Question

6(area of minor sector-area of triangle)
6(60/360×22/7×28×28-1/2×28×28 sin60)
My teacher gave 464.8 total answer in my notes but stillI have a doubt on this pls explain in easy way and need full explanation

Answer 1

427.2 (Approximate)

Step-by-step explanation:

The required area = 6(area of minor sector-area of triangle)

= $6[(\frac{60}{360}\times \frac{22}{7} \times 28 \times 28) - (\frac{1}{2}\times 28 \times 28 \times \sin60^{\circ})]$

= $6[(\frac{60 \times 22 \times 28 \times 28}{360 \times 7}) - (\frac{1 \times 28 \times 28 \times \sqrt{3}}{2 \times 2} )]$ {Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$}

Now, first of all do the multiplication job first in numerator and then denominator and then divide them.

= $6[(\frac{60 \times 22 \times 28 \times 28}{360 \times 7}) - (\frac{1 \times 28 \times 28 \times 1.732}{2 \times 2} )]$ {Since √3 = 1.732}

= $6[(\frac{1034880}{2520}) - (\frac{1357.888}{4})]$

= $6[410.67 - 339.472]$

= 6 × 71.198

= 427.188

= 427.2 (Approximate)

Hence, the teacher gave the incorrect answer. (Answer)