A round table cover has six equal designs as shown in figure. If the radius of the cover is 56 cm, find
the cost of making the design at the rate of 3.5 per cm2. (Use root 3=1.7, pie(ll), =22/7
Answers
Answer:
here is your answer mate
Step-by-step explanation:
r=28cm,θ=
6
360
0
=60
0
Area OAMB=
360
0
θ
×πr
2
=
360
0
60
0
×
7
22
×28×28
=
3
1232
cm
2
=410.67cm
2
Now, in ΔONA and ΔONB
i) ∠ONA=∠ONB[each90
0
]
ii) OA=OB [Radii of common circle]
iii) ON=ON [common]
∴ΔONA≅ΔONB[ByRHScongruency]
Hence, AN=NB=
2
1
AB
& ∠AON=∠BON=
2
1
∠AOB=
2
60
0
=30
0
Now in ΔONA,cos30
0
=
OA
ON
⇒
2
3
=
28
ON
⇒ON=14
3
cm
& sin30
0
=
OA
AN
⇒
2
1
=
28
AN
⇒AN=14cm
& 2AN=14×2=28cm=AB
∴ arΔAOB=
2
1
×AB×ON=
2
1
×28×14
3
=196
3
=196×1.7 =333.2cm
2
∴ Area of one design =410.67−333.2=77.47 cm
2
∴ Area of six design= 6×77.47=464.82cm
2
Therefore, Cost of making the designs=Rs.(464.82×3.5$$)
=Rs.1626.8
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Answer:
r=28cm,θ=
6
360
0
=60
0
Area OAMB=
360
0
θ
×πr
2
=
360
0
60
0
×
7
22
×28×28
=
3
1232
cm
2
=410.67cm
2
Now, in ΔONA and ΔONB
i) ∠ONA=∠ONB[each90
0
]
ii) OA=OB [Radii of common circle]
iii) ON=ON [common]
∴ΔONA≅ΔONB[ByRHScongruency]
Hence, AN=NB=
2
1
AB
& ∠AON=∠BON=
2
1
∠AOB=
2
60
0
=30
0
Now in ΔONA,cos30
0
=
OA
ON
⇒
2
3
=
28
ON
⇒ON=14
3
cm
& sin30
0
=
OA
AN
⇒
2
1
=
28
AN
⇒AN=14cm
& 2AN=14×2=28cm=AB
∴ arΔAOB=
2
1
×AB×ON=
2
1
×28×14
3
=196
3
=196×1.7 =333.2cm
2
∴ Area of one design =410.67−333.2=77.47 cm
2
∴ Area of six design= 6×77.47=464.82cm
2
Therefore, Cost of making the designs=Rs.(464.82×3.5$$)
=Rs.1626.8