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
54.8% of 800 - √(?) = 33.98% of 400 – 12.42% of 300
What approximate value will come in place of the question mark (?) in the following question?(Note: You are not expected to calculate the exact value.)<...
³√? `xx` 32.87 + 59.83 `xx` 28.7665 – 48.8745 `xx` 21.642 = 1085.344
{(1799.89 ÷ 8.18) ÷ 9.09 + 175.15} = 25.05% of ?
124% of 620.99 + 11.65% of 1279.23 = ?
3.98 × 29.67 ÷ 11.90 of √24.89 = ?% of 199.79
11.06 2 – 7.12 × 4.88 + 9.96 = 12.22 × ?
( 14.99% of 549.99 ) × 17.02 = ? 2 + 26.02 × 3200 ÷ 800
What approximate value will come in place of question (?) in the following given expression? You are not expected to calculate the exact value.
...? * 8.21 = (520.12 ÷ 12.98) % of 6800 - 1350.75
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