Imibuzo Yezibonelo kanye Nengxoxo Yemikhakha Yesiyingi
Imikhakha yendilinga iyisihloko esibalulekile ezibalweni esivame ukuvela ezivivinyweni nasemibuzweni yokuzijwayeza. Imikhakha yingxenye yendilinga eboshwe ngama-radii amabili kanye nomugqa oxhumanisa wona. Kulesi sihloko, sizoxoxa ngezibonelo eziningana zezinkinga zemikhakha yendilinga, kanye nezincazelo ezinemininingwane, ukuze sijulise ukuqonda kwethu.
Incazelo Yomkhakha Wendilinga
Umkhakha wendilinga uwumkhakha wendilinga ozungezwe ama-radii amabili kanye ne-arc eyodwa. Indawo yomkhakha ibalwa ngokusekelwe engxenyeni yendawo iyonke yendilinga. Ifomula eyinhloko esetshenziswa ukubala umkhakha yilena elandelayo:
– Indawo Yokuhlela Ijaji: \[L_juring = \frac{\theta}{360^\circ} \times \pi \times r^2\]
– Ubude be-Arc: \[P_b = \frac{\theta}{360^\circ} \times 2\pi r\]
Di mana:
– \( \theta \) ubukhulu be-engeli yomkhakha ngamadigri,
– \( r \) irediyasi yesiyingi,
– \( \pi \) kuyinto engaguquki (cishe u-3.14159).
Imibuzo Eyisibonelo Nengxoxo
Umbuzo 1:
Uma unikezwe indilinga enobubanzi obuyi-10 cm kanye nomkhakha one-engeli ephakathi engu-90°. Bala indawo yomkhakha.
Ingxoxo:
Kuyaziwa:
– \( r = 10 \) cm
– \( \theta = 90^\circ \)
Sisebenzisa ifomula yendawo yomkhakha:
\[L_juring = \frac{\theta}{360^\circ} \times \pi \times r^2\]
\[L_juring = \frac{90^\circ}{360^\circ} \times \pi \times (10\text{ cm})^2\]
\[L_juring = \frac{1}{4} \times \pi \times 100\text{ cm}^2\]
\[L_juring = 25\pi\text{ cm}^2\]
Uma i-\( \pi \) ithathwa njenge-3.14, khona-ke:
\[L_juring = 25 \izikhathi 3.14\umbhalo{ cm}^2 = 78.5\umbhalo{ cm}^2\]
Ngakho-ke, indawo yalo mkhakha ingu-78.5 cm².
Umbuzo 2:
Umkhakha wendilinga une-radius engu-7 cm kanye nobude be-arc obungu-11 cm. Thola i-engeli ephakathi yomkhakha ngama-radians.
Ingxoxo:
Kuyaziwa:
– \( r = 7 \) cm
– Ubude be-Arc \( P_b = 11 \umbhalo{ cm} \)
Sisebenzisa ifomula yobude be-arc ukuthola i-engeli \( \theta \):
\[P_b = \frac{\theta}{360^\circ} \times 2\pi r\]
Njengoba sicelwa ukuthi sithole i-engeli kuma-radians, sithatha indawo yama-radians angu-360° ngama-radians angu-\(2\pi\):
\[P_b = \theta \times r\]
\[11 = \theta \izikhathi 7\]
\[\theta = \frac{11}{7}\]
\[\theta \cishe 1.57 \umbhalo{ rad}\]
Ngakho-ke, i-engeli ephakathi yomkhakha ingama-radian angu-1.57.
Umbuzo 3:
Isiyingi esinobubanzi obungu-16 cm sinomkhakha onendawo engu-200 cm². Bala i-engeli ephakathi yomkhakha.
Ingxoxo:
Kuyaziwa:
– \( r = 16 \) cm
– \( L_juring = 200 \umbhalo{ cm}^2 \)
Sisebenzisa ifomula yendawo yomkhakha ukuthola \( \theta \):
\[L_juring = \frac{\theta}{360^\circ} \times \pi \times r^2\]
\[200 = \frac{\theta}{360^\circ} \times \pi \times (16)^2\]
\[200 = \frac{\theta}{360^\circ} \times \pi \times 256\]
\[200 = \frac{\theta \times 256 \times \pi}{360^\circ}\]
\[200 \izikhathi ezingu-360^\circ = \theta \izikhathi ezingu-256 \izikhathi ezingu-3.14\]
\[72000 = \theta \izikhathi ezingu-256 \izikhathi ezingu-3.14\]
\[72000 = \theta \izikhathi 804.64\]
\[\theta = \frac{72000}{804.64}\]
\[\theta \cishe 89.45^\circ\]
Ngakho-ke, i-engeli ephakathi yomkhakha cishe ingu-89.45°.
Umbuzo 4:
Bala umjikelezo ophelele womkhakha onobubanzi obungu-12 cm kanye ne-engeli ephakathi engu-120°.
Ingxoxo:
Kuyaziwa:
– \( r = 12 \) cm
– \( \theta = 120^\circ \)
Okokuqala, sithola ubude be-arc:
\[P_b = \frac{\theta}{360^\circ} \times 2\pi r\]
\[P_b = \frac{120^\circ}{360^\circ} \izikhathi 2\pi \izikhathi 12\]
\[P_b = \frac{1}{3} \izikhathi 2\pi \izikhathi 12\]
\[P_b = 8\pi\text{ cm}\]
Bese sibala umjikelezo womkhakha (ubude be-arc + ama-radii amabili):
\[K = 2r + P_b\]
\[K = 2 \izikhathi ezingu-12\umbhalo{cm} + 8\pi\umbhalo{cm}\]
\[K = 24\umbhalo{ cm} + 8\pi\umbhalo{ cm}\]
Uma i-\( \pi \) ithathwa njenge-3.14, khona-ke:
\[K = 24\umbhalo{cm} + 8 \izikhathi 3.14\umbhalo{cm}\]
\[K = 24\umbhalo{ cm} + 25.12\umbhalo{ cm}\]
\[K = 49.12\umbhalo{ cm}\]
Ngakho-ke, isiyingi esiphelele somkhakha singama-49.12 cm.
Umbuzo 5:
Uma indilinga enobubanzi obungu-18 cm inomkhakha owenza i-engeli engu-45°, nquma ubude be-arc nendawo yomkhakha.
Ingxoxo:
Kuyaziwa:
– \( r = 18 \) cm
– \( \theta = 45^\circ \)
1. Ubude be-Arc:
\[P_b = \frac{\theta}{360^\circ} \times 2\pi r\]
\[P_b = \frac{45^\circ}{360^\circ} \times 2\pi \times 18\text{ cm}\]
\[P_b = \frac{1}{8} \izikhathi ezingu-36\pi\text{ cm}\]
\[P_b = 4.5\pi\text{ cm}\]
Uma i-\( \pi \) ithathwa njenge-3.14, khona-ke:
\[P_b = 4.5 \izikhathi 3.14\umbhalo{ cm} \cishe 14.13\umbhalo{ cm}\]
Ngakho-ke, ubude bomnsalo bungaba ngu-14.13 cm.
2. Indawo Yomkhakha:
\[L_juring = \frac{\theta}{360^\circ} \times \pi \times r^2\]
\[L_juring = \frac{45^\circ}{360^\circ} \times \pi \times (18\text{ cm})^2\]
\[L_juring = \frac{1}{8} \times \pi \times 324\text{ cm}^2\]
\[L_juring = 40.5\pi\text{ cm}^2\]
Uma i-\( \pi \) ithathwa njenge-3.14, khona-ke:
\[L_juring = 40.5 \izikhathi 3.14\umbhalo{ cm}^2 \cishe 127.17\umbhalo{ cm}^2\]
Ngakho-ke, indawo yomkhakha icishe ibe ngu-127.17 cm².
Isiphetho
Kulesi sihloko, sixoxe ngezinkinga eziningana zezibonelo mayelana nemikhakha yendilinga kanye nezixazululo zazo. Ingqikithi yokuqonda imikhakha yendilinga itholakala ekuqondeni amafomula ayisisekelo okubala indawo yomkhakha kanye nobude be-arc. Ukuzijwayeza njalo nokuqonda ukuthi ungawasebenzisa kanjani la mafomula ezinhlotsheni ezahlukene zezinkinga kuzosiza ekuthuthukiseni ikhono lakho lokuxazulula izinkinga ezifanayo.
