Ceistean Eisimpleir agus Deasbad mu Roinnean Cearcaill
Tha earrannan cearcaill nan cuspair cudromach ann am matamataig a nochdas gu tric ann an deuchainnean agus ceistean cleachdaidh. Is e earrann am pàirt de chearcall air a chuartachadh le dà radii agus an arc a tha gan ceangal. San artaigil seo, bruidhnidh sinn air grunn eisimpleirean de dhuilgheadasan air earrannan cearcaill, còmhla ri mìneachaidhean mionaideach, gus ar tuigse a dhoimhneachadh.
Mìneachadh air Roinn Cearcaill
Is e earrann de chearcall earrann de chearcall air a chuartachadh le dà radii agus aon bhogha. Tha farsaingeachd earrann air a thomhas a rèir bloigh de farsaingeachd iomlan a’ chearcaill. Seo am prìomh fhoirmle a thathar a’ cleachdadh airson earrann obrachadh a-mach:
– Raon Juring: [L_juring = \frac{\theta}{360^\circ} \times \pi \times r^2\]
– Fad a’ Bhog: [P_b = \frac{\theta}{360^\circ} \times 2\pi r]
Càite:
– ’S e meud ceàrn na roinne ann an ceumannan a th’ ann an (\theta\),
– \(r \) 's e radius a' chearcaill a th' ann,
– Tha \( \pi \) na luach cunbhalach (timcheall air 3.14159).
Ceistean Eisimpleir agus Deasbad
Ceist 1:
Ma tha cearcall ann le radius de 10 cm agus earrann le ceàrn meadhanach de 90°. Obraich a-mach farsaingeachd na h-earrainn.
Deasbad:
Tha fios air:
– \(r = 10 \) cm
– \( \theta = 90^\circ \)
Bidh sinn a’ cleachdadh na foirmle airson farsaingeachd roinne:
[L_juring = \frac{\theta}{360^\circ} \times π \times r^2\]
[L_juring = \frac{90^\circ}{360^\circ} \times π \times (10\text{ cm})^2\]
[L_juring = \frac{1}{4} \times \pi \times 100 \text{ cm}^2 \]
[L_juring = 25\pi\text{ cm}^2\]
Ma ghabhas sinn π a ghabhail mar 3.14, an uairsin:
[L_juring = 25 × 3.14 cm² = 78.5 cm²]
Mar sin, tha farsaingeachd na roinne 78.5 cm².
Ceist 2:
Tha radius 7 cm agus fad bogha 11 cm aig earrann de chiorcal. Obraich a-mach ceàrn meadhanach na h-earrainn ann an radianan.
Deasbad:
Tha fios air:
– \(r = 7 \) cm
– Fad a’ bhogha (P_b = 11 cm)
Bidh sinn a’ cleachdadh foirmle faid a’ bhogha gus an ceàrn _( \theta \) a lorg:
[P_b = \frac{\theta}{360^\circ} \times 2π r\]
Leis gu bheil sinn air iarraidh an ceàrn a lorg ann an radianan, bidh sinn a’ cur 2 pi radianan an àite 360°:
[P_b = θ uair r]
[11 = θ × 7]
[\theta = \frac{11}{7}\]
[\theta \approx 1.57 \text{ rad}\]
Mar sin, is e 1.57 radians ceàrn meadhanach na roinne.
Ceist 3:
Tha earrann aig cearcall le radius de 16 cm le farsaingeachd de 200 cm². Obraich a-mach ceàrn meadhanach na h-earrainn.
Deasbad:
Tha fios air:
– \(r = 16 \) cm
– \( L_juring = 200 \text{ cm}^2 \)
Bidh sinn a’ cleachdadh foirmle na sgìre airson roinn gus \( \theta \) a lorg:
[L_juring = \frac{\theta}{360^\circ} \times π \times r^2\]
[200 = \frac{\theta}{360^\circ} \times π \times (16)^2\]
200 = θ/θ/360 × π × 256]
[200 = \frac{\theta \times 256 \times π}{360^\circ}]
[200 × 360 × circumference = θ₀ 256 × 3.14]
[72000 = θ x 256 x 3.14]
[72000 = θ × 804.64]
[\theta = \frac{72000}{804.64}\]
[\theta \timcheall air 89.45^\circ\]
Mar sin, tha ceàrn meadhanach na roinne timcheall air 89.45°.
Ceist 4:
Obraich a-mach cearcall-thomhas iomlan earrainn aig a bheil radius 12 cm agus ceàrn meadhanach 120°.
Deasbad:
Tha fios air:
– \(r = 12 \) cm
– \( \theta = 120^\circ \)
An toiseach, lorg sinn fad a’ bhogha:
[P_b = \frac{\theta}{360^\circ} \times 2π r\]
[P_b = \frac{120^\circ}{360^\circ} \times 2π \times 12\]
[P_b = \frac{1}{3} \times 2π \times 12\]
\[P_b = 8\pi\text{ cm}\]
An uairsin, bidh sinn a’ tomhas cearcall-thomhas na roinne (fad a’ bhogha + dà radii):
\[K = 2r + P_b\]
[K = 2 × 12 cm + 8 pi cm]
[K = 24 cm + 8 pi cm]
Ma ghabhas sinn π a ghabhail mar 3.14, an uairsin:
[K = 24 cm + 8 × 3.14 cm]
[K = 24 cm + 25.12 cm]
\[K = 49.12\text{ cm}\]
Mar sin, tha cearcall-thomhas iomlan na roinne 49.12 cm.
Ceist 5:
Ma tha earrann aig cearcall le radius 18 cm a tha a’ dèanamh ceàrn 45°, obraich a-mach fad a’ bhogha agus farsaingeachd na h-earrainn.
Deasbad:
Tha fios air:
– \(r = 18 \) cm
– \( \theta = 45^\circ \)
1. Fad a' Bhogha:
[P_b = \frac{\theta}{360^\circ} \times 2π r\]
[P_b = \frac{45^\circ}{360^\circ} \times 2π \times 18\text{ cm}]
[P_b = \frac{1}{8} \times 36\pi\text{ cm}]
\[P_b = 4.5\pi\text{ cm}\]
Ma ghabhas sinn π a ghabhail mar 3.14, an uairsin:
[P_b = 4.5 × 3.14 cm = timcheall air 14.13 cm]
Mar sin, tha fad a’ bhogha mu 14.13 cm.
2. Raon na Roinne:
[L_juring = \frac{\theta}{360^\circ} \times π \times r^2\]
[L_juring = \frac{45^\circ}{360^\circ} \times π \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\]
Ma ghabhas sinn π a ghabhail mar 3.14, an uairsin:
[L_juring = 40.5 × 3.14 cm² timcheall air 127.17 cm²]
Mar sin, tha farsaingeachd na roinne timcheall air 127.17 cm².
Co-dhùnadh
San artaigil seo, tha sinn air grunn eisimpleirean de dhuilgheadasan a dheasbad a thaobh earrannan cearcaill agus na fuasglaidhean aca. Tha brìgh tuigse air earrannan cearcaill na laighe ann a bhith a’ maighstireachd nam foirmlean bunaiteach airson farsaingeachd earrainn agus fad bogha obrachadh a-mach. Tha sinn an dòchas gun cuidich cleachdadh tric agus tuigse air mar a chuireas tu na foirmlean seo an sàs ann an diofar sheòrsaichean dhuilgheadasan le bhith a’ leasachadh do chomas air duilgheadasan coltach riutha fhuasgladh.
