Problem 1 :
Find the area of a triangle whose vertices are (3, 0), (7, 0) and (8, 4).
Solution:
The vertices of the given triangle A(3, 0), B(7, 0) and C(8, 4).
x_{1} = 3, y_{1} = 0, x_{2} = 7, y_{2} = 0, x_{3} = 8, y_{3} = 4
Area of a triangle
Area can't be negative.
Therefore, the area of the triangle is 20 square units.
Problem 2 :
The area of a triangle whose vertices are (5, 0), (8, 0) and (8, 4) (in sq.units) is
A) 20 B) 12 C) 6
Solution:
The vertices of the given triangle A(5, 0), B(8, 0) and C(8, 4)
x_{1} = 5, y_{1} = 0, x_{2} = 8, y_{2} = 0, x_{3} = 8, y_{3} = 4
Area of a triangle
So, option (C) is correct.
Problem 3 :
The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is (7/2, y), find the value of y.
Solution:
Let A(2, 1), B(3, -2) and C(7/2, y)
x_{1} = 2, y_{1} = 1, x_{2} = 3, y_{2} = -2, x_{3} = 7/2, y_{3} = y
Area of a triangle
Problem 4 :
Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.
Solution:
The vertices of the given triangle A(1, -1), B(-4, 2k) and C(-k, -5)
x_{1} = 1, y_{1} = -1, x_{2} = -4, y_{2} = 2k, x_{3} = -k, y_{3} = -5
Area of a triangle
Problem 5 :
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).
Solution:
The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).
Hence, the area of the required triangle is 1 sq. unit.
Problem 6 :
For what type of k, (k > 0), is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) to 53 sq. units?
Solution:
The vertices of the given triangle A(-2, 5), B(k, -4) and C(2k + 1, 10)
x_{1} = -2, y_{1} = 5, x_{2} = k, y_{2} = -4, x_{3} = 2k + 1, y_{3} = 10
Area of a triangle
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