An isosceles triangle has sides that are #sqrt125#, #sqrt125#, and 10 units. What is its area?

Question

Jayden Jackson · Accepted Answer

#50" units"^2#

the area of a triangle is found through #A=1/2bh#. We have the base, but not the length of the height.

Imagine that an altitude is drawn down the center of the triangle, perpendicularly bisecting the base and bisecting the vertex angle. This creates two congruent right triangles inside the original isosceles triangle.

You know each of these triangles has a hypotenuse of #sqrt125# and a base of #5#, since they're half the original base.

We can use the Pythagorean theorem to figure out the length of the missing side, which is the height of the triangle.

#(5)^2+(h)^2=(sqrt125)^2#

#25+h^2=125#

#h^2=100#

#h=10#

So, we now know that the base is #10# units and the height is #10# units.

Thus, #A=1/2(10)(10)=50 " units"^2#

Sadie Allen · Answer

undefined To find the area of the isosceles triangle, we can use the formula:

Area = (1/4) * sqrt(4a^2 - b^2) * b

Where 'a' is the length of one of the equal sides and 'b' is the length of the base.

Given that the lengths of the equal sides are sqrt(125) units and the base is 10 units, we can substitute these values into the formula:

Area = (1/4) * sqrt(4 * (sqrt(125))^2 - (10)^2) * 10

Solving this equation:

Area = (1/4) * sqrt(4 * 125 - 100) * 10
Area = (1/4) * sqrt(500 - 100) * 10
Area = (1/4) * sqrt(400) * 10
Area = (1/4) * 20 * 10
Area = (1/4) * 200
Area = 50 square units

Therefore, the area of the isosceles triangle is 50 square units.