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Thread: Triangle ABC, AB=50, AC=10, area 120; D midpt of AB, E midpt of AC; angle bisector...

  1. #1

    Triangle ABC, AB=50, AC=10, area 120; D midpt of AB, E midpt of AC; angle bisector...

    Extremely difficult geometry question (not homework, just interesting):

    1) Triangle ABC with AB=50 and AC=10 has area 120. Let D be the midpoint of AB, and let E be the midpoint of AC. The angle bisector of angle BAC intersects DE and BC at F and G respectively. What is the area of quadrilateral FDBG?

    •Options are 60, 65, 70, 75, 80
    Last edited by stapel; 02-10-2018 at 05:32 PM. Reason: Creating useful subject line.

  2. #2
    Elite Member stapel's Avatar
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    Quote Originally Posted by wellskyw View Post
    Extremely difficult geometry question (not homework, just interesting):

    1) Triangle ABC with AB=50 and AC=10 has area 120. Let D be the midpoint of AB, and let E be the midpoint of AC. The angle bisector of angle BAC intersects DE and BC at F and G respectively. What is the area of quadrilateral FDBG?

    •Options are 60, 65, 70, 75, 80
    Since you're working through this problem set (with numbered questions and multiple-choice answers) for fun ("not homework, just interesting"), clearly you have basic geometric knowledge, so you've been able to make at least a good start on an answer. So, as a fellow math-lover who wants to provide good examples for struggling students:

    What are your thoughts? What have you tried? How far have you gotten? Where are you getting bogged down?

    Please be complete. Thank you!

  3. #3
    Quote Originally Posted by stapel View Post
    Since you're working through this problem set (with numbered questions and multiple-choice answers) for fun ("not homework, just interesting"), clearly you have basic geometric knowledge, so you've been able to make at least a good start on an answer. So, as a fellow math-lover who wants to provide good examples for struggling students:

    What are your thoughts? What have you tried? How far have you gotten? Where are you getting bogged down?

    Please be complete. Thank you!
    Thankyou for your thoughtful response. I completely agree that people should attempt to work things out before asking other people. And I did try but I was stumped. I had figured out the relationship between most of the parts of the shapes within triangle ABC, but I wasn’t sure how to use that information.

    That said, I figured it out if anyone is interested:
    Bytriangle midpoint theorem (I think that is what it is called) De is half the length of BC. Since triangle ADE and triangle ACE are similar, and because the ratio of 2 similar triangles’ areas is equal to the ratio of the sides squared, the area of ADE must be 1/4 the area of ABC. Thus, the area of quadrilateral BDEC is 120-30 which is 90.By angle bisector theorem, BG and BC are in a ratio of 5:6 and since heights of the two quadrilaterals ( BDEC and FDBG) are the same, the area of FDBG is equal to (5/6)*90 which is 75.

    i think this is correct but if anyone has any ideas on how better to solve it please share!

  4. #4
    Thankyou for your thoughtful response. I completely agree that people should attempt to work things out before asking other people. And I did try but I was stumped. I had figured out the relationship between most of the parts of the shapes within triangle ABC, but I wasn’t sure how to use that information.

    That said, I figured it out if anyone is interested:
    By triangle midpoint theorem (I think that is what it is called) De is half the length of BC. Since triangle ADE and triangle ACE are similar, and because the ratio of 2 similar triangles’ areas is equal to the ratio of the sides squared, the area of ADE must be 1/4 the area of ABC. Thus, the area of quadrilateral BDEC is 120-30 which is 90.By angle bisector theorem, BG and GC are in ratio of 5:6 and since heights of the two quadrilaterals ( BDEC and FDBG) are the same, the area of FDBG is equal to (5/6)*90 which is 75.

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