How to Calculate the Area of an Isosceles Triangle

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This article was co-written by David Jia. David Jia is a tutoring teacher and founder of LA Math Tutoring, a private tutoring facility based in Los Angeles, California. With over 10 years of teaching experience, David teaches a wide variety of subjects to students of all ages and grades, as well as college admissions counseling and prep for SAT, ACT, ISEE, etc. scoring 800 in math and 690 in English on the SAT, David was awarded a Dickinson Scholarship to the University of Miami, where he graduated with a bachelor’s degree in business administration. Additionally, David has worked as an instructor in online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math.

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An isosceles triangle is a triangle with two equal sides, which always form two equal angles with the base (third side) and intersect above the midpoint of the base. You can check this with a ruler and two pencils of equal length: if you try to tilt the triangle to either side, the ends of the two pencils cannot touch. These special properties of an isosceles triangle allow you to calculate its area with just a few facts.

Table of Contents

Find the area from the lengths of the sides

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See how to calculate the area of a parallelogram. Squares and rectangles are parallelograms because they have four sides with two pairs of parallel sides. All parallelograms have a simple area formula: area equal to base times height, or A = bh . ^{[1] X Research Source} If you place a parallelogram on a horizontal plane, the base edge is the side it is standing on. The altitude is the height of the parallelogram above the ground: the distance from the base edge to the opposite side. Always measure the length of the height at right angles (90 degrees) to the bottom edge.

For squares and rectangles, the height is equal to the length of the vertical side because the sides are perpendicular to the ground.

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Compare triangles with parallelograms. These two figures have a simple relationship with each other. Cut the parallelogram in half along the diagonal and separate it into two equal triangles. Similarly, if you have two identical triangles, you can combine them to form a parallelogram. That is, the area of any triangle will have the formula A = ½bh , which is exactly half the area of the corresponding parallelogram.

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Find the base side of an isosceles triangle. Now you have the formula, but where exactly are the “base” and “elevation” of the isosceles triangle? It’s easy to find the bottom edge: it’s the third side and not equal to the other two.

For example, if an isosceles triangle has sides 5 cm, 5 cm, and 6 cm, the side of length 6 cm is the base.
If the triangle has three equal sides (equilateral triangle), you can choose any side as the base. An equilateral triangle is a special type of isosceles triangle, but you can calculate the area in a similar way. ^{[2] X Research Source}

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Draw a straight line from the midpoint of the bottom edge to the opposite vertex. That line must be perpendicular to the base and is the altitude of the triangle, we call it h . After calculating the value of h , you will calculate the area.

In an isosceles triangle, this line always intersects the base at its midpoint.

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Observe one half of the isosceles triangle. Note that the altitude divides the isosceles triangle into two similar right triangles. Observe one of the two triangles and distinguish its three sides:

One of the two short sides is half the bottom edge: $b$ $2$ ${displaystyle {frac {b}{2}}}$ ${frac {b}{2}}$ .
The remaining short side is the height h .
The hypotenuse of a right triangle is one of two equal sides of an isosceles triangle. We call it s .

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Set up the formula of the Pythagorean Theorem . Whenever you know the two sides of a right triangle and want to find the third side, you can use the Pythagorean Theorem: (side 1) ² + (side 2) ² = (hype side) ² . Substitute the variables you are using for the problem to get

(

b

2

)

2

+

H

2

=

S

2

{displaystyle ({frac {b}{2}})^{2}+h^{2}=s^{2}}

({frac {b}{2}})^{2}+h^{2}=s^{2}

.

Have you ever learned the formula of the Pythagorean Theorem is $a$ $2$ $+$ $b$ $2$ $=$ $c$ $2$ ${displaystyle a^{2}+b^{2}=c^{2}}$ $a^{2}+b^{2}=c^{2}$ . Write “side” and “hypogonal” to avoid confusion with the variables of the triangle.

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Solve for h . Remember that the formula for area uses b and h , but you don’t know the value of h yet. Convert the formula to solve for h :

$($ $b$ $2$ $)$ $2$ $+$ $H$ $2$ $=$ $S$ $2$ ${displaystyle ({frac {b}{2}})^{2}+h^{2}=s^{2}}$ $({frac {b}{2}})^{2}+h^{2}=s^{2}$
$H$ $2$ $=$ $S$ $2$ $-$ $($ $b$ $2$ $)$ $2$ ${displaystyle h^{2}=s^{2}-({frac {b}{2}})^{2}}$ $h^{2}=s^{2}-({frac {b}{2}})^{2}$
$H$ $=$ $($ $S$ $2$ $-$ $($ $b$ $2$ $)$ $2$ $)$ ${displaystyle h={sqrt {(}}s^{2}-({frac {b}{2}})^{2})}$ $h={sqrt (}s^{2}-({frac {b}{2}})^{2})$ .

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Substitute in the values of the triangle to find h . Now you know this formula and can use it for any isosceles triangle with known side lengths. Just substitute the length of the base into b and the length of one of the two equal sides into s , then we get 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 values into the formula:
$H$ $=$ $($ $S$ $2$ $-$ $($ $b$ $2$ $)$ $2$ $)$ ${displaystyle h={sqrt {(}}s^{2}-({frac {b}{2}})^{2})}$ $h={sqrt (}s^{2}-({frac {b}{2}})^{2})$
$H$ $=$ $($ $5$ $2$ $-$ $($ $6$ $2$ $)$ $2$ $)$ ${displaystyle h={sqrt {(}}5^{2}-({frac {6}{2}})^{2})}$ $h={sqrt (}5^{2}-({frac {6}{2}})^{2})$
$H$ $=$ $($ $25$ $-$ $3$ $2$ $)$ ${displaystyle h={sqrt {(}}25-3^{2})}$ $h={sqrt (}25-3^{2})$
$H$ $=$ $($ $25$ $-$ $9$ $)$ ${displaystyle h={sqrt {(}}25-9)}$ $h={sqrt (}25-9)$
$H$ $=$ $($ $16$ $)$ ${displaystyle h={sqrt {(}}16)}$ $h={sqrt (}16)$
$H$ $=$ $4$ ${displaystyle h=4}$ $h=4$ cm.

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Substitute the base and altitude into the area formula. Now you have the data you need to use the formula mentioned at the beginning: Area = ½bh. Substitute the found values of b and h into this formula and find the answer. Remember to write your answer in units of area.

Continuing with the above example, a triangle with sides 5-6 cm has a base of 6 cm and a height of 4 cm.
A = bh
A = (6cm)(4cm)
A = 12cm ² .

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Solve a more difficult example. Most isosceles triangles are more difficult to solve than the example problem above. The altitude length is usually the square root instead of a simple integer. If this is the case, reduce the square root value. Here is an example:

What is the area of a triangle whose sides are 8 cm, 8 cm and 4 cm?
The side whose length is different from the other two is the base edge b and has a length of 4 cm.
High road $H$ $=$ $8$ $2$ $-$ $($ $4$ $2$ $)$ $2$ ${displaystyle h={sqrt {8^{2}-({frac {4}{2}})^{2}}}}$ $h={sqrt {8^{2}-({frac {4}{2}})^{2}}}$
$=$ $sixty four$ $-$ $4$ ${displaystyle ={sqrt {64-4}}}$ $={sqrt {64-4}}$
$=$ $60$ ${displaystyle ={sqrt {60}}}$ $={sqrt {60}}$
Simplify the square root by finding the coefficients: $H$ $=$ $60$ $=$ $4$ $*$ $15$ $=$ $4$ $15$ $=$ $2$ $15$ $.$ ${displaystyle h={sqrt {60}}={sqrt {4*15}}={sqrt {4}}{sqrt {15}}=2{sqrt {15}}.}$ $h={sqrt {60}}={sqrt {4*15}}={sqrt {4}}{sqrt {15}}=2{sqrt {15}}.$
Acreage $=$ $first$ $2$ $b$ $H$ ${displaystyle ={frac {1}{2}}bh}$ $={frac {1}{2}}bh$
$=$ $first$ $2$ $($ $4$ $)$ $($ $2$ $15$ $)$ ${displaystyle ={frac {1}{2}}(4)(2{sqrt {15}})}$ $={frac {1}{2}}(4)(2{sqrt {15}})$
$=$ $4$ $15$ ${displaystyle =4{sqrt {15}}}$ $=4{sqrt {15}}$
Leave the response as is or enter it into the calculator to find the approximate value as a decimal (approximately 15.49 square centimeters).

Using trigonometry

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Start with an edge and a corner. If you know trigonometry, you can find the area of an isosceles triangle even if you don’t know the lengths of any of the three sides. Here is an example problem with the following data: ^{[3] X Research Source}

The length s of two equal sides is 10 cm.
The angle θ between two equal sides is 120 degrees.

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Divide the isosceles triangle into two right triangles. Draw a line from the vertex between two equal sides and perpendicular to the base. Now you have 2 similar right triangles.

This line divides angle θ into two equal halves. Each right triangle has an angle ½θ, in this case (½)(120) = 60 degrees.

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Use trigonometry to find the value of h . You now have a right triangle and can use the sine, cosine, or tangent trigonometric functions. In the example above, you know the length of the hypotenuse and want to find the value of h , which is the side that makes up the known angle. Use the equation cos = adjacent / hypotenuse to solve for h :

cos(θ/2) = h / s
cos(60º) = h / 10
h = 10cos(60º)

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Find the length of the remaining side: There is only one unknown side of the right triangle, we call it x . Solve for this side length using the equation sin = opposite side / hypotenuse:

sin(θ/2) = x / s
sin(60º) = x / 10
x = 10sin(60º)

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The relationship between x and the base of an isosceles triangle. You can now zoom in to the original isosceles triangle. The sum of the lengths of the base b is 2 x , because the base is divided into two equal parts with each half being x .

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Substitute the values of h and b into the basic area formula. After knowing the base and altitude, you now use the standard formula A = ½bh:

$A$ $=$ $first$ $2$ $b$ $H$ ${displaystyle A={frac {1}{2}}bh}$ $A={frac {1}{2}}bh$
$=$ $first$ $2$ $($ $2$ $x$ $)$ $($ $ten$ $c$ $o$ $S$ $60$ $)$ ${displaystyle ={frac {1}{2}}(2x)(10cos60)}$ $={frac {1}{2}}(2x)(10cos60)$
$=$ $($ $ten$ $S$ $i$ $n$ $60$ $)$ $($ $ten$ $c$ $o$ $S$ $60$ $)$ ${displaystyle =(10sin60)(10cos60)}$ $=(10sin60)(10cos60)$
$=$ $100$ $S$ $i$ $n$ $($ $60$ $)$ $c$ $o$ $S$ $($ $60$ $)$ ${displaystyle =100sin(60)cos(60)}$ $=100sin(60)cos(60)$
You can enter numbers into the calculator (set to degrees) and get a result of about 43.3 square centimeters. Or you can use properties of trigonometry to simplify it to A = 50sin(120º).

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Convert to the general formula. Once you know how to solve this problem, you can rely on the general formula without going through each step each time. Here is the result if you repeat this solution method without using specific values (simplify the equation by trigonometric properties): ^{[4] X Research Source}

$A$ $=$ $first$ $2$ $S$ $2$ $S$ $i$ $n$ $θ$ ${displaystyle A={frac {1}{2}}s^{2}sintheta }$ $A={frac {1}{2}}s^{2}sintheta$
s is the length of one of the two equal sides.
θ is the angle between two equal sides.

Advice

If the problem is for an isosceles right triangle (two equal sides and one 90 degree angle), the area of the triangle will be much easier to find. If you choose one of the two short sides as the bottom edge, the other side is the high line. Now the formula A = ½ bh is simplified to ½s ² , where s is the length of one of the two short sides.
After taking the square root, we get two values, one positive and one negative, but in geometry you can ignore negative values. Of course, there is no such thing as a triangle whose altitude length is a “negative value”.
Some trigonometry problems can give other initial data, such as the length of the base and an angle (and indicate it is an isosceles triangle). The basic method remains the same: divide the isosceles triangle into two right triangles and solve for the altitude using trigonometric functions.

Steps

Find the area from the lengths of the sides

Using trigonometry

Advice

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