Question

Last question of the day!!

Answer 1

Answer:

Correct options are 2, 5 and 7.

Step-by-step explanation:

Consider the given vertices of triangle are A(-3,-3), B(-3,2) and C(1,2).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$AB=\sqrt{(-3-(-3))^2+(2-(-3))^2}$

$AB=\sqrt{(0)^2+(5)^2}$

$AB=\sqrt{25}$

$AB=5$

Similarly,

$BC=\sqrt{(1-(-3))^2+(2-2)^2}=4$

$AC=\sqrt{(1-(-3))^2+(2-(-3))^2}=\sqrt{16+25}=\sqrt{41}$

From the above calculation it is clear that AC>AB and AC>BC.

According to Pythagoras theorem, in a right angle triangle, the square of largest side is equal to the sum of squares of two small sides.

$hypotenuse^2=base^2+perpendicular^2$

$AC^2=(\sqrt{41})^2=41$

$AB^2+BC^2=(5)^2+4^2=24+16=41=AC^2$

So, given triangle is a right angle triangle and AC is its hypotenuse.

Therefore, the correct options are 2, 5 and 7.