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?

1 Answer
Feb 7, 2016

205.9sq.cm

Explanation:

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

Where a and b=parallel 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=hypotenuse (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