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Hint: To calculate the area of triangle, we use the formula,

Area = \[\dfrac{1}{2}\left[ {{x}_{1}}~\left( {{y}_{2-}}~{{y}_{3}}~ \right)+{{x}_{2}}~\left( {{y}_{3}}-{{y}_{1}}~ \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]-- (1)

Where, $({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$are the coordinates of A, B and C.

Complete step-by-step answer:

We can calculate the area by using the area formula. One should however remember that since area is a positive quantity, we take the absolute value of the area we get from formula (1).

Thus, we get,

Area = \[\dfrac{1}{2}\left[ {{x}_{1}}~\left( {{y}_{2-}}~{{y}_{3}}~ \right)+{{x}_{2}}~\left( {{y}_{3}}-{{y}_{1}}~ \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]

Now, we know the coordinates of A, B and C which are (-5, 7), (-4, -5) and (4, 5) respectively. These are put in place of $({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$. Here we have taken the order of$({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$ as A, B and C. However, one can also take any other order (say B, C and A). We should however be consistent while inserting the values. Now, we start solving,

Area = $\dfrac{1}{2}$[-5(-5-5) + {-4(5-7)} + 4{7-(-5)}]

Area = $\dfrac{1}{2}$[50+8+48]

Area = 53 square units.

Hence, the area of triangle ABC is 53 square units.

Note: While calculating the area of the triangle when the Cartesian coordinates are given, one can also proceed by first plotting the triangle on X-Y graph. This can help in identifying if the triangle is an equilateral triangle, isosceles triangle or right triangle. If we can identify that the triangle is one of them, we can easily calculate the area of the triangle, by using the respective formulas for these special types of triangles. This greatly reduces the time taken in calculating the area of the triangle. In case, the triangle is none of the above types of triangles, we can always use the normal formula in Cartesian coordinates to calculate the area.

Area = \[\dfrac{1}{2}\left[ {{x}_{1}}~\left( {{y}_{2-}}~{{y}_{3}}~ \right)+{{x}_{2}}~\left( {{y}_{3}}-{{y}_{1}}~ \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]-- (1)

Where, $({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$are the coordinates of A, B and C.

Complete step-by-step answer:

We can calculate the area by using the area formula. One should however remember that since area is a positive quantity, we take the absolute value of the area we get from formula (1).

Thus, we get,

Area = \[\dfrac{1}{2}\left[ {{x}_{1}}~\left( {{y}_{2-}}~{{y}_{3}}~ \right)+{{x}_{2}}~\left( {{y}_{3}}-{{y}_{1}}~ \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]

Now, we know the coordinates of A, B and C which are (-5, 7), (-4, -5) and (4, 5) respectively. These are put in place of $({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$. Here we have taken the order of$({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$ as A, B and C. However, one can also take any other order (say B, C and A). We should however be consistent while inserting the values. Now, we start solving,

Area = $\dfrac{1}{2}$[-5(-5-5) + {-4(5-7)} + 4{7-(-5)}]

Area = $\dfrac{1}{2}$[50+8+48]

Area = 53 square units.

Hence, the area of triangle ABC is 53 square units.

Note: While calculating the area of the triangle when the Cartesian coordinates are given, one can also proceed by first plotting the triangle on X-Y graph. This can help in identifying if the triangle is an equilateral triangle, isosceles triangle or right triangle. If we can identify that the triangle is one of them, we can easily calculate the area of the triangle, by using the respective formulas for these special types of triangles. This greatly reduces the time taken in calculating the area of the triangle. In case, the triangle is none of the above types of triangles, we can always use the normal formula in Cartesian coordinates to calculate the area.