Two corners of an isosceles triangle are at #(9 ,2 )# and #(1 ,7 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?
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"How do I change #int_0^1int_0^sqrt(1-x^2)int_sqrt(x^2+y^2)^sqrt(2-x^2-y^2)xydzdydx# to cylindrical or spherical coordinates?"
The length of three sides of triangle are #9.43 ,14.36 , 14.36# unit
Base of the isocelles triangle is #B= sqrt((x_1-x_2)^2+(y_1-y_2)^2)) = sqrt((9-1)^2+(2-7)^2)) =sqrt(64+25)=sqrt89 =9.43(2dp)#unit
We know area of triangle is #A_t =1/2*B*H# Where #H# is altitude.
#:. 64=1/2*9.43*H or H= 128/9.43=13.57(2dp)#unit.
Legs are #L = sqrt(H^2+(B/2)^2)= sqrt( 13.57^2+(9.43/2)^2)=14.36(2dp)#unit
The length of three sides of triangle are #9.43 ,14.36 , 14.36# unit [Ans]
The sides are #9.4, 13.8, 13.8#
The length of side #A=sqrt((9-1)^2+(2-7)^2)=sqrt89=9.4#
Let the height of the triangle be #=h#
The area of the triangle is
#1/2*sqrt89*h=64#
The altitude of the triangle is #h=(64*2)/sqrt89=128/sqrt89#
The mid-point of #A# is #(10/2,9/2)=(5,9/2)#
The gradient of #A# is #=(7-2)/(1-9)=-5/8#
The gradient of the altitude is #=8/5#
The equation of the altitude is
#y-9/2=8/5(x-5)#
#y=8/5x-8+9/2=8/5x-7/2#
The circle with equation
#(x-5)^2+(y-9/2)^2=(128/sqrt89)^2=128^2/89#
The intersection of this circle with the altitude will give the third corner.
#(x-5)^2+(8/5x-7/2-9/2)^2=128^2/89#
#(x-5)^2+(8/5x-8)^2=128^2/89#
#x^2-10x+25+64/25x^2-128/5x+64=16384/89#
#89/25x^2-178/5x+89-16384/89=0#
#3.56x^2-35.6x-95.1=0#
We solve this quadratic equation
#x=(35.6+-sqrt(35.6^2+4*3.56*95.1))/(2*3.56)#
#x=(35.6+-51.2)/7.12#
#x_1=86.8/7.12=12.2#
#x_2=-15.6/7.12=-2.19#
The points are #(12.2,16)# and #(-2.19,-7)#
The length of #2# sides are #=sqrt((1-12.2)^2+(7-16)^2)=sqrt189.4=13.8#
