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Find the area of a triangle ABC

If AC=9CM, AB=8CM, CB=10cm
Give your answer to one decimal place

User Camp
by
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1 Answer

3 votes

Answer:

Area = 34.2 cm²

Explanation:

The general formula of the area of the triangle is half the product of two sides multiplied by the sine the angle between them.

So, for the given triangle ABC

AC = 9 cm, AB = 8 cm, CB = 10 cm

Area = 0.5 AC * AB * sin A or 0.5 BC * BA * sin B or = 0.5 CA * CB * sin C

Using the first form, so we need the measure of angle A

Using cosine low: cos A = (b² + c² - a²)/(2bc)

Where: a = BC = 10 , b = AC = 9 and c = AB = 8

So. cos A = (9² + 8² - 10²)/(2 * 9 * 8 ) = 0.3125

∠A = cos⁻¹0.3125 = 71.79°

So, Area = 0.5 AC * AB * sin A = 0.5 * 9 * 8 * sin 71.79° = 34.2 cm²

So, Area = 34.2 cm² to the nearest one decimal place

User Walmik
by
4.9k points