Questions on Geometry: Length, distance, coordinates, metric length answered by real tutors!

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The distance between two towns on a map is 6 cm. The actual distance between two towns is 48 ikm. What is the scale on the map?
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6 cm to 48 km = 6 cm to 4800000 cm
Scale is 1:800,000

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The scale would be in 1 cm = km



The scale is 1 cm = 8 km

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this forms a geometric progression.
a reference for the sum of a geometric progression formula that i used is shown here:
http://www.jimloy.com/algebra/gseries.htm
since the geometric progression is infinite, the formula to be used is:
S = a / (1-r)
the trick in this problem is to first determine that you have a geometric series and then to find the common ratio.
the common ratio turns out to be(1/sqrt(2))
a is the perimeter of the square we are starting with, which is 6 * 4 = 24.
the formula becomes:
S = 24 / (1 - (1/sqrt(2))
this becomes:
S = 81.9411255
the logic used is as follows:
the side of the original square is 6.
the subtending square is found by forming another square whose corners are at the midpoint of the sides of the original square.
the corners of the original square form an isosceles right triangle with the side of the subtending square (attached diagram shows this).
the side of the subtending square is the hypotenuse of this isosceles right triangle.
the formula to find the hypotenuse of this isosceles right triangle uses the formula of c^2 = a^2 + b^2 where c is the hypotenuse of the right triangle and a and b are legs of the right triangle.
since the legs are equal to each other, the formula becomes:
c^2 = a^2 + a^2 which is then equal to 2 * a^2
the formula becomes:
c^2 = 2a^2
the length of the side of the subtending square is therefore equal to sqrt(2a^2)
half the length of the side of the subtending square is then used to find the length of the side of it's subtending square.
this process goes on indefinitely.
a table was created to capture this relationship.
this table is shown below:
columns are identified as follows:
roe = the particular row that you are on in the table.
A is the side of the original square.
B is the perimeter of the original square.
C is half the side of the original square.
D is the side of the subtending square.
E is the perimeter of the subtending square.
F is the ratio of the perimeter of the original square to the perimeter of the subtending square.

row     A         B          C         D         E          F
1	6.00000   24.00000   3.00000   4.24264   16.97056   1.41421
2	4.24264   16.97056   2.12132   3.00000   12.00000   1.41421
3	3.00000   12.00000   1.50000   2.12132   8.48528    1.41421
4	2.12132   8.48528    1.06066   1.50000   6.00000    1.41421
5	1.50000   6.00000    0.75000   1.06066   4.24264    1.41421
6	1.06066   4.24264    0.53033   0.75000   3.00000    1.41421
7	0.75000   3.00000    0.37500   0.53033   2.12132    1.41421

the side of subtending square in row 1 becomes the side of the original square in row 2 and so on down the line.
this means that column A row 2 is the same value as column D row 1, and that column A row 3 is the same value as column D row 2, etc. down the line.
the calculation for the side of the subtending square is based on the formula of:
value in column D = sqrt(2 * value in column C squared).
For example:
in row 3, value in column C is 1.5 and value in column D is 2.12132.
column C is half the length of the side of the original square.
column D is the length of the side of the subtending square.
in row 3, column D is calculated as sqrt(2*1.5^2) which equals 2.121320344 which was shown in the table to 5 decimal places of 2.12132

one the common ratio of the perimeters was found, it was just a matter of plugging the values into the formula to get the sum of the infinite series which resulted in the answer of 81.9411255.

the diagram is shown below:

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Number of diagonals in a pentagon: (5*2)/2 = 5

Measure of each exterior angle in regular decagon: 360/10 = 36

5:36

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The triangle is a right triangle if one of the angles is a right angle. If one of the angles is a right angle then two of the sides will be perpendicular. If two lines are perpendicular, then their slopes will be negative reciprocals of each other. That is to say that if one slope is then the slope of the perpendicular will be .

So your first step is to use the slope formula:



where and are the coordinates of the given points to find the slopes of the three lines that contain the segments that form the sides of the triangle.

Once you have the three slope numbers, determine if any pair of them are negative reciprocals of each other.

The point of intersection of the two sides that are perpendicular is the right angle vertex.

The hypotenuse is the side opposite the right angle, hence the endpoints of the hypotenuse are the two points that are NOT the right angle. Use the midpoint formulas to calculate the coordinates of the midpoint of the hypotenuse:

and



You can't write the equation of a median. The median is a line SEGMENT. You can write the equation of the line containing the median. The median from a vertex is the line segment that joins the vertex to the midpoint of the opposite side. You know the coordinates of the vertex from part b of this problem and you know the coordinates of the midpoint of the hypotenuse from part d of this question, so you know the coordinates of two points on the line containing the desired median segment. Use the two-point form of an equation of a line to derive the desired equation.



where and are the coordinates of the given points.

Simplify as necessary/desired.

The altitude from the vertex of the right angle to the hypotenuse is, by definition, perpendicular to the hypotenuse. Recall that perpendiculars have negative reciprocal slopes. You have the slope of the hypotenuse from the calculations you did for part a of this problem. Calculate the negative reciprocal of the slope of the hypotenuse to find the slope of any line perpendicular to the line containing the hypotenuse.

Then use the point-slope form of an equation of a line:



where are the coordinates of the right angle vertex (part b of this problem) and is the calculated slope.

Simplify as desired or required.

John

My calculator said it, I believe it, that settles it

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the acute angle equals 26 and the side acroos the 90 degree angle is b and the side across the 26 degree angle equals a and the last side aquals 80
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It's not clear what you mean.

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solve for y and z
round values to the nearest whole number.
a right triangle.
the side across the right angle 13
the length across the acute or the smallest angle "y" equals 5
and the other angle is z
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It's the 5, 12, 13 right triangle.
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The smallest angle is opposite the side of 5.
Its sine = 5/13
--> angle =~ 22.62 degs
The other angle is 90 - 22.62 = 67.38 degs

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Let the shadow at 45 deg be x
at 30 deg it is x+650 ft
let height of tower be h ft
Tan 45 = h/x
but tan 45 = 1
so h = x
Tan 30 = h/(h+650)
1/sqrt(3)= h/(h+650)
h+650 = h sqrt(3)
hsqrt(3)-h = 650
h(sqrt(3)-1)=650
h= 650/(sqrt(3)-1)
h= 888 ft

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Let the length be 'l' and the width be 'w.' Then (3/4)l=w implies (3/4)l=8, so l=32/3 [10.6666 repeating].

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they are travelling at right angles to each other.
At any given instant they form a right triangle with their starting point

after 2 hours one car is 100 km away from starting point and another car 140 km
leg1 = 100 km
Leg2 = 140 km

Hypotenuse^2 100 ^2 + 140 ^2
Hypotenuse^2 10000 + 19600
Hypotenuse^2 29600
take the square root
Hypotenuse^2 172.05 km

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The squares should be 7 cm x 7 cm
check:




OK

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what is the formula for finding the sqare root of 30 feet by 40 feet
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If it's a rectangle, it's area = L*W
area = 30*40 = 1200 sq ft
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Square root is something different.

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On the coordinate grid of a map Jeff's house is located at (9,5). Hannah's house is at (-5,-5), Kenya's house is located at the midpoint between Jeff and Hannah's houses. What is the distance between Hannah's house to Kenya's house (in mileage).
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d =~ 17.2 units from Jeff's house to Hannah's house
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--> 17.2/2 = 8.6 units to Kenya's house
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There's no info given on miles. No way to convert units to miles.

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using similar triangles ___ 4 / 8 = 20 / (d + 8)

"cross" multiplying ___ 4d + 32 = 160 ___ d = 32 ___ this is the starting distance from the lamppost

4 / s = 20 / (31 + s) ___ 4s + 124 = 20s ___ 31 / 4 = s

her shadow decreases 1/4 ft or 3 in

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The units conversion is ( 3 feet / 1 yard )
yds x ft / yd = ft
ft is less than ft

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This is not hard, write an equation for each statement
The distance around (perimeter) a certain rectangular playing field is 310 meters.
2L + 2W = 310
We can simplify this, divide by 2
L + W = 155
:
The length is 65 meters more than the width.
L = W + 65
now
Using the 1st simplified equation, we can replace L with (W+65), find W
(W+65) + W = 155
2W = 155 - 65; subtracted 65 from both sides
2W = 90
W = 90/2; divided both sides by 2
W = 45 meters is the width
then
L = W + 65
L = 45 + 65
L = 110 meters is the length
:
:
Confirm this by finding the perimeter using these dimensions
2(110) + 2(45) =
220 + 90 = 310
:
Was this step-by-step enough?

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a. use the slope formula:



where and are the coordinates of points P and Q.

b. Presuming you mean "what is AN equation of the line containing the segment PQ", use the point-slope form of an equation of a line:



where are the coordinates of either point P or point Q and is the slope calculated in part a.

c. Use the midpoint formulas:

and



where and are the coordinates of points P and Q. Then the midpoint is the point

d. Again, presuming you mean "AN equation": The perpendicular bisector of PQ is the line perpendicular to PQ that passes through the midpoint. The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line. Calculate the negative reciprocal of the slope that you calculated in part a, that is calculate . Then use the point-slope form of an equation of a line to derive the desired equation:



where are the coordinates of the midpoint of segment PQ and is the negative reciprocal of the slope calculated in part a.

John

My calculator said it, I believe it, that settles it

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A rectangle has a perimeter of 12 m. If each side is a whole number of meters, what arethe possible dimensions for the length and width
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Perimeter = 2(length + width)
12 = 2(length + width)
length + width = 6
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Length can be 1,2,3,4,5
thenwidth will be 5 or 4 or 3 or 2 or 1
------
Cheers,
Stan H.
=============

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Just solve for x and y.

The more intuitive approach is: to get from (6,0) to (4,2) you can subtract 2 from the x-coordinate and add 2 to the y-coordinate. Hence, to find the other point, "reverse" it: add 2 to the x-coordinate and subtract 2 from the y-coordinate to obtain (8,-2).

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1.25m or 1200cm?
Convert the meters to centimeters.
1.25m(100cm/m) = 125cm, so...
So which is bigger?
125cm or 1200cm?

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Something is wrong in this problem. If the (x, y) coordinate points at the ends of the line ab are at:
.
a = (-6,3) and
b = (2,-3)
.
then the midpoint of the line ab is not located at (2.5, 1)
.
So either one or both of the points a and b is wrong, or the midpoint is wrong. Or possibly all three are wrong.
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Here is the way to do this problem. Let's assume that the coordinates for a and b are correct. It will help you to visualize the following actions if you make a quick sketch of the coordinate axes and then plot the two points. You don't need to be accurate for this sketch. The purpose of the sketch is just to help you visualize what is going on. Notice that point a is in the second quadrant (the upper left quarter) of the coordinate axes where x is to the left of the y-axis by 6 units and y is 3 units above the x-axis. And point b is in the fourth quadrant (the lower right quarter) where b is 2 units to the right of the y-axis and 3 units below the x-axis.
.
Starting with these two locations and only moving in the horizontal and vertical directions, how can we get from point a to point b. Let's start at point a and move horizontally to the right until we are above point b. [For reference, let's call this location point c. And point c should be (2, 3) just to make sure you are doing this correctly.] At that point we switch directions and go vertically down until we arrive at point b.
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Look at your sketch. Can you see that figure acb is a right triangle? Its legs are ac and cb and its hypotenuse is ab.
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How long is leg ac? You should see that ac goes from -6 on the x axis to +2 on the x-axis. Therefore, ac is 8 units long. (Notice that this distance can be found by subtracting the x coordinate of point a from the x coordinate of point b. That is by subtracting 2 - (-6) and getting 2 + 6 = 8.)
.
Then how long is leg cb? This time you should be able to see that this vertical leg begins at 3 units above the x-axis and goes down to b at 3 units below the x-axis, a total distance of 6 units. (This time notice that this distance can be found by subtracting the y coordinate of point c from the y coordinate of point b. That is by subtracting -3 minus 3 and getting -3 -3 = -6 which is a length of 6 units downward.)
.
So we have right triangle acb with legs ac = 8 and cb = 6. The length ab is the hypotenuse of this triangle and it is the distance you are trying to find.
.
Since this is a right triangle shown on your sketch, you can use the Pythagorean theorem to find the length of the hypotenuse. Remember that the Pythagorean theorem says that the square of the hypotenuse is equal to the sum of the squares of the two legs. So letting H represent the hypotenuse (line ab), M represent line ac, and N represent the leg cb, we can write the Pythagorean theorem as:
.
H^2 = M^2 + N^2
.
Now we can substitute 8 for the leg M and 6 for the leg N to get:
.
H^2 = 8^2 + 6^2
.
8^2 is equal to +64 and 6^2 is equal to +36. So we can say:
.
H^2 = 64 + 36
.
and the right side of this equation totals to 100. So we have:
.
H^2 = 100
.
and we can solve for H by taking the square root of both sides to get:
.
H = 10
.
So the distance from point a to point b (the length of line ab) is equal to this hypotenuse and is 10 units in length. That's the answer you were to find. Notice that you did not need to know the midpoint of line ab to find this distance.
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Just for added measure, how do you find the midpoint of line ab? Not too tough.
.
We begin by finding the x-value of the midpoint of line ac on your sketch. Recall that line ac started at x = -6 and went horizontally to x = +2. Recall that we decided this line was 8 units long. So its midpoint will be half of 8 units which will make it 4 units away from each of its ends. We either go 4 units to the right of -6 or 4 units to the left of +2. Either way we find that the value of x at the midpoint is x = -2. So our midpoint of line ab has the x value of -2.
.
We continue by finding the y-value of the midpoint of line cb on your sketch. Recall that line cb started at y = +3 and went vertically to y = -3. Recall that we decided this line was 6 units long. So its midpoint will be half of 6 units or will be 3 units away from each end. We either go 3 units down from +3 or 3 units up from -3. Either way we find that the value of y at the midpoint is 0. So our midpoint of line cb has the y value of 0 and this is the y value for the midpoint of line ab.
.
Now we can say that the midpoint of line ab is at (-2, 0)
.
Hope this helps you to understand this problem. Maybe sketching out what is involved will help you to understand what you are doing when you work with the distance formula. You don't need to memorize the distance formula if you make a sketch such as we did in this problem and then apply the Pythagorean theorem to solve it.
.
Good luck. With a little practice this will become easier to understand what you need to do.
.

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let x equal the number of equal parts.
length of each piece = 15.5 feet divided by x.
if x = 5, then each piece is equal to 15.5/5 = 3.1 feet
3.1 * 5 = 15.5
if x = 742, then each piece is equal to 15/5/742 = .0208894879 feet.

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Let = the number of posts used when they are 2 m apart
There has to be a post at each end
so, the number of spaces between posts
is 1 less than the number of posts
given:
width = m
width = m




and
width =
width =
width =
The width of the property is 18 m
check:



OK

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First lets get everything in the same units. there are 100 cm for every 1 meter.
10m=10*100cm=1000cm
1000cm/75cm=13 1/3. So we can make 13 complete laces
To get the number of pairs take 13/2=6 with a remainder of 1 so we get 6 complete pairs.

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If the minute hand of a big clock is 1.2 cm long, how far does the tip of the minute hands move in 20 minutes?
Since circumference of a circle made by the tip of the minute hand is:
2(pi)r
2(3.14)1.2
7.536 cm
.
if it travels 20 minutes, that is:
20/60 of the circle
or
1/3 of circle
.
Distance travel is then:
(1/3)(7.536) = 2.512 cm

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cut it perfectly using a meter or anything that has measurement....

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1in:45yd write the scale without units
1:1620

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a cube has all sides equal in length.
this makes the volume of the cube equal to s^3 where s represents the length of a side.
your equation is:
s^3 = 80
take the cube root of both sides of the equation to get:
root(3,s^3) = s = root(3,80) = 4.30886938

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Her pace is ( distance ) / ( time )
miles/hour
miles/hour
mi/hr
Her pace is 2 1/4 miles/hour

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how would you solve (8,4) and (12,2) using the distance formula?
sqrt(x2-x1)^2+(y2-y1)^2
sqrt(12-8)^2+(2-4)^2
sqrt(4)^2+(-2)^2
sqrt16+4
sqrt20
=4.472135955

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sqrt(x2-x1)^2+(y2-y1)^2
sqrt4+1
sqrt(2-0)^2+(-1-1)^2
sqrt(2)^2+(-2)^2
sqrt(4+4)
sqrt8
=2.828427125

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Find the distance between point A(-3, 5) and point B(4, -6) in the coordinate plane
-----
d = sqrt[(-6-5)^2+(4--3)^2]
---
d = sqrt[121+49]
--
d = sqrt[170]
==================
Cheers,
Stan H.
==========

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1 mile = 1760 yards
2/3 mile =
2/3 *1760
=2*586.66
=1173.33 yards

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Points T and J are 7 units apart. What is the number of points 2 units from T and 4 units from J?
------------------
Zero.
The points 2 units from T are on a circle centered at T, radius 2 units.
Same for J, radius = 4 units.
The circles don't overlap or intersect.
--> zero

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Assuming the backyard is a rectangle, the perimeter of the yard is then P = 2*30 + 2*20 = 60+40 = 100 ft


So you need 100 ft of fencing.

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Boat speed = 40 km/h
current speed = x km/h
wind speed
speed against current 40 -x
speed with current 40 +x

Distance against current x km
Distance with current x km
time with current = 2 hours
y/(40+x) =2
y=2(40+x)
y=80+2x..........(1)
Time against current = 2 1/2 hours
y/(40-x)=5/2
y= 5(40-x)/2............(2)
equate equation (1) & (2)
80+2x=5(40-x)/2
multiply by 2
160+4x=200-5x
9x=40
/9
x=40/9
x=4.44 km/h speed of current
m.ananth@hotmail.ca

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let the distance he walked be x
t=d/r
time taken to walk = x/5
he went by jeep.
time taken by jeep = x/45
time to reach home = 1 hour
walking time + jeep time = 1
x/5+x/45 =1
LCD = 45
multiply the equation by 45
9x+x=45
10x=45
/10
x=4.5 km

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width =x
length = 4x-5
perimeter =110
Perimeter = 2(L+W)
110 = 2(x+4x-5)
110=2(5x-5)
110=10x-10
10x=110+10
10x=120
/10
x=12 m the width
length = 4x-5==> 4*12-5==> 48-5=43 m

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side of square = x
Area of square = x^2
width of rectangle = x-2
length of rectangle = x+4
Area of rectangle = (x-2)(x+4)
=x^2+2x-8
x^2+2x-8=x^2+6
2x-8=6
2x=8+6
2x=14
/2
x=7
square side = 7 cm

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North 45 km/h
south 60 km/h
opposite directions
so add up speeds
60+45 = 105 km/h
distance = 300 km
t=d/r
t= 300/105
t= 2.85 hours

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x,x+1,x+2 be the integers
5x-6 =x+2
5x-x=6+2
4x=8
/4
x=2
2,3,4 are the integers