Question-86528

Question Number 86528 by TawaTawa1 last updated on 29/Mar/20

Commented by TawaTawa1 last updated on 29/Mar/20

$$\mathrm{Student}\:\mathrm{sir}. \\ $$

Commented by jagoll last updated on 29/Mar/20

sin 60^o = (h/(12)) ⇒ h = 6(√3) cos 60^o = ((TS)/(12)) ⇒ TS = 6 TO = 6+14+6 = 26 the area of the remaining portion = (((14+26)/2))×6(√3) − (1/2)×((22)/7)×49 = 120(√3) −77 = 120×1.3 − 77 = 207.85−77 = 130.85

$$\mathrm{sin}\:\:\mathrm{60}^{\mathrm{o}} \:=\:\frac{\mathrm{h}}{\mathrm{12}}\:\Rightarrow\:\mathrm{h}\:=\:\mathrm{6}\sqrt{\mathrm{3}} \\ $$$$\mathrm{cos}\:\mathrm{60}^{\mathrm{o}} \:=\:\frac{\mathrm{TS}}{\mathrm{12}}\:\Rightarrow\:\mathrm{TS}\:=\:\mathrm{6} \\ $$$$\mathrm{TO}\:=\:\mathrm{6}+\mathrm{14}+\mathrm{6}\:=\:\mathrm{26} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{portion}\: \\ $$$$=\:\left(\frac{\mathrm{14}+\mathrm{26}}{\mathrm{2}}\right)×\mathrm{6}\sqrt{\mathrm{3}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{22}}{\mathrm{7}}×\mathrm{49} \\ $$$$=\:\mathrm{120}\sqrt{\mathrm{3}}\:−\mathrm{77}\: \\ $$$$=\:\mathrm{120}×\mathrm{1}.\mathrm{3}\:−\:\mathrm{77}\:=\:\mathrm{207}.\mathrm{85}−\mathrm{77} \\ $$$$=\:\mathrm{130}.\mathrm{85} \\ $$

Commented by TawaTawa1 last updated on 29/Mar/20