Ratio of the area of triangles

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Discussion Overview

The discussion revolves around the ratio of the areas of triangles in a geometric configuration, specifically comparing the areas of triangles $CFG$, $BEG$, and $CAB$. Participants explore the implications of given area relationships and the midpoint theorem.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants note that the area of triangle $ABC$ is twice that of triangle $BCD$ and seek to find the ratio of the area of triangle $CFG$ to triangle $BEG$.
  • One participant states that since $G$ is the midpoint of $BC$, the area of triangle $BEG$ can be expressed in terms of the areas of triangles $BCD$ and $ABC$.
  • Another participant claims that the area of triangle $CFG$ is half that of triangle $CAB$ based on their calculations.
  • However, a later reply disputes this, asserting that the area of triangle $CFG$ is actually one-fourth the area of triangle $CAB$, supporting this with the similarity of triangles $CFG$ and $CAB$ and their corresponding side ratios.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the areas of triangles $CFG$ and $CAB$, with some asserting a ratio of 1:2 and others claiming it to be 1:4. The discussion remains unresolved regarding the exact ratios.

Contextual Notes

The discussion relies on geometric properties and relationships that may depend on specific assumptions about the configuration of the triangles and their dimensions, which are not fully detailed in the posts.

mathlearn
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In the figure , the area of triangle $ABC$ is twice that of triangle $BCD$.USing the given information , find the ration of the area of the triangle $CFG$ to the area of triangle $BEG$

Hint- Use the midpoint theorem.

(Wave) Stuck in this problem & currently I have no workings to show.
 

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mathlearn said:
In the figure , the area of triangle $ABC$ is twice that of triangle $BCD$.USing the given information , find the ration of the area of the triangle $CFG$ to the area of triangle $BEG$

Hint- Use the midpoint theorem.

(Wave) Stuck in this problem & currently I have no workings to show.
Note that $G$ is the mid-point of $BC$. Thus we have
$$[BEG]=(1/2)[BGD]=(1/4)[BCD]=(1/8)[ABC]=(1/2)[CFG]$$
 
caffeinemachine said:
Note that $G$ is the mid-point of $BC$. Thus we have
$$[BEG]=(1/2)[BGD]=(1/4)[BCD]=(1/8)[ABC]=(1/2)[CFG]$$

Many Thanks caffeinemachine (Happy)

Is the area of $\triangle CFG \frac{1}{2}$ the area of $\triangle CAB$?
 
mathlearn said:
Many Thanks caffeinemachine (Happy)

Is the area of $\triangle CFG \frac{1}{2}$ the area of $\triangle CAB$?
No. The area of $CFG$ is 1/4-th the area of $CAB$. This is easy to show. One way to show it is use the fact that $CFG$ and $CAB$ are similar triangles with corresponding side ratio's $1/2$.
 
caffeinemachine said:
No. The area of $CFG$ is 1/4-th the area of $CAB$. This is easy to show. One way to show it is use the fact that $CFG$ and $CAB$ are similar triangles with corresponding side ratio's $1/2$.

Thank you very much again (Sun)
 

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