Question

A (0, 5), B (3, 0) and C (3, 5) are vertices of a/an .........
triangle.​

Answer 1

The vertices of A(0, 5), B(3, 0) and C(3, 5) are "an acute angled triange".

Step-by-step explanation:

Given, A(0, 5), B(3, 0) and C(3, 5) are the vertices of a triangle.

$AB^{2} = (3 - 0)^{2}$ + (0 - 5)^{2}[/tex]  

= ( 3)^{2}[/tex] + (- 5 )^{2}[/tex]  

= 9 + 25 = 41

$BC^{2} = (3 - 3)^{2}$ + (5 - 0)^{2}[/tex]  

= (0 )^{2}[/tex] + (25 )^{2}[/tex]  

= 0 + 25 = 25

$CA^{2} = (3 - 0)^{2}$ + (5 - 5)^{2}[/tex]  

= (3)^{2}[/tex] + (0)^{2}[/tex]  

= 9 + 0 = 9

∴ [tex]AB^{2} >[tex]BC^{2} + [tex]CA^{2}

⇒ 41 > 25 + 9

⇒ 41 > 34, it is an acute angled triangle.

Hence, the vertices of A(0, 5), B(3, 0) and C(3, 5) are "an acute angled triange".

Answer 2

Answer:

RIGHT ANGLED OR SCALENE TRIANGLE

Step-by-step explanation:

I HOPE IT HELP YOU GYUS

BY GUYS