Question

In ΔWXY, w = 4.8 cm, x = 1.5 cm and ∠Y=52°. Find the area of ΔWXY, to the nearest 10th of a square centimeter.

Answer 1

Area = 3.0 square cm

Explanation:

By cosine formula,

WX² = WY² + XY² - 2(WY)(XY)cos(∠W)

WX² = (1.5)² + (4.8)² - 2(1.5)(4.8)cos(52°)

WX² = 2.25 + 23.04 - 8.866

WX = √(16.424)

WX = 4.05 cm

Since, area of a triangle with known sides =
`√(s(s-a)(s-b)(s-c))`

Here, s =
`(a+b+c)/(2)`

And a, b, c are the sides of the triangle

By this formula,

s =
`(1.5+4.8+4.05)/(2)`

s = 5.2

Therefore, area of the triangle XYZ =
`√(5.2(5.2-1.5)(5.2-4.8)(5.2-4.05))`

=
`√(5.2(3.7)(0.4)(1.15))`

=
`√(8.85)`

= 2.97

≈ 3.0 square cm