How to Find the Area of an Isosceles Triangle
Review the area of a parallelogram., Compare triangles and parallelograms., Find the isosceles triangle's base., Draw a line between the base to the opposite vertex., Look at one half of your isosceles triangle., Set up the Pythagorean Theorem...
Step-by-Step Guide
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Step 1: Review the area of a parallelogram.
Squares and rectangles are parallelograms, as is any four-sided shape with two sets of parallel sides.
All parallelograms have a simple area formula: area equals base multiplied by the height, or A = bh.If you place the parallelogram flat on a horizontal surface, the base is the length of the side it is standing on.
The height (as you would expect) is how high it is off the ground: the distance from the base to the opposite side.
Always measure the height at a right (90 degree) angle to the base.
In squares and rectangles, the height is equal to the length of a vertical side, since these sides are at a right angle to the ground. -
Step 2: Compare triangles and parallelograms.
There's a simple relationship between these two shapes.
Cut any parallelogram in half along the diagonal, and it splits into two equal triangles.
Similarly, if you have two identical triangles, you can always tape them together to make a parallelogram.
This means that the area of any triangle can be written as A = ½bh, exactly half the size of a corresponding parallelogram. , Now you have the formula, but what exactly do "base" and "height" mean in an isosceles triangle? The base is the easy part: just use the third, unequal side of the isosceles.
For example, if your isosceles triangle has sides of 5 centimeters, 5 cm, and 6 cm, use 6 cm as the base.
If your triangle has three equal sides (equilateral), you can pick any one to be the base.
An equilateral triangle is a special type of isosceles, but you can find its area the same way., Make sure the line hits the base at a right angle.
The length of this line is the height of your triangle, so label it h.
Once you calculate the value of h, you'll be able to find the area.
In an isosceles triangle, this line will always hit the base at its exact midpoint. , Notice that the height line divided your isosceles triangle into two identical right triangles.
Look at one of them and identify the three sides:
One of the short sides is equal to half the base: b2{\displaystyle {\frac {b}{2}}}.
The other short side is the height, h.
The hypotenuse of the right triangle is one of the two equal sides of the isosceles.
Let's call it s. , Any time you know two sides of a right triangle and want to find the third, you can use the Pythagorean theorem: (side 1)2 + (side 2)2 = (hypotenuse)2 Substitute the variables we're using for this problem to get (b2)2+h2=s2{\displaystyle ({\frac {b}{2}})^{2}+h^{2}=s^{2}}.
You probably learned the Pythagorean Theorem as a2+b2=c2{\displaystyle a^{2}+b^{2}=c^{2}}.
Writing it as "sides" and "hypotenuse" prevents confusion with your triangle's variables. , Rearrange the formula to solve for h: (b2)2+h2=s2{\displaystyle ({\frac {b}{2}})^{2}+h^{2}=s^{2}}h2=s2−(b2)2{\displaystyle h^{2}=s^{2}-({\frac {b}{2}})^{2}}h=(s2−(b2)2){\displaystyle h={\sqrt {(}}s^{2}-({\frac {b}{2}})^{2})}. , Just plug in the length of the base for b and the length of one of the equal sides for s, then calculate the value of h.
For example, you have an isosceles triangle with sides 5 cm, 5 cm, and 6 cm. b = 6 and s =
5.
Substitute these into your formula:h=(s2−(b2)2){\displaystyle h={\sqrt {(}}s^{2}-({\frac {b}{2}})^{2})}h=(52−(62)2){\displaystyle h={\sqrt {(}}5^{2}-({\frac {6}{2}})^{2})}h=(25−32){\displaystyle h={\sqrt {(}}25-3^{2})}h=(25−9){\displaystyle h={\sqrt {(}}25-9)}h=(16){\displaystyle h={\sqrt {(}}16)}h=4{\displaystyle h=4} cm. , Now you have what you need to use the formula from the start of this section:
Area = ½bh.
Just plug the values you found for b and h into this formula and calculate the answer.
Remember to write your answer in terms of square units.
To continue the example, the 5-5-6 triangle had a base of 6 cm and a height of 4 cm.
A = ½bhA = ½(6cm)(4cm)A = 12cm2. , Most isosceles triangles are more difficult to work with than the last example.
The height often contains a square root that doesn't simplify to an integer.
If this happens, leave the height as a square root in simplest form.
Here's an example:
What is the area of a triangle with sides 8 cm, 8 cm, and 4 cm? Let the unequal side, 4 cm, be the base b.
The height h=82−(42)2{\displaystyle h={\sqrt {8^{2}-({\frac {4}{2}})^{2}}}}=64−4{\displaystyle ={\sqrt {64-4}}}=60{\displaystyle ={\sqrt {60}}} Simplify the square root by finding factors: h=60=4∗15=415=215.{\displaystyle h={\sqrt {60}}={\sqrt {4*15}}={\sqrt {4}}{\sqrt {15}}=2{\sqrt {15}}.} Area =12bh{\displaystyle ={\frac {1}{2}}bh}=12(4)(215){\displaystyle ={\frac {1}{2}}(4)(2{\sqrt {15}})}=415{\displaystyle =4{\sqrt {15}}} Leave this answer as written, or enter it in a calculator to find a decimal estimate (about
15.49 square centimeters). -
Step 3: Find the isosceles triangle's base.
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Step 4: Draw a line between the base to the opposite vertex.
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Step 5: Look at one half of your isosceles triangle.
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Step 6: Set up the Pythagorean Theorem.
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Step 7: Solve for h. Remember
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Step 8: the area formula uses b and h
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Step 9: but you don't know the value of h yet.
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Step 10: Plug in the values for your triangle to find h. Now that you know this formula
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Step 11: you can use it for any isosceles triangle where you know the sides.
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Step 12: Plug the base and height into your area formula.
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Step 13: Try a more difficult example.
Detailed Guide
Squares and rectangles are parallelograms, as is any four-sided shape with two sets of parallel sides.
All parallelograms have a simple area formula: area equals base multiplied by the height, or A = bh.If you place the parallelogram flat on a horizontal surface, the base is the length of the side it is standing on.
The height (as you would expect) is how high it is off the ground: the distance from the base to the opposite side.
Always measure the height at a right (90 degree) angle to the base.
In squares and rectangles, the height is equal to the length of a vertical side, since these sides are at a right angle to the ground.
There's a simple relationship between these two shapes.
Cut any parallelogram in half along the diagonal, and it splits into two equal triangles.
Similarly, if you have two identical triangles, you can always tape them together to make a parallelogram.
This means that the area of any triangle can be written as A = ½bh, exactly half the size of a corresponding parallelogram. , Now you have the formula, but what exactly do "base" and "height" mean in an isosceles triangle? The base is the easy part: just use the third, unequal side of the isosceles.
For example, if your isosceles triangle has sides of 5 centimeters, 5 cm, and 6 cm, use 6 cm as the base.
If your triangle has three equal sides (equilateral), you can pick any one to be the base.
An equilateral triangle is a special type of isosceles, but you can find its area the same way., Make sure the line hits the base at a right angle.
The length of this line is the height of your triangle, so label it h.
Once you calculate the value of h, you'll be able to find the area.
In an isosceles triangle, this line will always hit the base at its exact midpoint. , Notice that the height line divided your isosceles triangle into two identical right triangles.
Look at one of them and identify the three sides:
One of the short sides is equal to half the base: b2{\displaystyle {\frac {b}{2}}}.
The other short side is the height, h.
The hypotenuse of the right triangle is one of the two equal sides of the isosceles.
Let's call it s. , Any time you know two sides of a right triangle and want to find the third, you can use the Pythagorean theorem: (side 1)2 + (side 2)2 = (hypotenuse)2 Substitute the variables we're using for this problem to get (b2)2+h2=s2{\displaystyle ({\frac {b}{2}})^{2}+h^{2}=s^{2}}.
You probably learned the Pythagorean Theorem as a2+b2=c2{\displaystyle a^{2}+b^{2}=c^{2}}.
Writing it as "sides" and "hypotenuse" prevents confusion with your triangle's variables. , Rearrange the formula to solve for h: (b2)2+h2=s2{\displaystyle ({\frac {b}{2}})^{2}+h^{2}=s^{2}}h2=s2−(b2)2{\displaystyle h^{2}=s^{2}-({\frac {b}{2}})^{2}}h=(s2−(b2)2){\displaystyle h={\sqrt {(}}s^{2}-({\frac {b}{2}})^{2})}. , Just plug in the length of the base for b and the length of one of the equal sides for s, then calculate the value of h.
For example, you have an isosceles triangle with sides 5 cm, 5 cm, and 6 cm. b = 6 and s =
5.
Substitute these into your formula:h=(s2−(b2)2){\displaystyle h={\sqrt {(}}s^{2}-({\frac {b}{2}})^{2})}h=(52−(62)2){\displaystyle h={\sqrt {(}}5^{2}-({\frac {6}{2}})^{2})}h=(25−32){\displaystyle h={\sqrt {(}}25-3^{2})}h=(25−9){\displaystyle h={\sqrt {(}}25-9)}h=(16){\displaystyle h={\sqrt {(}}16)}h=4{\displaystyle h=4} cm. , Now you have what you need to use the formula from the start of this section:
Area = ½bh.
Just plug the values you found for b and h into this formula and calculate the answer.
Remember to write your answer in terms of square units.
To continue the example, the 5-5-6 triangle had a base of 6 cm and a height of 4 cm.
A = ½bhA = ½(6cm)(4cm)A = 12cm2. , Most isosceles triangles are more difficult to work with than the last example.
The height often contains a square root that doesn't simplify to an integer.
If this happens, leave the height as a square root in simplest form.
Here's an example:
What is the area of a triangle with sides 8 cm, 8 cm, and 4 cm? Let the unequal side, 4 cm, be the base b.
The height h=82−(42)2{\displaystyle h={\sqrt {8^{2}-({\frac {4}{2}})^{2}}}}=64−4{\displaystyle ={\sqrt {64-4}}}=60{\displaystyle ={\sqrt {60}}} Simplify the square root by finding factors: h=60=4∗15=415=215.{\displaystyle h={\sqrt {60}}={\sqrt {4*15}}={\sqrt {4}}{\sqrt {15}}=2{\sqrt {15}}.} Area =12bh{\displaystyle ={\frac {1}{2}}bh}=12(4)(215){\displaystyle ={\frac {1}{2}}(4)(2{\sqrt {15}})}=415{\displaystyle =4{\sqrt {15}}} Leave this answer as written, or enter it in a calculator to find a decimal estimate (about
15.49 square centimeters).
About the Author
Dorothy Bell
Creates helpful guides on DIY projects to inspire and educate readers.
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