This discussion is closed.
john !!
Badges: 14
Rep:
?
#1
Report Thread starter 16 years ago
#1
The area of the largest right angle triangle that will fit inside a circle is 18 square units. What is the area of the circle?

1!2!3!4!5!6! has how many factors?

Is the product of the cubes of the roots of the equation: x^2 + 7x + 22 = 0 a positive or negative number?

Is 9^n - 1 always divisible by 8? If it is, prove that it is, if it isn't, give a counterexample.
0
beauford
Badges: 0
Rep:
?
#2
Report 16 years ago
#2
(Original post by mik1a)
The area of the largest right angle triangle that will fit inside a circle is 18 square units. What is the area of the circle?

1!2!3!4!5!6! has how many factors?

Is the product of the cubes of the roots of the equation: x^2 + 7x + 22 = 0 a positive or negative number?

Is 9^n - 1 always divisible by 8? If it is, prove that it is, if it isn't, give a counterexample.
1!2!3!4!5!6! = 2^4.3^2.5, which has (4+1)(2+1)(1+1) = 5.3.2 = 30 factors (as in divisors)

The roots a,b satisfy ab = 22, so (ab)^3 = 22^3 > 0.

9^n = 1^n mod 8 = 1 mod 8, thus 9^n -1 = 0 mod 8.
0
Jonny W
Badges: 8
Rep:
?
#3
Report 16 years ago
#3
(Original post by mik1a)
1!2!3!4!5!6! has how many factors?
1!2!3!4!5!6!
= 2*3*(2*2)*5*(3*2)
= 2^4 * 3^2 * 5.

So the factors of 1!2!3!4!5!6! are all the numbers of the form

2^i * 3^j * 5^k,

where 0 <= i <= 4, 0 <= j <= 2 and 0 <= k <= 1. There are 5*3*2 = 30 such numbers.
0
beauford
Badges: 0
Rep:
?
#4
Report 16 years ago
#4
(Original post by mik1a)
The area of the largest right angle triangle that will fit inside a circle is 18 square units. What is the area of the circle?
Now any right angled triangled inscribed in a circle must have the hypotenuse as the diameter. Calling the diameter the base, the base is clearly fixed, and the height depends upon where the two other sides meet. so Area = bh/2, so we want to maximise h, this happens when the altitude to meet the base is halfway between the two ends of the diameter, i.e the two other sides are equal. it follows that base = 2r, h = r, and so r^2 = 18, and so Area of circle = 18pi.
0
lgs98jonee
Badges: 2
Rep:
?
#5
Report 16 years ago
#5
(Original post by mik1a)
The area of the largest right angle triangle that will fit inside a circle is 18 square units. What is the area of the circle?

1!2!3!4!5!6! has how many factors?

Is the product of the cubes of the roots of the equation: x^2 + 7x + 22 = 0 a positive or negative number?

Is 9^n - 1 always divisible by 8? If it is, prove that it is, if it isn't, give a counterexample.
does the middle q have any real roots?
0
beauford
Badges: 0
Rep:
?
#6
Report 16 years ago
#6
(Original post by lgs98jonee)
does the middle q have any real roots?
It doesn't matter
0
Jonny W
Badges: 8
Rep:
?
#7
Report 16 years ago
#7
(Original post by lgs98jonee)
does the middle q have any real roots?
The product of the roots is real (= 22).
0
john !!
Badges: 14
Rep:
?
#8
Report Thread starter 16 years ago
#8
IT has complex roots which can be multiplied... as the roots are conjunctuates the imaginary parts dissapear when you find their product or sum. (neat huh)

The sum of the roots of ax^2 + bx + c = 0 is always -b/a, and the product is always c/a, neat thing to know that will help avoid a lot of quadratic equationing.
0
JamesF
Badges: 1
Rep:
?
#9
Report 16 years ago
#9
If you didnt think of that, you could of course use the formula to get
(1/2[-7+sqrt(-39)])^3*(1/2[-7-sqrt(-39)])^3

Which is the difference of 2 squares giving
(88/4)^3 = 22^3.
0
john !!
Badges: 14
Rep:
?
#10
Report Thread starter 16 years ago
#10
A chord of length sqrt 3 divided a circle of radius 1 into two regions. Find the area of the largest rectangle which can be inscribed in the smaller region.

I tried this one about 6 months ago and didn't do too well, couldn't do the maths when it came to finding the constraints of the rectangle. Could see the calculus at the end of the tunnel, but couldn't get there!
0
Fermat
Badges: 8
Rep:
?
#11
Report 16 years ago
#11
(Original post by mik1a)
A chord of length sqrt 3 divided a circle of radius 1 into two regions. Find the area of the largest rectangle which can be inscribed in the smaller region.

I tried this one about 6 months ago and didn't do too well, couldn't do the maths when it came to finding the constraints of the rectangle. Could see the calculus at the end of the tunnel, but couldn't get there!
I get Amax = ¼
0
Fermat
Badges: 8
Rep:
?
#12
Report 16 years ago
#12
Naah, forget it. It's wrong!
0
Fermat
Badges: 8
Rep:
?
#13
Report 16 years ago
#13
OK, got a new result.

From Fig 1, and the dimensions given, it can be seen that the distance from the circle centre to the chord is ½.
In Fig 2, the dist h' = ½, and the height of the rectangle in the smaller region is h and the half-width of that rectangle is b.
From Fig 2, we can see that b² + (h+h')² = r², where r is the radius of the circle and r=1.

The maths bit

b² + (h+h')² = r²
b² + (h+½)² = 1
b² = 1 - (h+½)²
==========

Area of rectangle is A,

A = 2bh
A = 2h{1 - (h+½)²}^(½)
dA/dh = 2{1 - (h+½)²}^(½) - 2h.½.2(h+½){1 - (h+½)²}^(-½) = 0
{1 - (h+½)²}^(½) = h.(h+½){1 - (h+½)²}^(-½)
{1 - (h+½)²} = h.(h+½)
1 - h² - h - ¼ = h² + h/2
2h² + 1.5h - 0.75 = 0
h = -1.5 ± rt(1.5² + 4.2.0.75) / 4
h = {-1.5 ± rt(8.25) }/4
h = -0.375 + 0.718
h = 0.3443
=======

b² = 1 - (h+½)²
b² = 1 - 0.8443²
b² = 0.2892
b = 0.5378
=======

A = 2bh
A = 2*0.5378*0.3443
A = 0.37
======
Attached files
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

How are you feeling ahead of results day?

Very Confident (32)
8.08%
Confident (54)
13.64%
Indifferent (56)
14.14%
Unsure (102)
25.76%
Worried (152)
38.38%

Watched Threads

View All