We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

In triangle ABC, AB=13, AC=15, and BC=14. Let I be the incenter. The incircle of triangle ABC touches sides BC, AC, and AB at D, E, and F, respectively. Find the area of quadrilateral AEIF.

 Dec 16, 2018

Pretty tough one... I would wait for an answer.

I think we should use Heron's here for the area of the triangle, and we have sides 13, 14, and 15.

To use Heron's, we also need to find the semiperimeter, which is \(\frac{13+14+15}{2}=\frac{42}{2}=21.\)

Now, we have \(\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21(8)(7)(6)}=\sqrt{7056}=84.\)

Hmm, I don't know how to continue from here, though...this was my best shot.

 Dec 16, 2018

I think this is correct....

Based on what we have the area=84 and the semiperimeter=21. Therefore, the incircle(inradius)=84/21=4.

By tangent formulas, we see that the two sides of the quadrilateral are x. So, we have following from the top of the triangle, starting with the side length of 15: x, 15-x, Side length of 14: x-1 and 15-x, Side Length 13: x-1 and x. So, x-1+x=13, 2x-1=13, 2x=14, and x=7. Now, quadrilateral AEIF is made up of two triangles, both with areas of 14, since \(\frac{1}{2}*7*4=14\) . Thus, the answer is \(14+14=\boxed{28}.\)  

 Dec 16, 2018

Very impressive, tertre......A+++++   !!!!!!



cool cool cool

CPhill  Dec 16, 2018

14 Online Users