Q&A for How to Find the Area of an Isosceles Triangle

A=area, L=length of 1 equal side, b=base, θ=HALF of angle between 2 equal sides. Split the triangle in half down the middle. The middle line is h, the height. Analyze the left triangle, where L is the hypotenuse and the smallest angle is θ. The smallest side is b/2, and the last side is h. sinθ = (b/2) / L --> b/2 = Lsinθ. cosθ = h/L --> h = Lcosθ. A = (1/2)bh = (b/2)h = (Lsinθ)(Lcosθ)=(L^2)sinθcosθ. sin(2θ) = 2sinθcosθ (by trig identities) --> sinθcosθ = (1/2)sin(2θ). --> A = (L^2)sinθcosθ = (1/2)(L^2)sin(2θ). Because A and L are known, the above equation can be used to find sin(2θ). Arcsin of sin(2θ) gives 2θ, allowing you to find θ. Then, you can find b from the equation: b/2 = Lsinθ.

Coordinate proof: Given the coordinates of the triangle's vertices, to prove that a triangle is isosceles plot the 3 points (optional). Use the distance formula to calculate the side length of each side of the triangle. If any two sides have equal side lengths, then the triangle is isosceles.

You don't. You must be given certain information: perimeter, other sides, area, or height.

Let each of the two equal sides of the triangle be x meters.Then, the perimeter is 2x + 20 = 42. So x = 11. The area of the triangle is then 1/2*20*root(11^2 - 10^2) = 10root(21)

Use the perimeter to find the sides of the triangle (3x + 10 = 100). Then use half the largest side and one of the equal sides to find the height through the Pythagorean Theorem. Finally, use the newly found height and the largest side of the triangle as its base in the formula to find a triangle's area.

If you are told the length of the base (unequal side), then you know the other two sides are equal, so you know all three side lengths and can use the standard method. If you only know the lengths of the two equal sides, then you cannot find the area without more information (such as the perimeter or an angle).

The area of a triangle is the base times height divided by two (bh/2). Just plug in the numbers: (10)(8)/2 = 80/2 = 40. The area of your triangle is 40 cm².

In an isosceles right triangle, the two equal sides have a right angle between them. This means you can use one equal side as the base, and the other as the height. If these sides have length s, then the area is (1/2)s^2. To find the perimeter, use the Pythagorean theorem to find the length of the hypotenuse, and add it to the lengths of the other sides.

The perimeter is equal to the sum of all side lengths. Since there are two sides with length s, the perimeter of this triangle is 5 + s + s, which simplifies to 2s + 5 cm.

Since the angles of a triangle add up to 180 degrees, you can find the answer by adding the two known angles (60 and 60) and subtracting that total from 180. In this case, 60 plus 60 equals 120, and 180 minus 120 equals 60, so the third angle is also 60 degrees.

You don't. To find the area, you would also have to know angles and/or sides.

A = (b + h) / 2. Therefore, h = 2A / b. Use the area and base to find the height. Then use trigonometry to find either of the two equal sides. Add the two equal sides to the base.

The formula for any triangle is A = (1/2)bh, where b is the length of the base and h is the height.

It can't be done without trig.

It's the same formula as for any other triangle: one-half the base multiplied by the height.

Subtract the base from the perimeter: 20 - 6 = 14. Each of the equal sides is half of 14. Now knowing all three sides, use trigonometry to find the height. A = (bh) / 2.

You would have to know all three sides. Then you can use trigonometry to find the height (and then the area).

The area of a triangle is 1/2 x base x height. It can also be written as base x height divided by 2.

Yes, you can draw an isosceles triangle with one 80° angle and two 50° angles. The latter would be the angles opposite the two equal sides.

Let's assume the 6 cm dimension is the height of the triangle. Then the area is (1/2)(base)(height) = (1/2)(33)(6) = 99 square centimeters.