Tambayoyi Misali Game da Ma'anar Da'ira
Da'irar tana ɗaya daga cikin manyan siffofi na lissafi da muke fuskanta akai-akai a rayuwar yau da kullun. A cikin lissafi, da'irori suna da ma'anoni da halaye na musamman. Wannan labarin zai bincika ma'anar da'ira da abubuwan da ke da alaƙa da ita a cikin zurfi, tare da samar da misalai da mafita da yawa don zurfafa fahimtarmu game da da'ira.
Ma'anar Da'ira
Da'ira ita ce saitin dukkan maki a cikin jirgin sama wanda ke da nisan da ya yi daidai da wani wuri mai tsayayye da ake kira tsakiyar da'irar. Nisa tsakanin tsakiya da kowane wuri a kan da'irar ana kiransa da radius na da'irar. Jimlar daidaituwar da'ira mai tsakiya a wurin \((h, k)\) da radius \(r\) an bayar da ita ta hanyar:
\[ (x – h)^2 + (y – k)^2 = r^2 \]
A can:
– \((h, k)\) sune daidaitattun tsakiyar da'irar,
– \(r\) shine radius na da'irar,
– \(x\) da \(y\) sune daidaitattun kowane wuri akan da'irar.
Abubuwan Da'ira
Kafin mu shiga cikin tambayoyin misalai, yana da kyau mu san wasu muhimman abubuwan da ke cikin da'irar:
1. Tsakiyar Da'ira: Matsakaici mai tsayi wanda shine tsakiyar dukkan maki waɗanda suke da nisan iri ɗaya.
2. Radius (r): Nisa daga tsakiyar da'irar zuwa kowane wuri akan da'irar.
3. Diamita (d): Layi madaidaiciya wanda ke ratsa tsakiyar da'irar kuma ya haɗa maki biyu akan da'irar, yana da tsayi sau biyu sau biyu na radius (\(d = 2r\)).
4. Arc: Sashen da'irar da ke tsakanin wurare biyu a kan da'irar.
5. Chord: Layi madaidaiciya wanda ya haɗa maki biyu a kan da'ira amma bai ratsa ta tsakiya ba.
6. Apothem: Mafi gajarta tazara daga tsakiyar da'ira zuwa ga maƙallin waƙa.
7. Kusurwar Tsakiya: Kusurwar da aka samar ta hanyar radius biyu daga tsakiyar da'irar.
8. Kusurwar Gefen: Kusurwar da aka samar ta hanyar sarƙoƙi biyu waɗanda suka haɗu a wuri ɗaya a kan da'ira.
Tambayoyi da Tattaunawar Samfura
Misali Tambaya ta 1
Tambaya: An ba da da'ira mai tsakiya a wurin \(3, 4)\) da kuma wucewa ta wurin \(7, 4)\). Kayyade daidaiton da'irar.
Tattaunawa:
Domin nemo lissafin da'ira, muna buƙatar sanin radius ɗinsa da farko. Tunda da'irar ta ratsa ta wurin \(7, 4)\), za mu iya ƙididdige nisan da ke tsakanin wannan wurin da tsakiyar da'irar, wato \((3, 4)\).
\[
r = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}
\]
\[
r = \sqrt{(7 – 3)^2 + (4 – 4)^2}
\]
\[
r = \sqrt{4^2 + 0^2}
\]
\[
r = 4
\]
Tare da tsakiya a \((3, 4)\) da radius 4, lissafin da'irar shine:
\[
(x – 3)^2 + (y – 4)^2 = 4^2
\]
\[
(x – 3)^2 + (y – 4)^2 = 16
\]
Misali Tambaya ta 2
Tambaya: Kayyade yankin da kewayen da'irar da ke da radius na 5 cm.
Tattaunawa:
– Ana iya ƙididdige yankin da'ira (A) ta amfani da dabarar \(A = \pi r^2\),
\[
A = \pi \sau 5^2
\]
\[
A = 25\pi \rubutu{ cm}^2
\]
Idan \(\pi \approx 3.14\), to:
\[
A \kimanin sau 25 3.14 = 78.5 \rubutu{ cm}^2
\]
– Ana iya ƙididdige kewayen da'ira (C) ta amfani da dabarar \(C = 2\pi r\),
\[
C = sau 2 \pi \sau 5
\]
\[
C = 10\pi \rubutu{ cm}
\]
Idan \(\pi \approx 3.14\), to:
\[
C \kimanin 10 \sau 3.14 = 31.4 \rubutu{ cm}
\]
Misali Tambaya ta 3
Tambaya: Da'ira tana da tsakiya a wurin O da kuma radius na 7 cm. Idan aka zana igiyar tsayin 10 cm a kan da'irar, a ƙayyade mafi ƙarancin nisa daga tsakiyar O zuwa igiyar.
Tattaunawa:
Domin nemo mafi guntun nisa daga tsakiya zuwa ga chord, muna amfani da manufar apothem, wanda shine mafi guntun nisa daga tsakiya zuwa chord. Tare da radius na 7 cm da tsawon chord na 10 cm, za mu iya magance wannan matsalar ta amfani da alwatika mai daidai da aka samar.
Idan tsakiyar maƙallin shine maki C, to OC shine apothem ɗin da muke nema. Idan A da B sune ƙarshen maƙallin, to AC da BC kowannensu shine 5 cm (rabin santimita 10).
Daga alwatika OAC wanda aka yi masa kusurwar dama a C, muna amfani da ka'idar Pythagorean:
\[
OA^2 = OC^2 + AC^2
\]
\[
7^2 = OC^2 + 5^2
\]
\[
49 = OC^2 + 25
\]
\[
OC^2 = 49 – 25
\]
\[
OC^2 = 24
\]
\[
OC = \sqrt{24} = 2\sqrt{6} \text{ cm}
\]
Misali Tambaya ta 4
Tambaya: An ba da da'ira mai lissafin \(x^2 + y^2 + 6x – 8y + 9 = 0\). Kayyade tsakiya da radius na da'irar.
Tattaunawa:
Domin nemo tsakiya da radius, muna canza lissafin zuwa tsari na yau da kullun:
\[
x^2 + y^2 + 6x – 8y + 9 = 0
\]
Mun warware ta hanyar kammala siffar quadratic:
\[
x^2 + 6x + y^2 – 8y = -9
\]
Ƙara da cirewa (6/2)\(^2\) zuwa x-term da kuma (8/2)\(^2\) zuwa y-term:
\[
x^2 + 6x + 9 + y^2 – 8y + 16 = -9 + 9 + 16
\]
\[
(x + 3)^2 + (y – 4)^2 = 16
\]
Daga lissafin \((x + 3)^2 + (y – 4)^2 = 16\), mun gano cewa tsakiyar da'irar shine \((-3, 4)\) kuma radius shine \(r = \sqrt{16} = 4\).
Kammalawa
Da'irar wata muhimmiyar manufa ce ta lissafi. Ta hanyar misalan matsaloli da tattaunawa da ke sama, za mu iya samun fahimtar fannoni daban-daban da halayen da'irori. Kwarewar wannan batu zai sauƙaƙa fahimta da magance matsalolin lissafi masu rikitarwa.
