1. Find the exact value for area of the triangle with the side lengths 5, 9, and 10.

Question

1. Find the exact value for area of the triangle with the side lengths 5, 9, and 10.

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Eva 1 month 2021-10-20T20:47:55+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-20T20:49:08+00:00

    Answer:

    Step-by-step explanation:

    a = 5 units ; b = 9units ; c = 10 units

    Semi perimeter = (a+b+c)/2 = (5+9+10)/2 = 24/2 = 12 units

    s-a = 12 -5 = 7 units

    s -b = 12 – 9 = 3 units

    s-c = 12- 10 =2 units

    Area=\sqrt{s*(s-a)(s-b)(s-c)}\\\\=\sqrt{12*7*3*2}\\\\=\sqrt{2*2*3*7*3*2}\\\\=2*3\sqrt{2*7}\\\\=6\sqrt{14}\\

    =6*3.742 = 22.452

    = 22.45 square units

    0
    2021-10-20T20:49:15+00:00

    Answer:

    area = 6\sqrt{14}  units²

    Step-by-step explanation:

    Given the 3 sides of a triangle, we can use Hero’s formula to calculate the area (A)

    A = \sqrt{s(s-a)(s-b(s-c)}

    where s s the semi perimeter and a, b , c the sides of the triangle

    let a = 5, b = 9 and c = 10

    s = \frac{a+b+c}{2} = \frac{5+9+10}{2} = \frac{24}{2} = 12

    A = \sqrt{12(12-5)(12-9)(12-10)}

       = \sqrt{12(7)(3)(2)}

       = \sqrt{504}

       = \sqrt{36(14)}

       =  6\sqrt{14} units²

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45:7+7-4:2-5:5*4+35:2 =? ( )