A right triangle has base (x - 7) units and height (2x - 10) units.
Part A: What is the square of the length of the hypotenuse of the triangle? Show your work. (4 points)
Part B: What is the area of the triangle? (3 points)
Part C: Using the solution obtained in Part B, explain the closure property of multiplication of polynomials. (3 points)

- anonymous

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- schrodinger

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- anonymous

@ganeshie8 @Nnesha @hartnn

- anonymous

Please help
Anyone

- anonymous

@mathmate @welshfella Can you guys help?
I have a general idea of what I need to do for part one, but I get really lost afterwards.

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## More answers

- welshfella

square of hypotenuse , by Pythagoras theorem
= sum of squares on the other 2 sides
= (x - 7^2 + (2x - 10)^2

- anonymous

I got that much, but I need to simplify it I believe

- anonymous

@welshfella

- kobeni-chan

What do you have that you need to simplify?

- anonymous

I need the square length of the hypotenuse. so wouldn't I need to simplify (x-7)^2 + (2x+10)= C^2?

- anonymous

Like I know that much but I don't know what to do afterwards. do I square everything?

- kobeni-chan

You need to square (x-7) and (2x-10) right now. If you have a scientific calculator you can do this easily, but if you have a reg. calculator then do (x-7)(x-7) and use the distributive property. Then do the same with (2x-10)(2x-10)

- anonymous

|dw:1433177251700:dw|

- anonymous

|dw:1433177275411:dw|

- anonymous

|dw:1433177299845:dw|

- anonymous

|dw:1433177325031:dw| ?

- anonymous

Did I do that part correct?

- kobeni-chan

h/o

- anonymous

Oki cx

- kobeni-chan

Yes that's right :) sorry if I'm slow to respond I'm doing my exam

- anonymous

Okay, so now I do (2x-10)(2x-10)?

- kobeni-chan

Yep

- anonymous

|dw:1433177671889:dw|
Sorry for the bad writing ;-;~

- welshfella

now had these 2 expansions together
x&3 - 14x + 49 + 4x^2 - 40x + 100

- welshfella

**correction first term is x^2

- anonymous

add them?

- welshfella

yes

- anonymous

|dw:1433178085721:dw|

- welshfella

add like terms
eg x^2 + 4x^2 = 5 x^2

- welshfella

yep

- anonymous

is that the answer? for part A I mean

- welshfella

yes

- welshfella

Part B
the area = 1/2 * base * height
= 1/2 * (x - 7) (2x - 10)

- anonymous

how do I do that o.O

- anonymous

@kobeni-chan ? ;-;~

- welshfella

to make it easier 1/2 * (2x - 10) = (x - 5)
now work out (x - 7)(x - 5) and you gave the required area

- welshfella

* have

- anonymous

|dw:1433179051398:dw|

- anonymous

@amistre64 @Here_to_Help15

- anonymous

@mathstudent55 @paki @Luigi0210

- anonymous

@welshfella can you please help me

- kobeni-chan

I'm so sorry I got very busy and forgot about your question :( what part are you on?

- anonymous

Part B
I believe >.>

- anonymous

@Aureyliant >.>

- anonymous

Ok so Pt. B is asking the area.. what is your understanding of this Q?

- anonymous

That for the area, you need to do a=1/2*b*h
since we already have b and h, we just plug it in. Right?

- anonymous

Yup!

- anonymous

I got that much, but when it comes to trying to find said area, I get completely lost

- anonymous

\[1/2*(2x-10)(x-7)\] is your equation

- anonymous

I know that much, but I don't know how to apply the 1/2 to the (x-7)

- welshfella

you've worked it out correctly alice

- anonymous

so is that trinomial the Area?

- anonymous

You can either take it's inverse out, but since it's an expression & not an equation, it's like dividing (x-7) by 2.

- welshfella

apply the half to (2x -10) first makes it easier
1/2 * (2x - 10 ) = x - 5

- anonymous

^ Agreed

- welshfella

so it comes down to
(x -5)(x - 7) which you worked out correctly in an earlier post alice

- anonymous

I did (x-5)(x-7). Is there no other step afterwards?

- anonymous

Nope. You can check your work by multiplying again, and you should be getting the same answer you started with.

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