QUESTION IMAGE
Question
drag the tiles to the boxes to form correct pairs. match each set of vertices to the triangle they form. a(3,5),b(3,4),c(5,4) a(2,4),b(4,5),c(3,6) a(2,4),b(3,5),c(2,6) a(3,5),b(5,6),c(3,0) reset next obtuse scalene acute equilateral acute isosceles right isosceles right scalene
Step1: Recall distance - formula
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
Step2: For the first set of vertices $A(3,5),B(3,4),C(5,4)$
Calculate side - lengths:
$AB=\sqrt{(3 - 3)^2+(4 - 5)^2}=1$;
$BC=\sqrt{(5 - 3)^2+(4 - 4)^2}=2$;
$AC=\sqrt{(5 - 3)^2+(4 - 5)^2}=\sqrt{4 + 1}=\sqrt{5}$. Since $AB^2+BC^2=1 + 4 = 5=AC^2$ and all side - lengths are different, it is a right scalene triangle.
Step3: For the second set of vertices $A(2,4),B(4,5),C(3,6)$
$AB=\sqrt{(4 - 2)^2+(5 - 4)^2}=\sqrt{4 + 1}=\sqrt{5}$;
$BC=\sqrt{(3 - 4)^2+(6 - 5)^2}=\sqrt{1+1}=\sqrt{2}$;
$AC=\sqrt{(3 - 2)^2+(6 - 4)^2}=\sqrt{1 + 4}=\sqrt{5}$. Since two sides are equal ($AB = AC$) and all angles are acute (by using the cosine - law $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}>0$ for all angles), it is an acute isosceles triangle.
Step4: For the third set of vertices $A(2,4),B(3,5),C(2,6)$
$AB=\sqrt{(3 - 2)^2+(5 - 4)^2}=\sqrt{1 + 1}=\sqrt{2}$;
$BC=\sqrt{(2 - 3)^2+(6 - 5)^2}=\sqrt{1+1}=\sqrt{2}$;
$AC=\sqrt{(2 - 2)^2+(6 - 4)^2}=2$. Since $AB^{2}+BC^{2}=2 + 2=4=AC^{2}$ and two sides are equal, it is a right isosceles triangle.
Step5: For the fourth set of vertices $A(3,5),B(5,6),C(3,0)$
$AB=\sqrt{(5 - 3)^2+(6 - 5)^2}=\sqrt{4 + 1}=\sqrt{5}$;
$BC=\sqrt{(3 - 5)^2+(0 - 6)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}$;
$AC=\sqrt{(3 - 3)^2+(0 - 5)^2}=5$. Using the cosine - law, find the largest angle. Let $a = 2\sqrt{10}$, $b=\sqrt{5}$, $c = 5$. $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{5 + 25-40}{10\sqrt{5}}<0$, so the triangle is obtuse scalene.
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$A(3,5),B(3,4),C(5,4)$ - right scalene
$A(2,4),B(4,5),C(3,6)$ - acute isosceles
$A(2,4),B(3,5),C(2,6)$ - right isosceles
$A(3,5),B(5,6),C(3,0)$ - obtuse scalene