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Mathematics 8 Online
OpenStudy (anonymous):

Can anyone solve this? (still not solved)

OpenStudy (anonymous):

OpenStudy (anonymous):

Yes, but it is too cumbersome.

OpenStudy (anonymous):

i know -.-

OpenStudy (shayaan_mustafa):

hi ZyairaBoyd :) Is it regular pentagon?

OpenStudy (anonymous):

Pretty sure.....

OpenStudy (shayaan_mustafa):

it seems too. So just ignore EB and EC. As you know regular pentagon always have same sides and here one side is given that is 10cm. If it is regular pentagon then it must have all sides with 10cm. Right? And area of regular pentagon is\[\large A=1.72x^2\] substitute x=10cm\[\large A=1.72(10cm)^2\]\[\large A=1.72(100cm^2)\]\[\large A=172cm^2\]If and only if it is a regular pentagon. Otherwise it is wrong..

OpenStudy (anonymous):

It's not a regular pentagon.

OpenStudy (anonymous):

Ok.

OpenStudy (shayaan_mustafa):

OK. Let us see if it is not a regular pentagon. Be with me.

OpenStudy (anonymous):

Man got to go. See ya.

OpenStudy (shayaan_mustafa):

@ZyairaBoyd Is that enough for you?

OpenStudy (anonymous):

Huh? No....

OpenStudy (anonymous):

@Shayaan_Mustafa

OpenStudy (kinggeorge):

They want you to solve this with triangles. First, notice that solving for triangle ECB is the easiest. It's just \[{1 \over2}\cdot 10\cdot 13=65\]

OpenStudy (anonymous):

Yeah so i have to like do it one by one?

OpenStudy (katielong):

i'll be 2 mins.. i can do it though and il give you a good method!!

OpenStudy (anonymous):

I dont mind.

OpenStudy (kinggeorge):

Then, since EB=EC=4, the areas of triangles EBA and triangle ECD are the same. and are given by \[{1 \over 2}\cdot 15\cdot 4=30\]Since there are two of these triangles, we have an area of 60.

OpenStudy (kinggeorge):

Now we add \(60+65=125\) To get the overall area.

OpenStudy (anonymous):

oh ok.

OpenStudy (kinggeorge):

We have three triangles with area 65, 30, and 30. If we add them, we get 125.

OpenStudy (anonymous):

Ty for breaking it down...you saved me alot of time.

OpenStudy (anonymous):

theres another one just like that with 3 triangles

OpenStudy (anonymous):

Did i do this right so far?

OpenStudy (kinggeorge):

That looks correct to me.

OpenStudy (anonymous):

ok so now all i have to do is add them all together correct?

OpenStudy (kinggeorge):

Correct.

OpenStudy (anonymous):

Ok.

OpenStudy (katielong):

|dw:1334855745714:dw| here you see i worked out all of the sides by doing some halving. so... you can see the top triangles are 4cmx7.5cm to find the area times them which equals 30 if you notice, it now needs halving but 30 is the equivilant of the area of two of those triangles so you dont need to half it. its the same with the 13cmx5cm triangle 13x5=65 then you need to finally find the area of the last one... which is the same as the first... so to find the area of the shape you need to do... 30+65+30=........answer.....=125!!!!!!!!! hope it helped?

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