Are There Even Enough Information?

Geometry Level 2

Above shows a square $ABCD$ with side length $a=10$ . There is a point $E$ inside that square, that forms four triangles $F$ , $G$ , $H$ and $I$ with the corners of the square $A$ , $B$ , $C$ and $D$ , respectively as shown in the picture above.

Find the sum of areas, $F+H$ .

If you think there are not enough information to solve, type 3141 as your answer.

The answer is 50.

5 solutions

Darius B
May 14, 2016

Relevant wiki: Length and Area - Composite Figures

Let $h$ be the height of the triangle $F$ , so that $a-h$ is the height of triangle $H$ . Then $F+H$ is $\frac { a⋅h }{ 2 } +\frac { a⋅(a-h) }{ 2 } =\frac { a⋅h }{ 2 } +\frac { { a }^{ 2 }-a⋅h }{ 2 } =\frac { { a }^{ 2 } }{ 2 }$ and since $a=10$ , $F+H$ is $\frac{100}{2}=50$

Easy one.Shifting point E to the centre makes it a 5 second question!!

Deepak Kumar - 5 years, 1 month ago

But it istn in the centre. You would first have to proof that F+H is equal to G+I for every point E.

Darius B - 5 years, 1 month ago

The purpose of this question is to point out that F+H = G+I for every point E.

Consider if point E is in the very center of the square; obviously F+H = G+I.

Consider if point E is on any of the corners of the square. It drives two of the triangles to zero, and the other two to half of the square, so again, F+H = G+I.

Place point E on any of the lines of the square itself, or on any point inside the square, and you can see that as the area of one triangle increases, its counterpart decreases, always maintaining F+H = G+I.

It's a fun visualization problem.

Brian Egedy - 5 years, 1 month ago

true but u need proof i use thats method

Patience Patience - 5 years ago

Half of the proof that F+H=G+I is already done in my solution, just go through the same process with G+I as i did for F+H and you will also get a^2/2

Darius B - 5 years ago

Andy Pato
May 16, 2016

Make four rectangles, all going from each vertex of the square to point E. AE, BE, CE and DE are all diagonals of each rectangle, they bisect them. Therefore the ratio of F or H to G or I is 50:50. Ares of F+H is then half of 100. This will work for any position of E.

Dejan Stankovic
May 16, 2016

Ok,this is my solutions The area of triangles F is 10 h1/2 The area of triangles H is 10 h2/2 Where is h1+h2=a=10 After that,we have F+H=5 h1+5 h2=5(h1+h2)=5*10=50

Rishabh Tiwari
May 16, 2016

Clearly., F + H = (1/2)•(10)•(h1) + (1/2)•(10)•(h2) = 5 (h1 + h2) .....

Now since h1+h2=10...

Therefore area = 5•10=50..

Ganesh Ravisankar
May 15, 2016

Sum of triangles/total area = 10 x 10 =100

Since f+h forms half of the total area (shift the tris and you'll know),

Area = 100/2 = 50.

Are There Even Enough Information?

The answer is 50.

5 solutions

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