Right Triangles - High School Math
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Acute angles x and y are inside a right triangle. If x is four less than one third of 21, what is y?
Acute angles x and y are inside a right triangle. If x is four less than one third of 21, what is y?
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.
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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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?
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?
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.
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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In the figure above, what is the positive difference, in degrees, between the measures of angle ACB and angle CBD?
In the figure above, what is the positive difference, in degrees, between the measures of angle ACB and angle CBD?
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.
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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Which of the following sets of line-segment lengths can form a triangle?
Which of the following sets of line-segment lengths can form a triangle?
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.
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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In right 
, 
and 
.
What is the value of 
?
In right
What is the value of
There are 180 degrees in every triangle. Since this triangle is a right triangle, one of the angles measures 90 degrees.
Therefore, 
.
There are 180 degrees in every triangle. Since this triangle is a right triangle, one of the angles measures 90 degrees.
Therefore,
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If angle 
and angle 
, what is the value for angle 
?
If angle
For this problem, remember that the sum of the degrees in a triangle is 
.
That means that 
.
Plug in our given values to solve:
Subtract 
from both sides:
For this problem, remember that the sum of the degrees in a triangle is
That means that
Plug in our given values to solve:
Subtract
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A right triangle is defined by the points (1, 1), (1, 5), and (4, 1). The triangle's sides are enlarged by a factor of 3 to form a new triangle. What is the area of the new triangle?
A right triangle is defined by the points (1, 1), (1, 5), and (4, 1). The triangle's sides are enlarged by a factor of 3 to form a new triangle. What is the area of the new triangle?
The points define a 3-4-5 right triangle. Its area is A = 1/2bh = ½(3)(4) = 6. The scale factor (SF) of the new triangle is 3. The area of the new triangle is given by Anew = (SF)2 x (Aold) =
32 x 6 = 9 x 6 = 54 square units (since the units are not given in the original problem).
NOTE: For a volume problem: Vnew = (SF)3 x (Vold).
The points define a 3-4-5 right triangle. Its area is A = 1/2bh = ½(3)(4) = 6. The scale factor (SF) of the new triangle is 3. The area of the new triangle is given by Anew = (SF)2 x (Aold) =
32 x 6 = 9 x 6 = 54 square units (since the units are not given in the original problem).
NOTE: For a volume problem: Vnew = (SF)3 x (Vold).
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You have two right triangles that are similar. The base of the first is 6 and the height is 9. If the base of the second triangle is 20, what is the height of the second triangle?
You have two right triangles that are similar. The base of the first is 6 and the height is 9. If the base of the second triangle is 20, what is the height of the second triangle?
Similar triangles are proportional.
Base1 / Height1 = Base2 / Height2
6 / 9 = 20 / Height2
Cross multiply and solve for Height2
6 / 9 = 20 / Height2
6 * Height2= 20 * 9
Height2= 30
Similar triangles are proportional.
Base1 / Height1 = Base2 / Height2
6 / 9 = 20 / Height2
Cross multiply and solve for Height2
6 / 9 = 20 / Height2
6 * Height2= 20 * 9
Height2= 30
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In the figure above, line segments DC and AB are parallel. What is the perimeter of quadrilateral ABCD?
In the figure above, line segments DC and AB are parallel. What is the perimeter of quadrilateral ABCD?
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.
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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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?
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?
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.
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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A right triangle has a total perimeter of 12, and the length of its hypotenuse is 5. What is the area of this triangle?
A right triangle has a total perimeter of 12, and the length of its hypotenuse is 5. What is the area of this triangle?
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
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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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?
The perimeter of a right triangle is 40 units. If the lengths of the sides are
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.
.
The answer is 60 units squared.
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.
The answer is 60 units squared.
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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?
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?
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.
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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If the hypotenuse of a triangle is 5 meters, which of the following is the closest value to the area of the triangle?
If the hypotenuse of a triangle is 5 meters, which of the following is the closest value to the area of the triangle?
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).
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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The length of one leg of an equilateral triangle is 6. What is the area of the triangle?
The length of one leg of an equilateral triangle is 6. What is the area of the triangle?
The base is equal to 6.
The height of an quilateral triangle is equal to 
, where 
is the length of the base.
The base is equal to 6.
The height of an quilateral triangle is equal to
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Triangle ABC is drawn between the points A(4, 3), B(4, 8), and C(7, 3). What is the area of ABC?
Triangle ABC is drawn between the points A(4, 3), B(4, 8), and C(7, 3). What is the area of ABC?
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 
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
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If 
, 
, and 
, what is the area of this triangle?
If
The formula for area of a triangble is 
.
Since 
, 
will be the base and 
will be the height.
Plug in our given values to solve:
The formula for area of a triangble is
Since
Plug in our given values to solve:
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You are given a right triangle with legs of length 
feet and 
feet. What is the area of this triangle?
You are given a right triangle with legs of length
Since we have a right triangle, and are given the lengths of the legs, we are by definition given the base and height. Recall that the legs of the triangle are those next to the right angle, and the hypotenuse is across from the right angle.
Recall the formula for the area of a triangle:
Since we know the base and height we plug them into the equation above:
We are given units of feet, so the area will be in terms of square feet. Then, our answer is 
square feet.
Since we have a right triangle, and are given the lengths of the legs, we are by definition given the base and height. Recall that the legs of the triangle are those next to the right angle, and the hypotenuse is across from the right angle.
Recall the formula for the area of a triangle:
Since we know the base and height we plug them into the equation above:
We are given units of feet, so the area will be in terms of square feet. Then, our answer is
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Find the area of the following triangle:
Find the area of the following triangle:
The formula for the area of a right triangle is:
where 
is the length of the base and 
is the length of the height.
Use the formulas for a a 
triangle to find the length of the base. The formula for a 
triangle is 
.
Our 
triangle is: 
Plugging in our values, we get:
The formula for the area of a right triangle is:
where
Use the formulas for a a
Our
Plugging in our values, we get:
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Find the area of the following right triangle:
Find the area of the following right triangle:
The formula for the area of a right triangle is:
where 
is the length of the base and 
is the length of the height.
Use the formulas for a 
triangle to find the length of the base and height. The formula for a 
triangle is 
Our 
triangle is: 
Plugging in our values, we get:
The formula for the area of a right triangle is:
where
Use the formulas for a
Our
Plugging in our values, we get:
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