# quadrilateral that one of its diagonals splits and the areas

A quadrilateral has consecutive sides of length 10, 4, and 6 (in that order), and one diagonal divides the quadrilateral into two isosceles triangles. If A1 and A2 are the smallest and next smallest areas of quadrilaterals which satisfy these conditions, in which interval below does A2 - A1 lie? A. [0, 1] B. [3, 4] C. [7, 8] D. [8, 9] E. [12, 13] I was able to draw it and use Pythagoras to find sides of both internal triangles ( there are two ) then was able to find both areas . The next step would be what ? I think to use the radicals in my expression for the areas to see if smaller quadrilaterals exist and what they would be . If they have say sides of 3,2, and 5 ? This would be simply dividing each side of our ' grand quadrilateral' by 2 . Hmm maybe someone can help and show me an easier way or show me what I'm not seeing that I think would make this problem a lot easier ! Thanx in advanced

• Do you know Heron's formula? Commented Oct 30, 2018 at 23:21
• yes i do its an ugly formula though and id rather not use it but maybe u know a way to use it Commented Oct 30, 2018 at 23:34

Two possible ways to divide quadrilateral into two isosceles triangles are:
1) triangles with sides $$10,10,4$$ and $$4,4,6$$;
2) triangles with sides $$6,6,4$$ and $$6,6,10$$.

Using Heron'a formula to find area in both cases:
$$A_1=\sqrt{12 \cdot 4 \cdot 8}+\sqrt{7 \cdot 9}=8\sqrt{6}+3\sqrt{7}$$;
$$A_2=\sqrt{8 \cdot 16}+\sqrt{11 \cdot 25}=8\sqrt{2}+5\sqrt{11}$$.

Now all you need to do is to estimate the difference.

• nice vasya , but my main problem was finding the triples that u did ..how did u find them ? Commented Oct 31, 2018 at 1:09
• @Randin: One of the diagonals should match one of the sides. If diagonal is $4$, the fourth side can only be $10$ (based on the triangular inequality); if the diagonal is $6$ or $10$, the remaining side can be $6$ or $10$. So there are only five possible quadrilaterals. . Commented Oct 31, 2018 at 1:58
• Ah 🤔👍 so i did do it right on the contest day and think i got it right on the exam Commented Oct 31, 2018 at 2:07
• what is the least ? Commented Oct 31, 2018 at 2:07
• @Randin: the smallest one has diagonal $4$ and sides $10, 4, 6, 10$ Commented Oct 31, 2018 at 2:13