Question 1149604
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Given image
<img src = "https://i.imgur.com/BBREqMU.png">
Let point E be the midpoint of segment BC. The red triangle AEC has unknown height h and unknown base x
<img src = "https://i.imgur.com/UFIdzie.png">
Focusing solely on triangle AEC, we can use the pythagorean theorem to say


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


{{{h^2 + x^2 = 17^2}}}


{{{h^2 + x^2 = 289}}}


{{{h^2 = 289-x^2}}} subtract x^2 from both sides


We'll use the last equation later on


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Mark triangle AED with another color. I'll use green. This triangle also has height h, but the base is now 3x
<img src = "https://i.imgur.com/tUFe5yf.png">
The 3x is from the fact that


EC = x
CD = BC = 2x
ED = EC+CD = x+2x = 3x


Once again, use the pythagorean theorem, but focus solely on triangle AED.


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


{{{h^2 + (3x)^2 = 27^2}}}


{{{h^2 + 9x^2 = 729}}}


{{{289-x^2 + 9x^2 = 729}}} Replace {{{h^2}}} with {{{289-x^2}}} (we're using {{{h^2 = 289-x^2}}} found earlier)


{{{289+8x^2 = 729}}}


{{{8x^2 = 729-289}}} Subtract 289 from both sides


{{{8x^2 = 440}}}


{{{x^2 = 440/8}}} Divide both sides by 8


{{{x^2 = 55}}}


{{{x = sqrt(55)}}} Apply square root to both sides


{{{4x = 4sqrt(55)}}} Multiply both sides by 4


The base of triangle ABD is exactly {{{4*sqrt(55)}}} units long


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Use that info to find the following value for h


{{{h^2 = 289-x^2}}}


{{{h^2 = 289-55}}}


{{{h^2 = 234}}}


{{{h = sqrt(234)}}}


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Now compute the area of triangle ABD


{{{Area = (1/2)*(base)*(height)}}}


{{{Area = (1/2)*4sqrt(55)*sqrt(234)}}}


{{{Area = 2*sqrt(55)*sqrt(234)}}}


{{{Area = 2*sqrt(55*234)}}}


{{{Area = 2*sqrt(12870)}}}


{{{Area = 2*sqrt(9*1430)}}}


{{{Area = 2*sqrt(9)*sqrt(1430)}}}


{{{Area = 2*3*sqrt(1430)}}}


{{{Area = 6*sqrt(1430)}}} <font color=red size=4>Exact area</font>


{{{Area = 226.892}}} <font color=red size=4>Approximate area</font>


Units are in square cm, often written as {{{cm^2}}}
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