In the triangles, we have XY = 2AX = 2YB and AZ/ZC = 2/3. Find the ratio of the area of triangle YBC to the area of triangle ZYC.
You must be having an identity crisis -- will, lily, jake, Mona, Harry.
Please use the same name for your posts.
Also --
Homework Posting Tips
Please show your work. Tutors will not do your homework for you. Please show your work for any question that you are posting.
Please show your work. Tutors will not do your homework for you. Please show your work for any question that you are posting.
In the diagram below, we have XY = 2AX = 2YB and AZ/ZC = 2/3. Find the ratio of the area of triangle YBC to the area of triangle ZYC.
Does anyone know the answer
To find the ratio of the area of triangle YBC to the area of triangle ZYC, we need to determine the relationship between the lengths of their corresponding sides.
Let's start by considering triangle ABC, where XY represents a line segment connecting points X and Y on side AB. We are given that XY = 2AX = 2YB.
This implies that AX = XY/2 and YB = XY/2.
Now, let's consider triangle AZC. We are given that AZ/ZC = 2/3.
To relate triangle YBC to triangle ZYC, we need to identify common sides. Since AZ and BZ are parts of triangle AZC, we can use the ratios AZ/ZC and AX/YB to determine the relationship between these sides.
Let's denote the length of AZ as a and the length of ZC as b. Then, the length of AX is XY/2, which is equal to a/3 since AX is part of XY and the ratio AX/XY is 1/3. Similarly, the length of YB is XY/2, which is equal to b/3.
Now, we have AZ = a and BZ = b, and we know that AX = a/3 and YB = b/3.
The ratio of the area of triangle YBC to the area of triangle ZYC is given by the ratio of the squares of their corresponding sides.
Therefore, to find the ratio of the areas, we compare the squares of the side lengths:
(YB^2)/(ZC^2) = (b/3)^2 / (b)^2
(YB^2)/(ZC^2) = 1/3^2 = 1/9
Hence, the ratio of the area of triangle YBC to the area of triangle ZYC is 1/9.