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ZyairaBoyd
 3 years ago
Can anyone solve this? (still not solved)
ZyairaBoyd
 3 years ago
Can anyone solve this? (still not solved)

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prealgebra
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, but it is too cumbersome.

Shayaan_Mustafa
 3 years ago
Best ResponseYou've already chosen the best response.0hi ZyairaBoyd :) Is it regular pentagon?

Shayaan_Mustafa
 3 years ago
Best ResponseYou've already chosen the best response.0it 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..

prealgebra
 3 years ago
Best ResponseYou've already chosen the best response.0It's not a regular pentagon.

Shayaan_Mustafa
 3 years ago
Best ResponseYou've already chosen the best response.0OK. Let us see if it is not a regular pentagon. Be with me.

ZyairaBoyd
 3 years ago
Best ResponseYou've already chosen the best response.0Man got to go. See ya.

Shayaan_Mustafa
 3 years ago
Best ResponseYou've already chosen the best response.0@ZyairaBoyd Is that enough for you?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1They want you to solve this with triangles. First, notice that solving for triangle ECB is the easiest. It's just \[{1 \over2}\cdot 10\cdot 13=65\]

ZyairaBoyd
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah so i have to like do it one by one?

katielong
 3 years ago
Best ResponseYou've already chosen the best response.1i'll be 2 mins.. i can do it though and il give you a good method!!

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Then, 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Now we add \(60+65=125\) To get the overall area.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1We have three triangles with area 65, 30, and 30. If we add them, we get 125.

ZyairaBoyd
 3 years ago
Best ResponseYou've already chosen the best response.0Ty for breaking it down...you saved me alot of time.

ZyairaBoyd
 3 years ago
Best ResponseYou've already chosen the best response.0theres another one just like that with 3 triangles

ZyairaBoyd
 3 years ago
Best ResponseYou've already chosen the best response.0Did i do this right so far?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1That looks correct to me.

ZyairaBoyd
 3 years ago
Best ResponseYou've already chosen the best response.0ok so now all i have to do is add them all together correct?

katielong
 3 years ago
Best ResponseYou've already chosen the best response.1dw: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?

katielong
 3 years ago
Best ResponseYou've already chosen the best response.1no probss =) do you understand it??

ZyairaBoyd
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1334856366881:dw

ZyairaBoyd
 3 years ago
Best ResponseYou've already chosen the best response.0@KingGeorge & @katielong

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1You're very welcome.

katielong
 3 years ago
Best ResponseYou've already chosen the best response.1youre welcommeeeee =)
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