Question 35071: In a triangle labeled "a", "b", "c" the length of ac is 12. Also, a line is drawn from the vertex of "b" to a point on line ac. This point is labled "d", and helps to form 2 triangles that are labeled abd and cbd. If the ratio of the area of trangle abd to the area of triangle cbd is 3:5, what is the length of segment ad?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! In a triangle labeled "a", "b", "c" the length of ac is 12. Also, a line is drawn from the vertex of "b" to a point on line ac. This point is labled "d", and helps to form 2 triangles that are labeled abd and cbd. If the ratio of the area of trangle abd to the area of triangle cbd is 3:5, what is the length of segment ad?
GOOD TO SEE THE PROBLEM NOW REWRITTEN IN FULL.ONE IMPROVEMENT NEEDED IS TO FOLLOW THE SATANDARD CONVENTIONS OF NAMING VERTICES AND ANGLES BY CAPITAL LETTERS , SIDES BY SMALL LETTERS (YOU FOLLOWED IT HERE)..OK SO WE HAVE ABC TRIANGLE BD IS A LINE WITH D ON AC.
AREA TRIANGLE ABD/AREA TRIANGLE CBD = 3/5...TO FIND AD=?
AREA OF A TRIANGLE IS PROPORTIONAL TO ITS BASE AS WELL AS ITS HEIGHT
WE HAVE TRIANGLE BCD AND BAD HAVE SAME HEIGHT SINCE HEIGHT IS PERPENDICULAR DISTANCE FROM B TO DC AND AD WHICH ARE SAME STRAIGHT LINE ADC.
SO AREAS OF THESE TRIANGLES ARE PROPORTIONAL TO THEIR BASES
HENCE
AREA TRIANGLE ABD/AREA TRIANGLE CBD = 3/5..=BASE OF TRIANLE BAD/BASE OF TRIANGLE BCD=AD/DC
HENCE AD/DC=3/5
AD/(AD+DC)=3/(3+5)=3/8
AD/AC=3/8
AC=b...HENCE
AD/b=3/8
AD=3b/8=3*12/8=4.5
|
|
|
