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3 PS Qs I couldn't solve... detailed explanation someone pls

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3 PS Qs I couldn't solve... detailed explanation someone pls

by sohailmbaprep » Sun May 13, 2012 5:04 am
couldn't solve below attached 3 question from MGMT flash cards...please someone provide me detailed explanations of these Qs...thanks in advance
also pls let me know if any triangle is drawn(inside a circle) with diameter of a circle as its one side will be a right angled triangle... is it a rule ??

there are 2 more Qs..will attach them in next post....
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by mathbyvemuri » Sun May 13, 2012 6:33 am
Angle made by diameter at any point on the circumference of a circle is aright angle.So The triangle formed with the diameter as a side is right triangle with diameter as diagonal of the triangle.

Q1)we can have a total of 8C5 five member starting-groups.
Out of these, if we want to find the possibilities of inclusions of Tom and Alex, let us assume that those two are included. The rest 3 positions can be filled out of the remaining 6 in 6C3 ways.
Probability = (Required outcomes)/(All possible outcomes)
= 6C3/8C5
option "A"

Q2)let CD=x
angle ABD = angle DAB = 30 (as 180-120 = 60 is the sum of the two)
angle ABC = 180-90-30 = 60
angle CAB = 180-angle(ABC)-angle(ACB)=180-60-90=30
angle CBD = angle ABC - angle ABD = 60-30 = 30
angle CDB = 180-angle(DCB)-angle(CBD) = 180-90-30 = 60
Now consider trig. ratios:
tan(CDB) = tan(60) = sqrt(3) = CB/CD => CB = x*sqrt(3)

tan(CAB) = tan(30) = 1/sqrt(3) = BC/AC
BC^2 = AC^2/3 => 3x^2 = (8+x)^2 / 3
solving, we get only positive value 2 for x

applying pythagoras at angle BCD,
BD^2 = BC^2 + CD^2 = 3x^2 + x^2 = 4x^2
BD = 2x = 4

Perimeter of triangle BCD = BC+CD+BD = 2sqrt3+2+4 = 6+2sqrt3

Q3)If a,b,c are the sides of a rectangular cuboid,
length of main diagonal = sqrt(a^2+b^2+c^2)
= 6
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by aneesh.kg » Sun May 13, 2012 6:41 am
Yes, A triangle drawn inside a circle so that one of its sides is the diameter will always be right-angled at the angle opposite to the diameter. Always!

Solution 1:
If a lengths of the edges of a rectangular solid are a, b and c, then the length of its main diagonal will be (a^2 + b^2 + c^2)^0.5 (see derivation in the image below)

Image

When a = 4, b = 4, c = 2
Length of the main diagonal = (16 + 16 + 4)^0.5 = 6
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by aneesh.kg » Sun May 13, 2012 6:58 am
Solution 2:


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by aneesh.kg » Sun May 13, 2012 7:01 am
Solution 3:

Required Probability
= (Number of 5-member teams with Tom and Alex)/(Total number of 5-member teams)
= (Number of ways of selecting Tom and Alex AND Number of ways selecting remaining 3 members from the 5 people available)/(Number of ways of selecting 5 members from 8 people available)
= (2C2)*(6C3)/(8C5) = [6!/(3!3!]/[8!/(5!3!)]

Expression [spoiler](A)[/spoiler] is correct.
Last edited by aneesh.kg on Sun May 13, 2012 7:18 am, edited 2 times in total.
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by sam2304 » Sun May 13, 2012 7:04 am
Yes it is a rule. A triangle drawn inside a circle with diameter as one of its side will be right angled at the side opposite to the diameter.

1.5 students out of 8, total ways = 8C3.
Tom and alex are always selected so 1*1*6C3 - we are selecting remaining 3 from 6 ppl.

So probablity = fav outcome/total outcome
= 6C3/8C3

IMO A.

2.Given |_ABD = |_DAB so the triangle is isosceles.

DB = DA = 8.

In triangle BCD,

|_BDC = 180-120 = 60 |_BCD = 90 |_CBD = 30.

This is a right triangle of form 2, 1, sqrt3. We have 2 equivalent to 8. so 1 is equivalent to 4 and sqrt3 equivalent to sqrt48 = 4*sqrt3

Perimeter of triangle BCD = 8 + 4 + 4sqrt3 = 12 + 4sqrt3


3.Diagonal of a cube/cuboid formula = sqrt(sum of square of all three sides)

so diagonal = sqrt(4^2 + 4^2 + 2^2) = sqrt(36) =6.
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by sohailmbaprep » Sun May 13, 2012 8:04 am
Thank you all guys...it helped
By the way can someone of you please list the "ratio of sides of triangle formulae"
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by sam2304 » Sun May 13, 2012 9:25 am
sohailmbaprep wrote:Thank you all guys...it helped
By the way can someone of you please list the "ratio of sides of triangle formulae"
30 60 90 -> 1 sqrt3 2
45 45 90 -> 1 1 sqrt2

The above two are the basic ones which have integer angles. The other ones shown below don't have integer angles, yet they are widely known in that particular order.

3:4:5
5:12:13
6:8:10
8:15:17
7:24:25
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by aneesh.kg » Sun May 13, 2012 9:53 am
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by sohailmbaprep » Mon May 14, 2012 9:31 am
sam2304 wrote:
sohailmbaprep wrote:Thank you all guys...it helped
By the way can someone of you please list the "ratio of sides of triangle formulae"
30 60 90 -> 1 sqrt3 2
45 45 90 -> 1 1 sqrt2

The above two are the basic ones which have integer angles. The other ones shown below don't have integer angles, yet they are widely known in that particular order.

3:4:5
5:12:13
6:8:10
8:15:17
7:24:25
couldn't understand... please elaborate how it can be used in problems
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by sam2304 » Mon May 14, 2012 9:36 pm
When you are given the angles say 30 60 90 then that triangles will have values in the ratio of 1 sqrt3 2. Its the same way, which we used to solve the problem 2 posted by you. Triangle BCD is a right triangle we calculated the value of angles, we knew length of one side of the triangle. With this ratio we calculated the other sides and then the perimeter.

For the 3:4:5, 5:12:13 ... - it might be useful to find the value of one of the sides or that the triangle is a right triangle.
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