Question

The diagram shows a logo.

Answer 1

Answer:

Step-by-step explanation:

From the picture attached,

Area of ΔECD = $\frac{1}{2}(\text{Base})(\text{Height})$    

                       = $\frac{1}{2}(EF)(CD)$

From ΔEFD,

sin(32°) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

sin(32°) = $\frac{EF}{ED}$

EF = 14 × sin(32°)

    = 7.42 cm

By cosine rule,

EC² = DE² + CD² - 2(DE)(CD)cos(32°)

EC² = 14² + 27² - 2(14)(27)cos(32°)

EC² = 196 + 729 - 641.12

EC² = 283.88

EC = 16.85 cm

Area of ΔECD = Area of ΔAEB = $\frac{1}{2}(7.42)(27)$

                                                   = 100.17

Area of ΔECD + Area of ΔAEB = 2(100.17)

                                                   = 200.34 cm²

Area of sector BEC = $\frac{\theta}{360}(\pi r^{2})$

Here, θ = central angle of the sector

Area of sector BEC = $\frac{105}{360}(\pi)( EC)^{2}$

                                = $\frac{105\pi}{360}(16.85)^2$

                                = 260.16 cm²

Area of the logo = Area of triangles AEB + Area of triangle ECD + Area of sector BEC

= 200.34 + 260.16

= 460.50

≈ 460 cm²