Propositions and their Converses (Propositions aur unke Converses): Class 9 Maths Chapter 9 Hinglish mein

Questions 1 se 12 mein har ek ek proposition deta hai. Kuch 'if … then …' form mein nahi likhe gaye: pehle unhe dobara likho (Section 9.6 dekho). Converse likho, phir tay karo ki dono mein se kaun se statements true hain. Har true statement ke saath reason do aur har false ke saath counterexample. Questions 18 se 22 extra practice ke liye hain.

1. 1
   Short answerEasyNCERT Exercise Set 9.1, Q1

   Agar ek transversal do parallel lines ko kaatta hai, to alternate interior angles equal hote hain.

   ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

   Check karo

   HintAI se poochho

   Vertically opposite angles ke zariye alternate angles ko corresponding angles se jodo.

   Solution dikhao

   AnswerConverse: agar ek transversal do lines ko kaatta hai aur equal alternate interior angles banata hai, to lines parallel hain.

   Converse: agar ek transversal do lines ko kaatta hai aur alternate interior angles equal hain, to dono lines parallel hain.

   Proposition: true. Parallel lines ke liye corresponding angles equal hote hain, aur har alternate interior angle ek corresponding angle ke barabar hota hai, kyunki vertically opposite angles equal hote hain.

   Converse: true. Agar alternate interior angles equal hain, to wahi vertically-opposite-angle wala step corresponding angles ke ek pair ko equal bana deta hai, aur jo lines transversal ke saath equal corresponding angles banati hain woh parallel hoti hain.

2. 2
   Short answerEasyNCERT Exercise Set 9.1, Q2

   Agar koi triangle equilateral hai, to woh isosceles hai (uski kam se kam do sides equal hain).

   ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

   Check karo

   HintAI se poochho

   Aise isosceles triangle ke baare mein socho jo equilateral na ho.

   Solution dikhao

   AnswerConverse: agar koi triangle isosceles hai, to woh equilateral hai.

   Converse: agar koi triangle isosceles hai (uski kam se kam do sides equal hain), to woh equilateral hai.

   Proposition: true. Equilateral triangle ki teeno sides equal hoti hain, to unmein se do to pakka equal hain.

   Converse: false. 5 cm, 5 cm aur 8 cm sides wale triangle ki do sides equal hain lekin woh equilateral nahi hai.

3. 3
   Short answerHardNCERT Exercise Set 9.1, Q3

   Ek triangle hai $\mathrm{△}ABC,$ jismein $BE$, $B$ se altitude hai (yaani $BE\perp AC$) aur $CF$, $C$ se altitude hai (yaani $CF\perp AB$). Proposition: agar $AB=AC$, to $BE=CF$.

   Ek triangle ABC jiska A upar hai aur B aur C neeche left aur right mein. B se ek dashed line side AC ko E par right angle par milti hai, aur C se ek dashed line side AB ko F par right angle par milti hai. Right-angle marks dono feet dikhate hain.

   ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

   Check karo

   HintAI se poochho

   Triangle ka area do alag tareekon se nikaalo.

   Solution dikhao

   AnswerConverse: agar $BE=CF$, to $AB=AC$.

   Converse: altitudes $BE$ aur $CF$ wale $\mathrm{△}ABC$ mein, agar $BE=CF$, to $AB=AC$.

   $\mathrm{△}ABC$ ka area do tareekon se likho: base $AC$ ke saath height $BE$ hai, aur base $AB$ ke saath height $CF$ hai.

   $$\begin{array}{rl}{\displaystyle \frac{1}{2}\cdot AC\cdot BE}& {\displaystyle =\frac{1}{2}\cdot AB\cdot CF}\\ {\displaystyle AC\cdot BE}& {\displaystyle =AB\cdot CF}\end{array}$$

   Proposition: true. Agar $AB=AC$, to dono sides ko is common length se divide karo, $BE=CF$ milega.

   Converse: true. Agar $BE=CF$, to dono sides ko is common length se divide karo, $AC=AB$ milega.

   Converse ke liye ek aur raasta: right triangles $BEC$ aur $CFB$ ka hypotenuse $BC$ common hai aur $BE=CF$ hai, isliye woh RHS se congruent hain. Tab $\mathrm{\angle }C=\mathrm{\angle }B,$ aur equal angles ke saamne equal sides hoti hain.

4. 4
   Short answerEasyNCERT Exercise Set 9.1, Q4

   Kisi bhi numbers $k$, $x$ aur $y$ ke liye: agar $x=y$, to $kx=ky$.

   ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

   Check karo

   HintAI se poochho

   Agar $k$ zero ho to kya hota hai?

   Solution dikhao

   AnswerConverse: agar $kx=ky$, to $x=y$.

   Converse: agar $kx=ky$, to $x=y$.

   Proposition: true. Equal numbers ko same number se multiply karne par results equal aate hain.

   Converse: false. $k=0$, $x=2$, $y=5$ lo: tab $kx=0=ky,$ lekin $x\mathrm{\ne }y$.

   Jab $k\mathrm{\ne }0$ ho, tab converse sahi hai: dono sides ko $k$ se divide karo. Isiliye equation solve karte waqt tum dono sides ko sirf aise number se divide kar sakte ho jo zero na ho.

5. 5
   Short answerMediumNCERT Exercise Set 9.1, Q5

   Positive integers $a$ aur $b$ ke liye: jab bhi dono perfect cubes hon, unka product $ab$ bhi perfect cube hota hai.

   ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

   Check karo

   HintAI se poochho

   Aise do non-cubes dhoondho jinka product 8 ho.

   Solution dikhao

   AnswerConverse: agar $ab$ perfect cube hai, to $a$ aur $b$ dono perfect cubes hain.

   'If … then …' form mein: agar $a$ aur $b$ perfect cubes hain, to $ab$ perfect cube hai. Converse: agar $ab$ perfect cube hai, to $a$ aur $b$ dono perfect cubes hain.

   Proposition: true. Agar $a=m^3$ aur $b=n^3$, to $ab=m^3{n}^3=(mn{)}^3\mathrm{.}$

   Converse: false. $a=2$ aur $b=4$ lo: $ab=8=2^3$ cube hai, lekin na $2$ cube hai na $4$.

6. 6
   Short answerEasyNCERT Exercise Set 9.1, Q6

   Real numbers $x$ aur $y$ ke liye: agar $x=y$, to $x^4=y^4$.

   ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

   Check karo

   HintAI se poochho

   Even powers number ka sign chhupa deti hain.

   Solution dikhao

   AnswerConverse: agar $x^4=y^4$, to $x=y$.

   Converse: agar $x^4=y^4$, to $x=y$.

   Proposition: true. Equal numbers ki fourth powers equal hoti hain.

   Converse: false. $x=2$ aur $y=-2$ lo: $x^4=16=y^4,$ lekin $x\mathrm{\ne }y$.

7. 7
   Short answerMediumNCERT Exercise Set 9.1, Q7

   Real numbers $x$ aur $y$ ke liye: agar $x=y$, to $x^5=y^5$.

   ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

   Check karo

   HintAI se poochho

   Even powers ke ulat, odd powers sign bachaa kar rakhti hain. Proof by contradiction try karo.

   Solution dikhao

   AnswerConverse: agar $x^5=y^5$, to $x=y$.

   Converse: agar $x^5=y^5$, to $x=y$.

   Proposition: true. Equal numbers ki fifth powers equal hoti hain.

   Converse: true. Maan lo $x^5=y^5$ hai lekin $x\mathrm{\ne }y$, jaise $x<y$. Agar $0\le x<y,$ to $x<y$ ko khud se paanch baar multiply karne par $x^5<y^5$ milta hai.

   Agar $x<0\le y,$ to $x^5$ negative hai aur $y^5$ nahi, isliye $x^5<y^5$. Agar $x<y<0$, to $0<-y<-x,$ isliye $(-y{)}^5<(-x{)}^5,$ jiska matlab hai $x^5<y^5$.

   Har case $x^5<y^5$ deta hai, jo $x^5=y^5$ ke against hai. Isliye $x=y$.

8. 8
   Short answerEasyNCERT Exercise Set 9.1, Q8

   Yahan se Question 12 tak $n$ ek positive integer hai. 18 ka har multiple 3 se aur 6 se bhi divisible hai.

   ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

   Check karo

   HintAI se poochho

   Woh sabse chhota number try karo jo 3 aur 6 dono se divisible ho.

   Solution dikhao

   AnswerConverse: agar $n$ 3 aur 6 dono se divisible hai, to $n$ 18 se divisible hai.

   'If … then …' form mein: agar $n$ 18 se divisible hai, to woh 3 aur 6 dono se divisible hai. Converse: agar $n$ 3 aur 6 dono se divisible hai, to $n$ 18 se divisible hai.

   Proposition: true. $18k=3\times 6k=6\times 3k\mathrm{.}$

   Converse: false. $6$, 3 se aur 6 se divisible hai lekin 18 se nahi. ($12$ bhi chalega.)

9. 9
   Short answerMediumNCERT Exercise Set 9.1, Q9

   Jab bhi $n$ 45 se divisible ho, woh 5 se aur 9 se divisible hota hai.

   ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

   Check karo

   HintAI se poochho

   Prime factorisation use karo; 5 aur 9 ka koi common prime factor nahi hai.

   Solution dikhao

   AnswerConverse: agar $n$ 5 aur 9 dono se divisible hai, to $n$ 45 se divisible hai.

   'If … then …' form mein: agar 45, $n$ ko divide karta hai, to 5 aur 9 dono $n$ ko divide karte hain. Converse: agar $n$ 5 aur 9 dono se divisible hai, to $n$ 45 se divisible hai.

   Proposition: true. $45k=5\times 9k=9\times 5k\mathrm{.}$

   Converse: true. 9 se divisible hone ka matlab hai ki $n$ ke prime factorisation mein $3\times 3$ hai. 5 se divisible hone ka matlab hai ki usmein $5$ hai, jo ek alag prime hai. Isliye usmein yeh sab hai: $3\times 3\times 5=45,$ aur $45$, $n$ ko divide karta hai.

10. 10
    Short answerMediumNCERT Exercise Set 9.1, Q10

    Jab bhi kisi prime $p$ ke liye $n=p^3$ ho, $n$ ke theek chaar factors hote hain.

    ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

    Check karo

    HintAI se poochho

    $p^3$ ke factors list karo. Phir 4 factors wale chhote numbers try karo.

    Solution dikhao

    AnswerConverse: agar $n$ ke theek 4 factors hain, to $n$ kisi prime number ka cube hai.

    'If … then …' form mein: agar kisi prime $p$ ke liye $n=p^3$, to $n$ ke theek chaar factors hain. Converse: agar $n$ ke theek 4 factors hain, to $n$ kisi prime ka cube hai.

    Proposition: true. Agar $n=p^3$ aur $p$ prime hai, to uska ek hi prime factor $p$ hai, isliye uske factors $1$, $p$, $p^2$ aur $p^3$ hain: theek 4.

    Converse: false. $6$ ke theek 4 factors ($1,2,3,6$) hain lekin woh cube nahi hai.

11. 11
    Short answerMediumNCERT Exercise Set 9.1, Q11

    $pqr$ form ka har number, jahan $p$, $q$ aur $r$ teen alag primes hain, theek aath factors rakhta hai.

    ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

    Check karo

    HintAI se poochho

    $pqr$ ka factor $p$, $q$, $r$ mein se kuch chun kar banta hai.

    Solution dikhao

    AnswerConverse: agar $n$ ke theek 8 factors hain, to $n$ teen alag primes ka product hai.

    'If … then …' form mein: agar teen alag primes ke liye $n=pqr$, to $n$ ke theek aath factors hain. Converse: agar $n$ ke theek 8 factors hain, to $n$ teen alag primes ka product hai.

    Proposition: true. Agar $n=pqr$, to har factor $p$, $q$, $r$ mein se kuch ka product hai: $1,p,q,r,pq,pr,qr,pqr\mathrm{.}$ Yeh 8 factors hain.

    Converse: false. $24=2^3\times 3$ ke theek 8 factors ($1,2,3,4,6,8,12,24$) hain, lekin woh teen alag primes ka product nahi hai.

12. 12
    Short answerMediumNCERT Exercise Set 9.1, Q12

    Jab $n$ aur $n+2$ ka ek hi common factor 1 ho, to number $n$ odd hota hai.

    ADono true hainBSirf proposition true haiCSirf converse true haiDKoi bhi true nahi hai

    Check karo

    HintAI se poochho

    Do numbers ka common factor unke difference ko bhi divide karta hai.

    Solution dikhao

    AnswerConverse: agar $n$ odd hai, to $n$ aur $n+2$ ka 1 ke alawa koi common factor nahi hai.

    'If … then …' form mein: agar $n$ aur $n+2$ ka ek hi common factor 1 hai, to $n$ odd hai. Converse: agar $n$ odd hai, to $n$ aur $n+2$ ka 1 ke alawa koi common factor nahi hai.

    Proposition: true. Agar $n$ even hota, to $n$ aur $n+2$ dono even hote, aur unka common factor 2 hota. Isliye $n$ odd hi hona chahiye.

    Converse: true. $n$ aur $n+2$ ka common factor $d$ unke difference $2$ ko bhi divide karta hai. Isliye $d$ ya to $1$ hai ya $2$. Kyunki $n$ odd hai, $d\mathrm{\ne }2$, isliye $d=1$.

13. 13
    Short answerMediumNCERT Exercise Set 9.1, Q13

    Har claim kehta hai ki ek formula hamesha prime number deta hai. Har claim ko ek counterexample se todo.

    HintAI se poochho

    $n$ ki chhoti values ek-ek karke rakho aur results ko factorise karo.

    1. (a)

       Har whole number $n\ge 1$ ke liye $6n+1$ prime hai.

       Solution dikhao

       Answer$n=4$ lene par milta hai $25=5\times 5.$

       $n=1,2,3$ se $7$, $13$, $19$ milte hain, sab prime, lekin $n=4$ lene par milta hai $25=5\times 5.$

    2. (b)

       Har whole number $n\ge 0$ ke liye $n^2+n+17$ prime hai.

       Solution dikhao

       Answer$n=16$ lene par milta hai $289=17\times 17.$

       $n=0$ se $15$ tak ki saari values prime hain, jisse claim safe lagta hai. Lekin $n=16$ lene par milta hai $256+16+17=289=17\times 17.$

    3. (c)

       Har whole number $n\ge 1$ ke liye $2^n+3$ prime hai.

       Solution dikhao

       Answer$n=5$ lene par milta hai $35=5\times 7.$

       $n=1,2,3,4$ se $5$, $7$, $11$, $19$ milte hain, sab prime, lekin $n=5$ lene par milta hai $32+3=35=5\times 7.$

14. 14
    Short answerMediumNCERT Exercise Set 9.1, Q14

    Har statement ka ek counterexample dhoondho.

    HintAI se poochho

    Pehle kuch cases order mein test karo.

    1. (a)

       Kisi prime ko double karke 1 jodne par hamesha prime milta hai.

       Solution dikhao

       Answer$p=7$ se $15$ milta hai.

       $p=2,3,5$ se $5$, $7$, $11$ milte hain, sab prime. Lekin $p=7$ lene par aata hai $15=3\times 5,$ jo prime nahi hai.

    2. (b)

       Har odd number $n$ ke liye $3^n+2$ prime hai.

       Solution dikhao

       Answer$n=5$ se $245$ milta hai.

       $n=1$ se $5$ aur $n=3$ se $29$ milta hai, dono prime. Lekin $n=5$ lene par aata hai $243+2=245=5\times 49,$ jo prime nahi hai.

15. 15
    Short answerMediumNCERT Exercise Set 9.1, Q15

    Yeh raha ek true statement: 12 se divisible har number 2 aur 6 dono se divisible hota hai.

    HintAI se poochho

    Divisibility ka shortcut asal mein bhes badla hua converse hai.

    1. (a)

       Samjhao ki yeh statement true kyun hai.

       Solution dikhao

       12 ka multiple hai $12k=2\times 6k=6\times 2k,$ isliye woh 2 se aur 6 se divisible hai.

    2. (b)

       Ravi kehta hai: 'Jo number 2 aur 6 ke tests paas kar le, woh 12 se divisible hoga hi.' Kya Ravi sahi hai? Samjhao.

       AHaan, woh sahi haiBNahi, woh galat hai

       Check karo

       Solution dikhao

       Nahi. Yeh test converse par tika hai, agar koi number 2 aur 6 se divisible hai, to woh 12 se divisible hai, jo false hai: $18$ 2 se aur 6 se divisible hai lekin 12 se nahi. Asal mein 6 ka har multiple pehle se even hota hai, isliye 2 check karne se kuch naya nahi milta.

    3. (c)

       Kaun se do chhote divisibility tests milkar 12 ke liye sach mein kaam karte hain?

       Solution dikhao

       Answer3 se aur 4 se divisibility.

       3 aur 4 use karo. 4 se divisible number ke prime factorisation mein $2\times 2$ hota hai, aur 3 se divisible mein $3$. Saath milkar unmein aata hai $2\times 2\times 3=12.$ Yeh isliye kaam karta hai kyunki 3 aur 4 ka 1 ke alawa koi common factor nahi hai.

16. 16
    Short answerMediumNCERT Exercise Set 9.1, Q16

    9 ka ek shortcut: koi number 9 se divisible theek tab hota hai jab uske digits ka sum 9 ka multiple ho. Is relationship ko do 'if … then …' sentences mein likho, batao ki har ek true hai ya nahi, aur unhe $7614$ par check karo.

    HintAI se poochho

    Ek sentence '9 se divisible' se shuru hota hai; doosra digits ke sum se.

    Solution dikhao

    AnswerDono sentences true hain; number 9 se divisible hai if and only if uske digits ka sum 9 ka multiple hai.

    Agar koi number 9 se divisible hai, to uske digits ka sum 9 ka multiple hai. Iska converse: agar kisi number ke digits ka sum 9 ka multiple hai, to woh number 9 se divisible hai.

    Dono true hain, kyunki $10=9+1$, $100=99+1,$ $1000=999+1$ aur aise hi aage. Isliye kisi number aur uske digit sum mein hamesha 9 ke multiple ka fark hota hai: ek 9 ka multiple theek tab hai jab doosra ho.

    Dono saath mein: koi number 9 se divisible hai if and only if uske digits ka sum 9 ka multiple hai. Check: $7+6+1+4=18,$ jo 9 ka multiple hai, aur sach mein $7614=9\times 846.$

17. 17
    Short answerHardNCERT Exercise Set 9.1, Q17

    Uttarayan ke liye Ahmedabad ki ek kite-maker do bamboo sticks ko saath baandhti hai aur unke chaaron ends ke around ek string ghumaati hai. Sticks quadrilateral ke diagonals ban jaati hain aur string uski sides. Maan lo 'type K' quadrilaterals ki koi class hai jo woh banana chahti hai.

    Do frames saath-saath. Har ek mein do bamboo sticks right angles par cross karti hain aur ek dashed string unke chaaron ends ko jodti hai. Left mein sticks chhoti stick ke beech mein lekin lambi stick ke beech se kaafi upar cross karti hain, jisse kite shape banti hai. Right mein sticks dono ke beech mein cross karti hain, jisse rhombus banta hai.

    HintAI se poochho

    Socho kaun si property batati hai ki har type-K shape mein kya hota hai, aur kaun si batati hai ki type K hone ke liye kya kaafi hai.

    1. (a)

       Maan lo agar koi quadrilateral type K ka hai, to uske diagonals perpendicular hain. Kya use sticks right angles par hi baandhni padengi?

       AHaanBNahi

       Check karo

       Solution dikhao

       Haan. Har type-K quadrilateral ke diagonals perpendicular hote hain. Agar sticks kisi aur angle par cross karti hain, to frame type K ka ho hi nahi sakta.

    2. (b)

       Isi property ke saath, kya unhe right angles par baandhna type K pakka karne ke liye kaafi hai?

       AHaanBNahi

       Check karo

       Solution dikhao

       Nahi. Property batati hai ki har type-K shape mein kya hota hai, yeh nahi ki perpendicular diagonals wali har shape type K ki hai; woh iska converse hota, jo humein diya hi nahi gaya. Jaise, agar type K ka matlab 'rhombus' hota, to sticks ko right angles par off-centre cross karne se kite banti hai, rhombus nahi (Fig. 9.10).

    3. (c)

       Ab iske bajaay maan lo: agar kisi quadrilateral ke diagonals perpendicular hain, to woh type K ka hai. Kya ab sticks ko right angles par baandhna kaafi hai?

       AHaanBNahi

       Check karo

       Solution dikhao

       Haan. Perpendicular diagonals wala koi bhi quadrilateral type K ka hai, chahe sticks ki lengths kuch bhi hon aur woh kahin bhi cross karein.

    4. (d)

       (c) wali property ke saath, kya sticks ka right angles par hona zaroori hai?

       AHaanBNahi, zaroori nahi

       Check karo

       Solution dikhao

       Zaroori nahi. Yeh property perpendicular diagonals ke liye type K ki guarantee deti hai, lekin doosri shapes ke baare mein kuch nahi kehti. Ho sakta hai kuch type-K quadrilaterals ke diagonals perpendicular na hon.

18. 18
    Quick answerEasy

    Agar Ravi 90 se zyada score karta hai, to use A grade milta hai ka converse kaun sa sentence hai?

    AAgar Ravi 90 se zyada score nahi karta, to use A grade nahi milta.BAgar Ravi ko A grade nahi milta, to woh 90 se zyada score nahi karta.CAgar Ravi ko A grade milta hai, to woh 90 se zyada score karta hai.DRavi 90 se zyada score karta hai aur use A grade milta hai.

    Check karo

    Answer dekho

    Answer (C) Agar Ravi ko A grade milta hai, to woh 90 se zyada score karta hai.

    Converse dono parts ko swap karta hai aur kuch nahi badalta. Pehla option bina swap kiye dono parts mein 'not' jodta hai, aur doosra swap karke 'not' jodta hai; dono alag statements hain.

    HintAI se poochho

    Parts swap karo; 'not' mat jodo.

19. 19
    Quick answerEasy

    Agar koi proposition false hai, to uska converse bhi false hi hoga.

    SahiGalat

    Check karo

    Answer dekho

    Answer Galat

    Example 4: 'equal perimeters $\Rightarrow$ congruent' false hai, lekin uska converse 'congruent $\Rightarrow$ equal perimeters' true hai.

    HintAI se poochho

    Chaar cases wali table phir se dekho.

20. 20
    Quick answerEasy

    Aisa akela case jo proposition ke 'if' part ko true aur 'then' part ko false bana de, kehlata hai.

    Check karo

    Answer dekho

    Answer counterexample

    Proposition ko false dikhane ke liye ek counterexample kaafi hai.

    HintAI se poochho

    Yeh Section 9.2 ka key word hai.

21. 21
    Short answerMedium

    Part 1 mein tumne yeh fact dekha tha ki semicircle mein bana angle right angle hota hai. Ise ek point $P$ aur diameter $AB$ wale circle ke baare mein 'if … then …' proposition ki tarah likho. Phir iska converse likho aur tay karo ki converse true hai ya nahi.

    HintAI se poochho

    Part 1, Chapter 5 mein, right triangle ka circumcentre kahan hota hai?

    Solution dikhao

    AnswerConverse: agar $\mathrm{\angle }APB={90}^{\circ },$ to $P$ diameter $AB$ wale circle par hai. Yeh true hai.

    Proposition: agar $P$ diameter $AB$ wale circle par koi point hai ($A$ aur $B$ ke alawa), to $\mathrm{\angle }APB={90}^{\circ }\mathrm{.}$

    Converse: agar $\mathrm{\angle }APB={90}^{\circ },$ to $P$ diameter $AB$ wale circle par hai.

    Converse true hai. Right triangle $APB$ mein circumcentre hypotenuse $AB$ ka midpoint hota hai (Part 1, Chapter 5), isliye $A$, $P$ aur $B$ se guzarne wala circle bilkul diameter $AB$ wala circle hi hai. Isliye $P$ us par hai.

22. 22
    Short answerMedium

    Rectangles ke baare mein aisa proposition do jo false ho aur jiska converse bhi false ho. Har verdict ko counterexample se support karo.

    HintAI se poochho

    Rectangles ka perimeter aur area compare karo.

    Solution dikhao

    Answer'Same perimeter $\Rightarrow$ same area' false hai, aur iska converse bhi.

    Proposition: agar do rectangles ka perimeter same hai, to unka area same hai. False: 2 cm by 8 cm rectangle aur 4 cm by 6 cm rectangle dono ka perimeter 20 cm hai, lekin unke areas 16 cm² aur 24 cm² hain.

    Converse: agar do rectangles ka area same hai, to unka perimeter same hai. False: 2 cm by 8 cm rectangle aur 4 cm by 4 cm square dono ka area 16 cm² hai, lekin unke perimeters 20 cm aur 16 cm hain.