Question

plz Solve 8th one... answer will be marked as brainliest...

Answer 1

Given: OA=21 cm, AB=14 cm, x=60°. Where, x is the central angle.

To find; The area and the perimeter of the region swept by the wiper.

To find area,

ar( region swept by the wiper )=ar(OBC)-ar(OAD)

=(πx)/360°{(r¹)²-(r²)²}
where, r¹ is the radius of sector OBC and r² is the radius of sector OAD.

={(22*60)/(7*360)}{(21+14)²-21²}

=(1320/2520)(35²-21²)

=(1320/2510)(1225-441)

=0.525*784

=411.6 cm²

To find the perimeter of the region swept,

In sector OAD, OA=OD…{radii}
similarly, in sector OABCD, OAB=ODC

Hence, AB=DC.

Now, perimeter of the region=AB+DC+ length of Arc OAD+ length of Arc OBC

=14+14+[{(2πx)/360°}{r¹+r²}]

=28+[{(2*22*60)/(360*7)}{21+14}]

=28+{(2640/2520)(35)}

=28+(92400/35)

=28+2640

=2668 cm.

Hence, the perimeter and the area of the region swept by the wiper is 2668 cm and 411.6 cm² respectively.

Answer 2

Answer:

2668 ans...

a+b=10 and a^2+b^2=58

a+b=10onsquaringbothsides(a+b)2=100\begin{lgathered}a + b = 10 \\ on \: squaring \: both \: sides \\ {(a + b) }^{2} = 100 \\\end{lgathered}

a+b=10

onsquaringbothsides

(a+b)

2

=100

a^2+b^2+2ab=100

58+2ab=100 (a^2+b^2=58)

2ab=100-58

2ab=42

ab=21

now according to the question,

a^3+b^3=(a+b)(a^2+b^2+2ab)

= (10)(58-21)

=(10)(37)

=370