Which of these triangles will be formed using the points $(a, a), (- a, - a)$ and $(- \sqrt{3} a, \sqrt{3} a)$ ?

Right triangle

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Scalene triangle

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Isosceles triangle

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Equilateral triangle

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D Equilateral triangle

$A B = \sqrt{(a - (- a))^{2} + (a - (- a))^{2}}$
$= \sqrt{(2 a)^{2} + (2 a)^{2}}$
$= \sqrt{8 a^{2}}$
$A B = 2 \sqrt{2} a$ units

$B C = \sqrt{(- a - (- \sqrt{3} a))^{2} + (- a - \sqrt{3} a)^{2}}$
$= \sqrt{a^{2} + 3 a^{2} - 2 \sqrt{3} a^{2} + a^{2} + 3 a^{2} + 2 \sqrt{3} a^{2}}$
$B C = \sqrt{8 a^{2}} = 2 \sqrt{2} a$ units

$A C = \sqrt{(a - (- \sqrt{3} a))^{2} + (a - \sqrt{3} a)^{2}}$
$= \sqrt{a^{2} + 3 a^{2} + 2 \sqrt{3} a^{2} + a^{2} + 3 a^{2} - 2 \sqrt{3} a^{2}}$
$A C = \sqrt{8 a^{2}} = 2 \sqrt{2} a$ units

$∴ A B = B C = A C = 2 \sqrt{2} a$ units
$∴ Δ A B C$ is an equilateral triangle.
So, option (d) is correct

Suggest Corrections

Similar questions

(- \sqrt{3} a, \sqrt{3} a)

The points $(a, a), (- a, - a)$ and $(- \sqrt{3} a, \sqrt{3} a)$ are the vertices of a/an _____ triangle.

The points $(a, a)$ , $(- a, - a)$ and $(- \sqrt{3} a, \sqrt{3} a)$ form the vertices of an

The points (a, a) (–a, -a) and $(- \sqrt{3} a, \sqrt{3} a)$ form the vertices of an

((a + 3) (a + 4), a + 3), ((a + 2) (a + 3), (a + 2)) and ((a + 1) (a + 2), (a + 1))

Join BYJU'S Learning Program