Question-87671

Question Number 87671 by TawaTawa1 last updated on 05/Apr/20

Commented by TawaTawa1 last updated on 05/Apr/20

I got 10.5 m^(2 ) but am not sure. Please help me check.

$$\mathrm{I}\:\mathrm{got}\:\:\mathrm{10}.\mathrm{5}\:\mathrm{m}^{\mathrm{2}\:} \:\:\mathrm{but}\:\mathrm{am}\:\mathrm{not}\:\mathrm{sure}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{check}. \\ $$

Commented by Tony Lin last updated on 05/Apr/20

7^2 ×((2π)/3)×(1/2)−7^2 ×((√3)/4) =49((π/3)−((√3)/4)) ≈30(cm^2 ) =0.003(m^2 ) 0.0049−0.003=0.0019(m^2 )

$$\mathrm{7}^{\mathrm{2}} ×\frac{\mathrm{2}\pi}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{7}^{\mathrm{2}} ×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$=\mathrm{49}\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right) \\ $$$$\approx\mathrm{30}\left({cm}^{\mathrm{2}} \right) \\ $$$$=\mathrm{0}.\mathrm{003}\left({m}^{\mathrm{2}} \right) \\ $$$$\mathrm{0}.\mathrm{0049}−\mathrm{0}.\mathrm{003}=\mathrm{0}.\mathrm{0019}\left({m}^{\mathrm{2}} \right) \\ $$

Commented by TawaTawa1 last updated on 05/Apr/20

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 05/Apr/20

A_(shade) =7×7−2×((π×7^2 )/6)+((7^2 ×(√3))/4) =49−((49π)/3)+((49(√3))/4) ≈18.9 cm^2 =0.00189 m^2

$${A}_{{shade}} =\mathrm{7}×\mathrm{7}−\mathrm{2}×\frac{\pi×\mathrm{7}^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{7}^{\mathrm{2}} ×\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$=\mathrm{49}−\frac{\mathrm{49}\pi}{\mathrm{3}}+\frac{\mathrm{49}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\approx\mathrm{18}.\mathrm{9}\:{cm}^{\mathrm{2}} =\mathrm{0}.\mathrm{00189}\:{m}^{\mathrm{2}} \\ $$

Commented by TawaTawa1 last updated on 05/Apr/20

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by TawaTawa1 last updated on 05/Apr/20

Sir, have been trying to understand − 2 × ((π × 7^2 )/6) + ((7^2 × (√3))/4) But since i did not understand. Please help me explain when you are chanced sir. Please.

$$\mathrm{Sir},\:\mathrm{have}\:\mathrm{been}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{understand}\:\:\:\:\:\:−\:\:\mathrm{2}\:×\:\frac{\pi\:×\:\mathrm{7}^{\mathrm{2}} }{\mathrm{6}}\:\:+\:\:\frac{\mathrm{7}^{\mathrm{2}} \:×\:\sqrt{\mathrm{3}}}{\mathrm{4}}\: \\ $$$$\mathrm{But}\:\mathrm{since}\:\mathrm{i}\:\mathrm{did}\:\mathrm{not}\:\mathrm{understand}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{explain}\:\mathrm{when}\:\mathrm{you}\:\mathrm{are}\:\mathrm{chanced}\:\mathrm{sir}. \\ $$$$\mathrm{Please}. \\ $$

Commented by mr W last updated on 05/Apr/20

Commented by mr W last updated on 05/Apr/20

Commented by mr W last updated on 05/Apr/20

A_1 =square=7×7 A_2 =sector=((circle)/6)=((π×7^2 )/6) A_3 =equilateral triangle=(1/2)×7×(((√3)×7)/2)=((7^2 (√3))/4) A_(shade) =A_1 −2A_2 +A_3 =7×7−2×((7^2 π)/6)+((7^2 (√3))/4) =49−((49π)/3)+((49(√3))/4)

$${A}_{\mathrm{1}} ={square}=\mathrm{7}×\mathrm{7} \\ $$$${A}_{\mathrm{2}} ={sector}=\frac{{circle}}{\mathrm{6}}=\frac{\pi×\mathrm{7}^{\mathrm{2}} }{\mathrm{6}} \\ $$$${A}_{\mathrm{3}} ={equilateral}\:{triangle}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{7}×\frac{\sqrt{\mathrm{3}}×\mathrm{7}}{\mathrm{2}}=\frac{\mathrm{7}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${A}_{{shade}} ={A}_{\mathrm{1}} −\mathrm{2}{A}_{\mathrm{2}} +{A}_{\mathrm{3}} \\ $$$$=\mathrm{7}×\mathrm{7}−\mathrm{2}×\frac{\mathrm{7}^{\mathrm{2}} \pi}{\mathrm{6}}+\frac{\mathrm{7}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$=\mathrm{49}−\frac{\mathrm{49}\pi}{\mathrm{3}}+\frac{\mathrm{49}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$

Commented by TawaTawa1 last updated on 05/Apr/20

Hahaha. Am happy. I understand now, God will be helping you too sir.

$$\mathrm{Hahaha}.\:\:\mathrm{Am}\:\mathrm{happy}. \\ $$$$\mathrm{I}\:\mathrm{understand}\:\mathrm{now},\:\:\mathrm{God}\:\mathrm{will}\:\mathrm{be}\:\mathrm{helping}\:\mathrm{you}\:\mathrm{too}\:\mathrm{sir}. \\ $$

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