Question

The lenght of a rectangle is 2 inches greater than the width. The lenght of the base of an isosceles triangle is 12 inches, and the lenghts of the other two sides are 1 inch greater than the width of the rectangle. Draw a picture of each figure and label the dimensions

Answer 1

@jim_thompson5910 HELP ME

Answer 2

Does it give you the perimeter of either one?

Answer 3

Nope it actually wants me to figure out the perimeter as well-.-

Answer 4

ok one sec

Answer 5

ok the best we can do is find some expression for the perimeter in terms of the width w

we can't find the actual number for the perimeter (of either one)

Answer 6

Let's say

w = width of rectangle
L = length of rectangle

we would have something like this

Answer 7

we're told that the length is 2 inches more than the width, so

L = w+2

Answer 8

we can then replace L with w+2 to get this

Answer 9

the opposite sides are congruent, which means we can add this on

Answer 10

the perimeter is found by adding up all 4 sides


(side1)+(side2)+(side3)+(side4)

(w)+(w+2)+(w)+(w+2)

w + w + 2 + w + w + 2

4w + 4

Answer 11

so the perimeter of the rectangle is 4w + 4 inches

Answer 12

with me so far?

Answer 13

yeah kinda>.<

Answer 14

where are you stuck?

Answer 15

the last part>.<

Answer 16

where I added up all the sides?

Answer 17

yeah>.< i mean i kinda get it but im getting fustrated i trying to solve it but i keep getting stuck

Answer 18

which line below (1, 2, 3, or 4) is the line where you're getting stuck



(side1)+(side2)+(side3)+(side4)

(w)+(w+2)+(w)+(w+2)

w + w + 2 + w + w + 2

4w + 4

Answer 19

dude i get stuck in all the steps im not even close to the answer>.< im stuck in all the lines i just got myself confused>.<

Answer 20

well we're getting closer to the answer, I'm just curious where you're stuck

Answer 21

so I can help you understand all this

Answer 22

all of it>.<

Answer 23

alright you get the idea that to find the perimeter we add up all the sides right?

Answer 24

dude this is hard im trying to solve for it but im trying to make sense what you are telling me but i cant understand it at all

Answer 25

well the problem is that we can't solve for w, so that's why we must leave the perimeter as some algebraic expression of w

Answer 26

if we had more info connecting the length and width, then we could find the numeric value of the perimeter (and solve for w)

Answer 27

oh so we first make a n algebraic expression then solve for w right? i mean i know how to solve for w but i dont know what to set it equal to

Answer 28

so anyways, the perimeter of the rectangle is 4w + 4

if we knew more info, we could find the value for w and use it to find the actual perimeter

Answer 29

that's the thing, we don't have anything to set it equal to

Answer 30

if we knew the perimeter (say it was 20), then we could say

P = 4w + 4

20 = 4w + 4

but the perimeter wasn't given

Answer 31

so how can i solve for w then? i mean if there was a value i could probably set it equal to eachother but i dont know how to though

Answer 32

does it say anywhere that the rectangle and the triangle have the same perimeter?

Answer 33

The lenght of a rectangle is 2 inches greater than the width. The lenght of the base of an isosceles triangle is 12 inches, and the lenghts of the other two sides are 1 inch greater than the width of the rectangle. thats all it says

Answer 34

ok then there's no way of solving for w

Answer 35

the best we can do is come up with 2 expressions for the perimeter of the rectangle and triangle

Answer 36

so we already did that with the rectangle, we found the perimeter to be 4w + 4

Answer 37

i need to look for 2 expressions to find the perimeters of the rectangle and triangle and find the width of the rectangle if the perimeter of each side is equal>.<

Answer 38

oh so each perimeter is equal?

Answer 39

i think thats with it says thugh>.<

Answer 40

can you post a screenshot of the entire problem?

Answer 41

i cant i dont have a camera but i think it is the both perimeter equstions you made

Answer 42

oh it's not an online problem but a problem in your book?

ok let's assume the perimeters are equal

Answer 43

the perimeter of the triangle is

(side1)+(side2) + (side3)

(w+1)+(w+1)+12

w+1+w+1+12

2w+14