Question #259023

Two sides of a square are on the line x + 10y – 10 = 0 and x + 10y – 5 = 0. Find the area of the square.



1
Expert's answer
2021-11-01T19:25:38-0400

x+10y10=0y=10x10y=x10+1x+10y-10=0\to y={\frac {10-x} {10}}\to y=-{\frac x {10}}+1

x+10y5=0y=5x10y=x10+12x+10y-5=0\to y={\frac {5-x} {10}}\to y=-{\frac x {10}}+{\frac 1 2}

The picture is presented below



ABCD is the square, then CD is perpendicular to both BC and AD, then it's a distance between them

The area of the ABCD is equal to CD2CD^{2}

So, the goal is to find the distance between two given lines (CD).

Since two of the given lines has equal slopes, then they are parallel, and DK equal to b1b2|b{\scriptscriptstyle 1}-b{\scriptscriptstyle 2}| , where b1,b2b{\scriptscriptstyle 1},b{\scriptscriptstyle 2} - free terms

In the given case DK=b1b2=112=0.5DK=|b{\scriptscriptstyle 1}-b{\scriptscriptstyle 2}|=|1-{\frac 1 2}|=0.5

СDK=MDA∠СDK = ∠MDA

MD=DCMD=DC

From two statements above we can conclude that triangles MDA and CDK is equals, so AD = 0.5

From triangle MDA(A=π2∠A={\frac \pi 2}): AM=ADtan(MDA)AM=AD*tan(∠MDA)

MDA=πMDFtan(MDA)=tan(πMDF)=tan(MDF)=k=∠MDA=\pi - ∠MDF \to tan(∠MDA)=tan(\pi -∠MDF)=-tan(∠MDF) = -k =

=(0.1)=0.1= -(-0.1)=0.1 (k is the slope of the given lines)

AM=AD0.1=0.05AM=AD*0.1=0.05

Now from the triangle MDA using the Pythagorian theorem we receive:

AD2=MA2+AD2=0.2525=CD2AD^{2}=MA^{2}+AD^{2}=0.2525=CD^{2} = area of ABCD

The area of the square is 0.2525


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