Question-86379

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Question Number 86379 by jagoll last updated on 28/Mar/20

Commented by jagoll last updated on 28/Mar/20

$$\mathrm{to}\:\mathrm{miss}\:\mathrm{tawa} \\ $$

Commented by TawaTawa1 last updated on 28/Mar/20

$$\mathrm{Sir},\:\mathrm{am}\:\mathrm{here} \\ $$

Answered by jagoll last updated on 28/Mar/20

tan 30^o = (9/(AS)) ⇒ AS = 9(√3) tan 60^o = (9/(DP)) ⇒ DP = 3(√3) perimeter PQRS = 2×(18(√3) + 6(√3)) = 48(√3) cm

$$\mathrm{tan}\:\mathrm{30}^{\mathrm{o}} \:=\:\frac{\mathrm{9}}{\mathrm{AS}}\:\Rightarrow\:\mathrm{AS}\:=\:\mathrm{9}\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\:\mathrm{60}^{\mathrm{o}} \:=\:\frac{\mathrm{9}}{\mathrm{DP}}\:\Rightarrow\:\mathrm{DP}\:=\:\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{perimeter}\:\mathrm{PQRS}\:=\:\mathrm{2}×\left(\mathrm{18}\sqrt{\mathrm{3}}\:+\:\mathrm{6}\sqrt{\mathrm{3}}\right) \\ $$$$=\:\mathrm{48}\sqrt{\mathrm{3}}\:\mathrm{cm} \\ $$

Commented by TawaTawa1 last updated on 28/Mar/20