Warm up problem 21: triangle area medium

Feel free to draw a diagram as you read along .......

Using a diameter $AB$ as a base, in each hemisphere inscribe a $15/20/25$ right triangle so that the two triangles taken together form a kite.

Now the second diagonal of this kite will have length $24$ , (as it cuts the other diagonal, a diameter, in a $2:1$ ratio). Let this length $24$ diagonal be $CD$ , labeled so that $AC = AD = 15$ and $BC = BD = 20$ .

Now draw a diameter from $C$ to a point $E$ . Then $\Delta CDE$ will be a $7/24/25$ right triangle with $DE = 7$ . Now with $AE || CB$ , (as $AE = CB = 20$ and have endpoints at opposite ends of diameter $CE$ ), we must have $EB = AC = 15$ . Then, given that $BD = 20$ , we have that $\Delta BED$ is an inscribed triangle with sides of lengths $7, 15$ and $20$ , i.e., we have constructed the desired $\Delta XYZ$ .

Now we just need to apply Heron's formula to find this triangle's area. With

$s = \dfrac{7 + 15 + 20}{2} = 21$

we have that the area is

$\sqrt{21*(21 - 7)*(21 - 15)*(21 - 20)} = \sqrt{21*14*6} = \sqrt{7^{2}*3^{2}*2^{2}} = \boxed{42}$ ,

the ultimate answer to everything. :)

For now if no solutions are posted, refer to the third problem on this wiki under the section "Heron's formula"