https://clackamasweb.owschools.com/media/g_geo_ccss_2014/8/groupi31.gif Find the area of the trapezoid shown. 43.5 sq. units 43.5√3 sq. units
tan (60) = h/3 \[\sqrt{3} = h/3\] \[3\sqrt{3} = h\] area of trapezoid = \[Area = ((a+b)\div 2 ) h\] a = 10 b =16 h = 3sqrt3
so which one is it? @sangya21
neither of these \[\tan 60^0 = \sqrt{3}\] \[h = 3\sqrt{3}\] \[area = ((10+16)/2)* 3\sqrt{3}\] \[area = (26/2)* 3\sqrt{3}\] \[area = 13* 3\sqrt{3}\] \[area = 39\sqrt{3}\]
its not 39
thats what my calculation says. let me ask someone for help @StudyGurl14 @Compassionate
@TheSmartOne
What d'ya need help with, friends?
@StudyGurl14 although we have applied the correct formula, we are not able to get desirable answer. Is there any other way to do this?
Hm...It appears you used tan to get the height. Couldn't you have just used the information known about 30-60-90 special right triangles?
Although, it probably would result in the same height...let me check.
no, the height still comes out to \(\large 3\sqrt{3}\)...
Oh, I see what the problem is...yo forgot to add 3 to 16 to find the length of base b
So \[29 \times 3\sqrt{3} \times \frac{ 1 }{ 2 }= Area\]
That is what I am getting so far...
Area of trapezoid: \(\large(\frac{b_1+b_2}{2})h\)
Exactly ^^
Oops.
\(\large(\frac{10+(16+3)}{2})(3\sqrt{3})\)
thanks @StudyGurl14
Don't worry @sangya21 , you had the right idea and everything. Just one small mistake that I've made many times over. :)
Anytime. :)
\(\large(\frac{29}{2})(3\sqrt{3})\rightarrow\frac{29(3)\sqrt{3}}{2}\rightarrow\frac{87\sqrt{3}}{2}\)
and 87/2=43.5
\(87\div2=43.5\)
So \[43.5\sqrt{3}\]
Exactly. :)
so.... \(\large43.5\sqrt{3}\tt=the~answer\)
you keeping beating me to it @TheSmartOne , lol :)
It makes me so happy when my friends get to the answer after me...
No @StudyGurl14 Why are you at 86 SS. Now I have to work harder. :(
When I overtake you in SS I will change my profile pic to whatever yours is then. :)
Once I overtake Cam i will change my profile pic to hers Ok? @camerondoherty And then you are next. :)
Thank you all:))
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