Maths Olympiad Prep

Library / /279 of 520

Geometry Difficulty 3.1 AMC 10/12 Find the answer

Square ABCDABCD has sides of length 3. Segments CMCM and CNCN divide the square's area into three equal parts. How long is segment CMCM?

Pick one

Solution

Since the square has side length 33, the area of the entire square is 99.
The segments divide the square into 3 equal parts, so the area of each part is 9÷3=39 \div 3 = 3.
Since CBM\triangle CBM has area 33 and base CB=3CB = 3, using the area formula for a triangle:
Atri=12bhA_{tri} = \frac{1}{2}bh
3=123h3 = \frac{1}{2}3h
h=2h = 2
Thus, height BM=2BM = 2.
Since CBM\triangle CBM is a right triangle, CM=BM2+BC2=22+32=(C) 13CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \boxed{\text{(C)}\ \sqrt{13}}.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.