Imibuzo Eyisibonelo Exoxa Ngemibuthano Nemiphetho
Indilinga iyisimo esiyisisekelo sejiyometri esivame ukufundwa emazingeni ahlukahlukene emfundo. Lo mqondo awusebenzi nje kuphela kwezemfundo kodwa futhi unezinhlelo zokusebenza ezibanzi empilweni yansuku zonke, njengokuklama izakhiwo, ubunjiniyela bemigwaqo, ngisho nobuciko. Lesi sihloko sizoxoxa ngezinkinga ezahlukahlukene zezibonelo mayelana nezindilinga nama-arcs, kanye nezixazululo zazo.
Ukuqonda Izindilinga kanye nama-Arc Ajikelezayo
Indilinga yiqoqo lawo wonke amaphuzu endizeni aqhelelene nendawo ethile ebizwa ngokuthi isikhungo sendilinga. Ibanga ukusuka enkabeni yendilinga kuya kunoma yiliphi iphuzu elisendilinga libizwa ngokuthi i-radius. I-arc yendilinga yingxenye yesiyingi esivalwe amaphuzu amabili endilinga.
Amafomula Ayisisekelo Okudingeka Uwazi
1. Ukuzungeza Kwendilinga (K):
\[
K = 2 \pi r
\]
lapho \( r \) kuyirediyasi yesiyingi kanye \( \pi \approx 3.14 \) noma \( \pi \approx \frac{22}{7} \).
2. Indawo Yesiyingi (A):
\[
A = \pi r^2
\]
3. Ubude be-Arc:
\[
s = \frac{\theta}{360^\circ} \izikhathi 2 \pi r
\]
lapho \( \theta \) kuyi-engeli ephakathi ngamadigri.
4. Indawo Yomkhakha (L):
\[
L = \frac{\theta}{360^\circ} \times \pi r^2
\]
Imibuzo Eyisibonelo Nengxoxo
Umbuzo 1: Ukuzungeza Kwendilinga
Umbuzo:
Indilinga inobubanzi obungu-14 cm. Bala umjikelezo wendilinga.
Ingxoxo:
Ukusebenzisa ifomula yokujikeleza kwendilinga:
\[
K = 2 \pi r
\]
Lapho \(r = 14 \) cm,
\[
K = 2 \izikhathi \frac{22}{7} \izikhathi 14 = 2 \izikhathi 22 \izikhathi 2 = 88 \, \umbhalo{cm}
\]
Ngakho-ke, umjikelezo wendilinga ungama-88 cm.
Umbuzo 2: Indawo Yendilinga
Umbuzo:
Uma unikezwe indilinga enobubanzi obuyi-10 cm. Bala indawo yendilinga.
Ingxoxo:
Okokuqala, sithola irediyasi yendilinga:
\[
r = \frac{d}{2} = \frac{10}{2} = 5 \, \text{cm}
\]
Ukusebenzisa ifomula yendawo yendilinga:
\[
A = \pi r^2
\]
\[
A = \pi \izikhathi 5^2 = \pi \izikhathi 25 \cishe 3.14 \izikhathi 25 = 78.5 \, \umbhalo{cm}^2
\]
Ngakho-ke, indawo yendilinga ingu-78.5 cm².
Umbuzo 3: Ubude be-Arc Ejikelezayo
Umbuzo:
Indilinga enobubanzi obungu-21 cm inomugqa owakha i-engeli ephakathi engu-60°. Bungakanani ubude bomugqa?
Ingxoxo:
Ukusebenzisa ifomula yobude be-arc:
\[
s = \frac{\theta}{360^\circ} \izikhathi 2 \pi r
\]
Lapho \( \theta = 60^\circ \) kanye \( r = 21 \, \text{cm} \),
\[
s = \frac{60^\circ}{360^\circ} \izikhathi 2 \izikhathi \frac{22}{7} \izikhathi 21
\]
\[
s = \frac{1}{6} \izikhathi ezi-2 \izikhathi \frac{22}{7} \izikhathi ezingama-21
\]
\[
s = \frac{1}{6} \izikhathi 132 = 22 \, \umbhalo{cm}
\]
Ngakho-ke, ubude be-arc buyi-22 cm.
Umbuzo 4: Indawo Yomkhakha
Umbuzo:
Bala indawo yesigaba sendilinga ene-engeli ephakathi engu-90° kanye ne-radius engu-7 cm.
Ingxoxo:
Ukusebenzisa ifomula yendawo yomkhakha:
\[
L = \frac{\theta}{360^\circ} \times \pi r^2
\]
Lapho \( \theta = 90^\circ \) kanye \( r = 7 \, \text{cm} \),
\[
L = \frac{90^\circ}{360^\circ} \times \pi \times 7^2
\]
\[
L = \frac{1}{4} \izikhathi \pi \izikhathi 49
\]
\[
L = \frac{49 \pi}{4}
\]
\[
L \cishe \frac{49 \izikhathi 3.14}{4} \cishe \frac{153.86}{4} \cishe 38.465 \, \text{cm}^2
\]
Ngakho-ke, indawo yalo mkhakha ingu-38.465 cm².
Umbuzo 5: Inhlanganisela Yemibuzo Yokuzungeza Neyendawo
Umbuzo:
Indilinga inomjikelezo ongu-44 cm. Bala indawo yendilinga.
Ingxoxo:
Okokuqala, sithola irediyasi yendilinga sisebenzisa ifomula yokujikeleza:
\[
K = 2 \pi r
\]
Lapho \( K = 44 \, \text{cm} \),
\[
44 = 2 \izikhathi \frac{22}{7} \izikhathi r
\]
\[
44 = \frac{44}{7} \times r
\]
\[
r = \frac{44 \izikhathi 7}{44} = 7 \, \umbhalo{cm}
\]
Okulandelayo, bala indawo yendilinga:
\[
A = \pi r^2
\]
\[
A = \pi \izikhathi 7^2 = \pi \izikhathi 49 \cishe 3.14 \izikhathi 49 \cishe 153.86 \, \umbhalo{cm}^2
\]
Ngakho-ke, indawo yendilinga ingu-153.86 cm².
Umbuzo 6: Ukuqhathanisa Phakathi Kwemibuthano
Umbuzo:
Izindilinga ezimbili zine-radii engu-5 cm kanye no-10 cm ngokulandelana. Thola isilinganiso somjikelezo kanye nendawo yezindilinga ezimbili.
Ingxoxo:
Eduze:
Kwindilinga yokuqala \( r_1 = 5 \, \text{cm} \):
\[
K_1 = 2 \pi r_1 = 2 \pi \izikhathi 5 = 10 \pi \, \umbhalo{cm}
\]
Kwendilinga yesibili \( r_2 = 10 \, \text{cm} \):
\[
K_2 = 2 \pi r_2 = 2 \pi \izikhathi 10 = 20 \pi \, \umbhalo{cm}
\]
Ukuqhathaniswa kwe-Perimeter:
\[
\frac{K_1}{K_2} = \frac{10 \pi}{20 \pi} = \frac{1}{2}
\]
Okubanzi:
Kwindilinga yokuqala:
\[
A_1 = \pi r_1^2 = \pi \izikhathi 5^2 = 25 \pi \, \umbhalo{cm}^2
\]
Kwendilinga yesibili:
\[
A_2 = \pi r_2^2 = \pi \izikhathi 10^2 = 100 \pi \, \umbhalo{cm}^2
\]
Ukuqhathaniswa kwendawo:
\[
\frac{A_1}{A_2} = \frac{25 \pi}{100 \pi} = \frac{1}{4}
\]
Ngakho-ke, isilinganiso sezindilinga ezimbili siyi-1:2 kanti isilinganiso sezindawo zazo siyi-1:4.
Isiphetho
Ukuqonda imiqondo eyisisekelo kanye namafomula ezindilinga kanye nama-arcs kubalulekile ekuxazululeni izinkinga ezahlukahlukene ze-geometry. Lesi sihloko sinikeza izibonelo eziningana zezinkinga kanye nezingxoxo zokuqinisa ukuqonda kwakho kwalokhu okubalulekile. Izinkinga zokuzijwayeza kanye nokuqonda okujulile kuzokusiza ukuthi usebenzise le miqondo emikhakheni ehlukahlukene yokufunda kanye nezimo zangempela.
