Questions on Geometry: Pythagorean theorem answered by real tutors!

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Is the expression ???


If so, then


Or....


Is the expression ???


If so, then


Note:

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The iagonal, the eight, and the idth form a Right Triangle. And we all know in a Right riangle, the 3 sides are in a relation into equation:

where-----
Then:



, ANSWER
Thank you,
Jojo

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Using the law of tan we have the opposite/adjacent:
t(40)=x/200
.839=x/200
x=.839*200
x=167.82 meters is the height of the cliff.

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Use the pythagorean theorem then find the exact length of the hypotenuse in the following right triangle
5 and 7
-----------------------
hypoteuse^2 = 5^2 + 7^2
h^2 = 25+49
h^2 = 74
hypotenuse = sqrt(74)
===========================
Cheers,
Stan H.

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a radio tower located 18 miles north of a straight highway has a transmission
radius of 33 miles. find x the length of the section of highway that is within
the transmission radius of the tower.
:
If you draw this out, you can see that two right triangles are formed by the
radii, the highway, and 18 mi. The radius will be the hypotenuse, the third
side will be half the distance of the highway inside the transmission circle.
:
Let a = .5x
:
a^2 + 18^2 = 33^2
:
a^2 + 324 = 1089
:
a^2 = 1089 - 324
:
a^2 = 765
:
a =
a = 27.6586
:
x = 2*27.6586
:
x = 55.317 mi is the highway

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Because the "diagonal", "base" and "height" of the rectangle forms a "right triangle" we can apply Pythagorean's theorem.
.
Let h = height
then
h^2 + 4^2 = 5^2
h^2 + 16 = 25
h^2 = 9
h = 3 inches
.
Area is then:
"height" * "base"
= 3 * 4 = 12 square inches

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a^2+b^2=c^2
1^2+x^2=h^2
h=sqrt(1+x^2)

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a^2+b^2=c^2
1^2+2^2=3^2
1+4=9
5=9 this says there is no right triangle with sides=1,2 & 3.
However there is a right triangle that has sides of 3 consecutive natural numbers.
3,4 & 5.
Proof:
3^2+4^2=5^2
9+16=25
25=25

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If this is a length, then only use the positive value.

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1^2+1.7^2=x^2
1+3.24=x^2
x^2=4.24
x=sqrt4.24
x=2.06 meters.
Looks like a fit.

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You're missing some information. Please repost the entire problem (along with the given equations)

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The 2 sides and the diagonal form a right triangle. Thus we have the formula: a^2+b^2=c^2
x^2+(x+2)^2=10^2
x^2+x^2+4x+4=100
2x^2+4x+4-100=0
2x^2+4x-96=0
2(x^2+2x-48)=0
2(x+8)(x-6)=0
x-6=0
x=6 answer for the shorter side.
6+2=8 for the longer side.
Proof:
6^2+8^2=10^2
36=64=100
100=100

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One side of a rectangular stage is 2 meters longer than the other.
If the diagonal is 10 meters, then what are the lengths of the sides?
:
Actually it's:
a^2 + b^2 = c^2
:
In this problem call one side x:
Let x = a
then
(x+2) = b
and
c = 10
:
Substituting in our formula:
x^2 + (x+2)^2 = 10^2
:
FOIL (x+2)(x+2)
x^2 + (x^2 + 4x + 4) = 100
:
Combine like terms, arrange as a quadratic equation
x^2 + x^2 + 4x + 4 - 100 = 0
2x^2 + 4x - 96 = 0
:
Factor
(2x - 12)(x + 8) = 0
;
We want the positive solution here
2x = 12
x = 6 meters is the 1st side
then
6 + 2 = 8 meters is the 2nd side
:
:
We can check that: 6^2 + 8^2 = 10^2
:
Was this understandable? Any questions?

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This web site explains it in detail:
http://www.purplemath.com/modules/distform.htm
.
If you still have trouble, write back.

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here goes .......................
A = 4B
C = B-20
solve for B
pythagorean formula is C^2 = A^2 + B^2
substituting to get all unknowns in terms of B, equation becomes
(B-20)^2 = (4B)^2 + B^2
multiplying out, formula becomes
B^2 - 40B + 400 = 4B^2 + B^2
combining like terms, formula becomes
-4B^2 - 40B + 400 = 0
multiplying left hand side of equation and right hand side of equation by -1 and formula becomes
4B^2 + 40B - 400 = 0
dividing both sides of the equation by 4 and the equation becomes
B^2 + 10B - 400 = 0
answer is not evident by inspection so use quadratic formula to get the roots.
quadratic formula is
and
where
-b = -10
b^2-4*a*c = (10^2 - 4*1*(-100) = 100 - (-400) = 500
2a = 2
solving we get B = 6.180339888 or B = -16.180339888.
since B can't be negative, the answer is B = 6.180339888.
substituting for B in the original equation derived from the pythagorean formula, we get
B^2 + 10*B - 100 = (6.180339888)^2 + 10*(6.180339888) - 100
= 38.19660113 + 61.80339888 - 100 = 0 which becomes
100 - 100 = 0
which becomes 0 = 0 proving formula is correct.
answer is B = 6.180339888 = 6.18 rounded to nearest hundredth.

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width = w
length = 3w + 3 (3 times the width plus 3 centimeters)
diagonal = 3w + 4 (length is 1 cm less than diagonal which means diagonal is 1 cm more then length (3w + 3 is the length + 1 = 3w + 4 is the diagonal)).
solve for w using pythagorean theorem (length squared plus width squared = diagonal squared)since the diagonal is the hypotenuse of a right triangle formed by the length and the width and the diagonal.
equation becomes:
length (3w+3) squared plus width (w) squared = hypotenuse (3w+4) squared.
becomes (3w+3)^2 + w^2 = (3w+4)^2.
becomes 9w^2 + 18w + 9 + w^2 = 9w^2 + 24w + 16
becomes w^2 - 6w - 7 = 0
becomes (w-7) * (w+1) = 0
w = 7 or w = -1.
can't be -1 so w = 7
w = 7 then length = 3w + 3 = 24 and diagonal = 3w + 4 = 25.
width = 7
length = 24
diagonal = 25
check pythagorean theorem 7^2 + 24^2 = 25^2
becomes 49 plus 576 = 625
becomes 625 = 625.

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how will you factor 2x^2-178x-14280=0
(2x ) (x ) << I cant find the factors
-------------------------------------------
(x-140)(2x+102)
--------------------
How to do these?
Consider b^2-4ac = 178^2-4*2*-14280 = 145924
Take the sqrt to get 382
Add -b = 178 to get 560
Divide by 2a = 4 to get 140
So, one of the factors is (x-140)
Divide 14280 by 140 to get 102
----------
The other factor must be 2x+102
Check it out to see if it is correct.
========================================
Cheers,
Stan H.

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Let x=length of first leg and y=length of second leg


Since the "length of a leg of a right triangle is two times the length of the other", this means that . So in this case "y" is the longer leg.


So with the use of Pythagoreans theorem, we get




Square 25 to get 625


Plug in


Square to get


Add


Divide both sides by 5.


Take the square root of both sides. Note: only the positive square root is considered.


Simplify the square root.



So the length of one leg is (which approximates to ) and the length of the longer leg is (which approximates to )

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Are there any other shapes that can be formed of the side of the side of the triangle besides squares that can complete the formula for example a hexagon? I have use a circle and that works if you use the length of the sides as the diameters. I think this is because of the square in the r2 is essentially the same as Pythagoras.

All you need is that all three figures be similar
plane figures, that is, have the exact same shape.

Edwin

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x=height
2x+4=base
3x-4=incline

Therefore
x=a
2x+4=b
3x-4=c

These are the values to plug into a^2+b^2=c^2.

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3600= 2x^2 + 10x + 25
then
2x^2+10x+25-3600=0
2x^2+10x-3575=0
=
=

then solutions are and
positive solution is
=39.85 (aprox)
So legs are 39.85 and 44.85 aprox

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yes, you just solve for x (which is one side) and figure out what x+5 is to find the other side. put those two values back in the formula to test your answer and make sure that the pythagorean theorum still works.

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Let cos(beta) =a/1; this implies that x = a and r=1
Therefore y = sqrt(1-a^2)
Therefore sin(beta)= [sqrt(1-a^2)]/r
-------------------------------------
Find the expression for cos(2beta) and sin(2beta) in terms of a:
cos(2beta) = cos^2(beta)-sin^2(beta
= a^2 - [(1-a^2)/r^2]
------------------------------------ ------
sin(2beta) = 2*sin(beta)*cos(beta)
= 2 * sqrt(1-a^2)/r]
--------------------------
and hence confirm that - cos^2(2beta) + sin^2(2beta) = 1
------------------
Subsitute to confirm that statement is true.
====================
Cheers,
Stan H.

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c=2.5
a=b+1.8
a^2+b^2=c^2
(b+1.8)^2+b^2=(2.5)^2
b^2+3.6b+3.24+b^2=6.25
2b^2+3.6b-3.01=0
b=.621512
a=2.42151
c=2.5
.
Ed
.

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=37.04 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 0.621512405470294, -2.42151240547029. Here's your graph:

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We have a right triangle of height h, width h+2, and hypotenuse sqrt(6).
a^2+b^2=c^2
h^2+(h+2)^2=sqrt(6)^2
h^2+h^2+4h+4=6
2h^2+4h-2=0
2(h^2+2h-1)=0
h^2+2h =1
h^2+2h+1=1+1 completing the square.
(h+1)^2=2
h+1=+-sqrt(2) Square root of each side.
h= -1+sqrt(2) m
w= 1+sqrt(2) m
.
Ed

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Edwin's solution:
I need an explanation of why the Pythatgorean theorem can be used only for right triangles that an 8th grader can understand. What I know is that the Pythagorean Theorem is a special case of the Law of Cosines where the Cos(c) is equal to zero which only happens when c=90 degrees. But I don't think an 8th grader has had the law of cosines yet. I'd really appreciate some help with this. Thanks.

Suppose you had this triangle, which is not necessarily a right triangle:



You only know that  but you don't know necessarily 
that angle  is 90°.

Now you construct another triangle, this time a RIGHT triangle 
with the SAME SIZE legs,  and , but with its hypotenuse
 not given to be equal to :



But we know that the Pythagorean theorem DOES hold true for triangle
DEF because we constructed it so that it is a right triangle with
angle E being a right angle. Therefore



And we are given that



So  because both are equal to .

and we can take positive square roots of both sides:



or 

Now we have all three sides of triangle  equaling 
all three corresponding sides of triangle , so the
two triangles are congruent, and therefore angle 
equals angle E which equals 90°, for they are corresponding
parts of congruent triangles. Therefore  is a
right triangle.

Therefore all we need know is that  to be
able to conclude that triangle  is a right triangle.

So the only time the Pythagorean theorem works is when we
have a right triangle.

Edwin

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Here is why the Pythagorean Theorem can only be used to solve right triangles. The Pythagorean Theorem is with a and b the length of the legs of the triangle and c as the hypotenuse, or the longest side. It has variations too, like and . These three are used to figure out, with given any angles, whether a triangle is right, obtuse or acute (respectively) by plugging in the lengths given. If or holds true, than the triangle is obtuse or acute. But if holds true, than the triangle is right, and it has to be right. You are probably more familiar with this version: . It is only true for right triangles, because of the unique relationship among the three side lengths, which never holds true for any other type of triangle. That is why this variation of it (with the = sign) can only be used for right triangles and right triangles only. Right triangles are the only triangles that make this true.

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The web already has many proofs of Pythagorean theorem.
I would suggest you go to:
www.google.com
and enter "proof pythagorean"
you will find many examples along with explanations.
.
Here is one example:
http://mathforum.org/isaac/problems/pythagthm.html

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An overhead wire stretches from the service drop that is 20ft from the ground,on a building to an insulator on a pole 35 ft from the ground. How long is the wire if the pole and building are 48 ft apart?(Assume ground is level.)Round answer to tenths.
.
The first thing you need to do is to draw a diagram!
.
From your diagram, you'll notice that the ground, building and pole forms a "right triangle". Because it is a right triangle (one of the interior angle of the triangle is 90 deg), we can apply Pythagorean theorem.
.
Let x = length of the overhead wire
then
x^2 = 48^2 + (35-20)^2
x^2 = 48^2 + 15^2
x^2 = 2304 + 225
x^2 = 2529
x = 50.3 ft

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Solve for the unknown side in the following right triangle. (side AC is the hypotenuse.) Round your answer to two places, where applicable.Side AB ? Side BC12 Side AC 19
From Pythagorean theorem we know:
(AC)^2 = (AB)^2 + (BC)^2
plugging in what was given:
(19)^2 = (AB)^2 + (12)^2
(19)^2 = (AB)^2 + (12)^2
361 = (AB)^2 + 144
361 - 144 = (AB)^2
217 = (AB)^2
14.73 = (AB)

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Apply Pythagorean theorem
.
Let x = length of conduit
then
x^2 = 200^2 + 60^2
x^2 = 40000 + 3600
x^2 = 43600
x = 208.8 feet

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The sum of the lengths of the diagonals of a rhombus of side 61 cm is 142 cm. What is (a) the difference in the lengths of the diagonals, (b) the area of the rhombus?
i can't get the solution out of it as i think i do not have enough informantion so i do not know how to solve it.

Yes, there's enough information.  First we draw a rhombus with 61cm sides:







Now draw in the longer diagonal and label its length x cm.







Now take away the bottom half of the rhombus,

and you have an isosceles triangle. Label the

angle :  







We use the law of cosines:



 





That's one equation we will use.



Now let's go back to the original rhombus.







Since two angles of a parallelogram

which are next to each other are 

supplementary, we label the angle

° 

 

Now we draw the other diagonal and label its length y cm:







Take away the right half and we have 

another isosceles triangle.







We use the law of cosines again:



 





Now we use the fact that 







or







Take that equation with the other one:









Now we add equals to equals.  Adding

the left sides gives  and

adding the right sides, the cosine terms

cancel out.  So we have:







Now we are told that the sum of the diagonals

is 142 cm., so we have the equation







So we have this system of equations:









Can you solve that by solving the second for

one of the letters and substituting into the

first equation?  If not post again asking

how.



Solution to that system of equations: , 



Those are the lengths of the diagonals. 



The difference is just .



Now since the diagonals of a rhombus are 

perpendicular bisectors of each other, the

shorter diagonal, which is 22 cm is bisected 

by the long diagonal, and so each half is

11 cm. 



 



To find the area, take away the bottom half, and

we have this triangle to find the area of:







The base of that triangle is , and its

height is 







So we double that to find the total area of the rhombus.



Answer = 



Edwin

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If your triangle looks like what I have imagined it to be from your description then this is the solution :
According to the Pythagoras theorem : a^2+ b^2 = c^2
Here a = 8, b = 6
Which means 8^2 + 6^2 = c^2
64 + 36 = c^2
100 = c^2
c = sqrt(100) = 10
Hence, the third side of the triangle is 10
The above question is solved by one of the experts from 24HoursTutor.com
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AC is the hypotenuse of the right triangle ACD
Formula for hypotenuse = ^((Length)*(length)+(breadth)*(breadth))
=^(5*5+5*5)
=^50
=7.07 units

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Draw a diagram!
From it, you'll see that the diagonal and two sides form a right triangle.
Let x = length of a side of the square
then
x^2 + x^2 = 4.24^2
2x^2 = 4.24^2
2x^2 = 17.9776
x^2 = 8.9888
x = 3 feet

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The key is that an "equilateral triangle" implies that ALL three sides are of equal lengths. Once you know that (if you draw a diagram), you can see that the "altitude" will bisect (divide it in half) one side of the triangle (at exactly 90 degrees). Now you can apply Pythagorean's theorem:
Let x = length of a side of the equilateral triangle
17.32^2 + (x/2)^2 = x^2
Solving for 'x' above will yield your answer.

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When you see the term "equilateral triangle" it should immediately register that all three sides are equal!
And, that the altitude (h) bisects one side in half.
Since x = 6 feet (given in problem)
Draw the diagram above...
Now, you can apply Pythagorean's theorem:
h^2 + 3^2 = 6^2
h^2 + 9 = 36
h^2 = 27
h = sqrt(27)
h = sqrt(3*3*3)
h = 3sqrt(3)

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the triangles are similar, which means that the ratio between corresponding sides is constant

smallest is to smallest as longest is to longest __ let x="longest side"

288/96=x/111 __ 3=x/111 __ 333=x

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We basically have this triangle set up:


Since we can see that the triangle has legs of 2 and x with a hypotenuse of 8, we can use Pythagoreans theorem to find the unknown side.


Pythagoreans theorem:

where a and b are the legs of the triangle and c is the hypotenuse



Plug in a=2, b=x, and c=8. Now lets solve for x


Square each individual term



Subtract 4 from both sides


Combine like terms


Take the square root of both sides


Simplify the square root



So you are correct

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We basically have a triangle like this set up:


Since we can see that the triangle has legs of 10 and x with a hypotenuse of 16, we can use Pythagoreans theorem to find the unknown side.


Pythagoreans theorem:

where a and b are the legs of the triangle and c is the hypotenuse



Plug in a=10, b=x, and c=16. Now lets solve for x


Square each individual term



Subtract 100 from both sides


Combine like terms


Take the square root of both sides


Simplify the square root



So our answer is which is approximately


So the value of b is which is approximately

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The answer is 5. A pythagorean triple a^2+b^2=c^2 checks out very nicely.

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If (x,12,13) is a Pythagorean triple, then x =
------------
13^2 = 12^2 + x^2
x^2 = 13^2 - 12^2
x^2 = 169 - 144
x^2 = 25
x = 5
==========
Cheers,
Stan H.

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C^2=(8x)^2+(15x)^2
then
C^2= 64x^2+225x^2
C^2=289x^2
then
C= sqrt(289)x=17x

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A) sqrt75=5sqrt3=5*1.732=8.66 answer
B) sqrt6=2.45 answer.

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the altitude divides the large triangle into two smaller triangles, both of which are SIMILAR the the large triangle (and each other)

using ratios of similarity __ 4:8::8.9:b and 4:8::8:a

b=17.8 and a=16


trying to check the results using Pythagoras shows that the initial values have been rounded
__ a 4-8-8.9 triangle is NOT a right triangle
__ the 8.9 is actually 4sqrt(5)