Su'aalo Tusaale ah oo Ka Hadlaya Goobo iyo Waxyaabaha La Xiriira
Goobooyinku waa mowduuc muhiim ah oo ku jira joomatari-ga matrix-ka, halkaas oo lagu muujiyo fikradaha qoto dheer ee ku saabsan masaafada, xaglaha, iyo qaabka. Hal fikrad oo si joogto ah looga hadlo mowduucan waa xariiqda togan ee goobo. Maqaalkan, waxaan ka hadli doonnaa dhowr dhibaato oo tusaale ah oo ku lug leh goobo iyo toganto.
Fahamka Aasaasiga ah ee Goobo iyo Tangents
Goobo
Goobo waa qaab joomatari ah oo ay sameeyeen dhammaan dhibcaha ku jira diyaarad oo ah masaafo go'an oo u jirta meel la bixiyay oo loo yaqaan bartamaha goobada. Masaafadan go'an waxaa loo yaqaan gacanka goobada.
Tangent
Taangent-ka goobada waa xariiq taabanaysa goobada hal dhibic. Bartan waxaa loo yaqaan barta taangent-ka. Tangent-ku waxay leeyihiin dhowr astaamood oo muhiim ah, oo ay ku jiraan:
– Xariiqda taangent-ku had iyo jeer waxay ku toosan tahay gacanka goobada marka ay timaado meesha taangent-ku.
– Dhererka tangent-ka laga bilaabo barta ka baxsan goobada ilaa goobada waa isku mid haddii laba tangent laga soo jiido bartaas.
Su'aalo iyo Doodo Tusaale ah
Hoos waxaan ku soo bandhigi doonaa dhowr su'aalood oo tusaale ah oo si faahfaahsan uga hadlaya fikradda wareegyada iyo tangents-ka.
Su'aal Tusaale ah 1: Helitaanka Dhererka Xariiqda Tangent
Su'aal:
Waxaa la siiyay goobaabin leh xarun \(O\) iyo gacan \(r = 6 \, \text{cm}\). Laga bilaabo barta \(P\) ee ka baxsan goobaabka oo 10 cm u jirta bartamaha goobada, laba tangents \(PA\) iyo \(PB\) ayaa lagu sawiray goobada. Xisaabi dhererka tangent \(PA\).
Dood:
Dhibaatadan, waxaan isticmaali karnaa aragtida Pythagorean. Sawir saddexagal \(\saddexagal OAP\):
– \(OP = 10 \, \text{cm}\) (masaafada laga bilaabo barta dibadda ilaa bartamaha goobada)
– \(OA = 6 \, \text{cm}\) (cabir goobada)
– \(PA\) waa xariiqda tangent-ka ee la heli karo
\[
OP^2 = OA^2 + PA^2
\]
\[
10^2 = 6^2 + PA^2
\]
\[
100 = 36 + PA^2
\]
\[
PA^2 = 64
\]
\[
PA = \sqrt{64} = 8 \, \qoraal{cm}
\]
Markaa, dhererka xariiqda tangent \(PA\) waa 8 cm.
Su'aal Tusaale 2aad: Helitaanka Barta Isku-dhafka ah
Su'aal:
Waxaa la siiyay goobaabin leh isle'egta \((x – 3)^2 + (y – 4)^2 = 25\) iyo xariiq \(y = 2x + 1\). Go'aami barta isku-dhafka ah ee u dhaxaysa goobada iyo xariiqda.
Dood:
Marka hore, waxaan aqoonsannaa bartamaha iyo gacanka goobada:
– Xarunta \(O(3, 4)\)
– Gacanka \(r = \sqrt{25} = 5\)
Si aan u helno barta isku-dhafka, aan u qaadanno in barta isku-dhafka ay tahay \(T(x_1, y_1)\) taas oo sidoo kale ku taal xariiqda \(y = 2x + 1\). Kadib:
\[
y_1 = 2x_1 + 1
\]
\(T(x_1, y_1)\) sidoo kale waa inay buuxisaa isla'egta goobada:
\[
(x_1 – 3)^2 + (y_1 – 4)^2 = 25
\]
Ku beddel \(y_1 = 2x_1 + 1\) isle'egta goobada:
\[
(x_1 – 3)^2 + ((2x_1 + 1) – 4)^2 = 25
\]
\[
(x_1 – 3)^2 + (2x_1 – 3)^2 = 25
\]
Waxaan u baahanahay inaan xisaabino laba jibbaaran.
\[
(x_1 – 3)^2 = x_1^2 – 6x_1 + 9
\]
\[
(2x_1 – 3)^2 = 4x_1^2 – 12x_1 + 9
\]
Isku dar labada natiijo:
\[
x_1^2 – 6x_1 + 9 + 4x_1^2 – 12x_1 + 9 = 25
\]
\[
5x_1^2 – 18x_1 + 18 = 25
\]
Ka jar 25 labada dhinac:
\[
5x_1^2 – 18x_1 – 7 = 0
\]
Xalli isle'egta labajibbaaran:
\[
x_1 = \frac{18 \pm \sqrt{18^2 + 4 \jeer 5 \jeer 7}}{2 \jeer 5}
\]
\[
x_1 = \frac{18 \pm \sqrt{324 + 140}}{10}
\]
\[
x_1 = \frac{18 \pm \sqrt{464}}{10}
\]
\[
x_1 = \frac{18 \pm 2\sqrt{116}}{10}
\]
\[
x_1 = \frac{18 \pm 2\sqrt{4 \jeer 29}}{10}
\]
\[
x_1 = \frac{18 \pm 4\sqrt{29}}{10}
\]
\[
x_1 = 1.8 \pm 0.4 \ sqrt{29}
\]
Xisaabi qiimaha \(y_1\):
Taas oo qancisa y = 2x + 1:
– Haddii \(x_1 = 1.8 + 0.4\sqrt{29}\), markaa \(y_1 = 2(1.8 + 0.4\sqrt{29}) + 1\)
– Haddii \(x_1 = 1.8 – 0.4\sqrt{29}\), markaas \(y_1 = 2(1.8 – 0.4\sqrt{29}) + 1\)
Qiimaynta:
Markaa waxaan helnaa laba dhibcood oo isgoys ah oo isle'egta goobada ee xariiqdaas.
Su'aal Tusaale ah 3: Go'aaminta Isle'egta Xariiqda Tangent
Su'aal:
Si loo siiyo goobaabin leh isla'egta \((x – 2)^2 + (y – 3)^2 = 20\). Go'aami isle'egta xariiqda taangent-ka ee goobada dhex marta barta \(6, 7)\).
Dood:
Taangent-ka goobada oo leh xarun \((h, k)\) iyo radius \(r\) laga bilaabo bar dibadeed oo la yaqaan waxaa lagu heli karaa isla'egta:
Xariiqda tangent-ka waxay dhex martaa barta dibadda \((x_1, y_1)\):
\[
(x - 2) (x_1 - 2) + (y - 3) (y_1 - 3) = 20
\]
Ku beddel barta dibadda \(6, 7)\):
\[
(x - 2) (6 - 2) + (y - 3) (7 - 3) = 20
\]
\[
4(x – 2) + 4(y – 3) = 20
\]
\[
4(x – 2 + y – 3) = 20
\]
\[
4x + 2y -20 = 20
\]
\[
4x + 4y -20 = 20
\]
\[
x + y = 5
\]
Isle'egta xariiqda tangent-ka waa:
\[
x + y = 9
\]
Markaa, kala duwanaanshaha isle'egta xariiqda ee dhex marta barta xariiqda goobada waa mid aad u weyn waxayna isbeddeli kartaa iyadoo ku xiran natiijada ama matalaadda muuqaalka.
Gabagabo
Doodda wareegyada iyo tangent-ku waxay daboolaysaa dhowr dhinac oo aasaasi ah oo xisaabta ah, laga bilaabo adeegsiga qaacidooyinka aasaasiga ah sida aragtida Pythagorean ilaa xallinta isla'egyada labajibbaaran. Tusaalooyinkan, waxaan ku horumarin karnaa faham wanaagsan oo ku saabsan sida loo dabaqo fikradahan xaaladaha qaar adag. Waxaan rajeyneynaa, maqaalkani wuxuu gacan ka geystay bixinta sawir cad oo ku saabsan sida loo wajaho loona xalliyo dhibaatooyinka ku lug leh wareegyada iyo tangent-ka.
