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?

Answer 1

The length of three sides of triangle are `9.43 ,14.36 , 14.36` unit

Explanation:

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]

Answer 2

The sides are `9.4, 13.8, 13.8`

Explanation:

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`