 2x + \frac{x}{2}= 180$ .

$90=2.5x$

$x=36$

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Question

Refer to the above diagram. Which of the following gives a valid alternative name for $\Delta AGF$ ?

Answer

A triangle can be named after its three vertices in any order, so all of the choices given are valid.

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Question

You are given triangles $\Delta MNO$ and $\Delta PQR$ ,with $\angle N$ and $\angle Q$ both right angles, and . Which of these statements, along with what you are given, is not enough to prove that $\Delta MNO \cong \Delta PQR$ ?

I)

II) $\angle O \cong \angle R$

III) $\Delta MNO$ and $\Delta PQR$ have the same area.

Answer

, and the right angles are $\angle N$ and $\angle Q$ , so we have two right triangles with congruent legs.

If we also know that , then the hypotenuses of the right triangles are also congruent, and this sets up the conditions of the Hypotenuse-Leg Theorem.

If we also know that $\angle O \cong \angle R$ , then, along with the fact that $\angle N \cong \angle Q$ (both being right angles) and nonincluded sides , the conditions of the Angle-Angle-Side Theorem are set up.

If we also know $\Delta MNO$ and $\Delta PQR$ have the same area, we can demonstrate that the other legs are congruent. The area of a right triangle is half the product of its legs, and since we have the same areas,

$\frac{1}{2} (MN)(NO) = \frac{1}{2} (PQ)(QR)$

$2 \times \frac{1}{2} (MN)(NO) =2 \times \frac{1}{2} (PQ)(QR)$

Since ,

$\frac{(MN)(NO)}{MN} = \frac{(PQ)(QR)}{PQ}$

The legs and the included angles (the right angles) are congruent, thus setting up the conditions for the Angle-Side-Angle Postulate.

In all three cases, congruence follows, so the correct response is "Any of the three statements is enough to prove congruence."

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Question

In the figure above, line segments DC and AB are parallel. What is the perimeter of quadrilateral ABCD?

Answer

Because DC and AB are parallel, this means that angles CDB and ABD are equal. When two parallel lines are cut by a transversal line, alternate interior angles (such as CDB and ABD) are congruent.

Now, we can show that triangles ABD and BDC are similar. Both ABD and BDC are right triangles. This means that they have one angle that is the same—their right angle. Also, we just established that angles CDB and ABD are congruent. By the angle-angle similarity theorem, if two triangles have two angles that are congruent, they are similar. Thus triangles ABD and BDC are similar triangles.

We can use the similarity between triangles ABD and BDC to find the lengths of BC and CD. The length of BC is proportional to the length of AD, and the length of CD is proportional to the length of DB, because these sides correspond.

We don’t know the length of DB, but we can find it using the Pythagorean Theorem. Let a, b, and c represent the lengths of AD, AB, and BD respectively. According to the Pythagorean Theorem:

_a_2 + _b_2 = _c_2

152 + 202 = _c_2

625 = _c_2

c = 25

The length of BD is 25.

We now have what we need to find the perimeter of the quadrilateral.

Perimeter = sum of the lengths of AB, BC, CD, and DA.

Perimeter = 20 + 18.75 + 31.25 + 15 = 85

The answer is 85.

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Question

$\Delta GHJ \sim \Delta KLM$ and $\angle G$ is a right angle.

Which angle or angles must be complementary to $\angle H$ ?

I) $\angle G$

II) $\angle J$

III) $\angle K$

IV) $\angle L$

V) $\angle M$

Answer

$\angle G$ is a right angle, and, since corresponding angles of similar triangles are congruent, so is $\angle K$ . A right angle cannot be part of a complementary pair so both can be eliminated.

$\angle L$ can be eliminated, since it is congruent to $\angle H$ ; congruent angles are not necessarily complementary.

Since $\angle G$ is right angle, $\Delta GHJ$ is a right triangle, and $\angle H$ and $\angle J$ are its acute angles. That makes $\angle J$ complementary to $\angle H$ . Since $\angle M$ is congruent to $\angle J$ , it is also complementary to $\angle H$ .

The correct response is II and V only.

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Question

Note: Figures NOT drawn to scale.

Refer to the above figure. Given that $\Delta ABC \sim \Delta DE F$ , evaluate .

Answer

By the Pythagorean Theorem, since $\overline{AC}$ is the hypotenuse of a right triangle with legs 6 and 8, its measure is

$AC = \sqrt{6^{2} + 8 ^{2}} = \sqrt{36 + 64 } = \sqrt{100} = 10$ .

The similarity ratio of $\Delta DE F$ to $\Delta ABC$ is

$\frac{DF}{AC} = \frac{70}{10} = 7$ .

Likewise,

$\frac{DE}{AB} = 7$

$\frac{DE}{8} = 7$

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure. Given that $\Delta ABC \sim \Delta DE F$ , give the area of $\Delta DE F$ .

Answer

By the Pythagorean Theorem,

$BC = \sqrt{(AC)^{2} - (AB)^{2}}$

$= \sqrt{13^{2}- 5^{2}}$

$= \sqrt{169- 25}$

$= \sqrt{144}$

The similarity ratio of $\Delta DE F$ to $\Delta ABC$ is

$\frac{EF}{BC} = \frac{90}{12} = 7.5$ ,

This can be used to find :

$\frac{DE}{AB} = 7.5$

$\frac{DE}{5} = 7.5$

$DE = 5 \cdot 7.5 = 37.5$

The area of $\Delta DE F$ is therefore

$\frac{1}{2} \cdot DE \cdot EF = \frac{1}{2} \cdot 37.5 \cdot 90 =1,687.5$

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Question

Refer to the above figure. Given that $\Delta ABC \sim \Delta DE F$ , give the perimeter of $\Delta DE F$ .

Answer

By the Pythagorean Theorem,

$BC = \sqrt{(AC)^{2} - (AB)^{2}}$

$= \sqrt{13^{2}- 5^{2}}$

$= \sqrt{169- 25}$

$= \sqrt{144}$

The similarity ratio of $\Delta DE F$ to $\Delta ABC$ is

$\frac{EF}{BC} = \frac{90}{12} = 7.5$ ,

which is subsequently the ratio of the perimeter of $\Delta DE F$ to that of $\Delta ABC$ .

The perimeter of $\Delta ABC$ is

,

so the perimeter of $\Delta DE F$ can be found using this ratio:

$\frac{\textup{Perim } \Delta DEF}{\textup{Perim } \Delta ABC} = 7.5$

$\frac{\textup{Perim } \Delta DEF}{30} = 7.5$

$\textup{Perim } \Delta DEF = 7.5 \times 30 = 225$

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Question

The ratio for the side lengths of a right triangle is 3:4:5. If the perimeter is 48, what is the area of the triangle?

Answer

We can model the side lengths of the triangle as 3x, 4x, and 5x. We know that perimeter is 3x+4x+5x=48, which implies that x=4. This tells us that the legs of the right triangle are 3x=12 and 4x=16, therefore the area is A=1/2 bh=(1/2)(12)(16)=96.

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Question

A right triangle has a total perimeter of 12, and the length of its hypotenuse is 5. What is the area of this triangle?

Answer

The area of a triangle is denoted by the equation 1/2 b x h.

b stands for the length of the base, and h stands for the height.

Here we are told that the perimeter (total length of all three sides) is 12, and the hypotenuse (the side that is neither the height nor the base) is 5 units long.

So, 12-5 = 7 for the total perimeter of the base and height.

7 does not divide cleanly by two, but it does break down into 3 and 4,

and 1/2 (3x4) yields 6.

Another way to solve this would be if you recall your rules for right triangles, one of the very basic ones is the 3,4,5 triangle, which is exactly what we have here

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Question

The perimeter of a right triangle is 40 units. If the lengths of the sides are , , and units, then what is the area of the triangle?

Answer

Because the perimeter is equal to the sum of the lengths of the three sides of a triangle, we can add the three expressions for the lengths and set them equal to 40.

Perimeter:

Simplify the x terms.

Simplify the constants.

Subtract 8 from both sides.

Divide by 4

One side is 8.

The second side is

.

The third side is

.

Thus, the sides of the triangle are 8, 15, and 17.

The question asks us for the area of the triangle, which is given by the formula (1/2)bh. We are told it is a right triangle, so we can use one of the legs as the base, and the other leg as the height, since the legs will intersect at right angles. The legs of the right triangle must be the smallest sides (the longest must be the hypotenuse), which in this case are 8 and 15. So, let's assume that 8 is the base and 15 is the height.

The area of a triangle is (1/2)bh. We can substitute 8 and 15 for b and h.

$\text{Area} = \frac{1}{2}(8)(15) = 4(15) = 60$ .

The answer is 60 units squared.

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Question

Figure not drawn to scale.

In the figure above, rays PA and PB are tangent to circle O at points A and B, respectively. If the diameter of circle O is 16 units and the length of line segment PO is 17 units, what is the area, in square units, of the quadrilateral PAOB?

Answer

Because PA and PB are tangent to circle O, angles PAO and PBO must be right angles; therefore, triangles PAO and PBO are both right triangles.

Since AO and OB are both radii of circle O, they are congruent. Furthermore, because PA and PB are external tangents originating from the same point, they must also be congruent.

So, in triangles PAO and PBO, we have two sides that are congruent, and we have a congruent angle (all right angles are congruent) between them. Therefore, by the Side-Angle-Side (SAS) Theorem of congruency, triangles PAO and PBO are congruent.

Notice that quadrilateral PAOB can be broken up into triangles PAO and PBO. Since those triangles are congruent, each must comprise one half of the area of quadrilateral PAOB. As a result, if we find the area of one of the triangles, we can double it in order to find the area of the quadrilateral.

Let's determine the area of triangle PAO. We have already established that it is a right triangle. We are told that PO, which is the hypotenuse of the triangle, is equal to 17. We are also told that the diameter of circle O is 16, which means that every radius of the circle is 8, because a radius is half the size of a diameter. Since segment AO is a radius, its length must be 8.

So, triangle PAO is a right triangle with a hypotenuse of 17 and a leg of 8. We can use the Pythagorean Theorem in order to find the other leg. According to the Pythagorean Theorem, if a and b are the lengths of the legs of a right triangle, and c is the length of the hypotenuse, then:

a2 + b2 = c2

Let us let b represent the length of PA.

82 + b2 = 172

64 + b2 = 289

Subtract 64 from both sides.

b2 = 225

Take the square root of both sides.

b = 15

This means that the length of PA is 15.

Now let's apply the formula for the area of a right triangle. Because the legs of a right triangle are perpendicular, one can be considered the base, and the other can be considered the height of the triangle.

area of triangle PAO = (1/2)bh

= (1/2)(8)(15) = 60

Ultimately, we must find the area of quadrilateral PAOB; however, we previously determined that triangles PAO and PBO each comprise half of the quadrilateral. Thus, if we double the area of PAO, we would get the area of quadrilateral PAOB.

Area of PAOB = 2(area of PAO)

= 2(60) = 120 square units

The answer is 120.

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Question

If the hypotenuse of a triangle is 5 meters, which of the following is the closest value to the area of the triangle?

Answer

The answer is 12. In this circumstance, the area of the triangle cannot be smaller than its hypotenuse length, and cannot be bigger than its hypotenuse squared (that would be the area of a square).

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Question

The length of one leg of an equilateral triangle is 6. What is the area of the triangle?

Answer

$A=\frac{1}{2}(B)(H)$

The base is equal to 6.

The height of an quilateral triangle is equal to $\frac{B\sqrt{3}}{2}$ , where is the length of the base.

$A=\frac{1}{2}(6)(\frac{6\sqrt{3}}{2})=\frac{1}{2}(6)(3\sqrt{3})=9\sqrt{3}$

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Question

Triangle ABC is drawn between the points A(4, 3), B(4, 8), and C(7, 3). What is the area of ABC?

Answer

Drawing a quick sketch of this triangle will reveal that it is a right triangle. The lines AB and AC form the height and base of this triangle interchangeably, depending on how you look at it.

Either way the formula for the area of the triangle is the distance from A to B multiplied by the distance from A to C, divided by 2.

This is $\dpi{100} \small \frac{5\times 3}{2}=\frac{15}{2}$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. In terms of area, $\Delta ADB$ is what fraction of $\Delta ABC$ ?

Answer

The area of a triangle is half the product of its base and its height.

The area of $\Delta ADB$ is

$\frac{1}{2} (AC)(DB) = \frac{1}{2} \cdot 30 (DB) = 15\left ( DB \right )$

The area of $\Delta ABC$ is

$\frac{1}{2} (AD)(DB) = \frac{1}{2} \cdot 5(DB) = 2.5\left ( DB \right )$

Therefore, $\Delta ADB$ is $\frac{2.5 \cdot DB}{15 \cdot DB} = \frac{2.5}{15} = \frac{2.5 \div 2.5}{15 \div 2.5} = \frac{1}{6}$ of $\Delta ABC$ .

Note that actually finding the measure of $\overline{DB}$ is not necessary.

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