Class 7

Haryana State Board HBSE 7th Class Hindi Solutions Hindi Vyakaran Varn-Vichar : Uchchaaran Aur Vartani वर्ण-विचार : उच्चारण और वर्तनी Exercise Questions and Answers.

Haryana Board 7th Class Hindi Vyakaran वर्ण-विचार : उच्चारण और वर्तनी

इस अध्याय के अंतर्गत निम्नलिखित का अध्ययन किया जाएगा :

• स्वर एवं स्वरों के भेद
• व्यंजन एवं व्यंजनों के भेद
• अक्षर
• व्यंजन गुच्छ 00 बलाघात
• अनुतान
• संगम
• उच्चारण संबंधी अशुद्धियां और उनमें सुधार

वर्ण :
वर्ण क्या है ? : भाषा की सबसे छोटी इकाई ध्वनि है। इसके लिखित रूप को वर्ण कहते हैं। वर्ण शब्द का प्रयोग ध्वनि और ध्वनि-चिह्न (लिपि-चिह्न) दोनों के लिए होता है। इस प्रकार वर्ण भाषा के मौखिक और लिखित दोनों रूपों के प्रतीक हैं। इसे हम अक्षर भी कह सकते हैं। अक्षर का अर्थ है-उसके और टुकड़े नहीं किए जा सकते।
वर्ण या अक्षर वह छोटी-से-छोटी ध्वनि है, जिसके टुकड़े नहीं किए जा सकते।

वर्णमाला (Alphabet) : वर्गों के व्यवस्थित समूह को वर्णमाला कहते हैं।
मानक देवनागरी वर्णमाला (Standard Hindi Alphabet) :

अनुस्वार : ं अं
विसर्ग : : अ:

हल् चिह्न ( ) : सभी व्यंजन वर्णों के लिपि चिह्नों में ‘अ’ स्वर रहता है, जैसे – क – क् + अ । जब स्वर रहित व्यंजन का प्रयोग करना हो तो उसके नीचे हल चिह्न लगाया जाता है। वर्णों के भेद (Kinds of Alphabet) : वर्णों को दो भागों में बाँटा जाता है :
1. स्वर (Vowels)
2. व्यंजन (Consonants)

स्वर (Vowels) :
जिन ध्वनियों के उच्चारण में श्वास-वायु बिना किसी रुकावट के मुख से निकलती है, उन्हें स्वर कहते हैं।
हिंदी में निम्नलिखित स्वर हैं :

यद्यपि ‘ऋ’ को लिखित रूप में स्वर माना जाता है, परंतु आजकल हिंदी में इसका उच्चारण ‘रि’ के समान होता है।
आजकल अंग्रेजी प्रभाव के कारण ‘ऑ’ ध्वनि हिंदी में अपनी जगह बना चुकी है। यह ‘आ’ और ‘ओ’ के बीच की ध्वनि है। इसका लिपि-चिह्न (ऑ) है।
जैसे – बॉल, डॉक्टर, हॉकी आदि।

स्वर के भेद (Kinds of Vowels) :
(क) उच्चारण में लगने वाले समय के आधार पर स्वरों को दो भागों में बाँट सकते हैं :
1. हस्व स्वर (Short vowels)।
2. दीर्घ स्वर (Long vowels)

आइए, अब हम इनके बारे में जानें :
1. ह्रस्व स्वर : जिन स्वरों को सबसे कम समय (एक मात्रा) में उच्चरित किया जाता है, उन्हें हस्व स्वर कहते हैं। ये हैं – अ इ उ (ऋ)
हस्व ‘ऋ’ का प्रयोग केवल संस्कृत के तत्सम शब्दों में होता है, जैसे-
ऋषि, ऋतु, कृषि आदि।
ह्रस्व स्वरों को ‘मूल स्वर’ भी कहते हैं।

2. दीर्घ स्वर : जिन स्वरों के उच्चारण में हस्व स्वरों से अधिक (लगभग दुगुना) समय लगता है, उन्हें दीर्घ स्वर कहते हैं। ये स्वर हैं : आ ई ऊ ए ऐ ओ औ
ये स्वर हस्व स्वरों के दीर्घ रूप नहीं हैं, वरन् स्वतंत्र ध्वनियाँ हैं। इन स्वरों में ए तथा औ का उच्चारण संयुक्त स्वर रूप में भी है, जैसे-‘ऐ’ में ‘अ+इ’ दो स्वरों का संयुक्त रूप है। यह उच्चारण तब होता है जब बाद में क्रमशः ‘य’ और ‘व’ आएँ ; जैसे- भैया = भइया, कौआ – कउवा
शेष स्थिति में ‘ऐ’ और ‘औ’ का उच्चारण शुद्ध स्वर की भाँति होता है । जैसे- मैल, कैसा, औरत, कौन आदि।

व्यंजन (Consonants) :
जिन ध्वनियों के उच्चारण में वायु रूकावट के साथ मुँह से बाहर आती है, उन्हें व्यंजन कहते हैं।
हिंदी वर्णमाला में मूलतः 33 व्यंजन हैं। दो व्यंजन ‘ड’ और ‘ढ़’ क्रमशः ‘ड’ ‘ढ’ से विकसित हुए हैं।
हिंदी में अरबी, फारसी, तुर्की आदि के शब्द आ जाने के कारण आने वाली ध्वनियों के लिए जो व्यंजन बनाए गए हैं, वे हैं-क, ख, ग, फ, ज़ ।

प्रयत्न और स्थान की विविधता के अनुसार हिंदी-व्यंजनों की तालिका

व्यंजनों का वर्गीकरण (Classification of Consonants) :
उच्चारण की दृष्टि से व्यंजन वर्णों को दो प्रकार से विभाजित किया जाता है :
(1) स्थान के आधार पर
(2) प्रयल के आधार पर।
1. स्थान के आधार पर (Places of Pronunciation) :
व्यंजनों का उच्चारण मुख के विभिन्न अवयवों-कंठ, तालु, मूर्धा आदि से किया जाता है। जो वर्ण मुख के जिस भाग से बोला जाता है, वही उस वर्ण का ‘उच्चारण स्थान’ कहलाता है। वर्गों के उच्चारण स्थान इस प्रकार हैं :

2. प्रयत्न के आधार पर (Manner of Articulation) :
व्यंजन ध्वनियों के उच्चारण में श्वास का कंपन, श्वास की मात्रा तथा जिह्वा आदि अवयवों द्वारा श्वास के अवरोध की प्रक्रिया का नाम प्रयत्न है।
प्रायः यह प्रयत्न तीन प्रकार से होता है :
1. स्वरतंत्री में साँस के कंपन के रूप में।
2. श्वास (प्राण) की मात्रा के रूप में।
3. मुख-अवयवों द्वारा श्वास रोकने के रूप में।
अब हम इन तीनों रूपों के बारे में विस्तारपूर्वक जानकारी प्राप्त करेंगे।

1. स्वरतंत्री में श्वास का कंपन : हमारे गले में दो झिल्लियाँ होती हैं, जो वायु के वेग से काँपकर बजने लगती हैं, इन्हें स्वरतंत्री कहते हैं। स्वर-तंत्रियों में होने वाले कंपन के आध र पर व्यंजन वर्णों के दो भेद हैं- अघोष और सघोष।
(क) अघोष (Non-wavering Sound) : जिन ध्वनियों के उच्चारण में स्वरतंत्रियों में कंपन नहीं होता, उन्हें अघोष कहते हैं। हिंदी की अघोष ध्वनियाँ ये हैं :

प्रथम तथा द्वितीय व्यंजन तथा

(ख) सघोष (Wavering Sound) : जिन ध्वनियों के उच्चारण में स्वरतंत्रियों में कंपन होता है, उनको सघोष कहते हैं। हिंदी के सघोष व्यंजन हैं :

ब भ म – वर्गों के तीसरे, चौथे और पाँचवें व्यंजन
तथा
सभी स्वर सघोष होते हैं।

2. श्वास की मात्रा : इस आधार पर व्यंजनों के दो भेद किए जाते हैं :
(क) अल्पप्राण (Non-Aspirated)
(ख) महाप्राण (Aspirated)।

(क) अल्पप्राण (Non-Aspirated) : जिन ध्वनियों के उच्चारण में श्वास (प्राण वायु) कम मात्रा में बाहर निकलती है, उन्हें अल्पप्राणं व्यंजन कहते हैं। ये हैं :
← वर्गों के पहले, तीसरे और पाँचवें वर्ण
तथा

(ख) महाप्राण (Aspirated) : जिन ध्वनियों के उच्चारण में श्वास वायु अधिक मात्रा में बाहर निकलती है, उन्हें महाप्राण कहते हैं। ये हैं :
← वर्गों के दूसरे और चौथे वर्ण तथा

व्यंजन-गुच्छ :
जब दो या दो से अधिक व्यंजन एक साथ एक श्वास के झटके में बोले जाते हैं, तब उनको ‘व्यंजन-गुच्छ’ कहा जाता है। जैसे-प्यास। शब्द के आदि में प्राय: दो प्रकार के व्यंजन-गुच्छ मिलते हैं :
1. व्यंजन + य, र, ल, व
2. स + य र ल से भिन्न व्यंजन

शब्द के मध्य तथा अंत में भी अनेक व्यंजन-गुच्छ मिलते हैं ; जैसे- न् + त = अंत, र + म – मार्ग, प् + त – लुप्त आदि।

व्यंजन-संयोग :
जब एक व्यंजन के साथ दूसरा व्यंजन आता है और दोनों का उच्चारण अलग-अलग किया जाए तो व्यंजन-संयोग होता है। व्यंजन-संयोग में व्यंजनों को अलग-अलग लिखना चाहिए।
जैसे- जनता = जन् + ता में न् + त का संयोग है।
उलटा = उल्+ टा में ल् + ट का संयोग है।

व्यंजन-द्वित्व :
जब कोई व्यंजन अपने समरूप व्यंजन से मिलता है तो ऐसे रूप को व्यंजन-द्वित्व कहते हैं। जैसेइक्का, पक्का, बच्चा, कट्टर, लड्डू, दिल्ली, लटू, उद्देश्य, थप्पड़, उत्तेजित आदि। Is इन्हें भी समझो : ‘र’ व्यंजन युक्त।

• जब ‘र’ (स्वर रहित) किसी व्यंजन के पूर्व हो, जैसे – कर्म, धर्म, वर्ष आदि।
• जब ‘र’ (स्वर सहित) किसी व्यंजन के बाद हो, जैसे – प्रेम, क्रम आदि।
• यही स्वर सहित ‘र’ ट और ड के साथ वर्तनी के साथ कुछ भिन्न रूप ले लेता है, जैसे-ट्रेन, ट्रक, ड्रम आदि।

दो व्यंजनों के त और श के साथ इसके विशिष्ट रूप बन जाते हैं-
त् + र =त्र त्रिशूल, त्रिभुज, यंत्र
श् + र = श्र श्रम, श्री, आश्रय अन्य

संयुक्त व्यंजन :
क् + ष = क्ष क्षमा, क्षेत्र, क्षत्रिय
ज् + अ = ज्ञ ज्ञान, यज्ञ, विज्ञान
(‘ज्ञ’ का उच्चारण प्रायः ग् + य – ग्य के रूप में किया जाता है।)

वर्ण-विच्छेद :
जब किसी शब्द में प्रयुक्त वर्णों को अलग-अलग किया जाता है, तो उसे ‘वर्ण-विच्छेद’ कहते हैं।
उदाहरण-

बलाघात (Stress):
किसी शब्द के उच्चारण में किसी अक्षर पर जो बल दिया जाता है, उसे बलाघात कहते हैं। किसी भी अक्षर के सभी शब्द समान बल से नहीं बोले जाते। जैसे- ‘राम’ शब्द में ‘रा’ पर बल है। – ‘कबीर’ शब्द में ‘बी’ पर बल है। * बलाघात शब्द स्तर पर भी देखा जाता है, जैसे –
रोको, मत जाने दो।
आज मैं रामायण पढूंगा।
मैं रामायण कल पढूँगा ।

अनुतान (Intonation) :
बोलने में जो सुर का उतार-चढ़ाव (आरोह-अवरोह) होता है, उसे अनुतान कहते हैं। इसका महत्व शब्द एवं वाक्य दोनों स्तरों पर है। ‘अच्छा’ शब्द की विभिन्न अनुतान से –
अच्छा – सामान्य कथन/स्वीकृति
अच्छा ? – प्रश्नवाचक
अच्छा ! – आश्चर्य

संगम (Juncture) :
पदों का सीमा-संकेत संगम कहलाता है। संगम अक्षरों के बीच के हल्के-से विराम को जानना है। दो भिन्न स्थानों पर संगम से दो भिन्न अर्थ निकलते हैं ; जैसे –
सिरका – एक प्रकार का तरल पदार्थ
सिर + का – सिर से संबद्ध
जलसा – उत्सव
जल + सा – जल की तरह
मनका – माला का दाना
मन + का – मन का (भाव)

उच्चारण संबंधी अशधियाँ और उनका निराकरण (Correction in Pronunciation):
शुद्ध भाषा लिखने-पढ़ने में शुद्ध उच्चारण का बहुत महत्त्व है। हिंदी में वर्तनी की जो अशुद्धियाँ पाई जाती हैं, उनका एक प्रधान कारण अशुद्ध उच्चारण है। आगे ऐसे शब्दों के उदाहरण दिए जा रहे हैं जिनके उच्चारण में प्रायः अशुद्धि होती है :

तालिका :
1. ह्रस्व स्वर के स्थान पर दीर्घ तथा दीर्घ स्वर के स्थान पर हस्व



2. नासिक्य व्यंजन संबंधी अशुद्धियाँ

3. अल्पप्राण-महाप्राण संबंधी अशुद्धियाँ

4. व-ब की अशुद्धियाँ

5. श, ष, स संबंधी अशुद्धियाँ

6. छ-क्ष संबंधी अशुद्धियाँ

7. ऋ-र संबंधी अशुद्धियाँ

8. चंद्रबिंदु और अनुस्वार की अशुद्धियाँ

Haryana State Board HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 7 Congruence of Triangles InText Questions

Discuss, Think & Write (Page No. 137):

Question 1.
When two triangles, say ABC and PQR are given, there are, in all, six possible matchings or correspondences. Two of them are:
(i) ABC ⇔PQR and
(ii) ABC ⇔ QRP
Find the other four correspondences by using two cut outs of triangles. Will all these correspondences lead to congruence ? Think about it.
Solution:
For better understanding of the correspondence let us use a diagram.

(i) The correspondence is ABC ⇔ PQR. This means A ⇔ P ; B ⇔ Q ; and C ⇔ R.
So; (i) \(\overline{\mathrm{QR}} \leftrightarrow \overline{\mathrm{BC}}\) (ii) ∠R ↔ ∠C and
(iii) \(\overline{\mathrm{PQ}} \leftrightarrow \overline{\mathrm{AB}}\)
Similarly we find the all possible matchings or correspondences, e.g.

(ii) ABC ⇔ QRP : ∠A ⇔ ∠Q, ∠B ⇔ ∠R,
∠C ⇔ ∠P
\(\overline{\mathrm{AB}} \leftrightarrow \overline{\mathrm{QR}}, \overline{\mathrm{BC}} \leftrightarrow \overline{\mathrm{RP}}, \overline{\mathrm{AC}} \leftrightarrow \overline{\mathrm{QR}}\)

(iii) ABC ⇔ XYZ,
(iv) ABC ⇔ YZX
(v) ABC ⇔ ZXY,
(vi) ABC ⇔ DEF.

Try These (Page No. 140-141):

Question 1.
In Fig. lengths of the sides of the triangles are indicated. By applying the SSS congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form:

Solution:
From fig(i);
In triangle ABC and PQR
AB = PQ = 1.5
BC = QR = 2.5
and AC = PR = 2.2
This shows that the three sides of one triangle are equal to the three sides of the other triangle. So, by SSS congruence rule, the two triangles are congruent. From the above three equality relations, it can be easily seen that A ⇔ P.B ⇔ Q and C ⇔ R.
So, we have ΔABC ≅ ΔPQR .
From fig. (ii);
In triangles DEF and NLM,
DE = MN = 3.2
EF = LM = 3
FD = NL = 3.5
This shows that the three sides of one triangle are equal to the three sides of the other triangle. So, by SSS congruence rule, the two triangles are congruent.
From the above three equality relations, it can be easily seen that
D ⇔ N, E ⇔ M and F ⇔ L.
So, we have ΔDEF ≅ ΔNLM.
From Fig. (iii);
In triangles ABC and PQR AB = 2, PQ = 5 ∴ AB ≠ PQ
AC = 5, PR = 4 ∴ AC ≠ PR
BC = 4, QR = 2.5 ∴ BC ≠ QR
Hence, ΔABC and ΔPQR are not congruent. From Fig. (iv);
In triangles ABD and ADC,
AB = AC = 3.5
BD = DC = 2.5
AD = AD = Common
This shows that the three sides of one triangle are equal to the three sides of the other triangle. So by SSS congruence rule, the two triangles are congruent.
From the above three equlity relations, it can be easily seen that
A ⇔ A,B ⇔ C and D ⇔ D.
So, we have ΔABC ≅ ΔADC.

Question 2.
In Fig., AB = AC and D is the mid-point of \(\overline{\mathbf{B C}}\)
(r) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC ? Give reasons.
(iii) Is ∠B = ∠C Why ?

Solution:
From Fig. ;
In triangles ADB and ADC,
(i) AB = AC
AD = AD = Common BD = DC
(ii) This shows that the three sides of one triangle are equal to the three sides of the other triangle. So, by SSS congruence rule, the two triangles are congruent.
From the above three equality relations, it can be easily seen that
A ⇔ A, B ⇔ C and D ⇔ D
So, we have ΔADB ≅ ΔADC.
(iii) By corresponding parts of congruent triangles (C.P.C.T) are equals
hence, ∠B = ∠C.

Question 3.
In Fig., AC = BD and AD = BC. Which of the following statements is meaningfully written ?
(i) ΔABC ≅ ΔABD.
(ii) ΔABC ≅ ΔBAD.

Solution:
From Fig.
(i) In triangles ABC and ABD
AD = BC
BD = AC
AB = AB = Common
This shows that the three sides of one triangle are equal to the three sides of the other triangle. So, by SSS congruence rule, the two angles are congruent.

From the above three equality relations, it can be easily seen that
A ⇔ B, D ⇔ C and ∠CAB = ∠DBA.
So, we have AABC s AABD
(ii) Similarly, above three equality relations ∠ADB = ∠ACB, ∠ABD = ∠BAC and ∠DAB = ∠ABC.
So, we have ΔABC ≅ ΔBAD.

Think, Discuss & Write (Page No. 141):

Question 1.
ABC is an isosceles triangle with AB = AC (Fig. 7.15)
Take a trace copy of ΔABC and also name it as ΔABC.
(i) State the three pairs of equal parts in ΔABC and ΔACB.
(ii) Is ΔABC ≅ ΔACB ? Why or why not ?
(iii) Is ∠B = ∠C ? Why or why not ?

Solution:
in Triangle ΔABC and ΔACB
AB = AC, AC = AB and BC = CB

But in isosceles triangle two sides are equal and base angles are equal.
Hence, ∠B = ∠C and ∠C = ∠B
From the above three equality rela-tions, it can be easily seen that, A ⇔ A, B ⇔ C and C ⇔ B
So, we have ΔABC ≅ ΔACB. (by SAS congruence rule)

(i) AB = C, AC = AB, BC = CB
(ii) Yes, ΔABC ≅ ΔACB
(iii) Yes, ∠B = ∠C. (by CPCT and by isosceles triangle).

Try These (Page No. 143-144):

Question 1.
Which angle is included between the sides DE and EF of ΔDEF ?
Solution:
∠DEF or ∠E is included between the sides DE and EF of ΔDEF.

Question 2.
By applying SAS congruence rule, you want to establish that ΔPQR s ΔFED. It is given that PQ – EF and RP = DE What additional information (pair of equal parts) is needed to establish the congruence ?
Solution:
In ΔPQR and ΔDEF,
PQ = EF =3 and RP = DF = 3.5
If ∠F = ∠P then the ΔPQR ≅ ΔFED
Since, ∠F = 40° but ∠P is not given.

Question 3.
In Fig., measures of some parts of the tirangles are indicated, by applying SAS congruence rule, state the pairs of congruent triangles, if any, in each case. In case of congruent triangles, write them in symbolic form.

Solution:
From Fig. 7.17, ΔACB ≅ ΔRPQ;
ΔPRS ≅ ΔRPQ
Now, In ΔACB and ΔRPQ
AC = RP =2.5,
BC = PQ = 3
and included ∠C = included
∠P =35°
Also, A ⇔ R , B ⇔ Q and C ⇔ P

Therefore, ΔACB = ΔRPQ (By SAS congruence rule)
Now, in a ΔPRS and ΔRPQ
PR = RP = Common
RS = PQ =3.5
and included ∠R = included ∠P = 30°
Also, R ⇔ P.S ⇔ Q and P ⇔ R
Thererefore, ΔPRS ≅ ΔRPQ (By SAS congruence rule)

Question 4.
In Fig. \(\overline{\mathbf{A B}}\) and \(\overline{\mathrm{CD}}\) bisect each other at O
(i) State the three pairs of equal parts in two triangles AOC and BOD.
(ii) Which of the following statements are true ? (Fig.)
(a) ΔAOC ≅ ΔDOB
(b) ΔAOC ≅ ΔBOD

Solution:
(i) From Fig.
AB and CD bisect each other at O.
Therefore, AO = OB, CO = OD,
AC = DB
and ∠AOC = ∠DOB = Vertically opposite angles
= 30°
∠ACO = ∠BDO = alternate interior angle
= 70°
Hence, ∠CAO = ∠DBO = alternate interior angle

ii) (a) In ΔAOC and ΔDOB
AC = DB = 3,
CO = OD

and included ∠C = included
∠D =70°
Also, A ⇔ B,C ⇔ D and O ⇔ O
Therefore, ΔAOC ≅ ΔDOB
(6) In AAOC s ABOD
AO = OB,
CO = OD and
Included angle ∠AOC = included angle ∠BOD = 30°
Also, A ⇔ B, C ⇔ D and O ⇔ O
Therefore, ΔAOC ≅ ΔBOD
(By SAS congruence rule)

Try These (Page No. 145-146):

Question 1.
What is the side included between the angles M and N of MNP

Solution:
Side MN = MP
included angle = ∠NMP
and side NM = NP
included angle = ∠MNP

Question 2.
You want to establish ΔDEF = ΔMNP, using the ASA congruence rule. Your are given that ∠D = ∠M and ∠F = ∠P. What information is needed to establish the congruence ? (Draw a rough figure and then try !)
Solution:
In ΔDEF and ΔMNP,

∠D = ∠M, ∠F = ∠P and DF = MP i.e. D ⇔ M, F ⇔ P and DF ⇔MP
For ASA congruence rule, we need the two sides which the two angles ∠D = ∠M and ∠F = ∠P are included, so the addtional information are as follows:
DF = MP
Hence ΔDEF ≅ ΔMNP.

Question 3.
In Fig., measures of some parts are indicated. By applying ASA congruence rule, state which parts of triangles are congruent. In case of congruence, write the result in symbolic form.

Solution:
(i) In ΔABC and ΔEFD, we have
∠BAC=∠EFD = 40° (given)
∠ABD = ∠DEF = 60° (given)
AB = EF = 3.5 (given)
Hence ΔABC ≅ ΔEFD (By ASA congruence rule)

(ii) In ΔPQR and ΔDEF,
∠PQR = ∠EDF = 90° (given)
∠PRQ = ∠DEF = 50° (given)
PR = EF = 3.5 (given)
Hence ΔPQR ≅ ΔDEF (By ASA congruence rule)

(Hi) In ΔRQP and ΔLNM,
∠PRQ = ∠MLN = 60° (given)
∠PQR = ∠LMN = 30° (given)
RQ = LN = 6 (given)
Hence ΔRQP ≅ ΔLNM (By ASA congruence rule)

(iv) In ΔDAB and ΔCBA,
∠DAB = ∠CBA = (45°+ 30°)
= 75° (given)
∠ABD = ∠BAC = 30° (given)
AB = AR = common
Hence ΔDAB ≅ ΔCBA (By ASA congruence rule)

Question 4.
Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by ASA congruence rule. In case of congruence, write it in symbolic form.

Solution:

If under a correspondence
By applying ASA congruence rule, ∠D = ∠Q = 60° (given)
∠F = ∠R = 80° (given)
DF = QR = 5 cm (given)
Hence ADEF ≅ APQR

(ii) In ADEF and APQR
∠D = ∠Q = 60° (given)
∠F = ∠R = 80° (given)
DF = QR = 5 cm (given)
Hence ΔDEF ≅ ΔPQR

(iii) In ΔDEF and ΔPQR
∠D = ∠Q = 60° (given)
∠F = ∠R = 80° (given)

But included side DF ≠ QR
Hence ASA congruence rule not satisfied.
So, ASA congruence rule cannot be applied and we cannot conclude that two triangles are congruent.

(iii) In ΔDEF and ΔPQR
∠E = ∠P = 80° (given)
∠F = ∠R = 30° (given)
But included side EF ≠ PR

So, ASA congruence rule cannot be applied and we cannot conclude that two triangles are congruent.

Question 5.
In Fig., ray AZ bisects ∠DAB as well as ∠DCB.

(i) State the three pairs of equal parts triangles BAC and DAC.
(ii) Is ΔBAC ~ ΔDAC ? Give reasons.
(iii) Is AB = AD ? Justify your answer.
(iv) Is CD = CB ? Give reasons.
Solution:
Since ray AZ bisects ∠DAB as well as∠DCB.
Hence ∠DAC = ∠BAC and ∠DCA = ∠BCA

(i) In ΔBAC and ΔDAC,
∠BAC = ∠DAC (given)
∠BCA = ∠DCA (given)
AC = AC = Common
Hence ΔBAC = ΔDAC (by ASA congruence rule)

(ii) From solution (i), by ASA congruence rule, we have
ΔBAC s ΔDAC.

(iii) Since ΔBAC = ΔDAC
hence by CPCT (corresponding part of congruent triangle)
AD = AB
DC = CB
and ∠D ↔ ∠B
(iv) By CPCT, CD = CB
because ΔBAC = ΔDAC.

Try These (Page No. 148):

Question 1.
In Fig., measures of some parts of triangles are given. By applying RHS congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form.


Solution:
If under a correspondence the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right angled triangle, then the triangles are congruent.

(i) In ΔPQR and ΔDEF,
∠Q = ∠E = 90°
Hypotenuse PR = Hypotenuse DF = 6
Side PQ ≠ side DE
So, the triangles are not congruent.

(ii) In ΔCAB and ΔDAB,
∠C = ∠D = 90°
Hypotenuse AB = Hypotenuse
AB = Common
Side CA = side DB = 2
So, ΔCAB = ΔDAB (By RHS congruence rule)

(iii) In ΔBCA and DCA,
∠B = ∠D = 90°
Hypotenuse CA = Hypotenuse
CA = Common
Side BA = side DA = 3.6
So, ΔBCA = ΔDCA (By RHS congruence rule)

(iv) In ΔPQS and ΔPRS,
∠PSQ = ∠PSR = 90°
Hypotenuse PQ = Hypotenuse PR = 3
Side PS = side PS = Common
So, ΔPQS ≅ ΔPRS.

Question 2.
Is to be established by RHS congruence rule that ΔABC S ΔRPQ. What additional information is needed, if it is given that ∠B = ∠P = 90° and AB = RP ?

Solution:
In ΔABC and ΔRPQ
∠B = ∠P = 90°
Side AB = Side RP
The additional information are as follows :
If hypotenuse AC = hypote-nuse RQ
So, ΔABC ≅ ΔRPQ.

Question 3.
In Fig. , BD and CE are altitudes of ΔABC such that BD = CE.

(i) State the three pairs of equal parts in ΔCBD and ΔBCE.
(ii) Is ΔCBD ≅ ΔBCE ? Why or why not ?
(iii) Is ∠DCB ≅ ∠EBC ? Why or why not ?
Solution:
BD = CE,
(i) ∠BEC = ∠BDC = 90° (given)
BD = EC
BE = DC
i. e., ΔCBD = ΔBCE (SAS congruence)
(ii) Yes, because SAS congruency.
(tit) By CPCT, ∠DCB = ∠EBC. (Fig. 7.30)
hence ΔCBD ≅ ΔBCE.

Question 4.
ABC is an isosceles triangle with AB = AC and AD is one of its altitudes (Fig. 7.31)
(i) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC ? Why or why not ?
(iii) Is ∠B = ∠C ? Why or why not ?
(iv) Is BD = CD ? Why or why not ?
Sol. Since, ABC is an isosceles triangle.
Hence, AB = AC and AD is altitude of BC.

(i) In ΔADB and ΔADC
∠ADB = ∠ADC = 90° (given)
hypotenuse AB = hypotenuse AC (given)
SideBD = Side DC (given)

(ii) So, ΔADB ≅ ΔADC (By RHS congruence rule)

(iii) By CPCT, ∠B = ∠C
(Isosceles’s base angles are equal)

(iv) By CPCT, BD = CD (AD is one of its altutude of side BC, hence BD = CD)

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes InText Questions

Try These (Page 277)

Question 1.
Match the shape with the name:



Solution:
(i) (b) Cylinder
(ii) (d) Sphere
(iii) (a) Cuboid
(iv) (c) Cube
(v) (f) Cone
(vi) (e) Pyramid

Try These (Page 278)

Question 1.
Match the 2 dimensional figure with the names:


Solution:
(i) Rectangle
(ii) Circle
(iii) Triangle
(iv) Square
(v) Quadrilateral

Try These (Page 279)

Question 1.
Do this : Complete the following table :

Try These (Page 281)

Question 1.
Here you find four nets. There are two correct nets among them to make a tetrahedron. See if you can work out which nets will make a tetrahedron.


Solution:

Try These (Page 286)

Question 1.
Try to guess the number of cubes in the following arrangements (Fig.):

Solution:
(i) Numerofcubes = 24
Let the edge of cube = 2 cm
Height = 4 cm
Breadth = 6 cm
Length = 8cm
(ii) Number of cubes = 8
Let the edge of cube = 2 cm
Length = 6 cm
Height = 4cm
Breadth = 4 cm
(iii) Number of cubes = 9
Lettheedgeofcube = 2cm
Length = 6cm
Breadth = 6 cm
Height = 4 cm

Try These (Page 287)

Question 1.
Two dice are placed side by side as shown; Can you say what the total would be on the face opposite to
(а) 5 + 6
(b) 4 + 3
(Remember that in a dice sum of numbers on opposite faces is 7)

Solution:
(i) (a) 5 + 6 → The face opposite to 5 is 2 and 6 is 1.
So, 2 + 1 = 3
(b) 4 + 3 → The face opposite to 4 is 3 and 3 is 4.
So, 3 + 4 = 7.

Question 2.
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Try to make an oblique sketch and say what could be its length, breadth and height.
Sol. (Fig.)

Length = 6 cm
Breadth = 2 cm
Height = 2 cm.

Try These (Page 291)

Question 1.
For each solid, the three views (1), (2), (3) are given. Identify for each solid the corresponding top, front and side views.
Solid In views


Solution:

Question 2.
Draw a view of each solid as seen from the direction indicated by the arrow.

Solution:

Haryana Board HBSE 7th Class Sanskrit Solutions रुचिरा भाग 2

• Chapter 1 सुभाषितानि
• Chapter 2 दुर्बुद्धिः विनश्यति
• Chapter 3 स्वावलम्बनम्
• Chapter 4 हास्यबालकविसम्मेलनम्
• Chapter 5 पण्डिता रमाबाई
• Chapter 6 सदाचारः
• Chapter 7 सड.कल्पः सिद्धिदायकः
• Chapter 8 त्रिवर्णः ध्वजः
• Chapter 9 अहमपि विद्यालयं गमिष्यामि
• Chapter 10 विश्वबंधुत्वम्
• Chapter 11 समवायो हि दुर्जयः
• Chapter 12 विद्याधनम्
• Chapter 13 अमृतं संस्कृतम्
• Chapter 14 अनारिकायाः जिज्ञासा
• Chapter 15 लालनगीतम्

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Ex 15.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Exercise 15.1

Question 1.
Identify the nets which can be used to make cubes (cut out copies of the nets and try it) :

Solution:

Question 2.
Dice are cubes with dots on each face.
Opposite faces of a die always total to seven dots on them. Fig.

Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box.

Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7.
Solution:

Question 3.
Can this be a net for a die ? Explain your answer. (Shown Fig. 15.11.)

Solution:
(Shown in Fig. 15.12). No, the net is not for die, because the sum of the opposite laces is not 7.

Question 4.
Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here ? (Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation.)

Solution:

Question 5.
Match the nets with appropriate solids :

Solution:
(a) → (ii)
(b) → (iii)
(c) → (iv)
(d) → (i)

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Ex 15.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Exercise 15.2

Question 1.
Use isometric dot paper and make an isometric sketch for each one of the given shapes:
Solution:
Let us take m to be the number of pannit’s marbles.Fig. 15.16.

Question 2.
The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw three different isometric sketches of this cuboid.
Solution:

Question 3.
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Sketch an oblique or isometric sketch of this cuboid.
Solution:

Question 4.
Make an oblique sketch for each one of the given isometric shpaes:
Solution:

Question 5.
Give (i) an oblique sketch and (ii) an isometric sketch for each of the following :
(a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique ?)
(b) A cube with an edge 4 cm long.
An Isometric sheet is attached at the end of the book. You could try to make on it some cubes or cuboids of dimensions specified by your friend.
Solution:

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Ex 15.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Exercise 15.3

Question 1.
What cross-sections do you get when you give a
(i) vertical cut
(ii) horizontal cut to the following solids ?
(a) Abrick
(b) Around apple
(c) A die
(d) A circular pipe
(e) An ice cream cone
Solution:
(a) (i) Vertical cut: Here is a brick. When you give the vertical cut to brick you get rectangle. So the cross-section of brick is nearly rectangle in this case.

(ii) Horizontal cut :
When you give the horizontal cut to brick you get rectangle.
So the cross section of brick is nearly rectangle in this case.

(b) (i) Vertical cut : Here is a round apple. When you give the vertical cut to round apple you get nearly circle. So, the cross section is nearly circle in this case.

(ii) Horizontal cut: When you give the horizontal cut to round apple you get nearly circle.
So, the cross-section of round apple is nearly circle.

(c) (i) Vertical cut: Here is a dice. It is like a cube.
When you give a vertical cut, you get a square face. So the cross-section of dice is nearly square in this case.
(ii) Horizontal cut :
When you give the horizontal cut to the dice you get the rectangle. So the cross-section of the dice is nearly rectangle in this case.

(d) (i) Vertical cut: Here is a circular pipe. When you give the horizontal. Cut to circular pipe you get semi cylinder. So, the cross-section of circular pipe is semi cylinder in this case.

(ii) Horizontal cut: When you give the horizontal cut to circular pipe, its cross-section remains same.

(e) (i) Vertical cut: Here is an ice cream cone when you give the vertical cut to an ice cream cone you get nearly triangle. So the cross section of an ice cream cone is nearly triangle.
(ii) Horizontal cone: When you give the horizontal cut to an ice cream cone you get nearly triangle. So the cross section of an ice cream cone is nearly triangle in this case.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Ex 15.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Exercise 15.4

Question 1.
A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions).

Solution:
A Ball : The shape of the shadow obtained in case of ball is nearly semi-circle.

Cylindrical pipe : The shape of the shadow obtained in case of cylindrical Pipe is nearly rectangular.
A book: The shape of the shadow obtained in case a book in nearly rectangular.

Question 2.
Here are the shadows of some 3-D objects, when seen under the lamp of an Overhead projector. Identify the solid(s) that match each shadow. (There may be multiple answers for these !)

Solution:
(i) Sphere, iii) Cube, cuboid, (iii) Pyramid, cone, (iv) Cylinder.

Question 3.
Examine if the following are true statements :
(i) The cube can cast a shadow in the shape of a rectangle.
(ii) The cube can cast a shadow in the shape of a hexagon.
Solution:
(i) True, (ii) False.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 14 Symmetry InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 14 Symmetry InText Questions

Try These (Page 272)

Question 1.
(a) Can you now tell the order fo the rotational symmetry for an equilateral triangle?
(b) How many positions are there at which the triangle looks exactly the same, when rotated about its centre by 120°?


Solution:
The number of times a shape will fit onto itself in one complete turn is called the order of rotation symmetry.
(a) Thus, we say that an equilateral triangle has rotational symmetry of orders.
(b) We have seen in the figure that there are three positions where it appears not to have moved.

Question 2.
Which of the following shapes have rotational symmetry about the marked point ?

Solution:

Try These (Page 273)

Question 1.
Give the order of the rotational symmetry of the given figures :


Solution:
Fig. (i).
Look at the fig (i) and rotate the figure about the mark point (x). Rotate the figure (i) by 90°.
We get,

And again rotate the fig. (i) by another 90°, a total of 180° we get,
Look at the fig (ii). Rotate the figure about the mark point (x). Rotate the Fig. (i) by 90°, we get
Fig. (ii) And again rotate the fig. (ii) by another 90°, a total of 180° we get,

Fig. (iii). Look at the fig. (iii) and, rotate the figure about the mark point (x). Rotate the figure (iii) by 90°, we get,

And again rotate the fig. (iii) by another 90°, a total of 180°, we get,

Haryana State Board HBSE 7th Class Maths Solutions Chapter 14 Symmetry Ex 14.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 14 Symmetry Exercise 14.3

Question 1.
Name any two figures that have both line symmetry and rotational symmetry.
Solution:

The alphabet letters H and 0 both have line symmetry and rotational symmetry.

Question 2.
Draw, wherever possible, a rough sketch of:
(i) a triangle with both line and rotational symmetries of order more than 1.
(ii) a triangle with only line symmetry and no rotational symmetry of order more than 1.
(iii) a quadrilateral with a rotational symmetry of order more than 1 but not a line symmetry.
(iv) a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
Solution:

Question 3.
If a figure has two or more lines of symmetry, should it have rotational symmetry of order more than 17.
Solution:

It has two lines of symmetry and rotational symmetry of order 2.

Question 4.
Fill the blanks :
Solution:

Question 5.
Name the quadrilateral which have both line and rotational symmetry of order more than 1.
Solution:
Square is both line and rotational symmetry.

Hence, order of rotation 4.

Question 6.
After rotation by 60° about a centre, a figure looks exactly the same as its original position. At what other angles will this happen for the figure?
Solution:
This figure is circle.

Centre of Rotation = 1
Order of Rotation = Infinite
Angle of Rotation = Nil.

Question 7.
Can we have a rotational symmetry of order more than 1 whose angle of rotation is :
(i) 45° ? (ii) 17° ?
Solution:
(i) Yes, e.g., square.
(ii) No.