The points A (4,-2) B(7,2)C(0,9)D(-3,5) from a parallelogram . Find the length of the altitude on the base AB.

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Solution

Let the height of parallelogram taking AB as base be h.

Now, AB = √[(7-4)2 + (2+2)2]

AB = √(32 + 42) = 5 units

Area (Δ ABC) = ½[4(2 - 9) + 7(9 + 2) + 0(-2 - 2)]

Area (Δ ABC) = 49/2 sq units

Now, ½ × AB × h = 49/2

½ × 5 × h = 49/2

h = 49/5

h = 9.8 units

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The points A (4,-2) B(7,2)C(0,9)D(-3,5) from a parallelogram . Find the length of the altitude on the base AB.

Let the height of parallelogram taking AB as base be h. Now, AB = √[(7-4)2 + (2+2)2] AB = √(32 + 42) = 5 units Area (Δ ABC) = ½[4(2 - 9) + 7(9 + 2) + 0(-2 - 2)] Area (Δ ABC) = 49/2 sq units Now, ½ × AB × h = 49/2 ½ × 5 × h = 49/2 h = 49/5 h = 9.8 units

Let the height of parallelogram taking AB as base be h.

Now, AB = √[(7-4)2 + (2+2)2]

AB = √(32 + 42) = 5 units

Area (Δ ABC) = ½[4(2 - 9) + 7(9 + 2) + 0(-2 - 2)]

Area (Δ ABC) = 49/2 sq units

Now, ½ × AB × h = 49/2

½ × 5 × h = 49/2

h = 49/5

h = 9.8 units