A circular field has an area 1256{cm}^21256cm 2 . A horse is tied at the centre of the field with the help of a rope 15m15m long. How much area of the field will the horse be able to graze? Also find the area of the remaining field.
Answers
Step-by-step explanation:
Using Pythagoras theorem:
Using Pythagoras theorem:PS
Using Pythagoras theorem:PS 2
Using Pythagoras theorem:PS 2 =PO
Using Pythagoras theorem:PS 2 =PO 2
Using Pythagoras theorem:PS 2 =PO 2 +SO
Using Pythagoras theorem:PS 2 =PO 2 +SO 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r)
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r)
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R 2 −6Rr=0
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R 2 −6Rr=0⇒2R(2R−3r)=0
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R 2 −6Rr=0⇒2R(2R−3r)=0⇒R=
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R 2 −6Rr=0⇒2R(2R−3r)=0⇒R= 2
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R 2 −6Rr=0⇒2R(2R−3r)=0⇒R= 23r
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R 2 −6Rr=0⇒2R(2R−3r)=0⇒R= 23r
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R 2 −6Rr=0⇒2R(2R−3r)=0⇒R= 23r
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R 2 −6Rr=0⇒2R(2R−3r)=0⇒R= 23r Total grazed area =
Using Pythagoras theorem:PS 2 =PO 2 +SO 2 ⇒(R+r) 2 =R 2 +(2R−r) 2 ⇒R 2 +r 2 +2Rr=R 2 +4R+r 2 −4Rr⇒4R 2 −6Rr=0⇒2R(2R−3r)=0⇒R= 23r Total grazed area = 2
π
+
+ 2
π
+π(
+π( 3
+π( 32R
+π( 32R
+π( 32R )
+π( 32R ) 2
+π( 32R ) 2
+π( 32R ) 2 =
+
+ 9
+ 94
+ 94
+ 94
+ 94
+ 94 Ungad area = π
+ 94 Ungad area = π 2
+ 94 Ungad area = π 2(2R)
+ 94 Ungad area = π 2(2R) 2
+ 94 Ungad area = π 2(2R) 2
+ 94 Ungad area = π 2(2R) 2
+ 94 Ungad area = π 2(2R) 2 −π
+ 94 Ungad area = π 2(2R) 2 −π2
+ 94 Ungad area = π 2(2R) 2 −π2 −
+ 94 Ungad area = π 2(2R) 2 −π2 − 9
+ 94 Ungad area = π 2(2R) 2 −π2 − 94
+ 94 Ungad area = π 2(2R) 2 −π2 − 94
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 =
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 9
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area =
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2 9
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2 9
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2 9
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2 9 ×100
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2 9 ×100=
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2 9 ×100= 18
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2 9 ×100= 18500
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2 9 ×100= 18500
+ 94 Ungad area = π 2(2R) 2 −π2 − 94 π2 = 95πR 2 Percentage of gazed area = 2πR 2 9 ×100= 18500 =28%
Answer:
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