The base angles of an isosceles trapazoid are 60 degrees and
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You've drawn the picture, of course, noting the various angles. You have the area formula, and the relation of the two bases to the median, which is helpful. You only need to find the value of the height.
Using the median relation and the ratios formed by the sides of any 30-60-90 triangle, what equations can you form? How far can you get in working toward a solution?
Please reply showing all of your steps. Thank you.
Eliz.
Using the median relation and the ratios formed by the sides of any 30-60-90 triangle, what equations can you form? How far can you get in working toward a solution?
Please reply showing all of your steps. Thank you.
Eliz.
The small trapezoid on the top is proportional to the big one. The ratio of their areas is 1:4, because the ratio of their sides is 1:2.
If we let the measure of the small leg equal "x", then the measure of the height of the small trapezoid will be x/2sqrt[3], and the height of the large trapezoid will be x sqrt[3]. this will also mean that the leg of the small right triangle will equal x/2.
Ao the area of the top trapezoid will equal (30 - x)x / 2sqrt[3]. Since the area of the big trapezoid is four times that, then it will equal (30 - x)2x sqrt[3].
The area of the big trapezoid can also be displayed as 30(x sqrt[3]) because, since the median is 15, then the sum of the lengths of the two large legs equals 30. So if I set the two expressions for the area of the large trapezoid equal to each other, I get that x = 15.
Then the area of the trapezoid is 450 sqrt[3].
Please tell if this is right. Thank you.
____________________________________
Edited by stapel -- Reason for edit: spelling, capitalization, etc
If we let the measure of the small leg equal "x", then the measure of the height of the small trapezoid will be x/2sqrt[3], and the height of the large trapezoid will be x sqrt[3]. this will also mean that the leg of the small right triangle will equal x/2.
Ao the area of the top trapezoid will equal (30 - x)x / 2sqrt[3]. Since the area of the big trapezoid is four times that, then it will equal (30 - x)2x sqrt[3].
The area of the big trapezoid can also be displayed as 30(x sqrt[3]) because, since the median is 15, then the sum of the lengths of the two large legs equals 30. So if I set the two expressions for the area of the large trapezoid equal to each other, I get that x = 15.
Then the area of the trapezoid is 450 sqrt[3].
Please tell if this is right. Thank you.
____________________________________
Edited by stapel -- Reason for edit: spelling, capitalization, etc
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I don't believe this is true.Fouad1013 said:
Note: I get an answer that checks out in this "trapezoid" calculator, and my answer differs from yours. I used 30-60-90 triangles, not proportional areas.
Eliz.
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It would be really nice if you showed what you'd done. (Most legitimate tutors don't "do" students' work for them.) Help us help you by showing what you've done and where you're stuck.
Instead of using your ratio (which does not appear to be valid), try following the suggestions provided earlier.
You have named the shorter base as being "x". Now label the height (of the original trapezoid) as "h". What relations can you form, using h and the ratios for any 30-60-90 triangle? You have a sixty-degree triangle (cut off from one end of the trapezoid) and a thirty-degree triangle (formed by the diagonal). You can write expressions for the base of the sixty-degree triangle in terms of "h", using the standard ratios, and for the base of the thirty-degree triangle using similar methods. Then note that the base of the thirty-degree triangle is equal to the sum of "x" and the base of the sixty-degree triangle. What can you get from that?
Eliz.
Instead of using your ratio (which does not appear to be valid), try following the suggestions provided earlier.
You have named the shorter base as being "x". Now label the height (of the original trapezoid) as "h". What relations can you form, using h and the ratios for any 30-60-90 triangle? You have a sixty-degree triangle (cut off from one end of the trapezoid) and a thirty-degree triangle (formed by the diagonal). You can write expressions for the base of the sixty-degree triangle in terms of "h", using the standard ratios, and for the base of the thirty-degree triangle using similar methods. Then note that the base of the thirty-degree triangle is equal to the sum of "x" and the base of the sixty-degree triangle. What can you get from that?
Eliz.