In △ ABC, ab = AC = 13, BC = 10, then the area of △ ABC is () A. 30B. 60C. 65D. 120

In △ ABC, ab = AC = 13, BC = 10, then the area of △ ABC is () A. 30B. 60C. 65D. 120

As shown in the figure, make ad ⊥ BC at point D, then BD = 12bc = 5. In RT △ abd, ∵ ad2 = ab2-bd2, ∵ ad = 132 − 52 = 12, ∵ ABC area = 12bc · ad = 12 × 10 × 12 = 60

In RT triangle ABC, the angle c = 90 degrees, if AC: ab = 3:5, BC = 8, the area of △ ABC is obtained

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In the triangle ABC, if AB = √ 3, AC = 4, ∠ a = 60 °, the area of triangle ABC is

Area of triangle ABC = (1 / 2) AB * ac * sin ∠ a = (1 / 2) * √ 3 * 4 * (√ 3 / 2) = 3

In the triangle ABC, if the angle B = 60 degrees, AC = 3, ab = root 6, then the area of the triangle ABC

Sine theorem AB / sinc = AC / SINB
sinC=AB×sinB/AC
=√6×√3/2×1/3
=√2/2
C = 45 ° C = 135 ° (rounding off)
A=75°
The height on the edge of ABC, AC = ab × Sina
=AB×sin75°
=√6×(√6+√2)/4
=(3+√3)/2
S △ ABC = (1 / 2) AC × H
=1/2×3×(3+√3)/2
=（9+3√3）/4