Question
Area of a triangles with vertices at (2, 3), (-1, 0) and
(2, -4) is :Solution
Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃). The area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)] Let A(x₁, y₁) = (2, 3), B(x₂, y₂) = (- 1 , 0), and C(x₃, y₃) = (2, - 4) Area of a triangle is given by 1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)] By substituting the values of vertices, A, B, C in the formula. Area of the given triangle = 1/2 [2(0 - (-4)) + (-1)((- 4) - 3) + 2(3 - 0)] = 1/2 (8 + 7 + 6) = 21/2 square units
29% of 400 + 66% of 1100 - 50% of 1200 = ?
(47.5 ÷ 9.5) × (78.5 ÷ 15.7) + 475 = ? + 15% of 150
What will come in the place of question mark (?) in the given expression?
(17/27) of 162 + ?² = 632 - (73 - 12) X 5If x + y + 3 = 0, then find the value of x3 + y3 – 9xy + 9.
√324 + √576 = ?/ √9
1/6+ 999*53/54 ×9 = ?
22 + 60 × 3 ÷ 12 = ?
? = 60% of 2500 + 85% of 2000 – 5³
654.056 + 28.9015 × 44.851 – 43.129 = ?
4261 + 8234 + 2913 + 8217 + 6283 + 4172 =?