The area of triangle ABC is 154 square inches, and point P is marked on side AB so that AP:PB = 3:4. What are the areas of triangles CPA and CPB?

Question

CP is not an altitude. No other information is given.

Answer 1

Doug C. has a very nice solution. However, since I have been working on this problem I will give another solution. Draw a diagram. Segment AP will be shorter than segment PB. From point C, draw the altitude to triangle PBC. Call this altitude "h". This altitude will be inside triangle PBC but outside triangle APC. This altitude for triangle PBC will be the altitude for triangle APB also. Therefore the areas will be (1/2)(h)(4x) and

(1/2)(h)(3x). The ratios of the areas will be (1/2)(h)(3x)/(1/2)(h)(4x)=3/4. The ratio of the areas of the two triangles will be 3:4 also and the total area is 154 square inches. Therefore 3x+4x=154, 7x=154 and x=22. Therefore 3(22)=66 sq in and 4(22)=88 sq in. The area of triangle APB is 66 sq in and the area of triangle PBC is 88 sq in.

Answer 2

Let AP=3x and PB=4x (those segments are in the ratio 3:4.

That means side AB = 7x and this side can be treated as the base of triangle ABC.

Let h = the altitude from vertex C to base segment AB.

Then 1/2 (7x)h = 154. Solve for h to get a representation for h in terms of x.

h = 154(2)/7x = 44/x.

The area of triangle CPA is then: 1/2 (3x)(h) = 1/2 (3x)(44/x) = 66.

The area of triangle CPB is then 1/2 (4x)(44/x) = 88.

Check: 66 + 88 = 154 which is the area or triangle ABC.