Question

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Solve it very fast.

11. The perimeter of the triangle with vertices (0. 4),(10. 0) and (3. 0) is units.
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Answer 1

Answer:

23

Step-by-step explanation:

Given that the vertices of the triangle are (0, 4), (10, 0) and (3, 0). We need to find out the perimeter of the triangle.

Let's say that the triangle is ABC having AB (0, 4), BC (10, 0) and CA (3, 0).

Perimeter of triangle = Sum of it's all sides = AB + BC + CA

Now,

Distance between the points is given by (d) = √[(x2 - x1)² + (y2 - y1)²]

Distance between A and B = √[(10 - 0)² + (0 - 4)²]

Used identity: (a + b)² = a² + b² - 2ab

→ d = √(100 + 16)

→ d = √116

→ d = 10.77

Distance between B and C = √[(3 - 10)² + (0 - 0)²]

→ d = √(9 + 100 - 60)

→ d = √49

→ d = 7

Distance between C and A = √[(0 - 3)² + (4 - 0)²]

→ d = √(9 + 16)

→ d = √25

→ d = 5

Perimeter of triangle = 10.77 + 7 + 5

= 22.77

= 23 (approx.)

Therefore, the perimeter of triangle is 23.

Answer 2

Answer:

First find length of each sides of ∆

Let A( 4 , 0) B(0, 0) and C (0 , 3) be the vertices of the triangle.

Using distance formula we have

AB =

$\sqrt{4 ^{2} + 0} = 4$

$BC = \sqrt{0 \times {3}^{2} } = 3$

$CA = \sqrt{ {4}^{2} + {3}^{2} } = 5$

Clearly adding AB + BC + CA = perimeter of ∆

⇒ Perimeter of ∆ ABC = 3 + 4 + 5= 12 units

Step-by-step explanation:

hope it helps (~‾▿‾)~