Question 1137384: The area of triangle ABC is 231 square inches, and point P is marked on side AB so that AP:PB = 3:4. What are the areas of triangles APC and BPC?
Found 2 solutions by rk2019, ikleyn:
Answer by rk2019(3)   (Show Source):
Answer by ikleyn(52788)  (Show Source):
You can put this solution on YOUR website! .
Draw the perpendicular CD from the vertex C of the triangle ABC to its side AB (so the point D lies on the straight line AB).
Then CD is the common height in triangles ABC, APC and BPC.
Since the triangle area is half the product of the base length and the height drawn to the base, it is clear that
- the area of the triangle ABC is the sum of areas of triangles APC and BPC, and
- the ratio of the area of triangle APC to the area of triangle BPC is the same as the ratio of the segment
lengths |AP| to |BP|, i.e. 3:4.
Then it is clear that area of the triangle APC = 3x and the area of the triangle BPC = 4x, where 3x + 4x = 231.
Hence, 7x = 231, x =  = 33,
area of the triangle APC = 3x = 3*33 = 99 square inches and area of the triangle BPC = 4*33 = 132 square inches. ANSWER
Solved.
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In the post by the other tutor, I see the words "the solution is attached".
I do not understand their meaning, because I do not see "the attached solution" there.
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