Suppose triangle abc has vertices (r,s)(r+1,s+7)(r-3,s+4)Show that it is an isosceles right triangle?

Question

must also find the area

Answer 1

Dunno. Try the following:

1) Use the distance formula on all three segments to see if two of them (and ONLY two of them) are congruent. If so, it's isosceles.

2) Since you have all three sides, Pyhtagorize to see if it's a right triangle (square the two smaller sides, and see if their sum matches the square of the longest side).

3) On the assumption that is IS an isosceles right triangle, then the two congruent (and shorter) legs are the base and height. Do 1/2(base)(height) to get the area.


EDIT" Upon further inspection (i.e. lemme think for about 30 seconds), it IS an isosceles right triangle. Don't let the r's and s's bother you: when you use the distance formula, they all cancel really quickly.

Answer 2

(r,s) and (r+1,s+7)
D = sqrt(((r + 1) - r)^2 + ((s + 7) - s)^2)
D = sqrt((r + 1 - r)^2 + (s + 7 - s)^2)
D = sqrt(1^2 + 7^2)
D = sqrt(1 + 49)
D = sqrt(50)
D = 2sqrt(5)

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(r+1,s+7) and (r-3,s+4)

D = sqrt(((r - 3) - (r + 1))^2 + ((s + 4) - (s + 7))^2)
D = sqrt((r - 3 - r - 1)^2 + (s + 4 - s - 7)^2)
D = sqrt((-4)^2 + (-3)^2)
D = sqrt(16 + 9)
D = sqrt(25)
D = 5

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(r,s) and (r-3,s+4)

D = sqrt((r - (r - 3))^2 + (s - (s + 4))^2)
D = sqrt((r - r + 3)^2 + (s - s - 4)^2)
D = sqrt(3^2 + 4^2)
D = sqrt(9 + 16)
D = sqrt(25)
D = 5

So now we know that the legs are

5, 5, and 5sqrt(2)

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A = (ab)/2
A = (5 * 5)/2
A = 25/2
A = 12.5

ANS : 12.5 square units.

Answer 3

let's call them p1,p2,p3
p1(r,s) p2(r+1,s+7) p3(r-3,s+4)

p1p2=Sqrt[(r+1-r)^2+(s+7-s)^2]
p1p2= Sqrt[1+49]=Sqrt[50]=5Sqrt[2]

p1p3=Sqrt(r-3-r)^2+(s+4-s)^2]
p1p3=Sqrt[9+16]=Sqrt[25]=5

p2p3=Sqrt[(r-3-r-1)^2+(s+4-s-7)^2
p2p3=Sqrt[16+9]=Sqrt[25]=5

so the sides are 5,5,5Sqrt[2]

5^2+5^2=(5Sqrt[5])^2
so the triangle is right and isosceles

the area is bh/2 = (5)(5)/2= 12.5

Answer 4

since this is independent of the values of r & s, lets set them to 0
a (0,0)
b (1,7)
c (-3,4)
ab=sqrt(7^2+1^2)=sqrt(50)
ac=sqrt(3^2+4^2)=sqrt(25)
bc=sqrt((1-(-3))^2+(7-4)^2)=sqrt(4^2+3^2)=sqrt(25)

sides are
bc=5
ac=5
ab=5sqrt(2)
it is isosceles

Answer 5

CAB opposite side angle of ADC we know that sum of any two sides of a triangle is greater than the third side so when CAB=20 so ADC will be also 20.........