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Minchanka [31]
3 years ago
15

Find the area of a segment formed by a chord 8" long in a circle with radius of 8"

Mathematics
1 answer:
Marta_Voda [28]3 years ago
7 0
Draw 2 radii such that their endpoints on the circle are 8 inches apartconnect these points; this is the chord and it is 8 inches longthe 2 radii and the chord form an equilateral triangleeach side is 8 inches and each angle is 60 degreesfind the area of the sector formed by the 2 radiifind the area of the equilateral trianglesubtract the area of the triangle from the area of the sector and you have the area of the segment of the circlearea of the sector: A=(q/360)*∏r2A=(60/360)*∏*82A=(1/6)(∏)(64)A=(64/6)(∏)A=(32/3)(3.14159)A=33.5103 square inchesarea of the triangle:A=(√3/4)a2 where a=length of a sideA=(1.73205/4)(82)A=(1.73205/4)(64)A=(1.73205)(16)A=27.7128 square inchessubtract33.5103-27.7128=5.7975 square inchesthe area of the segment is 5.7975 square inches
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Answer:

\frac{BE}{EC} =\frac{1}{3}

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

\frac{BK}{CK} =\frac{1}{4}

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE                ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE     ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK  and Δ EBK are similar

so we have

\frac{AD}{BE} =\frac{DK}{BK}

\frac{AD}{BE} =\frac{4}{1}

\frac{BC}{BE}=\frac{4}{1}     ( ABCD is parallelogram so AD=BC)

\frac{BE+EC}{BE}=\frac{4}{1}         ( BC= BE+EC)

\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}

1+\frac{EC}{BE}=4

\frac{EC}{BE}=3  ( subtracting 1 from both side )

\frac{EC}{BE}=\frac{3}{1}

taking reciprocal both side

\frac{BE}{EC} =\frac{1}{3}


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