As isosceles triangle is a triangle that has two equal sides. Find AB, BC and AC. Determine if the triangle???

Question

Could someone please explain to me how to do this problem step by step, I tried to draw the triangle example the best i could. i could not copy it into here.
As isosceles triangle is a triangle that has two equal sides. Find AB, BC and AC. Determine if the triangle is isoseles or not based on the lengths of the three sides. Show all your work.


B (1,5) .
| .
| . C (5,3)
| .
A (1,1).

Answer 1

To find the lenghts of the sides use the distance between 2 points formula.
Hence BC = ((5-1)^2 + (3-5)^2)^(1/2) =20^(1/2)
Similarly CA=20^(1/2)
Since BC = AC The triangle is isoceles.

Answer 2

QUESTION IS NOT CLEAR.PL. TRY FOLLOWING:
Isosceles Triangle :

An isosceles triangle is a triangle with (at least) two equal sides. In the figure , the two equal sides have length and the remaining side has length . This property is equivalent to two angles of the triangle being equal. An isosceles triangle therefore has both two equal sides and two equal angles. The name derives from the Greek iso (same) and skelos (leg).

A triangle with all sides equal is called an equilateral triangle, and a triangle with no sides equal is called a scalene triangle. An equilateral triangle is therefore a special case of an isosceles triangle having not just two, but all three sides and angles equal. Another special case of an isosceles triangle is the isosceles right triangle.

The height of the isosceles triangle illustrated above can be found from the Pythagorean theorem as :
h=sqrt(b^2-1/4(a^2).WHERE a IS THE BASE,h=HEIGHT

Let us see the whole sequence happening at once:

1. Consider a triangle abc with |ab| = |bc|
2. Draw a line from a perpendicular to the line segment [bc] meeting it at the point d.
3. Consider the two triangles abd and acd
4. The side |ab| = the side |ac| (given)
5. The side |ad| = itself
6. The angles adb and adc are both right angles. (We constructed them that way in step 2.
7. This forces the two triangles to be congruent by RHS
8. So the two angles abd and acd must be equal.

Let us see the whole sequence happening at once:

Answer 3

This Problem can be solved by using the distance formula

d = √(x₂- x₁)² + (y₂ - y₁)²

The ordered Pair of side AC

AC = (1, 1) (5, 3)

d = √(1 - 5)² + (1 - 3)²

d = √(-4)² + (-2)²

d = √16 + 4

d = √20

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The ordered Pair of side BC

BC = (1,5) (5, 3)

d = √(1 - 5)² + (5 - 3)²

d = √(- 4)² + (2)²

D = √16 + 4

d = √20

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The two sides are equal

AC = BC

d = √20 for both

This is a Isosceles Triangle

Answer 4

I think it is..
Let's try the phytagoras Theorem...
The base from BA's difference is 4 units ...
Okay...when you divide using 2 to divide the triange...you'll find the BC line...
So...it'll be...
2 at the top triangle height...the middle line would have a 4 units base area.
Than you find the line BC, hypotenuse which is square root of 20 units...
Down to the second triangle...
The bottom triangle's height is only two units...you have four units as the middle line, remember?
Find the hypotenuse again for AC which is the square root of 20 units...
So, according to my explaination, it is an isosceles triangle!

Answer 5

This triangle could be graphed on a Cartesian plane obviously. The points are its vertices...

So what you have to do is get the distance between the points using the distance formula:

AB=sqrt[(x1-x2)^2+(y1-y2)^2]
AB=sqrt[(1-1)^2+(1-5)^2]
AB=sqrt[0+(-4)^2]
AB=sqrt[0+16]
AB=4

AC=sqrt[(x1-x2)^2+(y1-y2)^2]
AC=sqrt[(1-5)^2+(1-3)^2]
AC=sqrt[(-4)^2+(-2)^2]
AC=sqrt[16+4]
AC=sqrt20
AC=2sqrt5

BC=sqrt[(1-5)^2+(5-3)^2]
BC=sqrt[(-4)^2+(2)^2]
BC=sqrt[16+4]
BC=sqrt20
BC=2sqrt5

So the length of the sides are 4, 2sqrt of 5 and 2sqrt5, thus the triangle is isosceles. BC=AC

Answer 6

Yes. BC and AC both have lengths of 4.4721.
Subtract the x and y values and square them, add the squares then take the square root.
Do this for each line defined by two of the points at a time.

Answer 7

Use the distance formula for each set of points and if two of them are equal then you have an isosceles triangle.

Answer 8

slope BC=-1/2

slope BA= infinite

slope CA= 1/2

m1.m2= -1
So triangle is a right angled triangle.

Answer 9

d(A,B)=sqrt{ (1-1)^2 + (5-1)^2}
=sqrt(16)=4
d(A,C)=sqrt{ 4^2+2^2)}
=sqrt{20}
d(B,C)=sqrt{ 4^2+(-2)^2}
=sqrt{20}
therefore the triangle is isosceles

Answer 10

we note the lengh of AB with c, the lengh of AC with b and the lengh of CB with a.

Xa=1, Ya=1, A(Xa,Ya)
Xb=1, Yb=5, B(Xb,Yb)
Xc=5, Yc=3, C(Xc,Yc)

a*a=(Xb-Xc)*(Xb-Xc)+(Yb-Yc)*(Yb-Yc)
=> a*a=4+16=20
a=sqrt(20) => BC=sqrt(20)

b*b=(Xa-Xc)*(Xa-Xc)+(Ya-Yc)*(Ya-Yc)
=> b*b=16+4=20
b=sqrt(20) => AC=sqrt(20)

c*c=(Xb-Xa)*(Xb-Xa)+(Yb-Ya)*(Yb-Ya) => c*c=0+16=16
c=4 => AB=4

because AC=BC=sqrt(20), the triangle is isosceles