Question

Find the perimeter of triangle formed by the points (0 0) (1 0) (0 1)

Answer 1

Answer:

${\pink{\green{\sf{\therefore Perimeter\:of\:triangle=2+\sqrt{2}\:units}}}}$

Step-by-step explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

• In the given question information given about vertices of triangle.

• We have to find the perimeter formed by joining these co-ordinates.

$\underline \bold{Given : } \\ \bold{Vertices \: of \: triangle } \\ \implies First \: vertices(A) = (0,0) \\ \implies Second \: vertices(B) = (1,0) \\ \implies Third \: vertices(C) = (0,1) \\ \\ \underline \bold{To \: Find : } \\ \implies perimeter \: of \: triangle = ?$

• According to given question :

$\bold{ According \: to \: perimeter \: of \: triangle \ }$

$\bold{in \: coordinate \: geometry}$

$\bold{By\:distance\:formula}$ $\implies AB=\sqrt{({0-1}^{2})+(0-0)^{2}}$

$\implies AB=\sqrt{1}$

$\bold{\implies AB=1\:units}$$\implies BC=\sqrt{(1-0)^{2}+(0-1)^{2}}$

$\implies BC=\sqrt{1+1}$

$\bold{\implies BC=\sqrt{2}\:units}$

$\implies CA=\sqrt{(0-0)^{2}+(1-0)^{2}}$

$\implies CA=\sqrt{1}$

$\bold{\implies CA=1\:units}$ $\bold{Perimeter\:of\:triangle=AB+BC+CA}$

$\implies Perimeter=1+\sqrt{2}+1$

$\bold{\implies perimter=2+\sqrt{2}}$

$\bold{\therefore Perimeter\:of\:triangle=2+\sqrt{2}\:units}$

Answer 2

ANSWER:-

Given:

A triangle formed by points (0,0),(1,0),(0,1).

To find:

Find the perimeter of ∆?

Solution:

Vertices of triangle;

⚫A (0,0)

⚫B (1,0)

⚫C (0,1)

According to this question:

Let the points be A(0,0), B(1,0)& C(0,1)

⚫The distance between two points always taken as positive.

Using the formula of distance between two points:

=) AB= √(0-1)² + (0-0)²

=) AB = √1 + 0

=) AB =√1

=) AB= 1

BC= √(1-0)² + (0-1)²

=) BC= √1 + 1

=) BC = √2

CA = √(0-0)² +(1-0)²

=) CA= √0 + 1

=) CA = √1

=) CA = 1

Therefore,

We know that perimeter of triangle is;

Side + side + side

=) AB + BC + AC

=) 1 + √2 + 1

=) 2+ √2

Hence,

The perimeter of ∆ is 2+√2.

Hope it helps ☺️