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?

Feb 7, 2016

$205.9 s q . c m$

Explanation:

Area of trapezium=$\frac{1}{2} \left(a + b\right) h = \frac{\left(a + b\right) h}{2}$

Where $a \mathmr{and} b = p a r a l l$$e$$l$ $s i \mathrm{de} s , h = h e i g h t$

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:

$a b = 23 , f d = 12 , \mathrm{dp} = f s = h$

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 = h y p o t e n$$u s e$ (longest side)

But we should know the length of $p b$ to know the height:

$\rightarrow p b = \frac{23 - 12}{2} = \frac{11}{2} = 5.5$

We divide it by $2$ because there is another side as $a s$ which equals $p b$ :

So,

$\rightarrow {h}^{2} + {5.5}^{2} = {13}^{2}$

$\rightarrow {h}^{2} + 30.25 = 169$

$\rightarrow {h}^{2} = 169 - 30.25$

$\rightarrow {h}^{2} = 138.75$

$\rightarrow h = \sqrt{138.25} = 11.77$

Now,

$A r e a = \frac{\left(23 + 12\right) 11.77}{2}$

$\rightarrow = \frac{\left(35\right) 11.77}{2}$

$\rightarrow = \frac{411.9}{2} = 205.95 s q . c {m}^{2}$

If we round it off to the nearest tenths we get $205.9 s q . c {m}^{2}$