an isoceles triangle w/ the base of 9 and legs measuring 13 each, what is the area??

Answer 1

Drop a perpendicular from the apex to the base. This will bisect the base into two lengths of 4.5 each. Now using pythagoras' theorem, work out the height of the triangle:

13² = h² + (4.5)²

Then the area of the triangle is half the base by the height.

Answer 2

Divide the triangle in two. So now you have two identical triangles each with measurments of 4.5, 13, and x.
to find x, use Pythagorean theorem.
a^2 + b^2 = c^2.
a is unknown
b is 4.5
c is 13
b^2=20.25
c^2=169
a^2+20.25=169
a^2=148.75
Then since you want to know the area of the original triangle, 1/2 base times height. base is 9. height is square root of 148.75
1/2 (9) * sqroot (148.75)
area is approx 54.8834

Answer 3

Heron's formula for the area of a triangle is
A = sqrt(s(s - a)(s - b)(s - c)) where s = (1/2)(a + b + c).

s = (1/2)(13 + 13 + 9)
s = 35/2

A = sqrt((35/2)(35/2 - 13)(35/2 - 13)(35/2 - 9))

The area of the triangle is approximately 54.9 square units.

Answer 4

The area of a triangle is 1/2bh. The height of this triangle is sqrt(13^2-81/4), which equals sqrt(555)/2. Therefore, your answer is 9/4sqrt555.

Answer 5

Let's divide this isosceles triangle into 2 equal right triangles with hypotenuse of 13 and one of the legs of 4.5.

Using the Pythagorean theorem, solve the other leg, y.

13^2 = 4.5^2 + y^2
y^2 = 169 - 20.25 = 148.75
y = 12.20

Now y = 12.20, which is the altitude of the original isosceles triangle.

Area = 1/2 b * a

Area = (1/2)(9)(12.20) = 54.90 square units
.

Answer 6

you could get the height through pythagoras h=sqrt(13^2-9^2)
then the area =2*(area of the right angled triangle= 0.5*4.5*h)