Asked by ben
Find the area of an equilateral triangle (regular 3-gon) with the given measurement.
9-inch perimeter
A = sq. in.
Answers
Answered by
Ms. Sue
Each side = 3 inches
By drawing a line from the center of the base to the top vertex you'll divide the triangle into two triangles.
The base = 1.5"
a^2 + b^2 = c^2
a^2 + 1.5^2 = 3^2
a^2 + 2.25 = 9
a^2 = 6.75
a = 2.6
A = 1/2bh
A = (1/2) * 1.5 * 2.6
A = ?
By drawing a line from the center of the base to the top vertex you'll divide the triangle into two triangles.
The base = 1.5"
a^2 + b^2 = c^2
a^2 + 1.5^2 = 3^2
a^2 + 2.25 = 9
a^2 = 6.75
a = 2.6
A = 1/2bh
A = (1/2) * 1.5 * 2.6
A = ?
Answered by
TutorCat
A=(1/2)*(base)*(height)
Side length=(1/3)*(9)=3
Split the triangle in half. This will give you the base, 1.5 in.
If you can image, you have two hypotenuse triangles. To find the height, you need to find the length of the line used to split the triangle in half. So, using Pythagorean Theorem, A^2 + B^2 = C^2 or (1.5)^2 + B^2 = (3)^2.
Side length=(1/3)*(9)=3
Split the triangle in half. This will give you the base, 1.5 in.
If you can image, you have two hypotenuse triangles. To find the height, you need to find the length of the line used to split the triangle in half. So, using Pythagorean Theorem, A^2 + B^2 = C^2 or (1.5)^2 + B^2 = (3)^2.