A) right angled
B) isosceles
C) equilateral
D) right angled isosceles
Correct Answer: D
Solution :
Vertices of\[\Delta \]ABC are \[A\equiv (0,7,10),\text{ }B=(-1,6,6)\]and \[C\equiv (-4,9,6)\] \[\therefore \] \[AB=\sqrt{{{(0+1)}^{2}}+{{(7-6)}^{2}}+{{(10-6)}^{2}}}\] \[=\sqrt{1+1+16}=\sqrt{18}\] \[BC=\sqrt{{{(-1+4)}^{2}}+{{(6-9)}^{2}}+{{(6-6)}^{2}}}\] \[=\sqrt{9+9+0}=\sqrt{18}\] \[CA=\sqrt{{{(0+4)}^{2}}+{{(7-9)}^{2}}+{{(10-6)}^{2}}}\] \[=\sqrt{16+4+16}\] \[=\sqrt{36}\] \[\because \] \[AB=BC\]and \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] \[\therefore \]Triangle is right angled isosceles.
You need to login to perform this action.
You will be redirected in
3 sec
