Question

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

Answer 1

Answer

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

Answer 2

Answer:

Joining BD, there are two triangles.

Area of quad ABCD = Ar △ABD + Ar △BCD

Ar △ABD = ½| (-5)(-5 - 5) + (-4)(5 - 7) + (4)(7 + 5) |

= 53 sq units

Ar △BCD = ½| (-4)(-6 - 5) + (-1)(5 + 5) + (4)(-5 + 6) |

= 19 sq units

Hence, area of quad ABCD = 53 + 19 = 72 sq units

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Explanation: