Questions on Geometry: Geometric formulas answered by real tutors!

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Length = x+3
Width = 7
Perimeter of a rectangle = 2*Length + 2*Width
Perimeter = 2(x+3) + 2*7
Perimeter = 2x+6+14 = 2x+20
Area = Length * Width = (x+3)*7 = 7x+21

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what is the perimeter and the area of a rectangle with length (x+3) and width 7?
Perimeter:
P = 2(x+3) + 2(7)
P = 2x + 6 + 14
P = 2x + 20
:
Area:
A = 7(x+3)
A = 7x + 21

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cut 2 inches of each 80 inch post

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The perimeter of a rectangle is 50 m. The width of the rectangle is 5m less than the length. Find the length and width of the rectangle.
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Perimeter = 2(L + W)
50 m = 2(L + L-5)
25 = 2L-5
30 = 2L
L = 15 m (length)
W = L-5 = 10 m (width)
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Cheers,
Stan H.

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half of a 16 in diameter circle fits snugly inside a rectangular box, how many square inches larger is the area of the box than the area of the circle?
.
Draw a diagram of the problem -- label everything you know -- it'll help you see the solution.
.
Area of "half circle"
(1/2)(pi)8^2
= (1/2)(pi)8^2
= 32(pi)
= 100.53 sq inches
.
box:
length is 16
width is 8
Area = width*length
Area = 8*16
Area = 128 sq inches
.
It is larger by:
128 - 100.53 = 27.47 sq inches

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Step 1) Draw a 10' x 10' square to represent the shed.




Step 2) Extend a line segment from the midpoint of one side of the square (I chose the top) 18 ft out. This is the goat's leash.



Step 3) Now extend a segment 18 ft to the left (you could go either way). Take note how there's 5 feet from the midpoint to the corner.



Step 4) Sweep the two 18' segments (or just one) to form the segment shown in blue below. We'll call this area




Step 5) Since the rope is eventually going to be caught on the shed, this means that we now have 13 feet (instead of 18') of rope. So just draw out an arbitrary 13' segment




Step 6) Sweep a circular segment through to get the red area shown below in which we'll call




Step 7) Since 10 ft of rope is stuck on the wall, we now have only 3 ft of rope left. So sweep out the remaining area that the goat can roam (for this side at least). We'll call this new area in green




Step 8) From the last drawing, we see that there are 3 quarter circles which have the radii 18, 13, and 3 feet.

Recall that the area of a circle is . Divide this by 4 to find the area of a quarter circle (since all of these segments are quarter circles) to get

Using the last formula, we then get the areas:











Now add the three individual areas:

So the area that the goat can graze on the entire left side is square feet.

To find the total area, simply multiply that last value by 2 (since the right side is a reflection of the left side) to get

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Answer:
So the total area that the goat can graze is square feet. If we use the approximation for , then the approximate area is square feet.

So the total area available for the goat to graze is 788.14 square feet.

Note: if you need better precision, then just use a better approximation of .

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a=(A/.5h)-b

To find the common ratio, r, we divide any term
by its preceding term.

So we can either divide the second term 12 by the 
first term 36 and get  which reduces to
.  So 

or

we can divide the third term 4 by the second term 12 
and get  which also reduces to .  So 
either way we get .

I'll do it two ways, the first way we'll do it will not 
be acceptable to your teacher, but it gets the right 
answer, so you'll know when you get it right.

The first way is to write the terms out till you get
to  by multiplying by  each time:

36, 12, 4, 

So to get the next term we multiply 4 by  and get .
So far we have:

36, 12, 4, , 

To get the next term we multiply 4/3 by  and get .
So far we have:

36, 12, 4, , 

To get the next term we multiply 4/9 by  and get .
So that's it, we now have

36, 12, 4, , , 

Then we count the terms and see that there are 6.

But your teacher doesn't want you to do that in case there
might have been 100 terms! But at least we know the answer
is 6.

Here's what your teacher wants to to do:

The formula for the nth term of a geometric sequence is









Cross multiply:



Simplify by dividing both sides by 4:



Write  and  and  as 



Add exponents on the right:



Use the principle: If  and  and 
                   then 




Edwin