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

$205.9 s q . c m$ $205.9sq.cm$

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

Where $a and b = p a r a l l$ $a and b=parall$ $e$ $e$ $l$ $l$ $s i d e s, h = h e i g h t$ $sides,h=height$

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

Now consider the diagram:
enter image source here

Now we will have a short sypnosis of the lengths:

$a b = 23, f d = 12, d p = f s = h$ $ab=23,fd=12,dp=fs=h$

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

$a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$

Where $a$ $a$ and $b$ $b$ are the two adjacent sides, $c = h y p o t e n$ $c=hypoten$ $u s e$ $use$ (longest side)

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

$\to p b = \frac{23 - 12}{2} = \frac{11}{2} = 5.5$ $rarrpb=(23-12)/2=11/2=5.5$

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

So,

$\to h^{2} + {5.5}^{2} = 13^{2}$ $rarrh^2+5.5^2=13^2$

$\to h^{2} + 30.25 = 169$ $rarrh^2+30.25=169$

$\to h^{2} = 169 - 30.25$ $rarrh^2=169-30.25$

$\to h^{2} = 138.75$ $rarrh^2=138.75$

$\to h = \sqrt{138.25} = 11.77$ $rarrh=sqrt138.25=11.77$

Now,

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

$\to = \frac{(35) 11.77}{2}$ $rarr=((35)11.77)/2$

$\to = \frac{411.9}{2} = 205.95 s q . c m^{2}$ $rarr=411.9/2=205.95sq.cm^2$

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