precalculus question.?

Question

Show that the triangle with the vertices A(6,-7),B(11,-3) &C(2,-2) is a right triangle by using the converse of the Pythagorean Theorem. Find the area of the triangle.

Answer 1

AB^2 = 5^2 + 4^2 = 25 + 16 = 41.

BC^2 = 9^2 + 1^2 = 81 + 1 = 82.

AC^2 = 4^2 + 5^2 = 41 again.

Thus you have a trangle such that AB = AC, and

AB^2 + AC^2 = 41 + 41 = 82 = BC^2.

Hence, by the converse of Pythagoras's Theorem, ABC is a right-angled triangle with BC the hypoteneuse. QED

Since AB = AC, the area = 1/2 AB*AC = 1/2 (AB)^2 = 20.5.

As you have no doubt observed by now, only the SQUARES of various lengths are ULTIMATELY NEEDED in this problem. It is therefore quite unnecessary, for the purposes of this particular problem, to actually solve for the individual lengths of the various line segments. Good teachers look out for students who can see such time-saving points.

LIVE LONG AND PROSPER.

Answer 2

The triangle has vertices A(6,-7), B(11,-3), and C(2,-2).

| AB |² = (11-6)² + (-3+7)² = 25 + 16 = 41
| AC |² = (2-6)² + (-2+7)² = 16 + 25 = 41
| BC |² = (2-11)² + (-2+3)² = 81 + 1 = 82

| AB |² + | AC |² = | BC |²
41 + 41 = 82

Therefore the triangle is a right triangle.

The area is (1/2)bh = (1/2)(√41)(√41) = 20.5

Answer 3

First you can solve for the length of the three sides of your triangle using the distance formula:

For side A(6,-7) & B(11,-3)

D² = (x1 - x2)² + (y1 - y2)² so

(AB)² = (6 - 11)² + [(-7) - (-3)]²
(AB)² = (-5)² + (-4)²
(AB)² = 25 + 16
(AB)² = 41
AB = √41

For B(11,-3) & C(2,-2)
(BC)² = (11 - 2)² + [(-3) - (-2)]²
(BC)² = (9)² + (1)²
(BC)² = 81 + 1
(BC)² = 82
BC = √82

For A(6,-7) & C(2,-2)
(AC)² = (6 - 2)² + [(-7) - (-2)]²
(AC)² = (4)² + (-5)²
(AC)² = 16 + 25
(AC)² = 41
AC = √41

Now to see if it's really a right triangle by using pythagorean therem, the square of the longest side is equal to the sum of the squares of the other two adjacent sides so by inspection:

AB = √41
BC = √82; longest side
AC = √41

to check if ABC is a right triangle by pythagorean theorem c² = a² + b²

BC² = AC² + AB²
(√82)² = (√41)² + (√41)²
82 = 41 + 41
82 = 82; correct

Now to find the area of the triangle (1/2) base time height, since the base is equal to the height for our triangle so the area is now:

Area of triangle = (1/2) (base) (height); base = height
Area of triangle = (1/2) (base) (base)
Area of triangle = (1/2) (base)²; base = √41
Area of triangle = (1/2) (√41)²
Area of triangle = (1/2) (41)
Area of triangle = 20.5 sq. units

So the Area bounded by the three vertices is 20.5 sq. units

Answer 4

i would suggest u to get a graph paper to draw the triangle out...
den take all sides as hypo, u'll be able to get the ans by drawing the adj and opp to form a right angle triangle with the hypo.
from there u can get the length of the 3 sides, den using pythagoras theorem, u can proof that by adding (AC)^2 and (AB)^2, you'll get (BC)^2..


AC=AB=6.4
BC=9.05
area of triangle=(1/2)(6.4)(6.4) =20.48

Answer 5

AB=sqrt((11-6)^2+(-3+7)^2) = sqrt(41)
BC=sqrt((2-11)^2+(-2+3)^2) = sqrt(82)
CA=sqrt((2-6)^2+(-2+7)^2) = sqrt(41)

since BC is the longest side, if ABC is a right angled tringle, BC^2=AB^2+CA^2
82 = 41+41 which is true.

Hence proved.
area = 1/2*sqrt(41)*sqrt(41) = 41/2 = 20.5 sq units.

Answer 6

u must state where is d hypotenuse !!