Since we know 6 and 8 are the lengths of the biggest triangle. We can say that the 3rd side is 10 in length. We are given that ED = 5; Therefore, AE = 5.
That is correct, knowing that ABD is a right triangle and its hypotenuse is unknown. Note, that if the hypotenuse was 8 and one side was 6, then the other side would be √(8² – 6²) = 2√7 , even though the triangle is a right triangle. Remember, that in 3-4-5 right triangle, 5 is the hypotenuse.
… Therefore, AE = 5. Triangles AFE and DCE are right angled; Therefore they are (3, 4, 5) in length.
This conclusion is groundless. Knowing just one side of a right triangle is not enough to calculate the other two.
If we know that a hypotenuse is 5, then there are infinitely many such triangles. For example the sides can be: 5, 5√2 / 2, 5√2 / 2 5, 4, 3 5, 2√6, 1
Plug in a value for x ( 0< x < 5 ) to make your own example: 5, x, √(25 – x²)
Indeed, in this problem these triangles are 5-4-3 triangles, but its a lucky coincidence. Some more reasoning is needed to prove that.
Users browsing this forum: No registered users and 1 guest
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum
GMAT(TM) and GMAT CAT (TM) are registered trademarks of the Graduate Management Admission Council(TM). The Graduate Management Admission Council(TM) does not endorse, nor is affiliated in any way with the owner or any content of this site.