Given the equilateral triangle inscribed in a square of side #s# find the ratio of #Delta BCR " to " DeltaPRD#?

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2 Answers
Nov 7, 2016

#(DeltaBCR)/(DeltaPRD)=1/2#

Explanation:

Considering the figure to be symmetric w.r.t. #BD#, #/_CBR=(90^o-60^o)/2=15^o#

and side #CR# of #DeltaBPR# is given by

#CR=sxxtan15^o=(sqrt3-1)/(sqrt3+1)s#, where #BC=s#.

Area of #DeltaBCR=(BCxxCR)/2=sxx(sqrt3-1)/(sqrt3+1)s/2=(sqrt3-1)/(sqrt3+1)s^2/2#

Area of #DeltaPRD=(DR^2)/2=((s-CR)^2)/2=(s-(sqrt3-1)/(sqrt3+1)s)^2/2#

#(1-(sqrt3-1)/(sqrt3+1))^2s^2/2=((sqrt3+1-sqrt3+1)/(sqrt3+1))^2s^2/2=4/(sqrt3+1)^2s^2/2#

Hence #(DeltaBCR)/(DeltaPRD)=((sqrt3-1)/(sqrt3+1)s^2/2)/(4/(sqrt3+1)^2s^2/2)=((sqrt3-1)/(sqrt3+1))/(4/(sqrt3+1)^2)=((sqrt3-1)(sqrt3+1)^2)/(4(sqrt3+1))#

= #((sqrt3-1)(sqrt3+1))/4=(3-1)/4=2/4=1/2#

Nov 8, 2016

#(DeltaBCR)/(DeltaPRD)=1/2#

Explanation:

As per given figure #DeltaBCR ~=DeltaABP#
Since #/_BCR=/_ABP=90^@->"angle of a square"#

#"hypotenuse "BR ="hypotenuse "BP#
( sides of equilateral #DeltaBRP#)

#AB=BC->"sides of square ABCD"#

So #/_CBR=/_ABP#

Using trigonometry

In triangle BCR
#/_CBR=/_ABP=1/2(/_ABC-/_PBR)#

#=(90^@-60^@)/2=15^@#

and #/_BRC=90^@-15^@=75^@#

If the length of each side of the equilareral triangle BPR be a,then
#BC=asin/_CBR=acos15^@#

#CR=asin/_CBR=asin15^@#

For #DeltaPRD#

#/_PRD=(180-/_BRC-/_BRP)#

#=180^@-75^@-45^@=45^@#

So #PD=DR=PRcos45^@=acos45^@=asin45^@#

Now

#(DeltaBCR)/(DeltaPRD)#

#=(cancel(1/2)xxCRxxBC)/(cancel(1/2)xxPDxxDR)#

#=(asin15^@xxcos15^@)/(acos45^@xxasin45^@)#

#=(2cos15^@xxsin15^@)/(2cos45^@xxsin45^@)#

#=sin30^@/sin90^@#

#=sin30^@=1/2#

Without using trigonometry

We have shown above #DeltaBCR ~=DeltaABP#

So CR=AP

Now

#DR=CD-CR=AD-AP=DP#

Applying Pythagoras theorem for #DeltaPRD# we get

#PR^2=DP^2+DR^2=2DR^2#

#=>BR^2=2DR^2->"since "PR=BR#

Niw applying Pythagoras theorem for #DeltaBRC# we get

#BR^2=BC^2+CR^2#

So #2DR^2=BC^2+CR^2#

#=>2DR^2=BC^2+CR^2#

#=>2DR^2=CD^2+CR^2#

#=>2DR^2=(CR+DR)^2+CR^2#

#=>2DR^2=CR^2+DR^2+2CR*DR+CR^2#

#=>2DR^2-DR^2=2CR^2+2CR*DR#

#=>DR^2=2CR^2+2CR*DR#

#=>1/2*DR^2=CR^2+CR*DR#

#=>1/2*DR^2=CR(CR+DR)#

#=>1/2*DR*PD=2*1/2*CR*CD#

#=>1/2*DR*PD=2*1/2*CR*BC#

#=>DeltaPRD=2*DeltaBCR#

#=>(DeltaBCR)/(DeltaPRD)=1/2#