Question

15. Find the area of triangle whose vertices are (2, 3), (– 1, 0) and (2, – 4).​

Answer 1

Answer :

• Area of the triangle is 10.5cm².

Given :

• Vertices of a triangle are (2 , 3), (-1 , 0), (2 , -4).

To Find :

• Area of the triangle.

Solution :

Let

ABC be the triangle in which

1. A = (2 , 3)
2. B = (-1 , 0)
3. C = (2 , -4)

Here

• x1 = 2
• x2 = -1
• x3 = 2
• y1 = 3
• y2 = 0
• y3 = -4

As we know that

Area of a triangle is

• ½ [x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)

Put the values in the formula

Now,

According to question :

→ 1/2 [ 2 {0 - (-4)} + (-1) {(-4) - 3} + 2 {3 - 0) ]

→ 1/2 [ 2 × 4 + (-1) × (-7) + 2 × 3 ]

→ 1/2 [ 8 + 7 + 6 ]

→ 1/2 × 21

→ 21/2

→ 10.5cm²

Hence, the area of the triangle is 10.5cm².

Answer 2

Given

• Vertices of a triangle are (2,3), (-1,0) and (2,-4).

⠀

To find

• Area of the given triangle.

⠀

Solution

• We have three vertices of the triangle. Now, we have to find the area of the triangle.

⠀

⠀⠀⠀⠀❍ We know that

$\small\boxed{\tt{\bigstar{Area_{(Triangle)} = \dfrac{1}{2} \bigg| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigg|\: \: {\bigstar}}}}$

⠀

Here

• $\sf{x_1 = 2\: and\: y_1 = 3}$
• $\sf{x_2 = -1\: and\: y_2 = 0}$
• $\sf{x_3 = 2\: and\: y_3 = -4}$

⠀

★ Putting the values

$\tt\longmapsto{Area = \dfrac{1}{2} \bigg| 2(0 + 4) + (-1)(-4 - 3) + 2(3 - 0) \bigg|}$

⠀

$\tt\longmapsto{Area = \dfrac{1}{2} \bigg| 2(4) + (-1)(-7) + 2(3) \bigg|}$

⠀

$\tt\longmapsto{Area = \dfrac{1}{2} \bigg| 8 + 7 + 6 \bigg|}$

⠀

$\tt\longmapsto{Area = \dfrac{1}{2} \bigg| 21 \bigg|}$

⠀

$\tt\longmapsto{Area = \dfrac{1}{2} \times 21}$

⠀

$\bf\longmapsto{Area = \dfrac{21}{2}}$

⠀

Hence,

• The area of the triangle is $\sf{\dfrac{21}{2}}$ square units.

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