Plane Geometry Problem

math9maniac · 2016-01-26 06:06:28

Hi. I'm putting up these questions so you point out where I've gone wrong. My answers are not what the book has. Thanks.

#1. The diagonals of a rhombus are 60 mm and 80 mm long. Find, to the nearest 0.1 degree, the acute angle between the sides of the rhombus. Also, find the perimeter of the rhombus.

math9maniac · 2016-01-26 06:09:09

Question : Since the sides of a rhombus are equal, Is equality also the case of the angle formed by a pair of sides of the rhombus? Are the four angles equal? Why or why not?

math9maniac · 2016-01-26 06:20:36

#2. For an examination, candidates have to be at least 1.5 m apart. What is the greatest number of candidates that can be fitted into a rectangular room 19 m × 11 m?

Bob · 2016-01-26 06:24:35

hi math9maniac

That perimeter is correct. There are two things wrong with your angle calculation (1) To get the acute angle you need arctan(3/4). (2) Then you need to double this as the diagonal splits the angle in half.

A rhombus is a special case of a parallelogram. Opposite angles are equal but only all four if it's a square. You can think of a square as a special case of a rhombus.
An explanation could come from the unequal nature of the diagonals.

Bob

math9maniac · 2016-01-26 06:29:24

#3. PQRS is a rhombus, and A, B, C, D are the midpoints of the sides PQ, RS and SP respectively. Given that ABCD is a square and that the diagonal PR of the rhombus is 80 mm long, calculate the length of the diagonal QS.

Haven't solved this yet. I've drawndiagram, but I'm a little confused somewhere.

Book provides answer as 80 mm.

Thanks for your usual cooperation. Regards.

Relentless · 2016-01-26 06:31:21

Hi,

I agree with you about the perimeter, but the book is correct about the angle. There is a rule that the area is equal to half the product of the diagonals (2400mm^2 = 24cm^2 in this case). There is another rule that the area is equal to the side length squared times the sine of either angle.

Which is close to 73.74 degrees.

To your second post, it is part of the definition of a rhombus that the side lengths are equal. Regarding the angles, unless the rhombus is a square (where all angles are equal), there will be two different angles because the opposite angles are corresponding. One pair of angles will be acute, the other obtuse, and since they all sum to 360 each type of angle must sum to 180

Last edited by Relentless (2016-01-26 06:38:18)

math9maniac · 2016-01-26 06:32:13

Thanks for this so far.

math9maniac · 2016-01-26 06:33:56

Alright, I'll look at it again and get back.

Relentless · 2016-01-26 07:07:12

Hello again!

I think the answer to problem three is to look at the quadrants as triangles, and draw a line connecting, say, B and D. The angle that line DQ makes with line DB will be 45, and so the angle on the other side of DQ will be 45 as well. But then, symmetrically, so will the angle opposite. Since the angle of the rhombus itself is the only one remaining of the "quadrant triangle", it must be 180 - 45 - 45 = 90 degrees; therefore, the rhombus is a square.

Also, for problem two I think there are two reasons you are getting less than expected: your calculations assume that the students must be at least 0.75m away from the walls, and most importantly, the area occupied by a student will be a circle of radius 0.75, not a square. The mathematics of these packing problems can be very difficult; I find it is best just to fiddle with them.

I got the answer in the book for problem 2 also. This must be some book! Usually every third answer must be corrected.

Start by putting candidates at the corners, and then evenly spaced along the sides. You will fit 38 candidates. Next just work out how many candidates will fit down and across: you get 6 by 11 for a total of 66 + 38 = 104. No fiddling required.

I haven't ruled out the possibility of leaving a gap beside two corner candidates and strategically placing people such that you could fit one or two more, but having explored the possibility, I am not sure it can be done in this case.

Last edited by Relentless (2016-01-26 07:21:12)

Bob · 2016-01-26 20:22:42

hi math9maniac

3). In triangle PQR, as A and B are midpoints, AB is parallel to PR and equal to half its length. Similarly BC, CD and DA. Also the diagonals of all rhombuses bisect each other at right angles.

So for any rhombus ABCD will always be a rectangle, with sides equal to half the diagonals of the rhombus.

So, if you are additionally told that ABCD is a square, then PR = QS and that makes the rhombus a square too.

Bob

math9maniac · 2016-01-29 19:56:03

Thanks a million for your kind assistance.

Have a nice day!

Math Is Fun Forum

#1 2016-01-26 06:06:28

Plane Geometry Problem

#2 2016-01-26 06:09:09

Re: Plane Geometry Problem

#3 2016-01-26 06:20:36

Re: Plane Geometry Problem

#4 2016-01-26 06:24:35

Re: Plane Geometry Problem

#5 2016-01-26 06:29:24

Re: Plane Geometry Problem

#6 2016-01-26 06:31:21

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#7 2016-01-26 06:32:13

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#8 2016-01-26 06:33:56

Re: Plane Geometry Problem

#9 2016-01-26 07:07:12

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#10 2016-01-26 20:22:42

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#11 2016-01-29 19:56:03

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