## AREA OF TRIANGLE WORKSHEETS

Area of Triangle Worksheet :

Worksheet given in this section will be much useful for the students who would like to practice problems on finding area of a triangle when its vertices are given.

## Area of Triangle Worksheet - Problems

Problem 1 :

Find the area of triangle whose vertices are (6, 7), (2, -9) and (-4, 1).

Problem 2 :

Find the area of triangle whose vertices are (3, 4), (2, -1) and (4, -6).

Problem 3 :

Find the area of triangle whose vertices are (5, 6), (2, 4) and (1, -3).

Problem 4 :

Find the area of triangle whose vertices are (1, 3), (-7, 6) and (5, -1).

Problem 5 :

Find the area of triangle whose vertices are (1, 1), (3, 4) and (5, -2).

Problem 6 :

Find the area of triangle whose vertices are (-3, -9), (-1, 6) and (3, 9).

Problem 7 :

Find the area of triangle whose vertices are (-3, -9), (3, 9) and (5, -8).

Problem 8 :

Find the area of triangle whose vertices are (4, 5), (4, 2) and (-2, 2).

Problem 9 :

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

Problem 10 :

Find the area of triangle whose vertices are (3, 1) (0, 4) and (-3, 1). ## Area of Triangle Worksheet - Solutions

Problem 1 :

Find the area of triangle whose vertices are (6, 7), (2, -9) and (-4, 1).

Solution : Now we have to take anticlockwise direction. So, we have to take the points in the order A (6,7) C (-4,1) and B (2,-9)

x₁ = 6     x₂ = -4      x₃ = 2

y₁ = 7     y₂ = 1       y₃ = -9 Area of the triangle ACB

=  (1/2) {(6 + 36 + 14) - (-28 + 2 - 54)}

=  (1/2) {56 - ( -82 + 2)}

=  (1/2) {56 - (-80) }

=  (1/2) {56 + 80)}

=  (1/2)  x 136

=  68 Square units.

So, the area of triangle ACB is  68 square units.

Problem 2 :

Find the area of triangle whose vertices are (3, 4), (2, -1) and (4, -6).

Solution : Now we have to take anticlockwise direction. So, we have to take the points in the order A (6,7) C (-4,1) and B (2,-9)

x₁ = 3     x₂ = 2         x₃ = 4

y₁ = 4     y₂ = -1        y₃ = -6 Area of the triangle ACB

=  (1/2) {(-3 - 12 + 16) - (8 - 4 - 18)}

=  (1/2) {(-15 + 16) - (8 - 22)}

=  (1/2) {1 - (-14) }

=  (1/2) x (1 + 14)

=  (1/2)  x 15

=  15/2

=  7.5

So, the area of triangle ABC is 7.5 square units.

Problem 3 :

Find the area of triangle whose vertices are (5, 6), (2, 4) and (1, -3).

Solution : Now we have to take anticlockwise direction. So we have to take the points in the order A (5,6) B (2,4) and C (1,-3)

x₁ = 5     x₂ = 2        x₃ = 1

y₁ = 6     y₂ = 4       y₃ = -3 Area of the triangle ABC

=  (1/2) {(20 - 6 + 6) - (12 + 4 - 15)}

=  (1/2) {20 - (16 - 15)}

=  (1/2) {20 - (1) }

=  (1/2) {20 - 1}

=  (1/2)  x 19

=  19/2

9.5

So, the area of triangle ABC is 9.5 square units.

Problem 4 :

Find the area of triangle whose vertices are (1, 3), (-7, 6) and (5, -1).

Solution : Now we have to take anticlockwise direction. So we have to take the points in the order A (1,3) B (-7,6) and C (5,-1)

x₁ = 1     x₂ = -7      x₃ = 5

y₁ = 3     y₂ = 6       y₃ = -1 Area of the triangle ACB

=  (1/2) {(6 + 7 + 15) - (-21 + 30 - 1)}

=  (1/2) {28 - ( -22 + 30)}

=  (1/2) {28 - (8) }

=  (1/2) {28 - 8}

=  (1/2)  x 20

=  10 Square units.

So, the area of triangle ABC is 10 square units.

Problem 5 :

Find the area of triangle whose vertices are (1, 1), (3, 4) and (5, -2).

Solution : Now we have to take anticlockwise direction. So we have to take the points in the order B (3,3) A (1, 1) and C (5,-2)

x₁ = 3     x₂ = 1      x₃ = 5

y₁ = 3     y₂ = 1       y₃ = -2 Area of the triangle BAC

=  (1/2) {(4 - 6 + 5) - (3 + 20 - 2)}

=  (1/2) {(9 - 6) - (23 - 2)}

=  (1/2) {3 - 21}

=  (1/2) {-18}

=  -18/2

=  9 Square units.

So, the area of triangle ABC is 9 square units.

Problem 6 :

Find the area of triangle whose vertices are (-3, -9), (-1, 6) and (3, 9).

Solution : Now we have to take anticlockwise direction. So we have to take the points in the order C (3,9) B (-1,6) and A (-3,-9)

x₁ = 3     x₂ = -1      x₃ = -3

y₁ = 9     y₂ = 6      y₃ = -9 Area of the triangle CBA

=  (1/2) {(18 + 9 - 18) - (-9 - 18 - 27)}

=  (1/2) {9 - ( -54)}

=  (1/2) {9 + 54}

=  (1/2) (63)

=  (63/2)

=  31.5 Square units.

So, the area of triangle CBA is 31.5 square units.

Problem 7 :

Find the area of triangle whose vertices are (-3, -9), (3, 9) and (5, -8).

Solution : Now we have to take anticlockwise direction. So we have to take the points in the order B (3,9) A (-3,-9) and C (5,-8)

x₁ = 3     x₂ = -3      x₃ = 5

y₁ = 9     y₂ = -9      y₃ = -8 Area of the triangle BAC

=  (1/2) {(-27 - 24 + 45) - (-27 - 45 - 24)}

=  (1/2) {(-51 + 45) - (-96)}

=  (1/2) {-6 + 96}

=  (1/2) (90)

=  (90/2)

=  45 Square units.

So, the area of triangle BAC is 45 square units.

Problem 8 :

Find the area of triangle whose vertices are (4, 5), (4, 2) and (-2, 2).

Solution : Now we have to take anticlockwise direction. So we have to take the points in the order A (4,5) C (-2,2) and B (4,2)

x₁ = 4     x₂ = -2      x₃ = 4

y₁ = 5     y₂ = 2       y₃ = 2 Area of the triangle ACB

=  (1/2) {(8 - 4 + 20) - (-10 + 8 + 8)}

=  (1/2) {(28 - 4) - ( -10 + 16)}

=  (1/2) (24 - 6)

=  (1/2) x 18

=  (18/2)

=  9 Square units.

So, the area of triangle ACB is 9 square units.

Problem 9 :

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

Solution : Now we have to take anticlockwise direction. So we have to take the points in the order B (2,2) C (2,0) and A (3,1)

x₁ = 2     x₂ = 2       x₃ = 3

y₁ = 2     y₂ = 0       y₃ = 1 Area of the triangle BCA

=  (1/2) {(0 + 2 + 6) - (4 + 0 + 2)}

=  (1/2) {8 - 6}

=  (1/2) (2)

=  (2/2)

=  1 Square units.

So, the area of triangle ACB is 1 square units.

Problem 10 :

Find the area of triangle whose vertices are (3, 1) (0, 4) and (-3, 1).

Solution : Now we have to take anticlockwise direction. So we have to take the points in the order B (0,4) C (-3,1) and A (3,1)

x₁ = 0     x₂ = -3      x₃ = 3

y₁ = 4     y₂ = 1       y₃ = 1 Area of the triangle BCA

=  (1/2) {(0 -3 + 12) - (-12 + 3 + 0)}

=  (1/2) {9 - ( -12 + 3)}

=  (1/2) {9 - (-9) }

=  (1/2) {9 + 9)}

=  (1/2)  x 18

=  9 Square units.

So, the area of triangle BCA is 9 square units. After having gone through the stuff given above, we hope that the students would have understood how to find area of a triangle when its vertices are given.

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