Trapeze

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Radit2125

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Could You help me also with these 2 problems, please?
1. The bisectors of the obtuse angles of the trapezoid intersect on the large base in point M. Find the sides of the trapezoid, if the bisector of angle D is equal to 13 cm, the bisector of angle C is equal to 15 cm, and the heights of the trapezoid are12 cm.
2. The trapezoid ABCD WITH BASES AB = 3 cm AND CD = 1 cm IS GIVEN. A CIRCLE CAN BE ENTERED AND DESCRIBEDI around IT. FIND THE FACE OF THE DESCRIBED CIRCLE.

**for the first problem I can find the base H2M and the base H1M with the Pythagorean theorem and the base CD=H2H1 which is equal to the sum of the bases H2M and H1M(I'm sorry that the second graph is not so accurate )

** for the second problem I assume that if a circle can be entered or described around a trapezoid maybe this means that the trapezoid is isosceles
 

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Radit2125 said:
Could You help me also with these 2 problems, please?
1. The bisectors of the obtuse angles of the trapezoid intersect on the large base in point M. Find the sides of the trapezoid, if the bisector of angle D is equal to 13 cm, the bisector of angle C is equal to 15 cm, and the heights of the trapezoid are12 cm.
2. The trapezoid ABCD WITH BASES AB = 3 cm AND CD = 1 cm IS GIVEN. A CIRCLE CAN BE ENTERED AND DESCRIBEDI around IT. FIND THE FACE OF THE DESCRIBED CIRCLE.

**for the first problem I can find the base H2M and the base H1M with the Pythagorean theorem and the base CD=H2H1 which is equal to the sum of the bases H2M and H1M(I'm sorry that the second graph is not so accurate )

** for the second problem I assume that if a circle can be entered or described around a trapezoid maybe this means that the trapezoid is isosceles
Click to expand...
1. Find necessary angles, then sides.
2. Please make a better graph. Label the vertices correctly, make sure the circles are inscribed and circumscribed: https://www.chegg.com/homework-help/definitions/inscribed-and-circumscribed-63
What's a face of a circle?
 
1. I assume that I know how to find them but was there a theorem that, for example, proves that angle MDH2 = 30 degrees and angle H2MD= 60 degrees.
2. It is the area of the circle.
 

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Radit2125 said:
1. I assume that I know how to find them but was there a theorem that, for example, proves that angle MDH2 = 30 degrees and angle H2MD= 60 degrees.
2. It is the area of the circle.
Click to expand...
1. Prove? Why do you need to prove that it's 30? You can just calculate it. You know 2 sides in a right triangle. You can calculate all angles in the diagram. Then use the triangles on the left and right to find the sides based on right triangle properties.
2. You probably need to draw a bunch of additional segments. Which points do you think should be connected so that we can look for relationships among the circles and the trapezoid sides?
 
In this way, we could find the midsegment of the trapezoid which is (AB+CD)/2= 2 cm which is equal to the diameter of the inscribed circle. So, the radius is equal to 1 cm. But then, what should I do?
 
Radit2125 said:
In this way, we could find the midsegment of the trapezoid which is (AB+CD)/2= 2 cm which is equal to the diameter of the inscribed circle. So, the radius is equal to 1 cm. But then, what should I do?
Click to expand...
Could you add this to the diagram?
 
Radit2125 said:
Here it is
Click to expand...
If you 'zoom in' you'll see that
1. Your diameter line is not straight
2. The points where the inscribed circle touches the trapezoid are NOT the points where the 'horizontal' line through O intersects BC and AD. Think about it - if FG is 'horizontal' (parallel to AB), then a line tangent to the circle at G would be vertical. It can theoretically happen on one of the sides, but NOT both, since AB and CD have different lengths.
 
Can I add this?
Because the trapezoid is inscribed =>AB+CD=BC+AD (BC=AD)
BC+AD= 4 => BC=AD=2 cm

Could You give me more instructions, because I still don't understand? Could I find out the radius of the circumscribed circle, if I build the heights?
 

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Radit2125 said:
Can I add this?
Because the trapezoid is inscribed =>AB+CD=BC+AD (BC=AD)
BC+AD= 4 => BC=AD=2 cm

Could You give me more instructions, because I still don't understand? Could I find out the radius of the circumscribed circle, if I build the heights?
Click to expand...
I see that AB+CD=BC+AD. Do you know why?
I don't see why BC=AD.
I don't see why BC=AD=2 cm.
I am not saying this is wrong - if you have an explanation, please post.

Regarding the radius of the circumscribed circle - to find it should you probably draw some segments whose length is the radius. Those segments should probably form some shapes together with the trapezoid sides. E.g. triangles. Then, using the known trapezoid sides and properties of the newly constructed shapes you can find the radius. I haven't done it myself, but that's what I would try.
 
Radit2125 said:
Could You help me also with these 2 problems, please?
1. The bisectors of the obtuse angles of the trapezoid intersect on the large base in point M. Find the sides of the trapezoid, if the bisector of angle D is equal to 13 cm, the bisector of angle C is equal to 15 cm, and the heights of the trapezoid are12 cm.

**for the first problem I can find the base H2M and the base H1M with the Pythagorean theorem and the base CD=H2H1 which is equal to the sum of the bases H2M and H1M(I'm sorry that the second graph is not so accurate )
Click to expand...
Radit2125 said:
1. I assume that I know how to find them but was there a theorem that, for example, proves that angle MDH2 = 30 degrees and angle H2MD= 60 degrees.
Click to expand...

This would be far easier to follow if you had stuck with one problem per thread, as requested. I can't be sure who is talking about what!

I'll stick with #1.

There is no angle equal to 30 or 60 degrees, and you don't need to use angles at all, though you could if you don't want an exact answer. Is there a reason you believed that?

The problem can be solved by drawing in radii MX and MY to the points of tangency on AD and BC and working with the triangles MAX and BMY, which are similar to another pair you've already used.
 
lev888 said:
I see that AB+CD=BC+AD. Do you know why?
I don't see why BC=AD.
I don't see why BC=AD=2 cm.
I am not saying this is wrong - if you have an explanation, please post.

Regarding the radius of the circumscribed circle - to find it should you probably draw some segments whose length is the radius. Those segments should probably form some shapes together with the trapezoid sides. E.g. triangles. Then, using the known trapezoid sides and properties of the newly constructed shapes you can find the radius. I haven't done it myself, but that's what I would try.
Click to expand...

I wrote that BC=AD because I think that only an isosceles trapezoid could be inscribed in a circle.
If we define the triangles DOC and AOB, the radius would be a height in these triangles.
 
Radit2125 said:
I wrote that BC=AD because I think that only an isosceles trapezoid could be inscribed in a circle.
If we define the triangles DOC and AOB, the radius would be a height in these triangles.
Click to expand...
BC=AD - agree.
Yes, the radius is the height, might be useful. But keep in mind which circle's area and which radius we need.
 
What if we define the triangles DMC and AMB? Would not we then get 2 isosceles triangles, because MA=MB=MC=MD= R?
 
Radit2125 said:
What if we define the triangles DMC and AMB? Would not we then get 2 isosceles triangles, because MA=MB=MC=MD= R?
Click to expand...
Yes. All 4 triangles you get are isosceles.

Something to look into: what is the length of the bottom leg of the 2 right triangles? If the hypotenuse is 2, what does it tell you about the top angle?
 
I don't know what the length of the bottom leg is but if the hypotenuse is 3 cm, doesn't that mean that the top angle (AMB) is obtuse?
 

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Radit2125 said:
I don't know what the length of the bottom leg is but if the hypotenuse is 3 cm, doesn't that mean that the top angle (AMB) is obtuse?
Click to expand...
You wrote above that BC is 2, why is it now 3cm?
Look at the diagram with heights. AB consists of 3 segments. We know that the middle one is what? That leaves how much for the other 2? And if their lengths are equal...
 
Last edited:
I made a mistake, BC is 2 cm. About the middle one we know that it is equal to CD= 1 cm and the other 2 are also 1 cm.
 
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