MCP๋กœ ์—ฐ๊ฒฐ โ†’

๊ณ„์‚ฐ ์ž…๋ ฅ

๊ณต์‹

๊ณต์‹: ๋งˆ๋ฆ„๋ชจ ๊ณ„์‚ฐ๊ธฐ
Show calculation steps (1)
  1. Diagonals from side and angle

    Diagonals from side and angle: ๋งˆ๋ฆ„๋ชจ ๊ณ„์‚ฐ๊ธฐ

    The two diagonals p and q in terms of the side a and interior angle A, with the parallelogram law.

๊ด‘๊ณ 

๊ฒฐ๊ณผ

๋งˆ๋ฆ„๋ชจ ๋„“์ด K
20
๊ผญ์ง“์  ๊ฐ A = C = 53.1301 ยฐ
๊ผญ์ง“์  ๊ฐ B = D = 126.87 ยฐ
๋ณ€ a ์˜ ๊ธธ์ด a = 5
๋Œ€๊ฐ์„  ๊ธธ์ด p = 8.94427
๋Œ€๊ฐ์„  ๊ธธ์ด q = 4.47214
๋†’์ด h = 4
๋‘˜๋ ˆ P = 20
๋„“์ด K = 20

๋งˆ๋ฆ„๋ชจ ๊ณ„์‚ฐ๊ธฐ๋ž€?

๋งˆ๋ฆ„๋ชจ๋Š” ๋„ค ๋ณ€์˜ ๊ธธ์ด๊ฐ€ ๋ชจ๋‘ ๊ฐ™์€(a) ์‚ฌ๊ฐํ˜•์ž…๋‹ˆ๋‹ค. ๋งˆ์ฃผ ๋ณด๋Š” ๊ฐ์€ ์„œ๋กœ ๊ฐ™๊ณ (A = C, B = D), ์ด์›ƒํ•œ ๊ฐ์€ ํ•ฉ์ด 180ยฐ๊ฐ€ ๋˜๋ฉฐ(A + B = 180ยฐ), ๋‘ ๋Œ€๊ฐ์„  p์™€ q๋Š” ์„œ๋กœ๋ฅผ ์ง๊ฐ์œผ๋กœ ์ด๋“ฑ๋ถ„ํ•ฉ๋‹ˆ๋‹ค. ์ด ๊ณ„์‚ฐ๊ธฐ๋Š” ์•„๋Š” ๋‘ ๊ฐ€์ง€ ์†์„ฑ๋งŒ ์ž…๋ ฅํ•˜๋ฉด ๋‚˜๋จธ์ง€ ๊ฐ’์„ ๋ชจ๋‘ ๊ณ„์‚ฐํ•ด ์ค๋‹ˆ๋‹ค โ€” ๊ฐ ๊ผญ์ง“์ ์˜ ๊ฐ๋„, ๋‘ ๋Œ€๊ฐ์„ , ๋†’์ด, ๋‘˜๋ ˆ, ๊ทธ๋ฆฌ๊ณ  ๋„“์ด๊นŒ์ง€ ํ•œ ๋ฒˆ์— ๊ตฌํ•  ์ˆ˜ ์žˆ์Šต๋‹ˆ๋‹ค.

๋ณ€, ๋Œ€๊ฐ์„ , ๊ฐ, ๋†’์ด๋ฅผ ํ‘œ์‹œํ•œ ๋งˆ๋ฆ„๋ชจ
๊ฐ™์€ ๋ณ€ a, ๋Œ€๊ฐ์„  p์™€ q, ๋‚ด๊ฐ A์™€ B, ๋†’์ด h๋ฅผ ํ‘œ์‹œํ•œ ๋งˆ๋ฆ„๋ชจ.

์‚ฌ์šฉ ๋ฐฉ๋ฒ•

"๊ณ„์‚ฐ ํ•ญ๋ชฉ ์„ ํƒ" ๋ฉ”๋‰ด์—์„œ ์ž์‹ ์ด ์•Œ๊ณ  ์žˆ๋Š” ๋‘ ๊ฐ’์— ๋งž๋Š” ํ•ญ๋ชฉ์„ ๊ณ ๋ฅด์„ธ์š” โ€” ์˜ˆ๋ฅผ ๋“ค์–ด "a, h๊ฐ€ ์ฃผ์–ด์ง„ ๊ฒฝ์šฐ" ๋˜๋Š” "p, q๊ฐ€ ์ฃผ์–ด์ง„ ๊ฒฝ์šฐ" ๋“ฑ์ด ์žˆ์Šต๋‹ˆ๋‹ค. ํ•ด๋‹น ๊ฐ’์„ ์ž…๋ ฅํ•˜๊ณ  ํ‘œ์‹œ ๋‹จ์œ„(์ˆซ์ž๋ฅผ ๋ณ€ํ™˜ํ•˜์ง€ ์•Š๊ณ  ๋‹จ์œ„ ํ‘œ๊ธฐ๋งŒ ๋ถ™๋Š” ๋ผ๋ฒจ์ž…๋‹ˆ๋‹ค)์™€ ์œ ํšจ์ˆซ์ž ์ž๋ฆฌ์ˆ˜๋ฅผ ์„ ํƒํ•˜๋ฉด ์ „์ฒด ๊ฒฐ๊ณผ๋ฅผ ํ•œ๋ˆˆ์— ํ™•์ธํ•  ์ˆ˜ ์žˆ์Šต๋‹ˆ๋‹ค. ๊ฐ๋„๋Š” ๋„(degree) ๋‹จ์œ„๋กœ ์ž…๋ ฅํ•˜๊ณ  ํ‘œ์‹œ๋ฉ๋‹ˆ๋‹ค.

๊ณต์‹ ์ •๋ฆฌ

๊ณ„์‚ฐ ์—”์ง„์€ ๋„ค ๊ฐ€์ง€ ๋„“์ด ๊ณต์‹๊ณผ ๋Œ€๊ฐ์„  ๊ด€๊ณ„์‹์„ ๊ธฐ๋ฐ˜์œผ๋กœ ํ•ฉ๋‹ˆ๋‹ค. ๋„“์ด๋Š” ๋‹ค์Œ๊ณผ ๊ฐ™์ด ๋‚˜ํƒ€๋‚ผ ์ˆ˜ ์žˆ์Šต๋‹ˆ๋‹ค.

$$K = a^2 \sin A = \frac{p \cdot q}{2} = a \cdot h$$

๋†’์ด๋Š” \(h = a \cdot \sin A\) ๋ฅผ ๋งŒ์กฑํ•ฉ๋‹ˆ๋‹ค. ๋Œ€๊ฐ์„ ์€ ๋‹ค์Œ๊ณผ ๊ฐ™์ด ๊ตฌํ•ด์ง‘๋‹ˆ๋‹ค.

$$p = 2a\cos\tfrac{A}{2}, \quad q = 2a\sin\tfrac{A}{2}$$

์ด ๋‘˜์„ ํ•ฉ์น˜๋ฉด ํ‰ํ–‰์‚ฌ๋ณ€ํ˜• ๋ฒ•์น™ \(p^2 + q^2 = 4a^2\) ๊ฐ€ ์„ฑ๋ฆฝํ•ฉ๋‹ˆ๋‹ค. ๋”ฐ๋ผ์„œ ๋ณ€์˜ ๊ธธ์ด๋Š” \(a = \tfrac{1}{2}\sqrt{p^2 + q^2}\) ๋กœ ์—ญ์‚ฐํ•  ์ˆ˜ ์žˆ์Šต๋‹ˆ๋‹ค. ๋‘˜๋ ˆ๋Š” ๊ฐ„๋‹จํžˆ \(P = 4a\) ์ž…๋‹ˆ๋‹ค.

์ง๊ฐ์œผ๋กœ ๋งŒ๋‚˜ ์„œ๋กœ ์ด๋“ฑ๋ถ„ํ•˜๋Š” ๋งˆ๋ฆ„๋ชจ์˜ ๋Œ€๊ฐ์„ 
๋งˆ๋ฆ„๋ชจ์˜ ๋Œ€๊ฐ์„ ์€ ์„œ๋กœ ์ง๊ฐ์œผ๋กœ ์ด๋“ฑ๋ถ„ํ•˜์—ฌ ๋„ค ๊ฐœ์˜ ์ง๊ฐ์‚ผ๊ฐํ˜•์„ ์ด๋ฃน๋‹ˆ๋‹ค.

๊ณ„์‚ฐ ์˜ˆ์‹œ

\(a = 5\), \(h = 4\) ๋ผ๊ณ  ๊ฐ€์ •ํ•ด ๋ด…์‹œ๋‹ค. ๊ทธ๋Ÿฌ๋ฉด \(\sin A = h/a = 0.8\) ์ด๋ฏ€๋กœ \(A = \arcsin(0.8) = 53.1301ยฐ\), \(B = 180 - 53.1301 = 126.870ยฐ\) ๊ฐ€ ๋ฉ๋‹ˆ๋‹ค. ๋Œ€๊ฐ์„ ์€ \(p = 2 \cdot 5 \cdot \cos(26.5651ยฐ) = 8.94427\), \(q = 2 \cdot 5 \cdot \sin(26.5651ยฐ) = 4.47214\) ์ž…๋‹ˆ๋‹ค. ๋‘˜๋ ˆ๋Š” \(P = 4 \cdot 5 = 20\), ๋„“์ด๋Š” \(K = a \cdot h = 20\) (์ด๋Š” \(p \cdot q / 2 = 20\) ๊ณผ๋„ ์ผ์น˜ํ•ฉ๋‹ˆ๋‹ค). ์„ธ ๊ฐ€์ง€ ๋„“์ด ๊ณต์‹ ๋ชจ๋‘ ๊ฐ™์€ ๊ฒฐ๊ณผ๋ฅผ ๋ณด์—ฌ ์ค๋‹ˆ๋‹ค.

์ž์ฃผ ๋ฌป๋Š” ์งˆ๋ฌธ

์™œ ๊ฐ๋„๋Š” ์˜ˆ๊ฐ์œผ๋กœ ํ‘œ์‹œ๋˜๋‚˜์š”? ๋ณ€๊ณผ ๋†’์ด(๋˜๋Š” ๋„“์ด)๋งŒ ์ฃผ์–ด์ง€๋ฉด ๋งˆ๋ฆ„๋ชจ์˜ ๋ชจ์–‘์€ ์˜ˆ๊ฐ๊ณผ ๊ทธ ๋ณด๊ฐ ์‚ฌ์ด์—์„œ ๋ชจํ˜ธํ•ด์ง‘๋‹ˆ๋‹ค. ๋ณธ ๊ณ„์‚ฐ๊ธฐ๋Š” A๋ฅผ ์˜ˆ๊ฐ ๊ฐ’์œผ๋กœ, B = 180ยฐ โˆ’ A๋ฅผ ๊ทธ ๋ณด๊ฐ์œผ๋กœ ํ‘œ์‹œํ•˜๋ฉฐ, ์ด๋Š” ์™„์ „ํžˆ ์ผ๊ด€๋œ ๊ฒฐ๊ณผ์ž…๋‹ˆ๋‹ค.

๋‹จ์œ„ ๋ฉ”๋‰ด๊ฐ€ ๊ฐ’์„ ๋ณ€ํ™˜ํ•˜๋‚˜์š”? ์•„๋‹™๋‹ˆ๋‹ค. ๋ชจ๋“  ๊ธธ์ด๋Š” ํ•˜๋‚˜์˜ ๋‹จ์œ„๋ฅผ ๊ณต์œ ํ•˜๋ฏ€๋กœ ๋ฉ”๋‰ด๋Š” ๋‹จ์œ„ ๋ผ๋ฒจ๋งŒ ๋ง๋ถ™์ž…๋‹ˆ๋‹ค. ๋„“์ด๋Š” ํ•ด๋‹น ๋‹จ์œ„์˜ ์ œ๊ณฑ์œผ๋กœ ํ‘œ์‹œ๋ฉ๋‹ˆ๋‹ค.

๋งˆ๋ฆ„๋ชจ๊ฐ€ ์กด์žฌํ•˜์ง€ ์•Š๋Š” ๊ฒฝ์šฐ๋Š”? ๋†’์ด๊ฐ€ ๋ณ€๋ณด๋‹ค ํฌ๊ฑฐ๋‚˜ ๋Œ€๊ฐ์„ ์ด 2a์— ๋„๋‹ฌํ•˜๋ฉด ๊ทธ ๋„ํ˜•์€ ๋ถˆ๊ฐ€๋Šฅํ•˜๊ฑฐ๋‚˜ ํ‡ดํ™”ํ•œ ํ˜•ํƒœ๊ฐ€ ๋ฉ๋‹ˆ๋‹ค. ์ด๋•Œ ๊ณ„์‚ฐ๊ธฐ๋Š” ์‚ผ๊ฐํ•จ์ˆ˜์˜ ์ž…๋ ฅ๊ฐ’์„ ์ œํ•œ(clamp)ํ•˜์—ฌ ๊ฒฐ๊ณผ๊ฐ€ ํ•ญ์ƒ ์œ ํšจํ•˜๋„๋ก ์œ ์ง€ํ•ฉ๋‹ˆ๋‹ค.

์ตœ์ข… ์—…๋ฐ์ดํŠธ: