The perimeter of an isosceles triangle is 40, and the length of the altitude to the base is 8. Find the area?

Answer 1

(Click on the link below for the diagram. Expand to get a clearer view)

Let the point of intersection of altitude and base be P

The length BP and PC is x.
The length AB = AC = (40-x-x)/2 = 20-x

Then we use pythagoras theorem to find x.
x^2 + 8^2 = (20 - x)^ 2
40x = 336
x = 336/40 = 8.4

Hence the area of the triangle
= 1/2(base*height)
= 1/2(16.8*8)
= 67.2

Answer 2

A = .5*b*h
h = 8
x = length of congruent side
y = length of base

2x + y = 40
y = 40 - 2x

[(40 - 2x)/2]^2 + 8^2 = x^2
(20 - x)^2 + 64 = x^2
400 - 40x + x^2 + 64 = x^2
464 = 40x
11.6 = x

b = y = 40 - 2x = 40 - 2(11.6) = 40 - 23.2 = 16.8

A = .5(16.8)(8) = 67.2 sq units

Answer 3

p = 2a + b
K = b sqrt(4a^2 - b^2)/4

40 = 2a + b
b = -2a + 40

(-2a + 40)/2 = -a + 20

The reason for this, is because the altitude and half the base form 2 right triangles.

(-a + 20)^2 + 8^2 = a^2
(-a + 20)(-a + 20) + 64 = a^2
a^2 - 20a - 20a + 400 + 64 = a^2
a^2 - 40a + 464 = a^2
-40a = -464
a = 11.6

b = -2a + 40
b = -2(11.6) + 40
b = -23.2 + 40
b = 16.8

K = b sqrt(4a^2 - b^2)/4
K = 16.8 * sqrt(4(11.6)^2 - (16.8)^2)/4
K = 4.2 * sqrt(4(134.56) - 282.24)
K = 4.2 * sqrt(538.24 - 282.24)
K = 4.2 * sqrt(256)
K = 4.2 * 16
K = 67.2

ANS : 67.2 square units

Answer 4

If we assume that the base is the one which is not same as other 2 as it is givel isosceles...
then if x,x,y r the sides of triangle....
given that 2x+y=40-------------(1)
h=8=sqrt(4x^2-y^2)/2
from this u get 4x^2-y^2=256
=>(2x-y)(2x+y)=256
=>2x-y=6.4(as 2x+y=40)------------(2)
after simplifying (1) n (2)
u get x=11.6 n y=16.8
as
AREA=h*y/2
=8*16.8/2
=67.2

Answer 5

I think it's 1536.
Check out the resource below

Answer 6

96????? its probably wrong, my geometry is a little rusty.