A tiny triangle

Geometry Level 2

There are three squares and one triangle in the figure above. The side lengths of squares are $AB, BC$ and $CA$ . The number within each square is the area of each square. Find the area of the triangle $ABC$ .

12 8 11 9 10

2 solutions

Chan Lye Lee
May 25, 2020

This problem can be solved using Heron's formula or Cosine rule.

I suggest another way to solve it here. Note that $\large\begin{array} { l l l } 74 & = & 5^2+7^2 \\116& = & 4^2+10^2 \\370& = & (5+4)^2+(7+10)^2 \\ \end{array}$

The area of triangle $ABC$ is $\frac{1}{2}(9)(17)- \left(\frac{1}{2}(5)(7)+\frac{1}{2}(4)(10)+(4)(7)\right)=11$

Good solve!!

Mahdi Raza - 1 year ago

Wow! How did you get the thought of expressing in terms of squares?

Vinayak Srivastava - 1 year ago

Yes, good thinking by Chan!

Mahdi Raza - 1 year ago

How do I develop such thinking? :)

Vinayak Srivastava - 1 year ago

Genius thinking

Kumudesh Ghosh - 1 year ago

Interesting problem. Simple enough to solve algebraically but this diagram illustrates why the answer is a simple integer.

Richard Desper - 1 year ago

I used Heron's formula and my Ti89 actually stalled for sometime to spit out 11. That's the best part! I was hoping to see some brilliant way to get the simplification of $s( s-a)(s-b)(s-c) = 121$ with little computational effort. Wasn't expecting to see this elegant workaround. Nice find!

Eric Roberts - 1 year ago

Rab Gani
May 30, 2020

Use cos rule : cos A = 164/(74 sqrt.(5)), then sin A = 22/(74 sqrt.(5)). The area [ABC] = sqrt(74). sqrt(5 . 74) .sin A / 2 = 11

A tiny triangle

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