ZyairaBoyd Group Title Can anyone solve this? (still not solved) 2 years ago 2 years ago

1. ZyairaBoyd Group Title

2. pre-algebra Group Title

Yes, but it is too cumbersome.

3. ZyairaBoyd Group Title

i know -.-

4. Shayaan_Mustafa Group Title

hi ZyairaBoyd :) Is it regular pentagon?

5. ZyairaBoyd Group Title

Pretty sure.....

6. Shayaan_Mustafa Group Title

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..

7. pre-algebra Group Title

It's not a regular pentagon.

8. ZyairaBoyd Group Title

Ok.

9. Shayaan_Mustafa Group Title

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

10. ZyairaBoyd Group Title

Man got to go. See ya.

11. Shayaan_Mustafa Group Title

@ZyairaBoyd Is that enough for you?

12. ZyairaBoyd Group Title

Huh? No....

13. ZyairaBoyd Group Title

@Shayaan_Mustafa

14. KingGeorge Group Title

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

15. ZyairaBoyd Group Title

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

16. katielong Group Title

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

17. ZyairaBoyd Group Title

I dont mind.

18. KingGeorge Group Title

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.

19. KingGeorge Group Title

Now we add $$60+65=125$$ To get the overall area.

20. ZyairaBoyd Group Title

oh ok.

21. KingGeorge Group Title

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

22. ZyairaBoyd Group Title

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

23. ZyairaBoyd Group Title

theres another one just like that with 3 triangles

24. ZyairaBoyd Group Title

Did i do this right so far?

25. KingGeorge Group Title

That looks correct to me.

26. ZyairaBoyd Group Title

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

27. KingGeorge Group Title

Correct.

28. ZyairaBoyd Group Title

Ok.

29. katielong Group Title

|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?

30. ZyairaBoyd Group Title

i got 22.

31. ZyairaBoyd Group Title

Ty @katielong

32. katielong Group Title

no probss =) do you understand it??

33. ZyairaBoyd Group Title

yes.

34. ZyairaBoyd Group Title

|dw:1334856366881:dw|

35. ZyairaBoyd Group Title

@KingGeorge & @katielong

36. KingGeorge Group Title

You're very welcome.

37. katielong Group Title

youre welcommeeeee =)