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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Observe that A₁B₁=a+b. Thus AB²=(a-b)²+(a+b)²=2(a²+b²)i.e. S=2(a²+b²)
Explanation for A₁B₁=a+b. Join OA, OB. OA = OB,< AOB is a right angle. <AOA₁ = complement of <BOB₁ = OBB₁So Right Δs OAA₁, BOB₁ are congruent Thus A₁O = b, OB₁ = a and A₁B₁ = a + bComplete rectangle AA₁B₁E (draw figure)Note AE = A₁B₁ = a + b, BE = EB₁ - BB₁ = AA₁ - BB₁ = a - bFrom rt triangle AEB, S = square area = AB^2 = (a + b)² + (a-b)² =2(a²+b²)
In fact, since angle AOB is a right angle,S = AB² = OA² + OB²= [(OA₁)² + (AA₁)²] + [(OB₁)² + (BB₁)²]= [b² + a²] + [a² + b²] = 2(a²+b²)
Triangle AA1O= triangle OB1B because: AO=OB; angle A1=B1; angle AOA1=OBB1; AOA1+BOB1+90=180A1O=B1B=b ; A1A=OB1=a ; AO= sq(a2+b2); OB= sq(a2+b2)2(AOxOB)=S= 2(a2+b2).(a2+b2) =2(a2+b2)
Tr.s AA1O and BB1O are congruent ASA so A1O = bHence AO^2 = a^2 + b^2 = S^2 /2 since Tr. AOB is isoceles right and the result followsSumith PeirisMoratuwaSri Lanka