Olympiad Maths Prep

Track / Stage 5 / 157 of 400 #757 of 2000

Problem 757

AIME late
Algebra Difficulty 5.3 Find the answer

B2. The difference in the perimeters of two squares is 8 cm8 \mathrm{~cm}, and the difference in their areas is 16 cm216 \mathrm{~cm}^{2}. Calculate the sum of their areas.

Official solution

B2. The difference in perimeters should be 4a4b=8 cm4 a-4 b=8 \mathrm{~cm}, and the difference in areas a2b2=16 cm2a^{2}-b^{2}=16 \mathrm{~cm}^{2}. From the first equation, we get ab=2a-b=2, and from the second, due to (ab)(a+b)=16(a-b)(a+b)=16, we also get a+b=8a+b=8. From this, we arrive at a=5 cma=5 \mathrm{~cm} and b=3 cmb=3 \mathrm{~cm}. The sum of both areas is 52+32=34 cm25^{2}+3^{2}=34 \mathrm{~cm}^{2}.

Writing the difference in perimeters 4a4b=84 a-4 b=8. 1 point

Writing the difference in areas a2b2=16a^{2}-b^{2}=16 1 point

Correct solving of the system 1 point

Solutions a=5 cma=5 \mathrm{~cm} and b=3 cmb=3 \mathrm{~cm}. 1+11+1 points

Calculated sum of areas: S1+S2=34 cm2S_{1}+S_{2}=34 \mathrm{~cm}^{2} 1 point

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