Calculate the area of triangle ABC with altitude CD, given A (6,0) B(1,5) C (2,0) and D (4,2)

2 Answer

  • Answer:10 square units.

    Step-by-step explanation:

    There are three vertices in this triangle: \rm A, \rm B, and \rm C. The three sides are \rm AB, \rm BC, and \rm AC.

    Among the two endpoints of altitude \rm CD, only \rm C is a vertex of this triangle. Hence, \rm AB, the side opposite to vertex \rm C\!, would be the base of this altitude.

    Apply the Pythagorean Theorem to find the length of \rm AB (the base) and \rm CD (the height).

    By the Pythagorean Theorem, the distance between points (x_0,\, y_0) and (x_1,\, y_1) is \sqrt{(x_1 - x_0)^{2} + (y_1 - y_0)^{2}}.

    The distance between \rm C (2,\, 0) and \rm D (4,\, 2) is:

    \sqrt{(4 - 2)^{2} + (2 - 0)^{2}} = \sqrt{8} = 2\, \sqrt{2}.

    Hence, the length of altitude \rm CD would be 2\sqrt{2} units.

    Similarly, the length of side \rm AB would be:

    \sqrt{(6 - 1)^{2} + (0 - 5)^{2}} = \sqrt{50} = 5\, \sqrt{2}.

  • Answer:

    10 square units

    Step-by-step explanation:

    Hope this helps :)