What's the length of the |AC| ?

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What's the length of the |AC| ?

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Answer 1 · 2017-09-17T03:47:08

$A C = 5 \sqrt{2}$ $AC=5sqrt2$

Explanation:

If $A B ⊥ B C$ $AB ⊥ BC$ , and $A B = 5$ $AB=5$ , then $B C = 5$ $BC=5$ . This is a right angled triangle, which means sides $a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$ . In this case $a = A B = 5,$ $a=AB=5,$ $b = B C = 5,$ $b=BC=5,$ and $c = A C$ $c=AC$ . Therefor
$5^{2} + 5^{2} = {(A C)}^{2}$ $5^2+5^2=(AC)^2$
$25 + 25 = {(A C)}^{2}$ $25+25=(AC)^2$
$50 = {(A C)}^{2}$ $50=(AC)^2$
$\sqrt{50} = A C$ $sqrt50=AC$

From here we can simplify $\sqrt{50}$ $sqrt50$ , since
$\sqrt{50} = \sqrt{5 \cdot 5 \cdot 2} = 5 \sqrt{2}$ $sqrt50=sqrt(5*5*2)=5sqrt2$

I hope I helped!

Answer 2 · 2017-09-17T05:57:10

$A C = 8 \frac{1}{3} c m .$ $AC=8 1/3cm.$

Explanation:

As $B H = 3$ $BH=3$ and $A B = 5$ $AB=5$ , as $Delta ABH$ is right angled at $H$ , using Pythagoras Theorem

$AH=sqrt(5^2-3^2)=sqrt(25-9)=sqrt16=4$

In a right angled triangle, right angled at $/_B$ , if we connect midpoint of hypotenuse to $B$ , then

$BD=AD=DC$ .

Ler this be $x$ and then $HD=x-3$ and using Pythagoras Theorem on $DeltaAHD$ , we have

$AH^2+HD^2=AD^2$ or $4^2+(x-3)^2=x^2$

i.e. $x^2-6x+9+16=x^2$

or $6x=25$ and $x=25/6=4 1/6$

Hence, $AC=25/6xx2=25/3=8 1/3cm.$

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What's the length of the |AC| ?

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