Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Monday, March 2, 2015

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

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Observe that A₁B₁=a+b.

ReplyDeleteThus AB²=(a-b)²+(a+b)²=2(a²+b²)

i.e. S=2(a²+b²)

Explanation for A₁B₁=a+b.

ReplyDeleteJoin 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)

In fact, since angle AOB is a right angle,

ReplyDeleteS = 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=180

ReplyDeleteA1O=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 = b

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

Sumith Peiris

Moratuwa

Sri Lanka