## 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. 1. Observe that A₁B₁=a+b.

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

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)

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)

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)

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