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?

Question

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?

Sophia Alfred · Accepted Answer

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]

Savannah Aguilar · Answer

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#

Evelyn Arboleda · Answer

undefined Use the formula for the area of a triangle: Area = 0.5 * base * height. Since the triangle is isosceles, the height can be found by drawing a perpendicular line from the vertex opposite the base to the midpoint of the base. Then, use the distance formula to find the length of the base, which is the distance between the two given points. Finally, use the Pythagorean theorem to find the lengths of the remaining sides, which are congruent.