I need some help.. I've already done these problems and i got them back and they need fixed.. will someone please help me

- anonymous

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- anonymous

what are they?

- anonymous

##### 3 Attachments

- anonymous

on these three i need to find the surface area

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## More answers

- myininaya

for the first one we have 4 triangles and a square. the area of a triangle is
.5(base)(height) and the area of a square is (base)^2

- myininaya

so we have 4(.5)(base)(height)+b^2

- anonymous

well i redid the first one and i got 103.47

- myininaya

=2(base)(height)+(base)^2=2(4.2)(10)+(4.2)^2 is this what you did?

- anonymous

yea

- myininaya

I got 101.64

- anonymous

oh yeah that's what i meant..

- myininaya

the second we have 4 triangles

- myininaya

so we have 4 times .5(b)(h) since there are all congruent triangles

- myininaya

=2bh

- myininaya

=2(15)(9)=270

- myininaya

it looks like you got a check mark on that one and the answer was 67.5?

- anonymous

yea.. all the ones i post need fixed

- myininaya

k

- myininaya

does everything make sense so far?

- anonymous

yeah

- myininaya

ok lets look at the last one

- anonymous

6 times .5bh

- myininaya

yes there are 6 triangles and then there is that hexagon hex means 6 right?

- anonymous

yea

- myininaya

lets look at the triangles first the base 6 and we need to find the height

- myininaya

thats if we look at one traingle

- myininaya

if we divide a triangle and half we can from a right triangle so we can use the Pythagorean thm the hyp is 12 and if we divide the base in half the base is 3 or our newly formed triangle so we can find the height of the triangle let's call it h. we have 3^2+h^2=12^2 implies h=sqrt(135). so the area of the whole traingle is .5(6)(sqrt(135)
since we have 6 trangles we have this area times 6 so this equals 18(sqrt(135))
but now we need to find the area of the hexagon

- myininaya

if we draw two vertical lines from one vertex to another and one horizontal line from one vertex to another we would form 2 rectangles and 4 traingles

- myininaya

looking at one triangle we see that the hyp is 6 and we know that the legs are equal. so we have b^2+b^2=6^2 imples 2b^2=36 implies b^2=18 implies b=sqrt(18) so the area of the triangle is .5(sqrt18)(sqrt(18)=.5(18)=9 since we have 4 of these triangles then we multiply the 9 by 4 =36
now we need to find the area of the rectangles we know 1 side =6 so we need find the length of the other side that runs perpendicular to it

- myininaya

but we found that when we were finding the legs of the triangle above
so the side running perpendicular to the 6cm side has length sqrt(18) so the area of the rectangle is (6)sqrt(18) but since we have 2 we multiply by 2 giving us 12sqrt(18)
now we add all of our areas together

- myininaya

so we have the area of the 6 triangles is 18sqrt(135)
and we have the area of the hexagon is 12sqrt(18)+36
the total surface area is 18sqrt(135)+12sqrt(18)+36

- anonymous

so 974.22

- myininaya

i got 296.05

- anonymous

i did my math wrong..

- anonymous

but thank u

- myininaya

that last problem was really long I hope i'm right lol

- anonymous

can help me with some more

- myininaya

more surface area?

- anonymous

no volume

- myininaya

ok i will look at one more and then i have to leave

- anonymous

the first three are finding volume

##### 3 Attachments

- myininaya

ok the first one we need to find the area of the base and multiply by the height of the solid

- myininaya

the area of the base is .5(b)(h)=.5(6)(8)=24
the height of the solid is 10
so 24(10)=240

- myininaya

the second one you shouldnt had divided by 3 it should had been 2

- myininaya

the last one you set it up right but you should had got 3451.04

- anonymous

okay thank you

- myininaya

np i hope i was a help

- anonymous

ur were

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