Stereometriya: murakkab masalalar

Bu mavzuda DTM nimani so'raydi

DTM imtihonlarida stereometriyadan 2–3 ta masala tushadi. Ular fazoviy tasavvur va bir nechta formulani uyg'un qo'llashni talab qiladi:

1. Ko'pyoqlar: Prizma va piramidaning sirt yuzi, hajmi, apofema va balandlik bog'liqligi.
2. Aylanma jismlar: Silindr, konus va shar sirti hamda hajmi, o'q kesimlari.
3. Fazoviy kombinatsiyalar: Silindr va shar, prizma va konus, kub ichiga shar chizish.
4. Kesimlar va o'xshashlik: Asosga parallel tekislik bilan kesganda yuzalar va hajmlar nisbati.

Asosiy formulalar va usullar

Fazoviy jism	Sirt yuzasi	Hajm	Bog'liqliklar
Parallelepiped	$S = 2(ab + bc + ac)$	$V = abc$	$d^2 = a^2 + b^2 + c^2$
Prizma	$S_{yon} = P_{asos} \cdot H$	$V = S_{asos} \cdot H$	$S_{tola} = S_{yon} + 2S_{asos}$
Muntazam piramida	$S_{yon} = \frac{1}{2}P_{asos}h_a$	$V = \frac{1}{3}S_{asos}H$	$h_a^2 = H^2 + r^2$; $S_{yon} = \frac{S_{asos}}{\cos\varphi}$
Silindr	$S_{yon} = 2\pi RH$	$V = \pi R^2 H$	$S_{tola} = 2\pi R(R + H)$
Konus	$S_{yon} = \pi R l$	$V = \frac{1}{3}\pi R^2 H$	$l^2 = H^2 + R^2$; $S_{tola} = \pi R(R + l)$
Shar va sfera	$S = 4\pi R^2$	$V = \frac{4}{3}\pi R^3$	Kesim: $r^2 = R^2 - d^2$

Yechilgan misollar

1-misol

To'g'ri burchakli parallelepipedning o'lchamlari $2\text{ sm}$, $3\text{ sm}$ va $6\text{ sm}$. Uning bosh diagonali va to'la sirti yuzini toping.

A) $d = 7\text{ sm}, S = 72\text{ sm}^2$
B) $d = 8\text{ sm}, S = 64\text{ sm}^2$
C) $d = 7\text{ sm}, S = 36\text{ sm}^2$
D) $d = 6\text{ sm}, S = 72\text{ sm}^2$

Yechim:

1. Bosh diagonal: $d = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\text{ sm}$.
2. To'la sirti: $S = 2(2 \cdot 3 + 3 \cdot 6 + 2 \cdot 6) = 2(6 + 18 + 12) = 72\text{ sm}^2$.

Tezkor yo'l:
$(a+b+c)^2 = d^2 + S_{tola} \implies 11^2 = 121 \implies S_{tola} = 121 - 49 = 72\text{ sm}^2$.
Javob: A

2-misol

Muntazam to'rtburchakli piramidaning asos tomoni $6\text{ sm}$, yon yoqlari asos tekisligi bilan $60^\circ$ burchak hosil qiladi. Uning hajmini hisoblang.

A) $18\sqrt{3}\text{ sm}^3$
B) $36\sqrt{3}\text{ sm}^3$
C) $54\sqrt{3}\text{ sm}^3$
D) $108\text{ sm}^3$

Yechim:

1. Asos yuzi: $S_{asos} = 6^2 = 36\text{ sm}^2$.
2. Asosga ichki radius: $r = \frac{a}{2} = 3\text{ sm}$.
3. Balandlik: $H = r \cdot \operatorname{tg} 60^\circ = 3\sqrt{3}\text{ sm}$.
4. Hajm: $V = \frac{1}{3} S_{asos} H = \frac{1}{3} \cdot 36 \cdot 3\sqrt{3} = 36\sqrt{3}\text{ sm}^3$.

Tezkor yo'l:
$V = \frac{a^3 \operatorname{tg}\varphi}{6} = \frac{216 \sqrt{3}}{6} = 36\sqrt{3}\text{ sm}^3$.
Javob: B

3-misol

Konusning yasovchisi $10\text{ sm}$ bo'lib, u asos tekisligi bilan $60^\circ$ burchak tashkil qiladi. Konusning to'la sirti yuzini hisoblang.

A) $50\pi\text{ sm}^2$
B) $60\pi\text{ sm}^2$
C) $75\pi\text{ sm}^2$
D) $100\pi\text{ sm}^2$

Yechim:

1. Asos radiusi: $R = 10 \cdot \cos 60^\circ = 5\text{ sm}$.
2. To'la sirt: $S_{tola} = \pi R(R + l) = \pi \cdot 5 \cdot 15 = 75\pi\text{ sm}^2$.

Tezkor yo'l:
$60^\circ$ burchakda $l = 2R$. Demak, $S_{tola} = 3\pi R^2 = 3\pi \cdot 25 = 75\pi\text{ sm}^2$.
Javob: C

4-misol

Silindr ichiga shar ichki chizilgan. Agar sharning hajmi $36\pi\text{ sm}^3$ bo'lsa, silindr hajmini toping.

A) $48\pi\text{ sm}^3$
B) $54\pi\text{ sm}^3$
C) $72\pi\text{ sm}^3$
D) $108\pi\text{ sm}^3$

Yechim:

1. Shar hajmi: $\frac{4}{3}\pi R^3 = 36\pi \implies R = 3\text{ sm}$.
2. Silindr balandligi $H = 2R = 6\text{ sm}$, radiusi $R = 3\text{ sm}$.
3. Hajm: $V_{silindr} = \pi R^2 H = \pi \cdot 9 \cdot 6 = 54\pi\text{ sm}^3$.

Tezkor yo'l:
Arximed qoidasi: $V_{silindr} = \frac{3}{2} V_{shar} = \frac{3}{2} \cdot 36\pi = 54\pi\text{ sm}^3$.
Javob: B

5-misol

Piramidaning balandligi $12\text{ sm}$, asosining yuzi $72\text{ sm}^2$. Uchidan boshlab balandlikni $1 : 2$ nisbatda bo'luvchi asosga parallel kesim yuzini toping.

A) $8\text{ sm}^2$
B) $16\text{ sm}^2$
C) $18\text{ sm}^2$
D) $24\text{ sm}^2$

Yechim:

1. Uchidan masofa: $h_1 = \frac{1}{3} \cdot 12 = 4\text{ sm}$.
2. Koeffitsiyent: $k = \frac{4}{12} = \frac{1}{3}$.
3. Kesim yuzi: $S_{kesim} = k^2 \cdot S_{asos} = \frac{1}{9} \cdot 72 = 8\text{ sm}^2$.

Tezkor yo'l:
Nisbat $1 : 2 \implies k = \frac{1}{3} \implies S = 72 \cdot \frac{1}{9} = 8\text{ sm}^2$.
Javob: A

Mashq

1. To'g'ri prizma asosi $6\text{ sm} \times 8\text{ sm}$ to'g'ri to'rtburchak, balandligi $10\text{ sm}$. To'la sirti yuzini hisoblang.

Javob va yechim

$S_{asos} = 6 \cdot 8 = 48\text{ sm}^2$. $S_{yon} = 2(6+8) \cdot 10 = 280\text{ sm}^2$. $S_{tola} = 2 \cdot 48 + 280 = 376\text{ sm}^2$. *Javob:* $376\text{ sm}^2$.

2. Muntazam uchburchakli piramida asos tomoni $6\text{ sm}$, balandligi $4\text{ sm}$. Hajmini hisoblang.

Javob va yechim

$S_{asos} = \frac{36\sqrt{3}}{4} = 9\sqrt{3}\text{ sm}^2$. $V = \frac{1}{3} \cdot 9\sqrt{3} \cdot 4 = 12\sqrt{3}\text{ sm}^3$. *Javob:* $12\sqrt{3}\text{ sm}^3$.

3. Silindrning o'q kesimi kvadrat bo'lib, yuzi $36\text{ sm}^2$. Silindr hajmini toping.

Javob va yechim

$2R = H = 6\text{ sm} \implies R = 3\text{ sm}, H = 6\text{ sm}$. $V = \pi \cdot 9 \cdot 6 = 54\pi\text{ sm}^3$. *Javob:* $54\pi\text{ sm}^3$.

4. Konusning balandligi $8\text{ sm}$, yasovchisi $10\text{ sm}$. Yon sirti yuzini toping.

Javob va yechim

$R = \sqrt{10^2 - 8^2} = 6\text{ sm}$. $S_{yon} = \pi \cdot 6 \cdot 10 = 60\pi\text{ sm}^2$. *Javob:* $60\pi\text{ sm}^2$.

5. Shar radiusi $3\text{ sm}$ dan $6\text{ sm}$ gacha oshirilsa, uning sirt yuzi necha marta ortadi?

Javob va yechim

Radiusi $k = 2$ marta oshganda, sirt yuzi $k^2 = 4$ marta ortadi. *Javob:* $4$ marta.

6. Muntazam to'rtburchakli piramidaning asos tomoni $6\text{ sm}$, apofemasi $5\text{ sm}$. Balandligini toping.

Javob va yechim

$r = \frac{6}{2} = 3\text{ sm}$. Balandlik: $H = \sqrt{5^2 - 3^2} = 4\text{ sm}$. *Javob:* $4\text{ sm}$.

7. Kubning diagonali $4\sqrt{3}\text{ sm}$. Uning hajmini hisoblang.

Javob va yechim

$d = a\sqrt{3} = 4\sqrt{3} \implies a = 4\text{ sm}$. $V = 4^3 = 64\text{ sm}^3$. *Javob:* $64\text{ sm}^3$.

8. To'g'ri prizma asosi katetlari $6\text{ sm}$ va $8\text{ sm}$ bo'lgan to'g'ri burchakli uchburchak. Yon qirrasi $5\text{ sm}$. Hajmini toping.

Javob va yechim

$S_{asos} = \frac{1}{2} \cdot 6 \cdot 8 = 24\text{ sm}^2$. $V = 24 \cdot 5 = 120\text{ sm}^3$. *Javob:* $120\text{ sm}^3$.

9. Konusning hajmi $96\pi\text{ sm}^3$, balandligi $8\text{ sm}$. To'la sirti yuzini hisoblang.

Javob va yechim

$R^2 = \frac{96 \cdot 3}{8} = 36 \implies R = 6\text{ sm}$. $l = 10\text{ sm}$. $S_{tola} = \pi \cdot 6 \cdot 16 = 96\pi\text{ sm}^2$. *Javob:* $96\pi\text{ sm}^2$.

10. Radiusi $5\text{ sm}$ bo'lgan shardan markazdan $3\text{ sm}$ masofada kesim o'tkazilgan. Kesim yuzini toping.

Javob va yechim

$r = \sqrt{5^2 - 3^2} = 4\text{ sm}$. $S = \pi \cdot 4^2 = 16\pi\text{ sm}^2$. *Javob:* $16\pi\text{ sm}^2$.

Keng tarqalgan tuzoqlar

1. Piramida apofemasi va balandligini adashtirish:

   • ✗ Xato: Yon sirtda apofema o'rniga balandlikni qo'yish ($S_{yon} = \frac{1}{2}P_{asos}H$).
   • ✓ To'g'ri: Yon sirtda apofema qo'llanadi: $S_{yon} = \frac{1}{2}P_{asos}h_a$. Balandlik esa hajmda ishlatiladi: $V = \frac{1}{3}S_{asos}H$.

2. Konus yon sirtida yasovchi o'rniga balandlikni yozish:

   • ✗ Xato: $S_{yon} = \pi R H$ deb hisoblash.
   • ✓ To'g'ri: Yon sirt $S_{yon} = \pi R l$ ga teng, bu yerda $l = \sqrt{H^2 + R^2}$ — yasovchi.

3. Shar sirti va hajmi formulalarida radius darajasini almashtirish:

   • ✗ Xato: Sirt yuzini $4\pi R^3$ yoki hajmni $\frac{4}{3}\pi R^2$ deb yozish.
   • ✓ To'g'ri: Yuza kvadratik ($S = 4\pi R^2$), hajm esa kubikdir ($V = \frac{4}{3}\pi R^3$).

4. Fazoviy jismlar o'xshashligida hajm va yuza nisbatini farqlamaslik:

   • ✗ Xato: Chiziqli o'lchamlar $k$ marta kattalashganda hajm ham $k^2$ marta oshadi deb o'ylash.
   • ✓ To'g'ri: Yuzalar nisbati $k^2$, hajmlar nisbati esa $k^3$ marta ortadi.