Su'aalo Tusaale ah oo Ka Hadlaya Mowqifka Barta La Xiriirta Goobo
Go'aaminta booska dhibic marka loo eego goobada waa mowduuc muhiim ah oo ku jira joomatari aasaasiga ah, gaar ahaan daraasadda wareegyada. Maqaalkan, waxaan kaga hadli doonnaa dhowr dhibaato oo tusaale ah oo ku lug leh booska dhibic marka loo eego goobada, iyo sharraxaadahooda. Tani waxay gacan ka geysan doontaa in la caddeeyo fikradda iyada oo loo marayo codsiyada wax ku oolka ah.
Pendahuluan
Kahor inta aan la gelin su'aalaha tusaalaha ah, aan xasuusanno saddexda boos ee suurtogalka ah ee dhibic ku taal goobada:
1. Gudaha goobada: Haddii masaafada barta u jirta bartamaha goobada ay ka yar tahay radius-ka goobada.
2. Goobo ka baxsan: Haddii masaafada barta ilaa bartamaha goobada ay ka weyn tahay gacanka goobada.
3. Goobo: Haddii masaafada u dhaxaysa barta ilaa bartamaha goobada ay la mid tahay gacanka goobada.
Xisaab ahaan, booska barta \(T(x_1, y_1)\) marka loo eego goobada xudunta u ah \(a, b)\) oo leh radius \(r\) waxaa lagu go'aamin karaa iyadoo la barbar dhigayo \(T(x_1, y_1)\) iyo isla'egta goobada, kuwaas oo kala ah:
\[
(x – a)^2 + (y – b)^2 = r^2
\]
Haddii natiijada beddelka \(x_1\) iyo \(y_1\) ee isla'egta ay bixiso qiimaha:
– Ka yar \(r^2\), bartu waxay ku jirtaa gudaha goobada.
– Ka weyn \(r^2\), bartu waxay ka baxsan tahay goobada.
– Sida \(r^2\), qodobku wuxuu ku yaal goobada.
Su'aalo iyo Doodo Tusaale ah
Su'aal 1aad
Go'aami booska barta \(T(3, 4)\) marka loo eego goobada leh isle'egta \( (x - 1)^2 + (y - 2)^2 = 25 \).
Dood:
Tallaabada ugu horreysa waa in la qiimeeyo isle'egta goobada oo la helo masaafada u dhaxaysa barta \(T(3, 4)\) ilaa bartamaha goobada \((1, 2)\).
1. Aqoonso bartamaha iyo gacanka goobada:
Isle'egta goobada: \( (x – 1)^2 + (y – 2)^2 = 25 \)
– Xarunta goobada (\(a, b\)): (1, 2)
– Gacanka goobada (\(r\)): \(\sqrt{25} = 5\)
2. Xisaabi masaafada u dhaxaysa barta \(T(3, 4)\) iyo bartamaha goobada \( (1, 2) \):
\[
D = \sqrt{(3 – 1)^2 + (4 – 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]
Qiimaha \( 2\sqrt{2} \qiyaastii 2 \jeer 1.414 = 2.828 \) (ka yar \(5\)).
3. Gunaanad:
Maadaama \( 2\sqrt{2} < 5 \), markaas barta \( T(3, 4) \) ay ku jirto gudaha goobada. Su'aasha 2aad Goobo waxay leedahay xarun barta \( (0, 0) \) iyo gacan 7 ah. Go'aami booska barta \(P(5, 6)\) marka loo eego goobada.
Su'aal 3aad
Go'aami booska barta \(M(2, -1)\) marka loo eego goobada leh isla'egta \( x^2 + y^2 = 5 \).
Dood:
1. Xisaabi masaafada laga bilaabo barta \(M(2, -1)\) ilaa bartamaha goobada \( (0, 0):
\[
D = \sqrt{(2 – 0)^2 + (-1 – 0)^2} = \sqrt{4 + 1} = \sqrt{5}
\]
2. Isbarbardhig masaafada \(D\) iyo gacanka goobada:
Radius-ka goobada (\(r\)) = \(\sqrt{5}\).
3. Gunaanad:
Maadaama \( \sqrt{5} = \sqrt{5} \), markaas barta \( M(2, -1) \) waxay ku taal goobada.
Su'aal 4aad
Goobo leh xarun ku taal \( (4, 3) \) waxay leedahay gacan \(\sqrt{10}\). Muuji booska barta \( N(7, 7) \) marka loo eego goobadan.
Dood:
1. Isla'egta Goobo:
Isle'egta goobada oo leh xarun \( (4, 3) \) iyo gacan \( \sqrt{10} \) waa:
\[
(x – 4)^2 + (y – 3)^2 = 10
\]
2. Xisaabi masaafada laga bilaabo barta \( N(7, 7) \) ilaa bartamaha goobada \( (4, 3) \):
\[
D = \sqrt{(7 – 4)^2 + (7 – 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
3. Gunaanad:
Maadaama \( 5 > \sqrt{10} \), markaas barta \( N(7, 7) \) waxay ka baxsan tahay goobada.
Xiritaanka
Marka la fahmo sida loo xisaabiyo masaafada dhibicda laga bilaabo bartamaha goobada oo loo barbar dhigo gacanka, waxaan si fudud u go'aamin karnaa booska dhibicda marka loo eego goobada. Doodda maqaalkan waxaa la filayaa inay bixiso faham cad oo ku saabsan fikradda iyo sida loo xalliyo dhibaatooyinka ku lug leh booska dhibicda marka loo eego goobada.
Dhab ahaantii, ogaanshaha booska qodobbadan aad ayey faa'iido ugu leedahay codsiyada xisaabta ee kala duwan, oo ay ku jiraan falanqaynta joomatari, naqshadeynta garaafka, iyo injineernimada. Sidaa darteed, barashada fikraddan waa aasaas muhiim ah oo u qalma fiiro gaar ah iyo faham qoto dheer.
