Question

Aspinner of radius 7.5 sum is divided into 6 equal sectors. Find the area
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Answer 1

Answer:

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60o

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,GO=OI  (Radius)

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,GO=OI  (Radius)∠GOI=60o

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,GO=OI  (Radius)∠GOI=60oTherefore triangle GOI ids equilateral .

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,GO=OI  (Radius)∠GOI=60oTherefore triangle GOI ids equilateral .Hence, GI=GO=a 

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,GO=OI  (Radius)∠GOI=60oTherefore triangle GOI ids equilateral .Hence, GI=GO=a Hence, GI=GH=HJ=JL=LK=KI=a

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,GO=OI  (Radius)∠GOI=60oTherefore triangle GOI ids equilateral .Hence, GI=GO=a Hence, GI=GH=HJ=JL=LK=KI=aNow area of the resulting figure is -

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,GO=OI  (Radius)∠GOI=60oTherefore triangle GOI ids equilateral .Hence, GI=GO=a Hence, GI=GH=HJ=JL=LK=KI=aNow area of the resulting figure is -area of circle with radius a+(  area of the equilateral triangle with side a)×6

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,GO=OI  (Radius)∠GOI=60oTherefore triangle GOI ids equilateral .Hence, GI=GO=a Hence, GI=GH=HJ=JL=LK=KI=aNow area of the resulting figure is -area of circle with radius a+(  area of the equilateral triangle with side a)×6=πa2+43a2×6

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,GO=OI  (Radius)∠GOI=60oTherefore triangle GOI ids equilateral .Hence, GI=GO=a Hence, GI=GH=HJ=JL=LK=KI=aNow area of the resulting figure is -area of circle with radius a+(  area of the equilateral triangle with side a)×6=πa2+43a2×6=πa2+233a2

In circle ∠GOI=∠GOH=∠HOJ=∠JOL=∠LOK=∠KOI=60o360o=60oNw consider triangle GOI4In ΔGOI,GO=OI  (Radius)∠GOI=60oTherefore triangle GOI ids equilateral .Hence, GI=GO=a Hence, GI=GH=HJ=JL=LK=KI=aNow area of the resulting figure is -area of circle with radius a+(  area of the equilateral triangle with side a)×6=πa2+43a2×6=πa2+233a2=2a2(2π+33)