How To Find The Area Of An Isosceles Triangle With Pictures

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.

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. [4] X Research source

In an isosceles triangle, this line will always hit the base at its exact midpoint. [5] X Research source

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.

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.

(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})}.

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.

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.

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).

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

This line divides θ perfectly in half. Each right triangle has an angle of ½θ, or in this case (½)(120) = 60 degrees.

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

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

A=12bh{\displaystyle A={\frac {1}{2}}bh}=12(2x)(10cos60){\displaystyle ={\frac {1}{2}}(2x)(10cos60)}=(10sin60)(10cos60){\displaystyle =(10sin60)(10cos60)}=100sin(60)cos(60){\displaystyle =100sin(60)cos(60)} You can enter this into a calculator (set to degrees), which gives you an answer of about 43. 3 square centimeters. Alternatively, use properties of trigonometry to simplify it to A = 50sin(120º).

A=12s2sinθ{\displaystyle A={\frac {1}{2}}s^{2}sin\theta } s is the length of one of the two equal sides. θ is the angle between the two equal sides.