http://prntscr.com/i1thga
@Vocaloid
just split the shape into a rectangle + a triangle|dw:1516213782841:dw|
|dw:1516213790254:dw| then find the area of the triangle + rectangle and add them together
How would I know where the area is? By counting the squares?
area of triangle = (1/2)(base)(height) area of rectangle = (width)(length)
good, keep going
the height is 6 not 5, base is 3, so calculate the area of the triangle
6*3?
area for a triangle is (1/2)(base)(height)
18
don't forget the 1/2
What do I do with that again?
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid area for a triangle is (1/2)(base)(height) \(\color{#0cbb34}{\text{End of Quote}}\)
9
?
good, the area of the triangle is 9 now find the area of the rectangle and add them together
24
re-count the base and height of the rectangle
6 and 5
(1/2) 6*5 = 15
the area of a rectangle is just (base)(height)
Oh
39
good, so 39 = your ans
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|dw:1516215007977:dw|
try to apply the same formulas from before area of triangle = (1/2)(base)(height) area of rectangle = (base)(height)
(1/2) 6 x 3 = 9
area of a rectangle is just base * height
18 then
good, keep going
I forgot what to do next, because it says to find the area of the Polygon
find the area of the other triangles
the triangle on the right hand side is 4, 3 not 4,5
Ohhh
anyway, using these values, calculate the area of all the other triangles remember that triangle = (1/2)(base)(height)
6, 1.5, and 9
All of it added with 9 = 25.5
the area of the rectangle is 18 not 9, so the final answer is 34.5
Ohhh, forgot, I was looking at the area I added for the width and height
the area of the rectangle is 28 and the circular parts are closer to 8 (just estimating) so I would go with answer choice 2
This I shouldn't mess up since they mention the shapes
area of a parallogram is also base*height for the rectangle you will have to use the pythagorean theorem to find the length of that slanted slide
|dw:1516216363603:dw|
|dw:1516216368705:dw|
find the two marked slanted slides
3?
3,8?
|dw:1516216509795:dw|
use the pythagorean theorem to find the hypotenuse leave your answer in radical form
For that triangle 2,8?
the hypotenuse is a number not a coordinate point
remember a^2 + b^2 = c^2 for a triangle hypotenuse
Sorry, I'm a bit rusty when it comes to this stuff
for a right triangle, a and b are the legs and c is the hypotenuse
|dw:1516216705869:dw|
a^2 + b^2 = c^2, find c (leave in radical form)
C=2 with the square root of 10
good, now repeat the same process with the smaller triangle
|dw:1516216861359:dw|
3,1-Ah you beat me to it
good, now find the hypotenuse
Square root \[\sqrt{10}\]
good, so the area of the rectangle is sqrt(10) * 2 * sqrt(10) = 2*10 = 20 now add the area of the parallelogram to this
\[\sqrt{10*2} =2*10=20\] Like this?
it's sqrt(10) * sqrt(10) * 2 the 2 needs to be outside the radical and there needs to be another square root 10 to cancel out the radicals
\[\sqrt{10} * \sqrt{10} *2\]
Like that?
yes
now add that to the area of the parallelogram
That times 24?
|dw:1516217524656:dw|
the base is 6 and the height is 1 for the parallelogram so 6*1 + 20 = 26 = your answer
Just bxh or 1/2bxh?
for a parallelogram it is base * height
AHa
Last one
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the base and height of the triangle are 6 and 3 the base and height of the rectangle are 7 and 6 calculate the total area
18+21=39
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid area of triangle = (1/2)(base)(height) area of rectangle = (width)(length) \(\color{#0cbb34}{\text{End of Quote}}\)
So I guess 39 is my answer
check your math again, you mixed up the formulas
42+9, excuse me. Didn't know how I didn't see 6 and 3 were a triangle. 51
good, 51 = your answer
Thanks
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