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?

##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:

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 :

##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##