Find area of a quadrilateral in a very simple way.

I have seen similar problems solved with fancy theorems. Nothing like that is needed!

[franshartman]

I simply love this channel! I wish more people find this and subscribe.

[mputuchimezie]

You are the champ, just exposed a secret, Please keep it coming.

[timc]

Thanks - that method is cool, and would not have occurred to me, but I think I have another (even simpler?) one! Instead of D and E, let G and H lie on AC where FG is parallel to EC and FH is parallel to AD.Triangles AFG and AEC are similar, so AG = 10/15 of AC; triangles CFH and CDA are similar, so HC = 10/14 of AC, ,so HG = (2/3 + 5/7 -1) = 8/21 of AC.But triangles FHG and BAC are similar so Area BAC = (21x10)/8 =26.25, as triangles BAC and FAC have the same base. So area x =26.25 - (10 +5 +4) = 7.25

[davidseed]

its worth noting that the solution method requires you to find…
1) A baseline with a point on the line. eg. BC with E
2) Two apexes from the base line to make two triangles. Eg D and A
3) then using the ratios to get equations equating areas.

ie a divided baseline BEC with apexes D and A
Other possibilities are
divided baseline ADB with apexes E and C
divided baseline DFC with apexes A and E
so that’s three possibilities, I thnk a solution can be found using any two possibilities.
However, baseline AFE directly gives z/4= 5/10 so z=2.
This result can be used for Baseline ADB y/6= (y+7)/14
14y=6y+42, 8y=42, y=21/4= 5.25
z+y= x= 2+5.25=7.25

[juhivarma]

This is good, thanks for sharing. What were the fancy theorems people used?