Circles: Class 10 Maths Chapter 10 Hinglish mein

Riya kehti hai: “Circle ke sirf do tangents hote hain, kyunki main apna ruler uske saath sirf do sides par hi laga paati hoon.” Kya woh sahi hai? Circle ke kitne tangents hote hain?

Khaali jagah bharo.

1. (a)

   Jis line ka circle ke saath exactly ek common point ho, use kehte hain

   Check karo

   Answer dekho

   Answer tangent

   Ek common point: tangent.
2. (b)

   Secant aur circle ke common points hote hain.

   Check karo

   Answer dekho

   Answer two

   Secant circle ko do points par kaatta hai.
3. (c)

   Ek di gayi line ke parallel saari lines mein se zyada se zyada lines circle ke tangents ho sakti hain.

   Check karo

   Answer dekho

   Answer two

   Sirf centre se distance $r$ par wali do lines, dono taraf ek-ek, circle ko chhooti hain.
4. (d)

   Ek tangent circle ko jis point par chhoota hai woh hai $P\mathrm{.}$ Point $P$ ko kehte hain

   Check karo

   Answer dekho

   Answer point of contact

   Yeh point of contact hai.

$O$ 9 cm radius wale circle ka centre hai. Ek point $Q$ se guzarti line circle ko chhooti hai, point of contact hai $P,$ aur $OQ=41$ cm. Length $QP$ hai

Ek circle banao jiska radius 3 cm ho aur centre ho $O,$ aur ek line $m$ jo 5 cm door ho centre se, yaani $O\mathrm{.}$ Phir banao (a) $m$ ke parallel ek line jo circle ka tangent ho, aur (b) $m$ ke parallel ek line jo secant ho. Tumhari har line $O$ se kitni door hai?

Ek point se, jo hai $P,$ circle par tangent 20 cm lamba hai, aur $P$ centre se 29 cm door hai. Circle ka radius hai

$P$ se tangents ek circle (centre $O$) ko do points par chhoote hain, $A$ aur $B,$ aur radii $OA$ aur $OB$ banate hain $\mathrm{\angle }AOB={125}^{\circ }\mathrm{.}$ To $\mathrm{\angle }APB$ hai

$P$ se kheenche gaye do tangents ek circle (centre $O$) ko do points par chhoote hain, $A$ aur $B,$ aur unke beech ka angle hai ${64}^{\circ }\mathrm{.}$ To $\mathrm{\angle }AOP$ hai

$AB$ ek circle ka diameter hai. Prove karo ki $A$ par tangent aur $B$ par tangent kabhi nahi milte.

Ek carpenter ek gol table top ke kinaare ke ek point $P$ par tangent banati hai. Phir woh $P$ se us tangent ke right angle par ek line banati hai.

1. (a)

   Prove karo ki yeh line table top ke centre $O$ se guzarti hai.

   Solution dikhao

   Theorem 10.1 se $OP$ tangent par perpendicular hai, aur tangent ka point of contact hai $P\mathrm{.}$ $P$ se sirf ek line tangent par perpendicular ho sakti hai, isliye carpenter ki line wahi line hai jo hai $OP\mathrm{.}$ Yeh line guzarti hai centre se, yaani $O\mathrm{.}$

2. (b)

   Do aisi lines se woh centre kaise dhoondh sakti hai?

   Solution dikhao

   Woh kinaare ke ek doosre point $Q$ par yahi dohraati hai ($P$ se guzarte diameter ke doosre end par nahi). Dono lines centre se guzarti hain, yaani $O,$ isliye centre wahi hai jahan dono lines cross karti hain.

Ek point circle ke centre se 37 cm door hai, aur usse tangent 35 cm lamba hai. Radius nikaalo, cm mein.

Do concentric circles ke radii 17 cm aur 8 cm hain. Bade circle ka ek chord chhote circle ko chhoota hai. Chord ki length nikaalo, cm mein.

Ek quadrilateral $PQRS$ ek circle ko circumscribe karta hai: uski har side circle ko chhooti hai.

1. (a)

   Prove karo ki $PQ+RS=QR+SP\mathrm{.}$

   Solution dikhao

   Maan lo corners hain $P,$ $Q,$ $R,$ $S$ aur unse tangent lengths hain $a,$ $b,$ $c,$ $d;$ Theorem 10.2 se har corner se do tangents barabar hain.

   Tab $PQ=a+b,$ $QR=b+c,$ $RS=c+d$ aur $SP=d+a\mathrm{.}$

   Isliye $PQ+RS=a+b+c+d=QR+SP\mathrm{.}$

2. (b)

   Agar $PQ=9$ cm, $QR=7$ cm aur $RS=6$ cm ho, to $SP$ cm mein nikaalo.

   cm

   Check karo

   Solution dikhao

   $9+6=7+SP,$ isliye $SP=8$ cm.

Do parallel lines $\mathrm{\ell }$ aur $m$ dono ek circle ko chhooti hain, jiska centre hai $O\mathrm{.}$ Ek teesra tangent circle ko $R$ par chhoota hai, $\mathrm{\ell }$ se $E$ par milta hai, aur $m$ se jis point par milta hai woh hai $F\mathrm{.}$

1. (a)

   Prove karo ki $\mathrm{\angle }EOF={90}^{\circ }\mathrm{.}$

   Solution dikhao

   Maan lo $\mathrm{\ell }$ circle ko $U$ par chhooti hai aur $m$ jis point par chhooti hai woh hai $V\mathrm{.}$ $E$ se tangents $EU$ aur $ER$ barabar hain, isliye $OE$ bisect karta hai $\mathrm{\angle }UEF$ ko (Theorem 10.2 ke baad ke remarks). Isi tarah $OF$ bisect karta hai is angle ko: $\mathrm{\angle }VFE\mathrm{.}$

   Kyunki $\mathrm{\ell }\parallel m,$ co-interior angles $\mathrm{\angle }UEF$ aur $\mathrm{\angle }VFE$ ka sum hai ${180}^{\circ }\mathrm{.}$ Unke halves ka sum hai ${90}^{\circ }:$ $\mathrm{\angle }OEF+\mathrm{\angle }OFE={90}^{\circ }\mathrm{.}$

   Isliye triangle lo, $\mathrm{△}EOF,$ to $\mathrm{\angle }EOF={180}^{\circ }-{90}^{\circ }={90}^{\circ }\mathrm{.}$

2. (b)

   Agar radius 6 cm hai aur $ER=4$ cm, to $RF$ cm mein nikaalo.

   cm

   Check karo

   Solution dikhao

   $OR\perp EF$ (Theorem 10.1). Right $\mathrm{△}EOF$ mein altitude $OR$ se $\mathrm{△}ERO\sim \mathrm{△}ORF$ (AA) milta hai, isliye $\frac{ER}{OR}=\frac{OR}{RF}\mathrm{.}$

   $RF=\frac{O{R}^2}{ER}=\frac{36}{4}=9$ cm.

External point $P$ se tangents centre $O$ wale circle ko do points par chhoote hain, $A$ aur $B\mathrm{.}$

1. (a)

   Prove karo ki $\mathrm{\angle }APB+\mathrm{\angle }AOB={180}^{\circ }\mathrm{.}$

   Solution dikhao

   Quadrilateral hai $OAPB,$ jismein $\mathrm{\angle }OAP=\mathrm{\angle }OBP={90}^{\circ }$ (Theorem 10.1). Chaaron angles ka sum hai ${360}^{\circ },$ isliye $\mathrm{\angle }APB+\mathrm{\angle }AOB={360}^{\circ }-{180}^{\circ }={180}^{\circ }\mathrm{.}$

2. (b)

   Isse nikaalo ki $O,$ $A,$ $P$ aur $B$ ek hi circle par hain.

   Solution dikhao

   $OAPB$ mein $A$ aur $B$ par opposite angles ka sum hai ${180}^{\circ }\mathrm{.}$ Class 9 se, jis quadrilateral mein opposite angles ke ek pair ka sum ${180}^{\circ }$ ho, woh cyclic hota hai. (Yeh wahi circle hai jiska diameter $OP$ hai, do tangents kyun hote hain wale box mein.)

Ek parallelogram $ABCD$ ek circle ko circumscribe karta hai.

1. (a)

   Prove karo ki $ABCD$ ek rhombus hai.

   Solution dikhao

   Tangent pairs se (jaise Question 8 mein), $AB+CD=BC+DA\mathrm{.}$

   Parallelogram mein $AB=CD$ aur $BC=DA,$ isliye $2AB=2BC,$ yaani $AB=BC\mathrm{.}$

   Jis parallelogram ki do adjacent sides barabar hon, uski chaaron sides barabar hoti hain: woh rhombus hai.

2. (b)

   Kya koi rectangle jo square nahi hai, circle ko circumscribe kar sakta hai? Samjhao.

   Solution dikhao

   Nahi. Rectangle ek parallelogram hai, isliye (a) se use rhombus hona padega, aur jis rectangle ki saari sides barabar hon woh square hota hai.

$\mathrm{△}PQR$ 3 cm radius wale circle ko circumscribe karta hai, jo $QR$ ko $S$ par chhoota hai, jahan $QS=4$ cm aur $SR=6$ cm.

1. (a)

   $PQ$ cm mein nikaalo.

   cm

   Check karo

   Solution dikhao

   Maan lo $P$ se tangent length hai $x\mathrm{.}$ Tangent pairs se $PQ=x+4,$ $PR=x+6$ aur $QR=10,$ isliye semi-perimeter hai $s=x+10.$

   Centre $I$ ko teeno corners se jodo: $P,$ $Q$ aur $R\mathrm{.}$ Triangles $IQR,$ $IRP$ aur $IPQ$ sabki height 3 cm hai, isliye area hai $\frac{1}{2}\times 3\times 2s=3(x+10)\mathrm{.}$

   Heron's formula se area aata hai $\sqrt{s\cdot x\cdot 4\cdot 6}=\sqrt{24x(x+10)}\mathrm{.}$

   Isliye $9(x+10{)}^2=24x(x+10),$ yaani $9(x+10)=24x,$ jisse milta hai $x=6.$ Isliye $PQ=10$ cm.

2. (b)

   $PR$ cm mein nikaalo.

   cm

   Check karo

   Solution dikhao

   $PR=x+6=12$ cm. Check: sides 10, 10 aur 12 se $s=16$ milta hai aur area $\sqrt{16\cdot 6\cdot 6\cdot 4}=48=3\times 16.$

Quadrilateral $ABCD$ ek circle ko circumscribe karta hai, jiska centre hai $O\mathrm{.}$

1. (a)

   Prove karo ki $\mathrm{\angle }AOB+\mathrm{\angle }COD={180}^{\circ }\mathrm{.}$

   Solution dikhao

   Maan lo circle sides $AB,$ $BC,$ $CD,$ $DA$ ko kram se in points par chhoota hai: $P,$ $Q,$ $R,$ $S\mathrm{.}$ $O$ ko saare aath points se jodo.

   Triangles $OAS$ aur $OAP$ congruent hain (RHS), isliye $\mathrm{\angle }AOS=\mathrm{\angle }AOP\mathrm{.}$ Isi tarah $\mathrm{\angle }BOP=\mathrm{\angle }BOQ,$ $\mathrm{\angle }COQ=\mathrm{\angle }COR$ aur $\mathrm{\angle }DOR=\mathrm{\angle }DOS\mathrm{.}$

   $O$ ke around aath angles ka sum ${360}^{\circ }$ hai aur woh chaar barabar pairs banate hain, isliye $\mathrm{\angle }AOP+\mathrm{\angle }BOP+\mathrm{\angle }COR+\mathrm{\angle }DOR={180}^{\circ }\mathrm{.}$

   Kyunki $\mathrm{\angle }AOP+\mathrm{\angle }BOP=\mathrm{\angle }AOB$ aur $\mathrm{\angle }COR+\mathrm{\angle }DOR=\mathrm{\angle }COD,$ humein milta hai $\mathrm{\angle }AOB+\mathrm{\angle }COD={180}^{\circ }\mathrm{.}$

2. (b)

   Agar $\mathrm{\angle }AOB={100}^{\circ },$ to $\mathrm{\angle }COD$ degrees mein nikaalo.

   °

   Check karo

   Solution dikhao

   $\mathrm{\angle }COD={180}^{\circ }-{100}^{\circ }={80}^{\circ }\mathrm{.}$

Ek point $P$ se 4 cm radius wale circle par bane do tangents ke beech ka angle hai ${60}^{\circ }\mathrm{.}$ To $OP$ barabar hai

Assertion (A): 3 cm radius wale circle ke centre se 5 cm door ke point se, har tangent 4 cm lamba hai. Reason (R): Point of contact tak kheencha gaya radius tangent se right angle par milta hai.

Assertion (A): Circle ke andar ke point se circle ka tangent kheencha ja sakta hai. Reason (R): Circle ke andar ke point se guzarne wali har line circle ko do points par kaatti hai.

Ek chord (diameter nahi) ke ends $A$ aur $B$ par bane tangents jis point par milte hain woh hai $T\mathrm{.}$ Prove karo ki $\mathrm{\angle }TAB=\mathrm{\angle }TBA\mathrm{.}$