Question

Find the area of triangle ABC whose vertices are A(4, 3),
B(12, 5) and C(4, 6).
ans?
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Answer 1

Answer:

ar. of triangle= 1/2× |4 12 5 4|

|3 5 6 3|

hence on futher solving, ar.= 1/2× {20-36+72-25+15-24}

ar.= 70/2 =35

Answer 2

Question:

Find the area of ∆ABC whose vertices are A(4, 3), B(12, 5) and C(4, 6).

Formula Used:

$\boxed{\bf{Area\: of \:triangle = \frac{1}{2} [(x_1)(y_2 - y_3) + (x_2)(y_3 - y_1) + (x_3)(y_1 - y_2)]}}$

Solution:

$\sf{Area = \frac{1}{2} ((4)(5 - 6) + (12)(6 - 3) + (4)(3 - 5))}$

$\sf{Area=\frac{1}{2} ((4)( - 1) + (12)(3) + (4)( - 2))}$

$\sf{Area =\frac{1}{2} ( - 4 + 36 - 8)}$

$\sf{Area = \frac{1}{2} ( 24)}$

$\sf{Area = 12sq.units}$

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