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I have four questions:
1. The area of trapezoid ABCD is 60. One base is 4 units longer than the other, and the height of the trapezoid is 5. Find the length of the median of the trapezoid.
2. In trapezoid ABCD, line BC is parallel to line AD, angle ABD = 105 degrees, angle A = 43 degrees, and angle C = 141 degrees. Find angle CBD, in degrees.
3. The bases of trapezoid ABCD are lines AB and CD. Let M be the midpoint of line AD. If the areas of triangles ABM and CDM are 5 and 17 respectively, then find the area of trapezoid ABCD.
4. Let ABCD be a parallelogram. Extend line BC past B to F, and let E be the intersection of lines AB and DF. If the areas of triangles BEF and ADE are 1 and 9, respectively, find the area of parallelogram ABCD.
hi JCHAN
http://www.mathsisfun.com/geometry/trapezoid.html
That should help you to get (1). If you post your answer I'll try to help with the rest.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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okay, I made a real account.
The answer to 1 should be 12, since the bases are 10 and 14.
Last edited by SPARKS_CHAN (2014-12-06 05:29:58)
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hi
Welcome to the forum!
That answer looks good to me. I'll have to make a diagram for Q2.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Ok Hint for Q2.
Extend CB to a point E somewhere on the line.
AB cuts two parallels, so DAB = ABE = 43. ( some call this 'alternate angles' but I've also seen it as 'Z-angles' )
Hope that helps.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Ok Hint for Q2.
Extend CB to a point E somewhere on the line.
AB cuts two parallels, so DAB = ABE = 43. ( some call this 'alternate angles' but I've also seen it as 'Z-angles' )
Hope that helps.
Bob
I got question 2!
The answer is 32.
I've solved 3 as well. The answer was 44.
Last edited by SPARKS_CHAN (2014-12-06 06:49:34)
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hi SPARKS_CHAN
That's excellent; both correct. Now for Q4.
As AD is parallel to FB, the triangles ADE and BFE are similar => one is an exact enlargement of the other.
The areas are in the ratio 9:1 so the lengths must be in the ratio 3:1
So let BF = a => AD = 3a (= BC)
So FC = 4a.
The triangles BFE and CFD are also similar, so you can work out the area of CFD, then the quadrilateral CBED, and then ABCD.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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