Geometry Problems

MathStudent1999

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3. The diagonals of the parallelogram measure 7 cm and 13 cm. What is the length in cm of the side of the parallelogram adjacent to the side that mesures 8 cm? Express your answer in simplest radical form.

4. The lengths of the sides of the triangle are 13cm, 17 cm, and 20 cm. What is the length in cm of the median to the longest side? Express your answer in the simplest radical form.

Using Herson Formula I managed to get the height of the triangle. Then using Heron's Fourmula again, I got the length of the median. I got: √129. Can someone verify this answer for me?

6. Circle A has a radius of 10 units and Circle B has a radius of 5 units. The center of Circle B is on the circumference of Circle A. What is the difference between the area which is within A but not B and the area which is within B but not A. Put your answer as a multiple of Pi.

8. A triangle whose verticies (0,0), (5,12) & (8,6) is seperated into two parts by a vertical line so that each part has the same area. What us the cordinates of the intersection of the deviding line and the x-axis? Express your answer in simplest radical form.

17. Define T to be a triangle with sides 3,4 & 5. Define D(T,p) to be the sum of the squares of the distances from the point p to the vertices of triangle T. What is the least value of D(,T,p)? Express your answer as a common fraction.

30. Marna baked a batch of brownies in a 9x12 pan. She then decided to created a giant ice cream sandwich by cutting two congruent circular brownies out of the pan and placing a wedge of chocolate chip ice cream between them. What is the radius of the largest circular brownies she can cut? Express you answer as a common fraction in simplest radical form


I got (15√2 - 15 / 2) for #30. Can someone verify this for me?

Please Help me on these six questions
 
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OK...I'll give you hints...one by one...(hoping someone else steps in!)

DRAW the darn thing:
Code:
     B              C


          M


A              D
M is crossing point of the diagonals; so BM = DM and AM = CM.

Let AB = 8 (given); BM = 3.5 and AM = 6.5
So you have all side lengths of triangle ABM.
From those, calculate angleAMB; then angleAMD = 180 - angleAMB.
Now using triangle AMD: you have angle AMD, AM = 6.5 and DM = 3.5;
use Law of Cosines to calculate AD...and that's the solution...

No trigonometry as I'm only in grade 7, but i do know trigonometry and understand what you are doing :)
 
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Correct; good work! however, you could make it simpler; formula:
a=13, b=17, c=20
median to c = SQRT(2a^2 + 2b^2 - c^2) / 2 = SQRT(516) / 2 = SQRT(129)
Similarly:
median to a = SQRT(2b^2 + 2c^2 - a^2) / 2
median to b = SQRT(2a^2 + 2c^2 - b^2) / 2

Wow! I never thought of that! Thanks!
 
This one answered at other site where you posted this.....still have questions?

Left is #8 and #17; show your work, then we'll see...

For #8, I tried using Heron's Formula before realizing that it doesn't have to be a triangle. Now I'm stuck, not knowing how to begin.

For #17, I don't get the question.

For #3, wouldn't Heron's formula also work?

Also could you help me on these questions?

http://www.freemathhelp.com/forum/threads/75058-Probability-Problems
 
USE GRAPH PAPER AND DRAW THE DARN THING!
Make the intersection points C and D; make the centers A and B;
look at triangle ABC (congruent to triangle ABD):
you have AB = AC = 10 and BC = 5;
so you're now able to calculate central angles at A and B.
If you're still stuck, look up "circle segment area"; like:
http://www.mathopenref.com/segmentarea.html

Using the cosine law, I found the central angle is 28.955 degrees, and i rounded it to 29. Then I multiplied 29/360 by 100pi(The area of the circle) And got 145pi/18. Using the Pythagoras therom and the multiplying by two, I found the area of the circle segment for circle A. What I got was {(145pi)-[SQRT(93.75) x 36]}/9. Using the exact same method for Circle B I got the area the Circle Segment as {(442.5pi)-[90 x SQRT(195)]}/72. So then add these two answers up I got: {(1602.5pi)-[228 x SQRT(93.75)]-[90 x SQRT(195)]}/72. After I got the area of Circle A(100pi) and subtract it from {(1602.5pi)-[228 x SQRT(93.75)]-[90 x SQRT(195)]}/72. I did the same with Circle B. Then I subtracted the two answers and got my final answer of: 75pi.

Is this the correct answer?

OK...I'll give you hints...one by one...(hoping someone else steps in!)

DRAW the darn thing:
Code:
     B              C


          M


A              D
M is crossing point of the diagonals; so BM = DM and AM = CM.

Let AB = 8 (given); BM = 3.5 and AM = 6.5
So you have all side lengths of triangle ABM.
From those, calculate angleAMB; then angleAMD = 180 - angleAMB.
Now using triangle AMD: you have angle AMD, AM = 6.5 and DM = 3.5;
use Law of Cosines to calculate AD...and that's the solution...

The two diagonals split the parallelgram into two differnt triangles. Using Heron's formula I got two equations to equal each other and got X equal to 3 x SQRT(5). Is this correct?
 
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