Jusitn's Surface Area Growing Post

TSA of rectangular prism

The rectangular prism has rectangular faces.

So, use the formula A = lw to find the area of the faces.

For both the front and back faces, l = 5 in. and w = 2 in.

So, the area of the front and back faces is,

5 2 2 = 20 in.²

For both the top and bottom faces, l = 14 in. and w = 5 in.

So, the area of the top and bottom faces is,

14 5 2 = 140 in.²

For both the sides, l = 14 in. and w = 2 in.

So, the area of the two sides is,

14 2 2 = 56 in.²

Add the area of the faces.

20 + 140 + 56 = 216

So, the total surface area of the given solid is 216 in.².

TSA of triangular prism

The triangular prism has triangular bases.

So, to find the area of the bases, use the formula:

Substitute 6 for b and 8 for h.

Therefore, the area of the two bases is,

2(24) = 48 cm²

The triangular prism has rectangular sides.

So, use the formula A = lw to find the area of the sides.

The area of the front side is,

13 8 = 104 cm²

The area of the bottom side is,

13 10 = 130 cm²

The area of the back side is,

13 6 = 78 cm²

Find the sum of the base areas and the area of the sides.

48 + 104 + 130 + 78 = 360

So, the total surface area of the given solid is 360 cm².

TSA of a cylinder

The radius of the circular base is,

he area of a circle is, A = πr² where r is the radius.

Replace π with 3.14, r with 2.4.

A (3.14)(2.4)²

A = 18.0864

So, the area of one base is 18.0864 cm².

The sum of the areas of the two bases is,

18.0864 + 18.0864 = 36.1728 cm²

This video is on how to find the tsa of a cylinder , rectangular prism, and triangular prism. It was created by Warren, Bruce and I.

Pythagoras (late)

this question is from page 110. Question 3

3. Walter walks across a rectangular fi eld in a
diagonal line. Maria walks around two sides
of the fi eld. They meet at the opposite corner.




a) How far did Maria walk?
300+120= 420m

b) How far did Walter walk?
a2+b2=c2
3002+1202=c2
300x300+120x120=c2
90000x144000=c2
c2=104400 or 323m

c) Who walked farther? By how much?
420m-323m=97
Maria walked farther by 97 m

Reuben's scribe

pg.110 Q.4
Find the height of the pole where the guy wire is attached to the nearest tenth of a metre.
Sorry there is no pic.


If the guy wire(c,10m) is the hypotenuse and the ground(a,2m) is one of the legs , then the pole(b) is the second leg.To do this we need the work.

c2 -a2=c2
10m2-2m2=b2
(10x10)-(2x2)=b2
100m-4m=96
square root of 96 =9.78 something,something,something,something
Round to the nearest meter to get 9.8m2
Again sorry about no pics I couldn't get onto my textbook (server lost) or something like that

Scribe Jan 6 2010!

7. - What is the height of the wheelchair ramp? Give your answer to the nearest tenth of a centimeter.

a² = c²- b²
a² = 80cm² - 79cm²
a² = (80x80) - (79x79)
a² = 6400cm - 6241cm
a² = 159cm
√a² = √159

a = 12.6cm

The height of the ramp is 12.6 cm tall.

Jasmine's Scribepost January 6, 2010

Waaaaaaazzzzzaaaaaaap.

7) What is the height of the wheelchair ramp? Give your answer to the nearest tenth of a centimeter.
(Sorry I didn't put the 'a,b,&c)
c² - b² = a²
80² - 79² = a²

(80x80) - (79x79) = a²

6400 - 6241 = a²

159 = a²

√159 = √a

12.6cm= a

Telisa's Scribepost for January 6, 2010

Question #3






Walter walks across a rectangular field in a diagonal line. Maria walks around two sides of the field. They meet at the opposite corner.

a) How far did Maria walk ?
I didn't really know how to find the length of what Maria walked so I added together the side lengths...

120 + 300 = 420
Maria walked 420m


~~~~~~~~~~~~~~~~~~~~~~~
b) How far did Walter walk ?

I used the Pythagorean relationship to find the answer...

a² + b² = c²

120² + 300² = c²

( 12o x 120) + ( 300 x 300) = c²

14400 + 90000 = c²

10 4400 = c²

√ 104400 = √ c²

323.1 = c

Walter walked 323.1m.


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


c) who walked farther ? by what distance ?

Maria walked farther by ... 96.9m

420 - 323.1 = 96.9m

~~~~~~~~~~~~~~~~~~~~~

Question : Purple

Work : Blue

Answer : Bold

Me : Grey




Sorry if the picture is hard to read.


Thank you for reading my scribe...

please criticize and comment

Alex's pythogorean scribe

OK. so for my scribe, I decided to do question 14 on page 113 so here it is.
14).the question for this is find the missing side length of each triangle.

a)c+b=d-each of these letters are supposed to be squared
12m+5m=d-each of these letters are supposed to be squared
(12mx12m)+(5mx5m)=d-the d is supposed to be squared
144+25=d-the d is supposed to be squared
169=d-the d is supposed to be squared
169=d-the d is supposed to be squared and their should be a square root over the 169 and d
13=d

b)w-t=v-each of these letters are supposed to be squared
15cm-9cm=v-each of these letters are supposed to be squared
(15cmx15cm)-(9cmx9cm)=v-the v is supposed to be squared
225-81=v-the v is supposed to be squared
144=v-the v is supposed to be squared
144=v-the v is supposed to be squared and their should be a square root over the 144 and v
12=v

So A's side length is 13m and B's side length is 12cm.

Jecelyn's Scribepost Jan 6 , 2010

I picked number 4 on page 110

The question was: Find the height of the pole where the guy wire is attached to the nearest tenth or a metre.

So the height of the pole is 9.79.

Well hope my scribe post made sense. Please comment :D

Brianna's Scribe Post January 6, 2010

Question 12



Sarah has a vegetable garden in the shape of a right triangle. She wants to put fencing around it to keep the rabbits away.



a) What total length of fencing does she need? Give your answer to the nearest hundredth of a metre.



c2- b2=a2

42-32 =a2

16-9=a2

7=a2

the square root of 7 is 2.6 which equals a2.

Sarah would need 9.6m of fencing.



b) If fencing costs $2/m, what would be the total cost of the fencing?

The fencing would cost $19.20, I got this answer by multiplying 9.6m by 2.

Gordons Scribepost Jan 6 2010

For my scribe post question i volenteered to do number 9.

(Sorry its a little late)

This question asked what the diagnol length of a square (or two triangles right next to each other) was if it had two sides (legs) with the measurments:

a=6m

b=12m

c=?

With c as the diagnol length.

So to discover the length we use the formula a²+b²+c²

And now we begin the long math equation:

a²+b²=c²

6²+12²=c²

(6x6)+(12x12)=c²

36+144=c²

√180=√c²

13.4=c

So now we simply write a simple sentince answer

The diagnol length of the rectangle is 13.4m.

Again I apologize for this post being a little late, and remember that tomorrow is the last day of pythagoras and we need 8 of the math books questions done.

Deniel's scribpost 8-16 January 6th 2010 Question 16#

QUESTION 16#

Hey guys okay I really don't know if I was to do this question or not today but ill do it anyway...

The question I chose was Question 16 and it was the hard question but i figured it out anyway so here is my solution:

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

What is the length of the red diagonal in the box? Express your answer to the nearest tenth of a millimetre.

-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

Here is my answer to this problem:

(Sorry if i didn't add any the letters a,b,c because it was hard putting them on the picture.)

When I was estimating I tried to figure out the problem by seeing what the question meant so here's what I did:

I multiplied the numbers by 2 making it 218 so I squared that number making it 14.8 because I rounded it by a tenth which makes this my solution but my answer is gonna be next.

__+__+__+__+__+__+____+__+__+__+__+__+____+__+__+__+__+__+__+

All I did was:

A²+B²+C² = ???

5mm² + 7mm² + 12mm² =???

So I multiplied them making an answer: (5x5)+(7x7)+(12x12)=218

So I squared the number making an answer which was: √218 = 14.7 rounding it making it 14.8

so that is my answer for Question 16#

~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=

Question 15

HAPPY NEW YEAR!

Here is question 15:

HOW LONG IS THE LINE?









To find the answer, we have to use the Pythagorean relationship. From looking at the graph we can tell that the legs, (green) have lengths of 2cm and 4cm. We have to find the length of the hypotenuse, (red). Our formula is: a squared + b squared = c squared.

a squared + b squared = c squared

2cm squared + 4cm squared = c squared

(2cm x 2cm) + (4cm x4cm) = c squared

4cm + 16cm = c squared

20cm= c squared

By doing this we have figured out the area of the hypothetical square attached to the hypotenuse. Now we have to find the square root of that number to find the length of the hypotenuse.





the square root of 20= 4.4 (approx.)





The length of the line is approximately 4.4cm.





Thanks for reading! Please comment :)





~Laura~

Pg.105-Question 12 1/5/10

For those that can't see whats written the question was.... 12. The hypotenuse of the triangle cuts the circle in half. What is the diameter of the circle? Express your answer to the nearest tenth of a centimetre.

And the right answer was 8.60 cm

Jomari's Scribepost, January 5, 2010

For this question, I am supposed to find the length of the red line(C). I can do this by making a right triangle with the red line as the hypotenuse. The legs would meet each other at 5,2. The vertical leg (A) would be 4 units long and the horizontal leg (B) would be 2 units long. I can use this information to first, find out the length of C² and then find the length of C.



A² + B² = C²
4 cm² + 2 cm² = C²
(4 cm x 4 cm)+( 2cm x 2cm) = C²
16 cm + 4cm = C²
20 cm = C²
√ 20 cm = √ C²
4.5 cm= C

The red line has a length of 4.5 centimeters.

Jennifer's ScribePost On Pythagoras :]

Hello Everyone! I'm going to do Question 11 on page 106!

Thanks For Looking At My Post Folks!

Althea's Scribepost, January 5, 2010

Question 13, page 105



How to get the answer:

Thanks for reading!!

Bruce's Scribepost, Pythagoras

Hello,
Today Mr Harbeck assigned a few of us to make a scribe on one of the seven question. My question was #6.


6. Determine the length of the leg for each right triangle.
A)
One leg=7cm
Second leg=?cm
Hypotenuse=25cm

c(sqrd) - a(sqrd) = b(sqrd)
25cm(sqrd) - 7cm(sqrd) = ?cm(sqrd)
625cm(sqrd) - 49cm(sqrd) = 576cm(sqrd)

square root of 576=24cm
b=24cm
B)
One leg=24cm
Second leg=?cm
Hypotenuse=26cm

t(sqrd) - s(sqrd) = r(sqrd)
26cm(sqrd) - 24cm(sqrd) = ?cm(sqrd)
676cm(sqrd) - 576cm(sqrd) = 100cm(sqrd)

square root of 100=10
r=10cm

If you think it's wrong please leave a comment, because I just did my scribe half asleep.

Kevin's Scribepost , January 4th, 2010

Sorry its late, I just didn't really understand how to find the answer .

7. What is the missing length of the leg for
each triangle? Give your answer to the
nearest tenth of a millimeter.

( CLICK THE IMAGE !! )

Page 104 Question 3 And 4

3. Determine the length of each hypotenuse









4. What is the length of each hypotenuse?

Give your answer to the nearest tenth of
a centimetre.


A:)

Reuben Cram pg 92 q.4

I know I didn't ask to do this scribe but I lost my other one and need to catch up.
4. what are the areas of the three squares shown.



e-30mm. To find the area of this square you need to times the number by itself to get it. 30 is 900mm2.
f-40mm. Now find the area of this square to get 1600 mm2
g-Now for the hypotenuse 50mm. 2500mm2. But now you have to know if it is a right triangle. Add the legs value together to get the hypotenuse. If you get the hypotenuse comment me that it was right and tell me how to paste a pic because I have no idea how to do it.