Intersecting Lines and Triangles

Geometry Level 2

Lines l 1 l_1 and l 2 l_2 intersect at O O . A A and C C are points on l 1 l_1 such that O A = 11 , O C = 9 OA = 11, OC = 9 . B B and D D are points on l 2 l_2 such that O B = 3 , O D = 22 OB = 3, OD = 22 . What is the ratio [ O C D ] [ O A B ] \frac{ [OCD] } { [OAB] } ?

Details and assumptions

[ P Q R S ] [PQRS] denotes the area of figure P Q R S PQRS .


The answer is 6.

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Arron Kau by Arron Kau

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1 solution

Arron Kau Staff
May 13, 2014

Solution 1: Since [ O C D ] = 1 2 O C × O D × sin ⁡ ∠ C O D [OCD] = \frac{1}{2} OC \times OD \times \sin \angle COD , [ O A B ] = 1 2 O A × O B × sin ⁡ ∠ A O B [OAB] = \frac{1}{2} OA \times OB \times \sin \angle AOB and sin ⁡ ∠ C O D = sin ⁡ ∠ A O B \sin \angle COD = \sin \angle AOB , thus [ O C D ] [ O A B ] = O C × O D O A × O B = 9 × 22 11 × 3 = 6. \frac{[OCD]}{[OAB]} = \frac{OC \times OD}{OA \times OB} = \frac { 9 \times 22 } { 11 \times 3 } = 6.

Solution 2: Triangles which share a common base line and opposing vertex have areas that are proportional to the length of the base. Hence, we have [ O C D ] [ O A D ] = O C O A , [ O A D ] [ O A B ] = O D O B . \frac{[OCD]} { [OAD]} = \frac { OC}{ OA } , \frac{ [OAD] } { [OAB] } = \frac { OD} { OB}. As such, this gives

[ O C D ] [ O A B ] = [ O C D ] [ O A D ] × [ O A D ] [ O A B ] = O C × O D O A × O B = 6. \frac{ [OCD] } {[OAB] } = \frac{ [OCD]} { [OAD]} \times \frac{ [OAD] } { [OAB] } = \frac { OC \times OD} { OA \times OB} = 6.

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