Konus

Bu nima va nega kerak

Hayotda konus shakliga ko'p duch kelamiz: yo'l to'siqlari, muzqaymoq naychalari va don bunkerlari.

Stereometriyada konus asosiy aylanish jismlaridan biridir. U to'g'ri burchakli uchburchakni kateti atrofida $360^\circ$ aylantirishdan hosil bo'ladi. Asosi doira bo'lgani uchun, piramida kabi konus hajmida ham $\frac{1}{3}$ koeffitsiyenti qatnashadi.

Ushbu darsda konus elementlari, sirt yuzalari, hajmi hamda kesik konus formulalarini o'rganamiz.

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Asosiy g'oya va ta'riflar

1. Konus va uning elementlari

To'g'ri burchakli uchburchakni kateti atrofida to'liq aylantirishdan hosil bo'lgan jismga to'g'ri doiraviy konus deyiladi.

          S (uchi)
         /|\
        / | \
       /  |  \ yasovchi (l)
      /   |H  \
     /    |    \
    (-----O-----)
       R   asos (doira)

• Uchi ($S$) — konusning yuqori uchi.
• Asosi — radiusi $R$ bo'lgan doira ($S_{asos} = \pi R^2$).
• Balandligi ($H = SO$) — uchidan asos markaziga tushirilgan balandlik.
• Yasovchisi ($l = SA$) — uchini asos aylanasi bilan tutashtiruvchi kesma.
• O'q kesimi — o'q orqali o'tuvchi kesim; u asosi $2R$, yon tomonlari $l$, balandligi $H$ bo'lgan teng yonli uchburchak. Agar $l = 2R$ bo'lsa, u teng tomonli konusdir.

2. Elementlar orasidagi bog'lanish

O'q kesimining yarmidan ko'rinadiki, $R, H, l$ to'g'ri burchakli uchburchak tashkil qiladi:

[!IMPORTANT] $l^2 = R^2 + H^2 \iff l = \sqrt{R^2 + H^2}$ $H = \sqrt{l^2 - R^2}, \quad R = \sqrt{l^2 - H^2}$

3. Sirt yuzi va yoyilmasi

Konusning yon sirti yoyilmasi radiusi $l$ bo'lgan doiraviy sektordir. Sektor yoyining uzunligi asos aylanasi uzunligiga ($2\pi R$) teng.

[!IMPORTANT]

• Yon sirt yuzi: $S_{yon} = \pi R l$
• To'la sirt yuzi: $S_{to'la} = S_{yon} + S_{asos} = \pi R l + \pi R^2 = \pi R (l + R)$
• Yoyilma burchagi: $\alpha = \dfrac{R}{l} \cdot 360^\circ$

4. Konusning hajmi

Konusning hajmi asosi va balandligi unga teng bo'lgan silindr hajmining uchdan biriga teng:

[!IMPORTANT] $V = \frac{1}{3} S_{asos} H = \frac{1}{3}\pi R^2 H$

5. Kesik konus

Konus asosiga parallel tekislik bilan kesilganda hosil bo'ladi. Radiuslari $R$ va $r$, balandligi $h$:

• $l = \sqrt{h^2 + (R - r)^2}$
• $S_{yon} = \pi (R + r) l$
• $S_{to'la} = \pi (R + r) l + \pi R^2 + \pi r^2$
• $V = \frac{1}{3}\pi h (R^2 + Rr + r^2)$

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Yechilgan misollar

1-misol. Asosiy parametrlar bo'yicha to'la sirt va hajm

Konus asosi radiusi $R = 6\text{ sm}$, balandligi $H = 8\text{ sm}$. Uning yasovchisi, to'la sirti va hajmini toping.

• Yasovchini Pifagor teoremasidan topamiz: $l = \sqrt{R^2 + H^2} = \sqrt{6^2 + 8^2} = 10\text{ sm}$
• To'la sirt yuzi: $S_{to'la} = \pi R (l + R) = \pi \cdot 6 \cdot (10 + 6) = 96\pi\text{ sm}^2$
• Hajmi: $V = \frac{1}{3}\pi R^2 H = \frac{1}{3}\pi \cdot 36 \cdot 8 = 96\pi\text{ sm}^3$

Javob: $l = 10\text{ sm}$, $S_{to'la} = 96\pi\text{ sm}^2$, $V = 96\pi\text{ sm}^3$.

2-misol. O'q kesimi orqali hisoblash

Konus o'q kesimi to'g'ri burchakli teng yonli uchburchakdan iborat bo'lib, gipotenuzasi $12\text{ sm}$. Konus hajmini toping.

• Gipotenuza asos diametridir: $2R = 12 \implies R = 6\text{ sm}$.
• To'g'ri burchak uchidan gipotenuzaga tushirilgan balandlik gipotenuzaning yarmiga teng: $H = 6\text{ sm}$.
• Hajmi: $V = \frac{1}{3}\pi R^2 H = \frac{1}{3}\pi \cdot 6^2 \cdot 6 = 72\pi\text{ sm}^3$

Javob: $72\pi\text{ sm}^3$.

3-misol. Yon sirt yoyilmasi burchagi

Konus yon sirtining yoyilmasi radiusi $10\text{ sm}$ va markaziy burchagi $216^\circ$ bo'lgan sektordir. Konus hajmini toping.

• Sektor radiusi yasovchidir: $l = 10\text{ sm}$.
• Yoyilma burchagidan asos radiusi $R$ ni aniqlaymiz: $\alpha = \frac{R}{l} \cdot 360^\circ \implies 216^\circ = \frac{R}{10} \cdot 360^\circ \implies R = 10 \cdot \frac{216}{360} = 6\text{ sm}$
• Balandlik: $H = \sqrt{l^2 - R^2} = \sqrt{10^2 - 6^2} = 8\text{ sm}$.
• Hajm: $V = \frac{1}{3}\pi R^2 H = \frac{1}{3}\pi \cdot 36 \cdot 8 = 96\pi\text{ sm}^3$

Javob: $96\pi\text{ sm}^3$.

4-misol. Kesik konusda yon sirt va hajm

Kesik konus radiuslari $3\text{ sm}$ va $7\text{ sm}$, balandligi $3\text{ sm}$. Yon sirti va hajmini toping.

• Yasovchi: $l = \sqrt{3^2 + (7 - 3)^2} = \sqrt{9 + 16} = 5\text{ sm}$.
• Yon sirti: $S_{yon} = \pi (7 + 3) \cdot 5 = 50\pi\text{ sm}^2$.
• Hajmi: $V = \frac{1}{3}\pi \cdot 3 \cdot (49 + 21 + 9) = 79\pi\text{ sm}^3$.

Javob: $S_{yon} = 50\pi\text{ sm}^2$, $V = 79\pi\text{ sm}^3$.

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Mashq

1. Konus asosi radiusi $5\text{ sm}$, balandligi $12\text{ sm}$. Yasovchisi va to'la sirtini toping.
2. Konusning yasovchisi $13\text{ sm}$, yon sirti $65\pi\text{ sm}^2$. Balandligini toping.
3. Konus o'q kesimi tomoni $12\text{ sm}$ bo'lgan teng tomonli uchburchakdir. Konus hajmini hisoblang.
4. Konus balandligi $12\text{ sm}$, asosi radiusi $9\text{ sm}$. Yon sirti yoyilmasi burchagini toping.
5. Konus balandligining o'rtasidan asosga parallel tekislik o'tkazilgan. Hosil bo'lgan kichik konus hajmining berilgan konus hajmiga nisbatini toping.
6. Kesik konus asoslarining radiuslari $2\text{ sm}$ va $5\text{ sm}$, yasovchisi $5\text{ sm}$. Uning balandligi va hajmini hisoblang.
7. Konus hajmi $100\pi\text{ sm}^3$, balandligi $12\text{ sm}$. To'la sirti yuzini toping.
8. Teng tomonli konusning to'la sirti $48\pi\text{ sm}^2$. Uning asos radiusini toping.
9. To'g'ri burchakli uchburchak katetlari $3\text{ sm}$ va $4\text{ sm}$. Uni katta kateti atrofida aylantirishdan hosil bo'lgan jismning to'la sirtini toping.

Javoblar va yechim yo'llari

1. $l = \sqrt{5^2 + 12^2} = 13\text{ sm}$. $S_{to'la} = \pi \cdot 5 \cdot (13 + 5) = 90\pi\text{ sm}^2$. Javob: $l = 13\text{ sm}$, $90\pi\text{ sm}^2$.
2. $S_{yon} = \pi R l \implies 65\pi = 13\pi R \implies R = 5\text{ sm}$. $H = \sqrt{13^2 - 5^2} = 12\text{ sm}$. Javob: $12\text{ sm}$.
3. $2R = 12 \implies R = 6\text{ sm}$, $l = 12\text{ sm}$. $H = \sqrt{12^2 - 6^2} = 6\sqrt{3}\text{ sm}$. $V = \frac{1}{3}\pi \cdot 36 \cdot 6\sqrt{3} = 72\sqrt{3}\pi\text{ sm}^3$. Javob: $72\sqrt{3}\pi\text{ sm}^3$.
4. $l = \sqrt{12^2 + 9^2} = 15\text{ sm}$. $\alpha = \frac{R}{l} \cdot 360^\circ = \frac{9}{15} \cdot 360^\circ = 216^\circ$. Javob: $216^\circ$.
5. O'xshashlik koeffitsiyenti $k = \frac{1}{2}$. Hajmlar nisbati $k^3 = \frac{1}{8}$. Javob: $1 : 8$.
6. Balandlik $h = \sqrt{5^2 - (5 - 2)^2} = 4\text{ sm}$. $V = \frac{1}{3}\pi \cdot 4 \cdot (25 + 10 + 4) = 52\pi\text{ sm}^3$. Javob: $h = 4\text{ sm}$, $V = 52\pi\text{ sm}^3$.
7. $V = \frac{1}{3}\pi R^2 \cdot 12 = 4\pi R^2 = 100\pi \implies R = 5\text{ sm}$. $l = \sqrt{5^2 + 12^2} = 13\text{ sm}$. $S_{to'la} = \pi \cdot 5 \cdot (13 + 5) = 90\pi\text{ sm}^2$. Javob: $90\pi\text{ sm}^2$.
8. Teng tomonli konusda $l = 2R$. $S_{to'la} = 3\pi R^2 = 48\pi \implies R^2 = 16 \implies R = 4\text{ sm}$. Javob: $4\text{ sm}$.
9. $H = 4\text{ sm}$, $R = 3\text{ sm}$, $l = 5\text{ sm}$. $S_{to'la} = \pi \cdot 3 \cdot (5 + 3) = 24\pi\text{ sm}^2$. Javob: $24\pi\text{ sm}^2$.

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Keng tarqalgan xatolar

1. Konus hajmida $\frac{1}{3}$ koeffitsiyentini unutish. ✗ $V = \pi R^2 H$ deb yozish (bu silindr hajmi). ✓ Konus hajmi mos silindr hajmining uchdan biriga teng: $V = \frac{1}{3}\pi R^2 H$.

2. Yon sirt yuzida $l$ o'rniga $H$ ni qo'yish. ✗ $S_{yon} = \pi R H$ deb hisoblash. ✓ Yon sirt yoyilmasining radiusi yasovchidir: $S_{yon} = \pi R l$.

3. Kesik konus hajmida asos yuzalari o'rta geometrigini adashtirish. ✗ $V = \frac{1}{3}\pi h (R + r)^2$ deb hisoblash. ✓ To'g'ri formulada $2Rr$ emas, balki $Rr$ bo'ladi: $V = \frac{1}{3}\pi h (R^2 + Rr + r^2)$.

4. Yoyilma burchagida nisbatni teskari olish. ✗ $\alpha = \frac{l}{R} \cdot 360^\circ$ deb yozish ($360^\circ$ dan oshib ketadi). ✓ Har doim $R < l$ bo'lgani uchun: $\alpha = \frac{R}{l} \cdot 360^\circ$.

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DTM uslubidagi test

Yuqoridagi frontmatter bo'limida keltirilgan 5 ta test savolini ishlab, javoblaringizni tekshiring.

Keyingi qadam

Keyingi darsda aylanish jismlarining yana bir muhim vakili bo'lgan Shar va sfera mavzusini o'rganamiz: sferaning sirt yuzi, shar hajmi hamda uning tekislik bilan kesimlari qoidalarini ko'rib chiqamiz.