Question

Please help i will fan u and give u a medal. questions number13-15 and 19-21
http://cdn.virtuallearningcourses.com/ivtcontent/images/10-4%20Perimeters%20and%20Areas%20of%20Similar%20Figures.pdf

Answer 1

@ybarrap you help me several days ago idk if u remember me but i made a new account can u help me now

Answer 2

Ok. Take a wild guess on how to approach 13.

Answer 3

i am guessing its area/area=side/side

Answer 4

what's the area of a parallelogram?

Answer 5

AREA=BASE TIMES HEIGHT

Answer 6

No, you're guess was better approach

Answer 7

It's \( \tt \Large{\sqrt{\dfrac{area_1}{area_2}}= \dfrac{side_1}{side_2}}\)

Answer 8

ohh ok

Answer 9

Give me the numbers for area1, side1 and side 2 and then you'll have the remaining area

Answer 10

Way better than my initial thought!

Answer 11

area1 is 6 side 1is3 and side 2 is 6

Answer 12

Yes! Now hard part is calculator work.

Answer 13

Do you know what to do next?

Answer 14

sqaure root of 18 right

Answer 15

$$
\tt \Large{\sqrt{\dfrac{6}{area_2}}= \dfrac{1}{3}}\\
\tt \Large{\dfrac{6}{area_2}=} \left ( \dfrac{1}{3} \right )^2 \\
\tt \Large{\dfrac{6}{1}=} \dfrac{area_2}{9} \\
\tt \Large{\dfrac{6\times 9}{1}=} \left ( \dfrac{area_2}{1} \right ) \\
\tt \Large{area_2=54}
$$

Answer 16

isnt side one 3 and side 2 is 6

Answer 17

Yep 3/6 = 1/2:

$$
\tt \Large{\sqrt{\dfrac{6}{area_2}}= \dfrac{1}{2}}\\
\tt \Large{\dfrac{6}{area_2}=} \left ( \dfrac{1}{2} \right )^2 \\
\tt \Large{\dfrac{6}{1}=} \dfrac{area_2}{4} \\
\tt \Large{\dfrac{6\times 4}{1}=} \left ( \dfrac{area_2}{1} \right ) \\
\tt \Large{area_2=24}
$$

Answer 18

O ok so thats for number 13 the ithe other one up there was wrong ok thank u very much may we please move on to number 14

Answer 19

Same exact idea, except, we are now trying to find area1 instead of area2.

Answer 20

what is area2, side 2 and side 1?

Answer 21

121,18,12

Answer 22

So what's our equation?

Answer 23

a1/121 = (2/3)^2
Does that sound right?

Answer 24

radical area one over 121=12 over 18

Answer 25

That's right. Then you should get what I got

Answer 26

so a1 = 121*(2/3)^2

Answer 27

ok that will be easy i can solve the equation alone can we please go on to number 15

Answer 28

Same exact thing. Solve for area 1 again.

Answer 29

ok i can do those can we go on to 19 please

Answer 30

So find the scale factor, take sqrt(area1/area2). This is also the ratio of the perimeter, because take the octagon. Perimeter, P1 of the first is 5x1 and perimeter of the second is 5 *x2, where x1 and x2 are the sides. Taking the ratio, we have 5x1 / 5x2 = x1/x2, which is our scale factor

Answer 31

I meant 8x1 / 8x1 not 5x1 / 5x2

Answer 32

One more time, I meant 8x1 / 8x2 not 5x1 / 5x2

Answer 33

8/16 =1/2

Answer 34

so thats our scale factor

Answer 35

sqrt(a1/a2) = sqrt(4/16)=2/4=1/2 - yes!

Answer 36

All are done the same.

Answer 37

thats our scale factor and how do we find the ratop pf the perimeter

Answer 38

It's the same as our scale factor.

This is because, the formula for a perimeter is sum of all sides. Now the sum of the sides of one figure divided by the sum of the sides of the second figure is the same as the ratio side of one figure to the other. Imaging a square with side = 1, it's perimeter is 4. Now take a second square with side =2, its perimeter is 8. The scale factor is 2/1 =2 using the sides and 8/4=2, using the perimeters. It's the same.

Answer 39

ooo ok may we please go on to 20

Answer 40

identical, just using different numbers

Answer 41

i dont get this for some odd reason how is equal to 1/2 in number 19

Answer 42

i am very dumb in math as u can see

Answer 43

I'm wondering if for the questions 19-21, when they ask you to find the scale factor, you need to keep just area1/area2. Then for the ratio of the perimeter, its the square root. I was just looking at some of the examples above. I think that's what you should do.

Answer 44

You're not dumb! You're doing fantastic!!

Answer 45

ok i get it so its basically like this a square over b square is = is side oveer that right

Answer 46

doesnt matter what shape it is

Answer 47

just do area1/area2 for the scale factor. Like for problem #19 it would be 4/16 for scale factor. That's your answer. And then for ratios of perimeter, take square root of that: 2/4=1/2

Answer 48

That's a great question. Shape doesn't matter. Because similar figures are related to each other only by the squares, regardless of how many sides. That's the beauty of this technique.

Answer 49

the ratio of the primeter is equal to the scale factor so the scale factor is equal to 1/2 and so is the permiter based on problem number 4 in the examples

Answer 50

I am correcting my self. When they ask for scale factors in questions 19-21, I think that they want the ratio of the areas. And when they ask for perimeter ratios, its the square root of that. Not much different that what we did.

Answer 51

o ok thank u very much for u help :))

Answer 52

Keep on chugging on these problems. You have the potential given your tenacity. I was the same way.

Answer 53

ok i will thank u and i voted u as best response but i wont be able to fan u as i cant verfiy my acount because i dont no know gow

Answer 54

@ybarrap not to say ur wrong or anything but what u said first that the scale factor was equal to the perimeter was correct:) i check online

Answer 55

great!

Answer 56

ok see you if i need any help i will contact u

Answer 57

ok