Monday, March 2, 2015

Geometry Problem 1091. Square, Perpendicular, Distance, Area, Center

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.

Online Math: Problem 1091. Square, Perpendicular, Distance, Area, Center.

5 comments:

  1. Observe that A₁B₁=a+b.

    Thus AB²=(a-b)²+(a+b)²=2(a²+b²)
    i.e. S=2(a²+b²)

    ReplyDelete
  2. Explanation for A₁B₁=a+b.
    Join OA, OB.
    OA = OB,
    < AOB is a right angle.
    <AOA₁ = complement of <BOB₁ = OBB₁
    So Right Δs OAA₁, BOB₁ are congruent
    Thus A₁O = b, OB₁ = a and A₁B₁ = a + b
    Complete rectangle AA₁B₁E (draw figure)
    Note AE = A₁B₁ = a + b, BE = EB₁ - BB₁ = AA₁ - BB₁ = a - b
    From rt triangle AEB,
    S = square area = AB^2 = (a + b)² + (a-b)² =2(a²+b²)

    ReplyDelete
  3. 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)

    ReplyDelete
  4. Triangle AA1O= triangle OB1B because: AO=OB; angle A1=B1; angle AOA1=OBB1; AOA1+BOB1+90=180
    A1O=B1B=b ; A1A=OB1=a ; AO= sq(a2+b2); OB= sq(a2+b2)
    2(AOxOB)=S= 2(a2+b2).(a2+b2) =2(a2+b2)

    ReplyDelete
  5. Tr.s AA1O and BB1O are congruent ASA so A1O = b

    Hence AO^2 = a^2 + b^2 = S^2 /2 since Tr. AOB is isoceles right and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete