Question

The diagram shows a logo.
ABE and DCE are congruent triangles.
BCE is a sector of a circle, centre E.
Show that the area of the logo is 460 cm?
to 2 significant figures.
B В.
C
+
105°
E
27 cm
14 cm
32°
A
D

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

Answer 2

The area of the logo will be 460 square cm.

What is Geometry?

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

The diagram shows a logo.

ABE and DCE are congruent triangles.

BCE is a sector of a circle, center E.

Then the area of the logo will be

The radius of the sector will be

r² = 27² + 14² - 2 x 27 x 14 x cos 32°

r = 16.85 cm

Then the height of the triangle will be

h = 14 x sin 32°

h = 7.42 cm

Then the area of the geometry will be

Area = area of sector + 2 x area of triangle

Area = (105°/360°) x π x 16.85² + 2 x 1/2 x 27 x 7.42

Area = 260.16 + 200.30

Area = 460.46

Area ≅ 460

More about the geometry link is given below.

https://brainly.com/question/7558603

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