Right Triangles - SAT 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

Figure is not drawn to scale

Refer to the provided figure. Evaluate $m \angle CAD$ .

Answer

$\bigtriangleup ABC$ is an isosceles right triangle with right $\angle B$ , so, by the 45-45-90 Triangle Theorem, $m \angle ACB =45^{ \circ}$ . This angle is an exterior angle to $\bigtriangleup DAC$ , so its measure is equal to the sum of those of its two remote interior angles, $\angle DAC$ and $\angle D$ . That is,

$m\angle DAC + m\angle D = m \angle ACB$

Setting $m \angle ACB =45^{ \circ}$ and $m \angle D = 26^{\circ }$ , solve for $m\angle DAC$ :

$m\angle DAC + 26^{\circ }= 45^{\circ }$

$m\angle DAC = 45^{\circ } - 26^{\circ }= 19 ^{\circ }$

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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

A traffic light hangs t feet from the ground, over a street. A man standing the shadow of the traffic light is h feet tall, and his shadow is s feet long. How far is the man standing from the spot on the street directly under the traffic light?

Answer

We can set this problem up like a set of similar triangles.

The first triangle is created by the three points: The Traffic light, the spot beneath the traffic light, and the spot where the man is standing (which is also the spot where the traffic light's shadow is).

The height of this Triangle is "T" as given in the question, and its base is the part that we are asked to solve for.

The second triangle is created by the top of the man's head, his feet, and the end of his shadow.

The height of this Triangle is "h" as given in the questions, and the base is "s".

We set up a proportion:

$\frac{height}{base} = \frac {T}{X} = \frac{h}{s}$

where X is the distance we are asked to find. Simply cross-multiply to solve.

$X = \frac{sT}{h}$

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Question

In the given diagram, $\bigtriangleup ABC \sim \bigtriangleup DCA$ . Give the area of $\bigtriangleup DCA$ to the nearest whole number.

Answer

By the Pythagorean Theorem,

$(AC)^{2}= (AB)^{2} + (BC)^{2}$

Set , and solve for :

$(AC)^{2}= (AB)^{2} + (BC)^{2}$

$(AC)^{2}= 5^{2} + 12^{2}$

$(AC)^{2}= 25+144$

$(AC)^{2}=169$

Take the positive square root of both sides:

$AC = \sqrt{169} = 13$

$\bigtriangleup ABC \sim \bigtriangleup DCA$ , so corresponding sides are in proportion; specifically,

$\frac{DC}{AB} = \frac{AC}{CB}$

Set , and solve for :

$\frac{DC}{5} = \frac{13}{12}$

$\frac{DC}{5} \cdot 5 = \frac{13}{12} \cdot 5$

$DC = \frac{65}{12} = 5 \frac{5}{12}$

A right triangle has as its area half the product of the length of its legs, so the area of $\bigtriangleup DCA$ is

$\frac{1}{2} \cdot AC \cdot DC$

$= \frac{1}{2} \cdot 13 \cdot 5 \frac{5}{12}$

$= 35 \frac{5}{24}$

To the nearest whole number, this rounds to 35.

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Question

An meteor crashed in the desert and created an oblong shaped crater. Scientists want to find the width of the crater as it is near their research facility. Line segments AC and DE intersect at B making the angles E and D the same. If AB is 2000 meters, BD is 1800 meters, DC is 600 meters and EB is 3600 meters, what is the width of the crater?

Answer

To calculate the width of the crater, use the given information to establish that the image draws similar triangles. When triangles that have corresponding angles and a ratio to their side lengths they are considered to be similar triangles.

Identify the known information.

$\ \angle E=\angle D \\angle B=\angle B$

therefore,

$\angle A=\angle C$

and the bases of the triangles are parallel.

Also,

$\\overline{AB}=1800 \text{ m} \\overline{EB}=3600\text{ m} \\overline{BD}=1800 \text{ m} \\overline{CD}=600\text{ m}$

Set up the side ratios for this particular problem.

$\frac{AB}{BC}; \frac{AE}{CD}; \frac{EB}{BD}$

$\frac{1800}{BC}; \frac{x}{600}; \frac{3600}{1800}$

Looking at the only full ratio that is given, the scalar multiplier can be found.

$\frac{3600}{1800}=2$

Therefore, to find the width of the crater multiply $\overline{CD}$ by two.

$\\frac{x}{600}=2 \\x=2\cdot 600 \x=1200\text{ m}$

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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

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

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

Figure NOT drawn to scale.

$\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . What percent of $\bigtriangleup ABC$ is shaded in?

Choose the closest answer.

Answer

The altitude of a right triangle from the vertex of its right angle - which, here, is $\overline{BX}$ - divides the triangle into two triangles similar to each other and to the large triangle. From the Pythagorean Theorem, the hypotenuse of $\bigtriangleup ABC$ has length

$AC = \sqrt{13 ^{2}+ 21^{2}} = \sqrt{169+441} = \sqrt{610}$ .

The similarity ratio of $\bigtriangleup AXB$ to $\bigtriangleup ABC$ is the ratio of the lengths of the hypotenuses:

$\frac{AB}{AC} = \frac{13}{\sqrt{610}}$

The ratio of the areas of two similar triangles is the square of their similarity ratio, which here is

$\left ( \frac{13}{\sqrt{610}} \right )^{2} = \frac{169}{610}$

Therefore, the area of $\bigtriangleup AXB$ is

$\frac{169}{610} \times 100 % \approx 27.7 %$

the overall area of $\bigtriangleup ABC$ . This makes the closest response.

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Question

The vertices of a right triangle on the coordinate axes are at the origin, , and . Give the area of the triangle.

Answer

The triangle in question can be drawn as the following:

The lengths of the legs of the triangle are 12, the distance from the origin to , and 8, the distance from the origin to . The area of a right triangle is equal to half the product of the lengths of the legs, so set in the formula:

$A = \frac{1}{2} ab= \frac{1}{2} \cdot 12 \cdot 8 = 48$

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Question

Angela drives 30 miles north and then 40 miles east. How far is she from where she began?

Answer

By drawing Angela’s route, we can connect her end point and her start point with a straight line and will then have a right triangle. The Pythagorean theorem can be used to solve for how far she is from the starting point: a2+b2=c2, 302+402=c2, c=50. It can also be noted that Angela’s route represents a multiple of the 3-4-5 Pythagorean triple.

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