Create a random point D on circle AC. Construct the radii AC and AD . Determine the intersection of AC with circle AB, and label this point E. Determine the intersection of AD with circle AB, and label this point F. Record the coordinates of points E and F on circle AB.

b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)