Kite ABCD is inscribed in circle O such that BD = 12 and AE = 4.

(a) Determine the length of CE.

(b) Determine the perimeter of ABCD to the nearest tenth.

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a) CE = 9

b) Perimeter = 36.1

Step-by-step explanation:

a)

Using the property of chords in a circle, we have that:

DE * BE = AE * CE

As the diameter is perpendicular to the chord, the diameter cuts the chord in half, so BE = DE = BD/2 = 6

So we have that:

6 * 6 = 4 * CE

4 * CE = 36

CE = 9

b)

As the triangle BCE is congruent to the triangle DCE (using side-angle-side case), we have that BC = CD, and we can find them using Pythagoras' theorem in the triangle BCE:

BC^2 = DE^2 + EC^2

BC^2 = 6^2 = 9^2

BC^2 = 36 + 81 = 117

BC = 10.8167

As the triangle BAE is congruent to the triangle DAE (using side-angle-side case), we have that AB = AD, and we can find them using Pythagoras' theorem in the triangle BAE:

AB^2 = AE^2 + BE^2

AB^2 = 4^2 = 6^2

AB^2 = 16 + 36 = 52

AB = 7.2111

So the perimeter of ABCD is:

10.8167 + 10.8167 + 7.2111 + 7.2111 = 36.0556

Rounding to the nearest tenth, the perimeter is 36.1

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step-by-step explanation:

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