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Let's find out whether [tex]$\triangle ABC$[/tex] is an isosceles triangle or a right triangle by computing the lengths of its sides and checking the necessary conditions for isosceles and right triangles.
First, let's determine the lengths of the sides [tex]\( \overline{AB} \)[/tex], [tex]\( \overline{BC} \)[/tex], and [tex]\( \overline{AC} \)[/tex].
### Length of [tex]\( \overline{AB} \)[/tex]
The points [tex]$A$[/tex] and [tex]$B$[/tex] are given as [tex]\( A(2,0) \)[/tex] and [tex]\( B(4,4) \)[/tex].
The distance formula is:
[tex]\[ AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \][/tex]
Substituting the coordinates of [tex]$A$[/tex] and [tex]$B$[/tex]:
[tex]\[ AB = \sqrt{(4 - 2)^2 + (4 - 0)^2} = \sqrt{(2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.472 \][/tex]
### Length of [tex]\( \overline{BC} \)[/tex]
The points [tex]$B$[/tex] and [tex]$C$[/tex] are given as [tex]\( B(4,4) \)[/tex] and [tex]\( C(6,3) \)[/tex].
The distance formula is:
[tex]\[ BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} \][/tex]
Substituting the coordinates of [tex]$B$[/tex] and [tex]$C$[/tex]:
[tex]\[ BC = \sqrt{(6 - 4)^2 + (3 - 4)^2} = \sqrt{(2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236 \][/tex]
### Length of [tex]\( \overline{AC} \)[/tex]
The points [tex]$A$[/tex] and [tex]$C$[/tex] are given as [tex]\( A(2,0) \)[/tex] and [tex]\( C(6,3) \)[/tex].
The distance formula is:
[tex]\[ AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} \][/tex]
Substituting the coordinates of [tex]$A$[/tex] and [tex]$C$[/tex]:
[tex]\[ AC = \sqrt{(6 - 2)^2 + (3 - 0)^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]
### Checking for Isosceles Triangle
To check if [tex]$\triangle ABC$[/tex] is isosceles, we need to see if any two sides are of equal length.
We have:
[tex]\[ AB \approx 4.472, \quad BC \approx 2.236, \quad AC = 5 \][/tex]
Since none of the lengths are approximately equal, [tex]\(\triangle ABC\)[/tex] is not isosceles.
### Checking for Right Triangle
To check if [tex]$\triangle ABC$[/tex] is a right triangle, we need to use the Pythagorean theorem and verify if any combination of sides satisfies:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
Let's test the following combinations:
1. [tex]\(AB^2 + BC^2 \approx AC^2\)[/tex]:
[tex]\[ (4.472)^2 + (2.236)^2 \approx 5^2 \][/tex]
[tex]\[ 20 + 5 \approx 25 \quad \text{(True)} \][/tex]
Since [tex]\(AB^2 + BC^2 = AC^2\)[/tex], we confirm that [tex]\(\triangle ABC\)[/tex] is a right triangle.
Therefore, Verna is correct because [tex]\(\overline{BC} \perp \overline{AC}\)[/tex].
Answer: Verna, because [tex]\(\overline{BC} \perp \overline{AC}\)[/tex]
First, let's determine the lengths of the sides [tex]\( \overline{AB} \)[/tex], [tex]\( \overline{BC} \)[/tex], and [tex]\( \overline{AC} \)[/tex].
### Length of [tex]\( \overline{AB} \)[/tex]
The points [tex]$A$[/tex] and [tex]$B$[/tex] are given as [tex]\( A(2,0) \)[/tex] and [tex]\( B(4,4) \)[/tex].
The distance formula is:
[tex]\[ AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \][/tex]
Substituting the coordinates of [tex]$A$[/tex] and [tex]$B$[/tex]:
[tex]\[ AB = \sqrt{(4 - 2)^2 + (4 - 0)^2} = \sqrt{(2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.472 \][/tex]
### Length of [tex]\( \overline{BC} \)[/tex]
The points [tex]$B$[/tex] and [tex]$C$[/tex] are given as [tex]\( B(4,4) \)[/tex] and [tex]\( C(6,3) \)[/tex].
The distance formula is:
[tex]\[ BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} \][/tex]
Substituting the coordinates of [tex]$B$[/tex] and [tex]$C$[/tex]:
[tex]\[ BC = \sqrt{(6 - 4)^2 + (3 - 4)^2} = \sqrt{(2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.236 \][/tex]
### Length of [tex]\( \overline{AC} \)[/tex]
The points [tex]$A$[/tex] and [tex]$C$[/tex] are given as [tex]\( A(2,0) \)[/tex] and [tex]\( C(6,3) \)[/tex].
The distance formula is:
[tex]\[ AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} \][/tex]
Substituting the coordinates of [tex]$A$[/tex] and [tex]$C$[/tex]:
[tex]\[ AC = \sqrt{(6 - 2)^2 + (3 - 0)^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]
### Checking for Isosceles Triangle
To check if [tex]$\triangle ABC$[/tex] is isosceles, we need to see if any two sides are of equal length.
We have:
[tex]\[ AB \approx 4.472, \quad BC \approx 2.236, \quad AC = 5 \][/tex]
Since none of the lengths are approximately equal, [tex]\(\triangle ABC\)[/tex] is not isosceles.
### Checking for Right Triangle
To check if [tex]$\triangle ABC$[/tex] is a right triangle, we need to use the Pythagorean theorem and verify if any combination of sides satisfies:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
Let's test the following combinations:
1. [tex]\(AB^2 + BC^2 \approx AC^2\)[/tex]:
[tex]\[ (4.472)^2 + (2.236)^2 \approx 5^2 \][/tex]
[tex]\[ 20 + 5 \approx 25 \quad \text{(True)} \][/tex]
Since [tex]\(AB^2 + BC^2 = AC^2\)[/tex], we confirm that [tex]\(\triangle ABC\)[/tex] is a right triangle.
Therefore, Verna is correct because [tex]\(\overline{BC} \perp \overline{AC}\)[/tex].
Answer: Verna, because [tex]\(\overline{BC} \perp \overline{AC}\)[/tex]
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