What is the area of a triangle whose sides have length and ?

Area of a triangle is 1/2 Base x Height.

This is an isosceles triangle with base sqrt(28), so you can split it in half to form 2 right angled triangles with base = sqrt(28) / 2  ,  hypotenuse = sqrt(70) and height = x.

Use pythagorus to get the height of the right angled triangle:

x = sqrt[70 - (28/4)] = sqrt(63)

Therefore area of Isosceles triangle = 2 x area of right angled triangle = 2[1/2 * (sqrt28)/2 * sqrt(63)]

                                                                                                    = 21 units ^2

Gahh CPhill beat me to it...

Close your eyes, AWESOMEEE, if you haven't had Trig, because that's the only wayI know to do this....there are at least two methods......one would be EXTREMELY messy.....I'll opt for the other....

This triangle is iscoceles......so I can use the Law of Sines to find 1/2 of the measure of the apex angle opposite the shortest side....after that....it'll be a breeze.....!!!

So we have

sin 90 /√70  = sin Θ /  (1/2)√28   where Θ is the angle we're after.  So.....

sin Θ  = (1/2) √28 / √70     and using the sine inverse, we have

sin^-1 ( √28 / 2 √70) =   Θ = about 18.43°   and this is 1/2 of the apex angle, so that angle  = about 36.87°

Now....we're ready to find the area...so we have

A = (1/2) 70 sin(36.87) = 35 sin (36.87)  =  21 sq units

ohhhhh ok thx

That's OK, Brodudedoodebrodude....your method is easier, anyway....