Geometry Problem

Level: Mathematics Education, High School, College.

Click the figure below to see the complete problem 615.

## Tuesday, June 7, 2011

### Problem 615: Right Triangle with a Square, Diagonals, Center, Area of Quadrilateral

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OABC is a cyclic quadrilateral

ReplyDeleteBy Ptolemy's Theorem:

AB.OC+BC.OA=OB.AC

(c+a)(b/√2)=6b, c+a=6√2

Required area

=(ABC)+(COA)

=(1/2)ca + (b²/4)

=(1/4)(2ca+c²+a²)by Pythagoras

=(1/4)(c + a)²

=18 sq units

Problem 615: This solution was submitted by Michael Tsourakakis from Greece

ReplyDeleteThanks Michael

http://s30.postimg.org/qiktgwn6p/pro_615.png

ReplyDeleteSolution of this problem depend on value b , length of side of square ACDE.

Case 1: b< 6, we have no solution

Case 2: 6< b< 6.sqrt(2) we have 2 solutions. Area of ABCO is 18

Case 3: b> 6.sqrt(2) , we have 2 solutions. Point B is inside the square ACDE

And the area of ACBO is undefined.

Area • ac/2 + b^2/4 = (a+c)^2 / 4 using Pythagoras

ReplyDeleteApplying Ptolemy to cyclic quad ABCO

ab/sqrt2 + cb/sqrt2 = 6b

Hence a+c = 6aqrt2 and so area = 36 X 2 /4 = 18

Sumith Peiris

Moratuwa

Sri Lanka

Append triangles congruent to ABC to the other sides of the square, thus creating a new square called BFGH. Connect BO up through G. Since 0 is the center of both squares, then BO is 12. The area of the BFGH is 72 = 12^2 / 2. Since quadrilateral AOCB is 1/4 of the area of the square BFGH, then the area of AOCB is 18.

ReplyDeletehttps://photos.app.goo.gl/UGeJgNArp2tU6dzJ8

ReplyDeleteLet the length of the square be x

ReplyDelete<AOC=90=<ABC

ABCD is a cyclic quad

By Ptolemy Theorem, (AO)(BC)+(AB)(CO)=(AC)(BO)

(x/sqrt2)*(BC)+(x/sqrt2)*(AB)=6x

AB+BC=6sqrt2

AB^2+BC^2=x^2

2(AB)(BC)=72-x^2

(AB)(BC)/2=18-x^2/4

(AO)(OC)/2=x^2/4

[OABC]=[ABC]+[AOC]=[(AB)(BC)/2]*[(AO)(OC)/2]=18