It is possible to make two different isosceles triangles with equivalent sides measuring 4 meters that have an area of 4 square meters. Find all the angle measures in each of these triangles.
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Let BD and AD be b and h  respectively.

Area of triangle ABD = 2 sq m ( by symmetry)

So, 1/2 * b * h = 2

or b*h = 4

or b = 4/h  ----------(1)

Also, b^2 + h^2 = 16  (pythagoras theorem)  ------------------------(2)

On putting the value of (1) in (2) we get:

(4/h)^2 + h^2 =16

or (16 + h^4)/h2 = 16

or h^4 -16h^2 +16 = 0

Put x = h^2 and solve for x we get:

x = 8 - 43 or x = 8 + 43

So, h = 2(2 - 3) or h = 2(2 + 3)  and neglecting other two negative values of h, since h cannot be negative.

Sin x = 2(2 - 3)/4 = or sinx = 2(2 + 3)/4, since sinx = perpendicular/hyp.

of sinx = (2 - 3)/2 = or sinx = (2 + 3)/2

if sinx = (2 - 3)/2

then x = arcsin((2 - 3)/2) = <B =<C = 15 degrees

and <A = 180 -2*15 = 150 degrees

if sinx = (2 + 3)/2 then

x = arcsin((2 + 3)/2) =<B=<C = 75 degrees

and <A = 180 - 2*75 = 30 degrees

So the angles of the two isosceles triangles are:

(15,15,150) and (75,75,30)

by Level 7 User (27.4k points)

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