How to find the area of a right triangle - ACT Math

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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 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

Find the area of the following right triangle to the nearest integer

Note: The triangle is not necessarily to scale

Answer

The equation used to find the area of a right triangle is: $A = \frac{1}{2}b\cdot h$ where A is the area, b is the base, and h is the height of the triangle. In this question, we are given the height, so we need to figure out the base in order to find the area. Since we know both the height and hypotenuse of the triangle, the quickest way to finding the base is using the pythagorean theorem, $a^{2}+b^{2}=c^{2}$ . a = the height, b = the base, and c = the hypotenuse.

Using the given information, we can write $8^{2} + b^{2} = 13^{2}$ . Solving for b, we get $b^{2} = 105$ or $b \approx 10.25$ . Now that we have both the base and height, we can solve the original equation for the area of the triangle. $A = \frac{1}{2}\cdot10.25\cdot8 = 40.99\approx41$

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Question

In the above triangle, if and what is the area of the triangle?

Answer

To find the area of a triangle use the formula:

$A = \frac{1}{2}bh$ , since the base is and the height is , plugging in yields:

$\frac{1}{2}84 = 42$

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Question

A right triangle has integer sides with a ratio of , measured in $\textup{inches}$ . What is the smallest possible area of this triangle?

Answer

The easiest way to find the smallest possible integer sides is to simply factor the ratio we are given. In this case, is already prime (since is a prime number), so the smallest possible sides which hold to this triangle are and . You may also recognize this number as a common Pythagorean triple.

The area of a triangle is expressed as $A\Delta = \frac{1}{2}lh$ , where is the length and is the height. Since our triangle is right, we know that two lines intersect at a $90^{\circ}$ angle and thus serve well as our length and height. We also know that the longest side is always the hypotenuse, so the other two sides must be and .

Applying our formula, we get:

$A\Delta = \frac{1}{2}lh = \frac{1}{2}(7)(24) = \frac{168}{2} = 84$

Thus, the smallest possible area for our triangle is $84\textup{ in}^{2}$ .

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Question

Right triangle $\Delta GHJ$ has hypotenuse cm and $\angle J = 16.26^{\circ}$ . Find the area of the triangle, in cm2, by using $\sin$ .

Round angles to four significant figures. Round side lengths to the nearest integer.

Answer

To find the area of a right triangle, find the lengths of the two perpendicular legs (since this gives us our length and height for the area formula).

In this case, we know that one angle is $\angle J = 16.26^{\circ}$ , and SOHCAHTOA tells us that $\sin = \frac{\textup{opposite}}{\textup{hypotenuse}}$ , so:

$\sin 16.26^{\circ} = \frac{GH}{GJ} = \frac{GH}{50}$ Substitute the angle measure and hypotenuse into the formula.

$50\sin 16.26^{\circ} = GH$ Isolate the variable.

Solve the left side (rounding to the nearest integer) using our Pythagorean formula:

---> Substitute known values.

---> Simplify.

Square root both sides.

So with our two legs solved for, we now only need to apply the area formula for triangles to get our answer:

$A\Delta = \frac{lh}{2} = \frac{(GH)(HJ)}{2} = \frac{672}{2} = 336$

So, the area of our triangle is $336\textup{ cm}^{2}$ .

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Question

Find the area of a right triangle whose height is 4 and base is 5.

Answer

To solve, simply use the formula for the area of a triangle given height h and base B.

Substitute

and into the area formula.

Thus,

$A=\frac{1}{2}hB=\frac{1}{2}45=10$

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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.

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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.

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

Compare your answer with the correct one above

Question

Find the area of the following right triangle to the nearest integer

Note: The triangle is not necessarily to scale

Answer

The equation used to find the area of a right triangle is: $A = \frac{1}{2}b\cdot h$ where A is the area, b is the base, and h is the height of the triangle. In this question, we are given the height, so we need to figure out the base in order to find the area. Since we know both the height and hypotenuse of the triangle, the quickest way to finding the base is using the pythagorean theorem, $a^{2}+b^{2}=c^{2}$ . a = the height, b = the base, and c = the hypotenuse.

Using the given information, we can write $8^{2} + b^{2} = 13^{2}$ . Solving for b, we get $b^{2} = 105$ or $b \approx 10.25$ . Now that we have both the base and height, we can solve the original equation for the area of the triangle. $A = \frac{1}{2}\cdot10.25\cdot8 = 40.99\approx41$

Compare your answer with the correct one above

Question

In the above triangle, if and what is the area of the triangle?

Answer

To find the area of a triangle use the formula:

$A = \frac{1}{2}bh$ , since the base is and the height is , plugging in yields:

$\frac{1}{2}84 = 42$

Compare your answer with the correct one above

Question

A right triangle has integer sides with a ratio of , measured in $\textup{inches}$ . What is the smallest possible area of this triangle?

Answer

The easiest way to find the smallest possible integer sides is to simply factor the ratio we are given. In this case, is already prime (since is a prime number), so the smallest possible sides which hold to this triangle are and . You may also recognize this number as a common Pythagorean triple.

The area of a triangle is expressed as $A\Delta = \frac{1}{2}lh$ , where is the length and is the height. Since our triangle is right, we know that two lines intersect at a $90^{\circ}$ angle and thus serve well as our length and height. We also know that the longest side is always the hypotenuse, so the other two sides must be and .

Applying our formula, we get:

$A\Delta = \frac{1}{2}lh = \frac{1}{2}(7)(24) = \frac{168}{2} = 84$

Thus, the smallest possible area for our triangle is $84\textup{ in}^{2}$ .

Compare your answer with the correct one above

Question

Right triangle $\Delta GHJ$ has hypotenuse cm and $\angle J = 16.26^{\circ}$ . Find the area of the triangle, in cm2, by using $\sin$ .

Round angles to four significant figures. Round side lengths to the nearest integer.

Answer

To find the area of a right triangle, find the lengths of the two perpendicular legs (since this gives us our length and height for the area formula).

In this case, we know that one angle is $\angle J = 16.26^{\circ}$ , and SOHCAHTOA tells us that $\sin = \frac{\textup{opposite}}{\textup{hypotenuse}}$ , so:

$\sin 16.26^{\circ} = \frac{GH}{GJ} = \frac{GH}{50}$ Substitute the angle measure and hypotenuse into the formula.

$50\sin 16.26^{\circ} = GH$ Isolate the variable.

Solve the left side (rounding to the nearest integer) using our Pythagorean formula:

---> Substitute known values.

---> Simplify.

Square root both sides.

So with our two legs solved for, we now only need to apply the area formula for triangles to get our answer:

$A\Delta = \frac{lh}{2} = \frac{(GH)(HJ)}{2} = \frac{672}{2} = 336$

So, the area of our triangle is $336\textup{ cm}^{2}$ .

Compare your answer with the correct one above

Question

Find the area of a right triangle whose height is 4 and base is 5.

Answer

To solve, simply use the formula for the area of a triangle given height h and base B.

Substitute

and into the area formula.

Thus,

$A=\frac{1}{2}hB=\frac{1}{2}45=10$

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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.

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

Compare your answer with the correct one above

Question

Find the area of the following right triangle to the nearest integer

Note: The triangle is not necessarily to scale

Answer

The equation used to find the area of a right triangle is: $A = \frac{1}{2}b\cdot h$ where A is the area, b is the base, and h is the height of the triangle. In this question, we are given the height, so we need to figure out the base in order to find the area. Since we know both the height and hypotenuse of the triangle, the quickest way to finding the base is using the pythagorean theorem, $a^{2}+b^{2}=c^{2}$ . a = the height, b = the base, and c = the hypotenuse.

Using the given information, we can write $8^{2} + b^{2} = 13^{2}$ . Solving for b, we get $b^{2} = 105$ or $b \approx 10.25$ . Now that we have both the base and height, we can solve the original equation for the area of the triangle. $A = \frac{1}{2}\cdot10.25\cdot8 = 40.99\approx41$

Compare your answer with the correct one above

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