An isosceles trapezoid has bases of length 23 and 12 centimeters and legs of length 13 centimeters. What is the area of the trapezoid to the nearest tenth?

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Answer 1 · 2016-02-07T14:06:39

#205.9sq.cm#

Area of trapezium=#1/2(a+b)h=((a+b)h)/2#

Where #a and b=parall##e##l# #sides,h=height#

Now in an isoceles trapezium the legs are equal ,and in this case they are in a length of #13#

Now consider the diagram:
enter image source here

Now we will have a short sypnosis of the lengths:

#ab=23,fd=12,dp=fs=h#

Now we need to find the height:In this case the height is in a right triangle.So,we use the pythagorean theorem:

#a^2+b^2=c^2#

Where #a# and #b# are the two adjacent sides,#c=hypoten##use# (longest side)

But we should know the length of #pb# to know the height:

#rarrpb=(23-12)/2=11/2=5.5#

We divide it by #2# because there is another side as #as# which equals #pb# :

So,

#rarrh^2+5.5^2=13^2#

#rarrh^2+30.25=169#

#rarrh^2=169-30.25#

#rarrh^2=138.75#

#rarrh=sqrt138.25=11.77#

Now,

#Area=((23+12)11.77)/2#

#rarr=((35)11.77)/2#

#rarr=411.9/2=205.95sq.cm^2#

If we round it off to the nearest tenths we get #205.9sq.cm^2#