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drag the tiles to the boxes to form correct pairs. match each set of ve…

Question

drag the tiles to the boxes to form correct pairs. match each set of vertices to the triangle they form. right isosceles acute equilateral right scalene acute isosceles obtuse scalene a(3,5),b(5,6),c(3,0) a(2,4),b(3,5),c(2,6) a(2,4),b(4,5),c(3,6) a(3,5),b(3,4),c(5,4)

Explanation:

Step1: Recall distance - formula

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Also, for a right - triangle, the Pythagorean theorem $a^{2}+b^{2}=c^{2}$ holds, for an isosceles triangle two sides are equal, for an equilateral triangle all sides are equal, for an acute - triangle $a^{2}+b^{2}>c^{2}$ for all sides $a,b,c$ (where $c$ is the longest side), and for an obtuse - triangle $a^{2}+b^{2}

Step2: Calculate distances for $A(3,5),B(5,6),C(3,0)$

$AB=\sqrt{(5 - 3)^2+(6 - 5)^2}=\sqrt{4 + 1}=\sqrt{5}$; $BC=\sqrt{(3 - 5)^2+(0 - 6)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}$; $AC=\sqrt{(3 - 3)^2+(0 - 5)^2}=5$. Since $AB^{2}+AC^{2}=5 + 25=30
eq BC^{2}$, and no two sides are equal, and one angle is right (because $x_A=x_C$, so $AC$ is vertical and $AB$ has a non - vertical slope, so $\angle A = 90^{\circ}$), it is a right scalene triangle.

Step3: Calculate distances for $A(2,4),B(3,5),C(2,6)$

$AB=\sqrt{(3 - 2)^2+(5 - 4)^2}=\sqrt{1+1}=\sqrt{2}$; $BC=\sqrt{(2 - 3)^2+(6 - 5)^2}=\sqrt{1 + 1}=\sqrt{2}$; $AC=\sqrt{(2 - 2)^2+(6 - 4)^2}=2$. Since $AB = BC$ and $AB^{2}+BC^{2}=2 + 2=4=AC^{2}$, it is a right isosceles triangle.

Step4: Calculate distances for $A(2,4),B(4,5),C(3,6)$

$AB=\sqrt{(4 - 2)^2+(5 - 4)^2}=\sqrt{4 + 1}=\sqrt{5}$; $BC=\sqrt{(3 - 4)^2+(6 - 5)^2}=\sqrt{1+1}=\sqrt{2}$; $AC=\sqrt{(3 - 2)^2+(6 - 4)^2}=\sqrt{1 + 4}=\sqrt{5}$. Since $AB = AC$ and $AB^{2}+AC^{2}=5 + 5 = 10>BC^{2}$, it is an acute isosceles triangle.

Step5: Calculate distances for $A(3,5),B(3,4),C(5,4)$

$AB=\sqrt{(3 - 3)^2+(4 - 5)^2}=1$; $BC=\sqrt{(5 - 3)^2+(4 - 4)^2}=2$; $AC=\sqrt{(5 - 3)^2+(4 - 5)^2}=\sqrt{4 + 1}=\sqrt{5}$. Since $AB^{2}+BC^{2}=1 + 4 = 5=AC^{2}$, and no two sides are equal, it is a right scalene triangle.

Answer:

$A(3,5),B(5,6),C(3,0)\to$ right scalene
$A(2,4),B(3,5),C(2,6)\to$ right isosceles
$A(2,4),B(4,5),C(3,6)\to$ acute isosceles
$A(3,5),B(3,4),C(5,4)\to$ right scalene