NCERT Solutions for Class 7 Hindi Vasant Chapter 1 हम पंछी उन्मुक्त गगन के

NCERT Solutions for Class 7 Hindi Vasant Chapter 1 हम पंछी उन्मुक्त गगन के are part of NCERT Solutions for Class 7 Hindi. Here we have given NCERT Solutions for Class 7 Hindi Vasant Chapter 1 हम पंछी उन्मुक्त गगन के.

Board CBSE
Textbook NCERT
Class Class 7
Subject Hindi Vasant
Chapter Chapter 1
Chapter Name हम पंछी उन्मुक्त गगन के
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 7 Hindi Vasant Chapter 1 हम पंछी उन्मुक्त गगन के

पाठ्यपुस्तक के प्रश्न-अभ्यास

(पृष्ठ 2-3)

कविता से

प्रश्न 1.
हर तरह की सुख सुविधाएँ पाकर भी पक्षी पिंजरे में बंद क्यों नहीं रहना चाहते?
उत्तर
हर तरह की सुख-सुविधाएँ पाकर भी पक्षी पिंजरे में बंद नहीं रहना चाहते क्योंकि उन्हें बंधन पसंद नहीं। वे तो खुले आकाश में ऊँची उड़ान भरना, नदी-झरनों का बहता जल पीना, कड़वी निबौरियाँ खाना, वृक्ष की सबसे ऊँची डाली पर झूलना, अनार के दानों रूपी तारों को चुगना और क्षितिज मिलन करना ही पसंद करते हैं।

NCERT Solutions for Class 7 Hindi will help you to score more marks in your CBSE board Examination.

प्रश्न 2.
पक्षी उन्मुक्त रहकर अपनी कौन-कौन सी इच्छाएँ पूरी करना चाहते हैं?
उत्तर
पक्षी उन्मुक्त रहकर अपनी निम्न इच्छाओं को पूरा करना चाहते हैं

  1. पक्षी खुले आकाश में उड़ना चाहते हैं।
  2. नदी-झरनों का बहता जल पीना चाहते हैं।
  3. नीम के वृक्ष की कड़वी निबौरियाँ खाना चाहते हैं।
  4. पेड़ की सबसे ऊँची टहनी पर झूलना चाहते हैं।
  5. आकाश में ऊँची उड़ान भरकर अनार के दानों रूपी तारों को चुगना चाहते हैं।
  6. क्षितिज मिलन करना चाहते हैं।

प्रश्न 3.
भाव स्पष्ट कीजिए
या तो क्षितिज मिलन बन जाता / या तनती साँसों की डोरी।
उत्तर-
प्रस्तुत पंक्तियाँ हमारी पाठ्यपुस्तक वसंत भाग-2 की कविता ‘हम पंछी उन्मुक्त गगन के’ से ली गई है। इस कविता के रचयिता प्रसिद्ध कवि श्री शिवमंगल सिंह ‘सुमन’ हैं। इस पंक्ति में कवि पक्षी के माध्यम से कहना चाहते हैं कि पक्षी स्वतंत्र रहकर क्षितिज की सीमा तक उड़ जाने की अपनी इच्छा व्यक्त करते हैं कि उड़ते-उड़ते या तो वह तो क्षितिज की सीमा हूँढ़ ही निकालेंगे या प्राण त्याग देंगे। पक्षियों के इस कथन से उनकी उन्मुक्त उड़ान के प्रति ललक व्यक्त हुई है।

कविता से आगे

प्रश्न 1.
बहुत से लोग पक्षी पालते हैं-
(क) पक्षियों को पालना उचित है अथवा नहीं? अपने विचार लिखिए।
उत्तर
पक्षियों का पालना बिल्कुल भी उचित नहीं है क्योंकि ईश्वर ने उन्हें उड़ने के लिए पंख दिए हैं, तो इच्छाएँ भी वैसी ही दी हैं। वे हरदम स्वच्छंद रहना चाहते हैं। अपनी इच्छा से ऊँची-से-ऊँची उड़ान भरना, पेड़ों पर घोंसले बनाकर रहना, नदी-झरनों का जल पीना, फल-फूल खाना ही उनकी प्रवृत्ति है। उनके बंधन हेतु पिंजरे भले ही सोने के क्यों न हों, बढ़िया से बढ़िया खाने के पदार्थ उन्हें परोसे जाएँ, लेकिन यही सत्य है कि बंधन में रहकर मिलने वाली सुविधाओं से कठिनाइयों से भरी आज़ादी श्रेष्ठ है।

(ख) क्या आपने या आपकी जानकारी में किसी ने कभी कोई पक्षी पाला है? उसकी देखरेख किस प्रकार की जाती होगी, लिखिए।
उत्तर
हाँ! हमने एक बार तोता पाला था। मेरे पिताजी उसे मेले से खरीदकर लाए थे। पक्षी की परवरिश एक छोटे बच्चे की भाँति ही की जाती है। मेरी माँ सुबह-सुबह उसे पिंजरे समेत घर के बाहर, बगीचे में लेकर आतीं। वह बाहर का वातावरण देखकर प्रसन्न हो जाता और ज़ोर-जोर से पंख फड़फड़ाता। हम उसे नहलाते भी थे। माँ उसे बाजरा देती व कटोरी में पानी। अमरूद, अंगूर व हरी मिर्च तो वह बहुत चाव से खाता। धूप थोड़ी सी तेज होने पर वह सिकुड़कर बैठ जाता। हम उसे अंदर कमरे में ले आते। बिल्ली या कुत्ते आदि जानवरों से उसे बहुत बचाना पड़ता था। रात को उसके पिंजरे पर कोई कपड़ा डाल देते तो वह बैठे-बैठे ही सो जाता।

प्रश्न 2.
पक्षियों को पिंजरे में बंद करने से केवल उनकी आज़ादी का हनन ही नहीं होता, अपितु पर्यावरण भी प्रभावित होता है। इस विषय पर दस पंक्तियों में अपने विचार लिखिए।
उत्तर-
पक्षियों को पिंजरे में बंद करके उसकी स्वतंत्रता का हनन होता है, क्योंकि उनकी प्रवृत्ति है ‘उड़ना’। अतः प्राकृतिक संतुलन बनाए रखने में पृथ्वी के सभी जीवों की समान रूप से महत्त्वपूर्ण भूमिका है। पृथ्वी के पर्यावरण संतुलन के लिए मनुष्य एवं पशु दोनों की आवश्यकता समान रूप से है। पक्षियों को कैदी बना लेने से उनकी उड़ने की आजादी समाप्त हो जाती है। इससे पर्यावरण प्रभावित होता है। पर्यावरण को शुद्ध और प्राकृतिक बनाए रखने के लिए पक्षियों को प्रकृति के मध्य रहना आवश्यक है। वे इस प्रकार पर्यावरण को शुद्ध एवं संतुलित बनाते हैं। पर्यावरण में पक्षियों का अपना विशेष महत्त्व होता है।

अनुमान और कल्पना

प्रश्न 1.
क्या आपको लगता है कि मानव की वर्तमान जीवन-शैली और शहरीकरण से जुड़ी योजनाएँ पक्षियों के लिए घातक हैं? पक्षियों से रहित वातावरण में अनेक समस्याएँ उत्पन्न हो सकती हैं। इन समस्याओं से बचने के लिए हमें क्या करना चाहिए? उक्त विषय पर वाद-विवाद प्रतियोगिता का आयोजन कीजिए।
उत्तर-
हाँ, इसमें कोई शक नहीं है कि मानव की वर्तमान जीवन-शैली और शहरीकरण से जुड़ी योजनाएँ पक्षियों के लिए सभी दृष्टिकोण से घातक हैं। अंधाधुंध शहरीकरण के कारण पक्षी प्रकृति से समाप्त होते चले जा रहे हैं। बड़ी संख्या में पेड़ काटे जा रहे हैं जिससे पक्षियों का आश्रय समाप्त हुआ है। कल-कारखानों के खुलने से वातावरण का प्रदूषण बढ़ गया है। इस कारण पक्षियों का आसमान में उड़ना भी कठिन हो गया है क्योंकि उनका आश्रय समाप्त होने के साथ-साथ पेड़ों से प्राप्त खाद्य पदार्थ, फल-फूल आदि उन्हें नहीं मिल पाते । ऐसा होने पर उन्हें मुश्किलों का सामना करना पड़ता है। पक्षियों के पलायन से सबसे बड़ा खतरा है। अनाज़ में कमी होने का है। पर्यावरण संतुलित नहीं रहेगा इससे पर्यावरण संतुलन पर गहरा असर पड़ेगा और तब मनुष्य को अपने भविष्य की चिंता सताने लगेगी। अतः आवश्यक है कि मनुष्य जागरूक हो जाए और पक्षियों के संरक्षण के लिए अधिक से अधिक संख्या में वृक्षारोपण करें। पक्षियों के लिए जलाशयों के साथ-साथ बाग-बगीचों का भी निर्माण करवाएँ। पक्षियों को पिंजरों में बंदी बना करके नहीं रखना चाहिए।

अन्य समस्याओं के बारे में छात्र स्वयं सोचे और विचार-विमर्श करें। इसके लिए विद्यालय में वाद-विवाद प्रतियोगिता का आयोजन करें।

प्रश्न 2.
यदि आपके घर के किसी स्थान पर किसी पक्षी ने अपना आवास बनाया है और किसी कारणवश आपको अपना घर बदलना पड़ रहा है तो आप उस पक्षी के लिए किस तरह के प्रबंध करना आवश्यक समझेंगे? लिखिए।
उत्तर
यदि हमारे घर में किसी पक्षी ने अपना घोंसला बनाया हो और किसी कारणवश हमें घर बदलना पड़ रहा हो तो हम संभवतः प्रयास तो यह करेंगे कि घोंसले को छेड़ा न जाए और उस पक्षी के बाहर आने-जाने का स्थान भी खुला रहे। लेकिन यदि ऐसा न हो पाए तो हम घोंसले को सावधानीपूर्वक उठाकर घर के बाहर किसी ऊँचे स्थान पर रखेंगे जहाँ उस पक्षी की नज़र पड़ सके।

भाषा की बात

प्रश्न 1.
स्वर्ण-श्रृंखला और लाल किरण-सी में रेखांकित शब्द गुणवाचक विशेषण हैं। कविता से ढूंढकर इस प्रकार के तीन और उदाहरण लिखिए।
उत्तर

  1. कनक-तीलियाँ
  2. कटुक-निबौरी
  3. तारक-अनार।

प्रश्न 2.
‘भूखे-प्यासे’ में द्वंद्व समास है। इन दोनों शब्दों के बीच लगे चिह्न को सामासिक चिह्न (-) कहते हैं। इस चिह्न से ‘और’ का संकेत मिलता है, जैसे-भूखे-प्यासे = भूखे और प्यासे।
• इस प्रकार के दस अन्य उदाहरण खोजकर लिखिए।
उत्तर

  1. सुख-दुख      –   सुख और दुख
  2. दिन-रात       –  दिन और रात ।
  3. सुबह-शाम    –  सुबह और शाम
  4. अन्न-जल       –  अन्न और जल
  5. अपना-पराया – अपना और पराया
  6. अमीर-गरीब  –   अमीर और गरीब
  7. तन-मन        –     तन और मन
  8. दूध-दही       –     दूध और दही
  9. खट्टा-मीठा    –    खट्टा और मीठा
  10. पाप-पुण्य      –    पाप और पुण्य

We hope the NCERT Solutions for Class 7 Hindi Vasant Chapter 1 हम पंछी उन्मुक्त गगन के help you. If you have any query regarding NCERT Solutions for Class 7 Hindi Vasant Chapter 1 हम पंछी उन्मुक्त गगन के, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 English Honeycomb Chapter 1 Three Questions

NCERT Solutions for Class 7 English Honeycomb Chapter 1 Three Questions are part of NCERT Solutions for Class 7 English. Here we have given NCERT Solutions for Class 7 English Honeycomb Chapter 1 Three Questions.

Board CBSE
Textbook NCERT
Class Class 7
Subject English Honeycomb
Chapter Chapter 1
Chapter Name Three Questions
Number of Questions Solved 47
Category NCERT Solutions

NCERT Solutions for Class 7 English Honeycomb Chapter 1 Three Questions

IMPORTANT PASSAGES FOR COMPREHENSION
Read the following extracts and answer the questions that follow by choosing the correct option :

[I]

NCERT Solutions for Class 7 English Honeycomb will help you to score more marks in your CBSE board Examination.

Question 1.
Many wise men came to the king, but they all answered his questions differently. The king gave no reward.      (Page 7)
Multiple Choice Questions
Question 1.
The wise men wished to get
(a) some amount of money
(b) a handsome sum of money
(c) the land owned by the king
(d) the treasure buried under the temple
Answer.
(b) a handsome sum of money

Question 2.
Their answers to his questions were
(a) inappropriate
(b) different
(c) irrelevant
(d) to the point
Answer.
(b) different

Question 3.
The king’s reaction was
(a) undesirable
(b) humiliating
(c) justified
(d) instant
Answer.
(c) justified

Question 2.
“You are tired,” said the king. “Let me take the spade and work in your place.”   (Page 9)
Question 1.
Who is ‘you’ in the above extract ?
Answer.
‘You’refers to the hermit.

Question 2.
Why did the king offer to work ?
Answer.
The king offered to work because he felt that the hermit was very tired.

Question 3.
What does it show about the king’s character ?
Answer.
It shows the king’s compassion.

Question 3.
“Here comes someone running, said the hermit.”          (Page 10)
Multiple Choice Questions
Question 1.
The hermit drew the king’s attention
(a) to the axe kept nearby
(b) towards the bearded man
(c) towards the evil bodyguard
(d) to the seeds to be sown
Answer.
(b) towards the bearded man

Question 2.
The bearded man was
(a) stabbed by the king
(b) God-fearing
(c) the king’s enemy
(d) wearing a long robe
Answer.
(c) the king’s enemy

Question 3.
Both the hermit and the king
(a) slept for long
(b) nursed the wounds of the bearded man
(c) told tales
(d) were not surprized to meet the man
Answer.
(b) nursed the wounds of the bearded man

[II]

Question 4.
“Now if I live, I will serve you as your most faithful servant and will order my sons to do the same. Forgive me !”          (page 12)
Question 1.
Who is the speaker of the above lines ?
Answer.
The speaker was an injured bearded man whom the king had helped. The king had saved his life.

Question 2.
Who was he speaking to ?
Answer.
He was speaking to the king.

Question 3.
Why was he making such a promise ?
Answer.
He felt greatly indebted to the king. In order to repay the king’s goodness, he was making that promise.

Question 5.
The most important business is to do that person good, because we were sent into this world for that purpose alone.    (Page 13)
Question 1.
The hermit’s advice was
(a) to do good to the people
(b) to do important work
(c) to do hard work
(d) to do good to the person who needs your help at the moment
Answer.
(d) to do good to the person who needs your help at the moment

Question 2.
No one is sure
(a) if he would get time to meet someone else
(b) if he will live
(c) what is in store for him
(d) whether the future is bright for him
Answer.
(a) if he would get time to meet someone else

Question 3.
To do good is
(a) the only work for you
(b) the main purpose of our life
(c) greatly admired by gods
(d) liked by even the devils
Answer.
(b) the main purpose of our life

TEXTUAL QUESTIONS

Comprehension Check (Page 10)
Question 1.
Why did the king want to know answers to three questions ?
Answer.
The king wanted to be always successful. For this purpose, he wanted to know the answers to three questions.

Question 2.
Messengers were sent throughout the kingdom
(i) to fetch wise men.
(ii) to find answers to the questions.
(iii) to look for the wise hermit.
(iv) to announce a reward for those who could answer the questions.
Mark your choice.
Answer.
(iv) to announce a reward for those who could answer the questions.

Comprehension Check (Page 14)
Complete the following sentences by adding the appropriate parts of the sentences given in the box.

  1. Many wise men answered the king’s questions, …………..
  2. Someone suggested that there should be a council of wise men ……………
  3. Someone else suggested that the king should have a timetable …………..
  4. The king requested the hermit …………
  5. The king washed and dressed the bearded man’s wound, ………………

NCERT Solutions for Class 7 English Honeycomb Chapter 1 Three Questions 1
Answers.

  1. Many wise men answered the king’s questions but their answers were so varied that the king was not satisfied.
  2. Someone suggested that there should be a council of wise men to help the king act at the right time.
  3. Someone else suggested that the king should have a timetable and follow it strictly.
  4. The king requested the hermit o answer three questions.
  5. The king washed and dressed the bearded man’s wound but the bleeding would not stop.

Answer the following questions :
Question 1.
Why was the king advised to go to magicians ? (Imp.)
Answer.
Some people said that the magicians could look into the future. If one knows the future it is easy to decide the right time for doing something. The king wanted to know the right time to begin something. So some people advised the king to go to magicians.

Question 2.
In answer to the second question, whose advice, did the people say, would be important to the king ?
Answer.
In answer to the second question the wise men had given different answers. Some said that the king’s councillors were the people whose advice was important. Others thought the priests were more important. A few others chose the doctors. Some said that the advice of the soldiers would be important to the king.

The hermit’s answer was the best. He said that the person present with the king was the only important person. His advice would be important.

Question 3.
What suggestions were made in answer to the third question ?
Answer.
In answer to the third question, some said science was the most important affair. Others chose fighting and yet others advocated religious worship.

The hermit’s answer was the best. He said that the most important affair is to do good to the person before you. The hermit believed that man was sent on the earth for this purpose alone.

Question 4.
Did the wise men win the reward ? If not, why not ?   (V. Imp.)
Answer.
No, the wise men did not win the reward. It was so because none of their answers could satisfy the king.

Question 5.
How did the king and the hermit help the wounded man ?   (V. Imp.)
Answer.
The king and the hermit removed the man’s clothing. They saw that there was a wound in his stomach. The king washed the wound. Then he covered it with his handkerchief. Yet the blood kept on flowing. So the king redressed the wound. At last the bleeding stopped. The man felt better. Now he wanted to drink something. The king gave him fresh water. After this, they carried him into the hut and put him on the bed.

Question 6.
(i) Who was the bearded man ?
(ii) Why did he ask for the king’s forgiveness ?
Answer.

  1. The bearded man was the king’s sworn enemy. The king had killed his brother and seized his property. He wanted to avenge his murdered brother by killing the king.
  2. The king had washed the bearded man’s wound. He had covered the wound with his handkerchief. He redressed his wound because the blood kept on flowing. The bearded man felt grateful to the king. Therefore he asked for his forgiveness.

Question 7.
The king forgave the bearded man. What did he do to show his forgiveness ? (Imp.)
Answer.
To show his forgiveness, the king promised to help the bearded man. He promised to send his servants and his own doctor to look after him. The king also promised to give him back his property.

Question 8.
What were the hermit’s answers to the three questions ? Write each answer separately. Which answer do you like most, and why ? (V. Imp.)
Answer.
The hermit’s answer was that ‘Now’ is the best time to begin work. For, nobody knows whether he will have any time in the future or not. The king had just put his questions, when he pitied the hermit. Later on it proved to be the best time. Had he not done so and gone back, his enemy would have killed him.

Answering the second question, the hermit said that the people present were the best. They alone should be listened to. For, these are the only people whom one can consult at once.

To do good to the people present is the most important thing. It was the hermit’s reply to the third question. According to the hermit, man was sent on the earth for this purpose alone.
I like the answer to the first question most. It is most logical.

Working with Language

Question 1.
Match items in List A with their meanings in List B.

NCERT Solutions for Class 7 English Honeycomb Chapter 1 Three Questions 2

Use any three of the above words change the form of the word.
Answer.
NCERT Solutions for Class 7 English Honeycomb Chapter 1 Three Questions 3

Question 2.
Each of the following sentences has two blanks. Fill in the blanks with appropriate forms of the word given in brackets.   (Imp.)
He has ………. to help me. Do you think he will remember his ……. ?  (promise)
He has promised to help me. Do you think he will remember his promise ?
(i) The ……….. said that only fresh evidence would make him change his ……. (judge)
(ii) I didn’t notice any serious ……….. of opinion among the debaters, although they ………. from one another over small points. (differ)
(iii) It’s a fairly simple question to ……….. , but will you accept my …….. as final ? (answer)
(iv) It isn’t ……… that …………… should always be the mother of invention. (necessary)
(v) Hermits are ………. men. How they acquire their ………… no one can tell.  (wise)
(vi) The committee ………….. has to make Jagdish captain of the team. The ……….. is likely to please everyone.  (decide)
(vii) Asking for ………. is as noble as willingness to …………. .  (forgive)
Answers.

  1. The judge said that only fresh evidence would make him change his judgement.
  2. I didn’t notice any serious difference of opinion among the debaters, although they differed from one another over small points.
  3. It’s a fairly simple question to answer, but will you accept my answer as final?
  4. It isn’t necessary that necessity should always be the mother of invention.
  5. Hermits are wise men. How they acquire their wisdom no one can tell.
  6. The committee has decided to make Jagdish captain of the team. The decision is likely to please everyone.
  7. Asking for forgiveness is as noble as willingness to forgive.

Speaking and Writing

Question 1.
Imagine you are the king. Narrate the incident of your meeting the hermit. Begin like this :
The wise men answered my questions, but I was not
satisfied with their answers. One day I decided to go
and meet the hermit…
Answer.
… I put on ordinary clothes. I started for the woods, in which the hermit lived. I left my horse with my bodyguard. Then, I went to the hermit’s hut alone. I saw the hermit digging the ground. The hermit greeted me but continued digging. I told him the purpose of my visit. The hermit listened to me. Without speaking anything he busied himself again in digging. I felt the hermit was tired. I offered to do his work for him. He agreed, I began to dig the ground.

Question 2.
Imagine you are the hermit. Write briefly the incident of your meeting the king. Begin like this : (Imp.)
One day I was digging in my garden. A man in
ordinary clothes came to see me. I knew it was the
king …
Answer.
I greeted the king. He asked me three questions. He wanted to know the right time to do the right thing. He wanted to know who were the people he needed most. He also wanted to know the affairs which were most important. I made no reply. Then the king saw that I was tired. He offered to do my work. I gave him my spade. He worked till sunset.

SHORT ANSWER TYPE QUESTIONS

Question 1.
Why was the king anxious to know the answers to his questions ?
Answer.
The king wanted to know the answers to his three questions. He felt that if he knew them, he would never fail in his attempts.

Question 2.
The hermit told the king the purpose of human life. What was it ? What was the king’s aim of life before meeting the hermit ?   (V. Imp.)
Answer.
The hermit told the king that the purpose of life was to help others. Success in whatever he did, was the king’s aim of life before meeting the hermit.

Question 3.
Why did the king go alone and in ordinary clothes to meet the hermit ?
Answer.
The king went alone and in ordinary clothes because he knew that the hermit saw no one but simple people.

Question 4.
Why did the king dig the ground ?
Answer.
The king dug the ground to help the hermit.

Question 5.
Why did the hermit let the king work till sunset ?
Answer.
The hermit let the king work till sunset to save his life.

LONG ANSWER TYPE QUESTIONS

Question 1.
What was the hermit’s answer to the second question : ‘Which people should be listened to’ ? Justify the answer.   (V. Imp.)
Answer.
The hermit’s answer to the second question was that the king should listen to the people who were with him at that particular moment. It was correct because according to the hermit man’s only duty is to help his fellowmen. So if there is someone before us who needs our help, our foremost duty is to help him.

OBJECTIVE TYPE QUESTIONS

Read the statements given below and choose the correct option :

Question 1.
The answers to the three questions will enable :
(i) the hermit to go to the king
(ii) the king to lead a peaceful life
(iii) the king to be always successful
(iv) the subjects to lead a good life
Answer.
(iii) the king to be always successful

Question 2.
The king was not satisfied with the answers as :
(i) they were lengthy
(ii) they all were different
(iii) they were dull
(iv) they served no purpose
Answer.
(ii) they all were different

Question 3.
The king went to see the hermit in ;
(i) his cage
(ii) simple clothes
(iii) his cave
(iv) a hurry
Answer.
(ii) simple clothes

Question 4.
The king dug the beds
(i) because he loved gardening
(ii) to help the hermit
(iii) to pass time
(iv) to get his answers
Answer.
(ii) to help the hermit

Question 5.
The wounded man was
(i) a friend of the hermit
(ii) a friend of the king
(iii) an enemy of the hermit
(iv) an enemy of the king
Answer.
(iv) an enemy of the king

Question 6.
The king dressed the wounds of :
(i) the hermit
(ii) his guard
(iii) the horse
(iv) the bearded man
Answer.
(iv) the bearded man

Question 7.
The bearded man was the :
(i) relative of the king
(ii) friend of the king
(iii) king’s sworn enemy
(iv) guard of the king
Answer.
(iii) king’s sworn enemy

Question 8.
The bearded man asked for :
(i) water
(ii) property
(iii) forgiveness
(iv) money
Answer.
(iii) forgiveness

Question 9.
The king ‘had done him’ wrong as :
(i) he had killed his brother
(ii) he had killed his sister
(iii) he had seized his property
(iv) Both (i) and (iii)
Answer.
(iv) Both (i) and (iii)

Question 10.
The king sent a/an to look after him.
(i) doctor
(ii) anaesthetic
(iii) quack
(iv) servant
Answer.
(i) doctor

Question 11.
The king was happy that
(i) he had won a friend
(ii) he had killed his enemy
(iii) his action had pleased the hermit
(iv) his enemy was wounded
Answer.
(i) he had won a friend.

Question 12.
The answers of the hermit
(i) satisfied the king
(ii) failed to satisfy the king
(iii) confused the king
(iv) amused the king
Answer.
(i) satisfied the king.

We hope the NCERT Solutions for Class 7 English Honeycomb Chapter 1 Three Questions help you. If you have any query regarding NCERT Solutions for Class 7 English Honeycomb Chapter 1 Three Questions, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1.

Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 4
Chapter Name Lines and Angles
Exercise Ex 4.1
Number of Questions Solved 6
Category NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1

Question 1.
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 1
Solution:
Here, ∠AOC and ∠BOD are vertically opposite angles.
∴ ∠AOC = ∠BOD
⇒ ∠AOC = 40° [∵ ∠BOD = 40°(Given)] …(i)
We have, ∠AOC + ∠ BOE = 70° (Given)
40°+ ∠BOE = 70° [From Eq. (i)]
⇒ ∠BOE = 30°
Also, ∠AOC + ∠COE + ∠BOE = 180° (Linear pair axiom)
⇒ 40° + ∠COE + 30° = 180°
⇒ ∠COE = 110°
Now, ∠COE + reflex ∠COE = 360° (Angles at a point)
110°+reflex ∠COE = 360°
⇒ Reflex ∠COE = 250°

Question 2.
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 2
Solution:
We have, ∠POY = 90°
⇒ ∠POY + ∠POX = 180° (Linear pair axiom)
⇒ ∠POX = 90°
⇒ a+b = 90°
Also, a : b = 2 : 3 (Given)
⇒ Let a = 2k,b = 3k
Now, from Eq. (j), we get
2k + 3k = 90°
⇒ 5k = 90°
⇒ k = 18°
∴ a = 2 x 18°=36°
and b=3 x 18°=54°
Now, ∠MOX + ∠XON = 1800 (Linear pair axiom)
b+ c = 180°
⇒ 540 + c= 180°
⇒ c = 126°

Question 3.
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 3
Solution:
∵ ∠PQS+ ∠PQR = 180° (Linear pair axiom) ,..(i)
and ∠PRT + ∠PRQ = 180° (Linear pair axiom).. .(ii)
From Eqs. (i) and (ii), we get
∠PQS + ∠PQR =∠PRT + ∠PRQ
∠PQS + ∠PRQ =∠PRT + ∠PRQ
[Given, ∠PQR = ∠PRQ]
⇒ ∠PQS = ∠PRT

Question 4.
In figure, if x + y = w + z, then prove that AOB is a line.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 4
Solution:
∵ x+ y+w+ z = 360° (Angle at a point)
x + y = w + z (Given)…(i)
∴ x+ y+ x+ y = 360° [From Eq. (i)]
2(x + y) = 360°
⇒ x + y = 180° (Linear pair axiom)
Hence, AOB is a straight line.

Question 5.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 5
Solution:
We have,
∠POR = ∠ROQ = 90° (∵ Given that, OR is perpendicular to PQ)
∴ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS
On adding ∠ROS both sides, we get
2 ∠ROS = 90° – ∠POS + ∠ROS
⇒ 2 ∠ROS = (90° + ∠ROS) – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS (∵ ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS)
⇒ ∠ROS = \(\frac { 1 }{ 2 }\) (∠QOS – ∠POS)
Hence proved.

Question 6.
It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Solution:
Here, YQ bisects ∠ZYP.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 img 6
Hence, ∠ZYQ = ∠QYP = \(\frac { 1 }{ 2 }\) ∠ZYP ……..(i)
Given, ∠XYZ = 64° ….(ii)
∵ ∠XYZ + ∠ZYQ + ZQYP = 180° (Linear pair axiom)
⇒ 64° + ∠ZYQ + ∠ZYQ = 180° [From Eqs. (i) and (ii)]
⇒ 2 ∠ZYQ = 180° – 64°
⇒ ∠ZYQ = \(\frac { 1 }{ 2 }\) x 116°
⇒ ∠ZYQ = 58°
∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°
Now, ∠QYP + reflex ∠QYP = 360°
58° + reflex ∠QYP = 360°
⇒ reflex ∠ QYP = 302°

We hope the NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1

NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1.

Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 9
Chapter Name Quadrilaterals
Exercise Ex 9.1
Number of Questions Solved 12
Category NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1

Question 1.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution:
Given: the ratio of the angles of quadrilateral are 3 : 5 : 9 : 13.
Let the angles of the quadrilateral are 3x, 5x, 9x and 13x.
We know that, sum of angles of a quadrilateral = 360°
∴ 3x + 5x+ 9x+13x = 360°
⇒ 30x = 360° ⇒ x = \(\frac { { 360 }^{ \circ } }{ 30 }\) = 12°
∴ Angles of the quadrilateral are 3x = 3 x 12 = 36°
5x = 5 x 12 = 60°
9x= 9×12 = 108°
and 13x = 13×12 = 156°

Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Let given parallelogram is ABCD whose diagonals AC and BD are equal, i.e., AC = BD.
Now, we have to prove that ABCD is a rectangle.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 1
Proof: In ∆ ABC and ∆ DCB, we have
AB = CD (Opposite sides of parallelogram)
BC = CB (Common in both triangles)
and AC = BD (Given)
∴ ∆ABC ≅ ∆DCB (By SSS rule)
∴ ∆ABC = ∠DCB …(i)
(Corresponding Part of Congruent Triangle)
But DC || AB and transversal CB intersect them.
∴ ∠ABC+ ∠DCB = 180°
(∵ Both are interior angles on the same side of the transversal)
⇒ ∠ABC + ∠ABC = 180° [From Eq. (i)]
⇒ 2 ∠ ABC = 180°
⇒ ∠ABC = 90°= ∠DCB
Thus, ABCD is a parallelogram and one of angles is 90°.
Hence, ABCD is a rectangle.
Hence proved.

Question 3.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
Given: a quadrilateral ABCD whose diagonals AC and BD bisect each other at right angles.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 2
i.e, OA = OC and OB = OD
and ∠AOD = ∠AOB = ∠COD = ∠BOC = 90°
To prove: ABCD is a rhombus.
Proof: In ∆OAB and ∆ODC, we have
OA = OC and OB = OD (Given)
∠AOB = ∠COD (Vertically opposite angles)
∴ ∆OAB ≅ ∆OCD (By SAS rule)
∴ AB = CD ,..(i)
(Corresponding part of congruent triangles)
Again, in ∆OAD and ∆OBC, we have
OA = OC and OD = OB (Given)
and ∠AOD = ∠BOC (Vertically opposite angle)
∴ A OAD = A OCB (By SAS rule)
∴ AD = BC …(ii) (Corresponding part of congruent triangles)
Similarly, we can prove that
AB = AD
CD = BC …(iii)
Hence, from Eqs. (i), (ii) and (iii), we get
AB = BC = AD = CD
Hence, ABCD is a rhombus.
Hence proved.

Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Given: A square ABCD whose diagonals AC and BD intersect at O.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 3
i. e, . AC = BD, OD = OB, OA = OC and AC ⊥ BD
Proof: In A ABC and A BAD, we have
AB = BA (Common)
BC= AD (Sides of a square)
∠ABC = ∠BAD = 90°
∴ ∆ABC ≅ ∆BAD (By SAS rule)
AC = BD
(Corresponding Parts of Congruent Triangle)
∴ ∆OAB and ∆OCD
AB = DC (Sides of a square)
∠ OAB = ∠DCO
(∵ AB || CD and transversal AC intersect)
and ∠ OBA = ∠BDC
(∵AB|| CD and transversal BD intersect)
∴ ∆OAB ≅ ∆OCD
OA = OC and OB = OD
(Corresponding Parts of Congruent Triangle)
Now, in ∆AOB and ∆AOD, we have
OB = OD (Prove in above)
AB = AD (Sides of a square)
AO = OA (Common)
∴ ∆ AOB = ∆AOD (By SSS)
∴ ∠AOB = ∠AOD (By CPCT)
But ∠AOB + ∠AOD-180° (Linear pair)
∴ ∠AOB = ∠AOD = 90°
Thus, AO ⊥ BDi.e.,AC ⊥ BD.
Hence, AC = BD, OA = OC, OB = OD and AC ⊥ BD
Hence proved.

Question 5.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:
Given: A quadrilateral ABCD in which AC = BD and AC ⊥ BD such that OA = OC and OB = OD. So, ABCD is a parallelogram.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 4
To prove: ABCD is a square.
Proof: Let AC and BD intersect at a point O.
In ∆ABO and ∆ADO, we have
BO = OD (Given)
AO = OA (Common)
∠AOB = ∠AOD = 90° (Given)
∴ ∆ABO ≅ ∆ADO (By SAS)
∴ AB = AD (By CPCT)
and AD = BC (Opposite sides of parallelogram)
∴ AB = BC = DC = AD …(i)
Again, in ∆ABC and ∆BAD, we have
AB = BA (Common)
AC = BD (Given)
BC= AD [From Eq. (i)]
∴ ∆ABC = ∆BAD (By SSS)
∠ABC= ∠ BAD
∠ABC+ ∠ BAD =180° …(ii)
But
∠ABC = ∠ BAD = 180° (Sum of interior angles of a parallelogram)
∴ ∠ABC = ∠BAD = 90° [From Eq. (ii)]
Thus, AB = BC = CD = DA and ∠A = 90°
∴ ABCD is a square.
Hence proved.

Question 6.
Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 5
Solution:
Given: diagonal AC of a parallelogram ABCD bisects ∠A
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 6
i.e., ∠DCA = ∠ BAC = \(\frac { 1 }{ 2 }\) ∠BAD ….(i)
Here, AB || CD and AC is transversal.
∴ ∠DCA = ∠CAB (Pair of alternate angle)…(ii)
and ∠BCA = ∠ DAC (Pair of alternate angle).. .(iii)
From Eqs. (i), (ii) and (iii), we get
∠DAC = BCA = ∠ BAC = ∠DCA
Now, ∠BCD = ∠BCA + ∠DCA
= ∠DAC + ∠CAB
= ∠BAD
∴ Diagonal AC also bisects ∠C.
Again, in ∆OAD and ∆ OCD, we have
OA = OC (∵ Diagonals bisect each other)
OD = OD (Common)
∴ ∠AOD = ∠ COD = 90°
∴ ∆ OAD ≅ ∆ OCD (By SAS)
∴ AD = CD (By CPCT)
Now AB = CD and AD = BC (Opposite sides of parallelogram)
∴ AB = CD = AD = BC
Hence, ABCD is a rhombus.
Hence proved.

Question 7.
ABCD is a rhombus. Show that diagonal AC bisects ∠Aas well as ∠C and diagonal BD bisects ∠B as well AS ∠D.
Solution:
Given: ABCD is a rhombus
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 7
∴ AD = AB = BC = CD ….(i)
To prove:
(i) Diagonal AC bisect ∠A as well as ∠C.
(ii) Diagonal BD bisects ∠B as well as ∠D.
Proof:
(i) Let AC and BD are the diagonals of rhombus ABCD.
In ∆ABC and ∆ADC,
AD = AB
CD = BC [From Eq,(i)]
and AC = CA (Common)
∴ ∆ABC ≅ ∆ADC ( By SSS rule)
∴ ∠DAC = ∠BAC (By CPCT)
and ∠DCA = ∠ BCA
Also, ∠DAC = ∠DCA
and ∠BAC = ∠ BCA
This shows that AC bisect ∠ A as well as ∠C.
(ii) Again, in ∆BDCand ∆BDA,
AB = BC
AD = CD
BD = BD (Common)
∴ ∆ BDC ≅ ∆ BDA (SSS rule)
∴ ∠BDA = ∠BDC
and ∠DBA = ∠DBC
Also, ∠BDA = ∠DBA
and ∠BDC = ∠DBC
This shows that BD bisect ∠B as well as ∠D.

Question 8.
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that
(i) ABCD is a square
(ii) diagonal BD bisects ∠ B as well as ∠D.
Solution:
Given: ABCD is a rectangle.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 8
∴ AB = CD and BC= AD …(i)
To Prove (i) ABCD is a square.
i.e., AB = BC = CD = DA
(ii) Dioagonal BDbisects ∠B as well as ∠D.
Proof (i) In ∆ADC and ∆ABC, we have
Since, AB || DC and AC transversal intersect
∠DAC = ∠BAC
∠DCA = ∠BCA
and AC = CA (Common)
∴ ∆ADC ≅ ∆ ABC (By ASA rule)
AD = AB (By CPCT)
and CD = BC …(ii)
Hence, from Eqs. (i) and (ii), we get
AB = BC- AD = CD
∴ ABCD is a square.
(ii) In ∆AOB and ∆COB, we have
AB = BC (Side od square)
BO = OB (Common)
OA = OC
(∵ Diagonal of square bisect each other)
∴ ∆AOB ≅ ∆COB
∴ ∠OBA = ∠OBC
This shows that BO or BD bisect ∠B.
Similarly, in ∆ AOD and ∆ COD, we have
AD = CD (Side od square)
OD = DO (Common)
and OA = OC
(∵ Diagonal of square bisect each other)
∴ ∆ AOD ≅ ∆ COD (By ASA rule)
∴ ∠ADO = ∠CDO
This shows that DO or DB bisect ∠D.
Hence proved.

Question 9.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 9
Solution:
Given: ABCD is a parallelogram and P and Q are lie on BD such that
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 10
DP = PQ ….(i)
(i) We have to show,
∆APD ≅ ∆CQB
Now, in ∆APD and ∆CQB, we have
DP = BQ (Given)
AD = BC
(Opposite sides are equal in parallelogram)
∵ AD || BC and BD is a transversal.
∴ ∠ADP = ∠QBC (Alternate interior angle)
∴ ∆APD ≅ ∆CQB (By SAS)
(ii) Since, ∆APD ≅ ∆CQB
∴ AP = CQ
(iii) Here, we have to show, ∆AQB ≅ ∆CPD
Now, in ∆AQB and ∆CPD, we have
BQ = DP (Given)
AB = CD (Opposite sides of parallelogram)
∵ AB || CD and BD is a transversal.
∴ ∠ABQ = ∠CDP (Alternate interior angle)
∴ ∆AQB s ∆ CPD
(iv) Since, ∆AQB ≅ ∆CPD
∴ AQ = CP
(v) Now, in ∆APQ and ∆PCQ, we have
AQ = CP [From part (iv)]
AP = CQ [From part (ii)]
PQ = QP (Common)
∴ ∆APQ = ∆PCQ (By SSS)
∴ ∠APQ = ∠PQC
and ∠AQP = ∠CPQ (Vertically opposite)
Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.
∴ AP || CQ and AQ || CP
Now, both pairs of opposite sides of quadrilateral APCQ are parallel.
Hence, APCQ is a parallelogram.
Hence proved.

Question 10.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 11
Solution:
Given: ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD.
∵ AB || CD and BD is a transversal.
∴ ∠CDB = ∠DBA …(i)
(i) Now, in ∆ APB and ∆ CQD, we have
CD = AB (Sides of parallelogram)
∠CQD = ∠APB = 90° (Given)
∠CDQ = ∠ABP [From Eq. (i)]
∴ ∆APB ≅ ∆CQD (By ASA rule)
(ii) ∵ ∆APB ≅ ∆CQU (By CPCT)
∴ AP = CQ
Hence proved.

Question 11.
In ∆ABC and ∆DEF, AB = DE, AB || DE, BC – EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 12
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ∆ABC ≅ ∆DEF
Solution:
Given: in ∆ABC and ∆DEF,
AB = DE, AB || DE
and BC = EF,BC||EF

(i)
Now, in quadrilateral ABED,
AB = DE and AB || DE (Given)
⇒ ABED is a parallelogram.
(∵ A pair of opposite sides is equal and parallel)

(ii)
In quadrilateral BEFC,
BC = EF and BC || EF.
⇒ BEFC is a parallelogram.
(∵ A pair of opposite sides is equal and parallel)

(iii)
Since, ABED is a parallelogram.
∴ AD || BE and AD = BE …(i)
Also, BEFC in a parallelogram.
∴ CF || BE and CF = BE …(ii)
From Eqs. (i) and (ii), we get
AD || CF and AD =CF

(iv)
In quadrilateral ACFD, we have
AD || CF and AD = CF [From part (iii)]
⇒ ACFD is a parallelogram.

(v)
Since, ACFD is a parallelogram.
∴ AC = DF and AC || DF

(vi)
Now, in ∆ABC and ∆DEF,
AB = DE (Given)
BC = EE (Given)
and AC = DF [From part (v)]
∴ ∆ABC ≅ ∆DEF (By SSS rule)

Question 12.
ABCD is a trape∠ium in which AB || CD and AD = BC (see figure). Show that
(i )∠A=∠B
(ii )∠C=∠D
(iii) ∆ABC ≅ ∆BAD
(iv) diagonal AC = diagonal BD
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 13
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].
Solution:
Given: ABCD is a trapezium.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 img 14
AB || CD and AD = BC
Now, extend AB and draw a line through C parallel to DA intersecting AB produced at E.
Now, ADCE is a parallelogram.
∴ AD || CE and AD = CE
But AD = BC
∴ AD = BC = CE

(i)
We know that, ∠A + ∠E = 180°
(∵ Interior angles on the same side of the transversal AE)
⇒ ∠E = 180° – ∠A
Since, BC = EC
∴ ∠E = ∠CBE = 180° – ∠A
Also, ∠ ABC =180° – ∠CBE (∵ ABE is straight line)
= 180°- 180°+ ∠A
⇒ ∠B = ∠A …(i)

(ii)
Now ∠A+ ∠D = 180°
(∵ Interior angles on the same side of the transversal AD)
⇒ ∠D=180°-∠A
⇒ ∠D = 180° -∠B [From Eq. (i)]…(ii)
Also, ∠C+ ∠B = 180°
(∵ Interior angles on the same side of the transversal BC)
⇒ ∠C = 180°-∠B …..(iii)
From Eqs. (ii) and (iii), we get
∠C = ∠D

(iii)
Now, in ∆ABC and ∆ BAD, we have
AB = BA (Common)
AD = BC (Given)
∠A=∠B [From Eq. (i)]
∴ ∆ABC ≅ ∆ BAD (By SAS)

(iv)
Since, ∆ABC ≅ ∆BAD
∴ AC = BD
Hence proved.
We hope the NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1

NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1.

Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 12
Chapter Name Constructions
Exercise Ex 12.1
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:
Steps of construction
NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 1

  1. Taking O as centre and some radius, draw an arc of a circle which intersects OA say at a point B.
  2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
  3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
  4. Draw the ray OE passing through C.
    Then, ∠EOA = 60°
  5. Draw the ray of passing through D. Then, ∠FOE = 60°.
  6. Next, taking Cand Das centres and with the radius more than \(\frac { 1 }{ 2 }\) CD, draw arcs to intersect each other, say at G.
  7. Draw the ray OG. This ray OG is the bisector of the ∠FOE i.e.,
    ∠FOG = ∠EOG = \(\frac { 1 }{ 2 }\) ∠FOE = \(\frac { 1 }{ 2 }\) (60°) = 30°
    Thus, ∠GOA = ∠GOE + ∠EOA
    = 30° + 60° = 90°

Justification
(i) Join BC.
Then, OC=OB = BC (By construction)
∴ ∆COB is an equilateral triangle.
∴ ∠COB = 60°
∴ ∠EOA = 60°

(ii) Join CD.
Then, OD=OC=CD (By construction)
∴ ∆DOC is an equilateral triangle.
∴ ∠DOC = 60°
∴ ∠FOE = 60°

(iii) Join CG and DG.
In ∆ODG and ∆OCG,
OD = OC (Radii of the same arc)
DG = CG (Arcs of equal radii)
OG = OG (Common)
∴ ∆ ODG ≅ ∆OCG (SSS rule)
∴ ∠DOG = ∠COG (CPCT)
∴ ∠FOG = ∠EOG = \(\frac { 1 }{ 2 }\) ∠FOE
= \(\frac { 1 }{ 2 }\) (60°) = 30°
Thus, ∠GOA = ∠GOE + ∠EOA = 30° + 60° = 90°

Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Solution:
Steps of construction

  1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.
  2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
  3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
  4. Draw the ray OE passing through C. Then, ∠EOA = 60°.
  5. Draw the ray OF passing through D. Then, ∠FOE = 60°.
  6. Next, taking C and D as centres and with radius more than \(\frac { 1 }{ 2 }\) CD, draw arcs to intersect each other, say at G.
  7. Draw the ray OG. This ray OG is the bisector of the ∠FOE,
    i.e., ∠FOG = ∠EOG = \(\frac { 1 }{ 2 }\) ∠FOE = \(\frac { 1 }{ 2 }\) (60°) = 30°.
    thus , ∠GOA = ∠GOE + ∠EOA
    = 30° + 60° = 90°
  8. Now taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at Hand /, respectively.
  9. Next, taking H and las centres and with the radius more than \(\frac { 1 }{ 2 }\) Hl, draw
    arcs to intersect each other, say at J.
  10. Draw the ray OJ. This ray OJ is the required bisector of the ∠GOA.
    ∠GOJ = ∠AOJ = \(\frac { 1 }{ 2 }\) ∠GOA
    = \(\frac { 1 }{ 2 }\) (90°) = 45°

Justification
(i) Join BC. (By construction)
Then, OC = OB = BC
∴ ∆COB is an equilateral triangle.
∴ ∠COB = 60°
∴ ∠EOA = 60°

Question 3.
Construct the angles of the following measurements
(i) 30°
(ii) 22 \(\frac { 1 }{ 2 }\)
(iii) 15°
Solution:
(i) Steps of construction

  1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.
  2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
  3. Draw the ray OE passing through C. Then, ∠EOA = 60°.
  4. Taking B and C as centres and with the radius more than \(\frac { 1 }{ 2 }\) BC, draw arcs to intersect each other, say at D.
  5. Draw the ray OD, this ray OD is the bisector of the ∠EOA, i.e.,
    ∠EOD = ∠AOD = \(\frac { 1 }{ 2 }\) ∠EOA = \(\frac { 1 }{ 2 }\) (60°) = 30°

(ii) Steps of construction

  1. Taking O as centre and some radius, draw an arc of a circle which, intersects OA, say at a point B.
  2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
  3. Taking C as centre and with the same radius as before, drawn an arc intersecting the arc drawn in step 1, say at D.
  4. Draw the ray OE passing through C. Then, ∠EOA = 60°.
  5. Draw the ray OF passing through D. Then, ∠FOE = 60°.
  6. Next, taking C and D as centres and with radius more than \(\frac { 1 }{ 2 }\)CD, draw arcs to intersect each other, say at G.
  7. Draw the ray OG. This ray OG is the bisector of the ∠FOE,
    NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 7
  8. Now, taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and l, respectively.
  9. Next, taking H and l as centres and with the radius more than \(\frac { 1 }{ 2 }\)Hl, draw arcs to intersect each other, say at J.
  10. Draw the ray OJ. This ray OJ is the bisector of the ∠GOA
    i. e., ∠GOJ = ∠AOJ = \(\frac { 1 }{ 2 }\) ∠GOA
    = \(\frac { 1 }{ 2 }\) (90°) = 45 °
  11. Now, taking O as centre and any radius, drawn an arc to intersect the rays OA and OJ, say at K and L, respectively.
  12. Next, taking K and L as centres and with the radius more than \(\frac { 1 }{ 2 }\)KL, draw arcs to intersect each other, say at H.
  13. Draw the ray OM. This ray OM is the bisector of the ∠AOJ, i.e., ∠JOM = ∠AOM
    = \(\frac { 1 }{ 2 }\) ∠AOJ = \(\frac { 1 }{ 2 }\) (45°) = 22 \(\frac { 1 }{ 2 }\) °

(iii) Steps of construction

  1. Taking O as centre and some radius, draw an arc of a circle which intersects OA say at a point B.
  2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C..
  3. Draw the ray OE passing through C. Then, ∠EOA = 60°.
    NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 8
  4. Now, taking 6 and Cas centres and with the radius more than \(\frac { 1 }{ 2 }\) BC, draw arcs to intersect each other, say at D.
  5. Draw the ray OD intersecting the arc drawn in step 1 at F. This ray OD is the bisector of the ∠EOA, i.e.,
    ∠EOD = ∠AOD = \(\frac { 1 }{ 2 }\) ∠EOA = \(\frac { 1 }{ 2 }\) (60°) = 30°
  6. Now, taking B and F as centres and with the radius more than \(\frac { 1 }{ 2 }\) BF, draw arcs to intersect each other, say at G.
    1. Draw the ray OG. This ray OG is the bisector of the ∠AOD,
      i. e., ∠DOG = ∠AOG = \(\frac { 1 }{ 2 }\) ∠AOD = \(\frac { 1 }{ 2 }\) (30°) = 15°

Question 4.
Construct the following angles and verify by measuring them by a protractor
(i) 75°
(ii) 105°
(iii) 135°
Solution:
(i) Steps of construction
NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 9

  1. Taking O as centre and some radius, draw an arc of a circle which intersects OA say at a point B.
  2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
  3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
  4. Join the ray OE passing through C. Then, ∠EOA = 60°.
  5. Draw the ray of passing through D. Then, ∠FOE = 60°.
  6. Next, taking C and D as centres and with the radius more than \(\frac { 1 }{ 2 }\)CD, draw arcs to intersect each other, say at G.
  7. Draw the ray OG intersecting the arc of step 1 at H. This ray OG is the bisector of the ∠FOE, i.e., ∠FOG = ∠EOG
    = \(\frac { 1 }{ 2 }\) ∠FOE = \(\frac { 1 }{ 2 }\)(60°) = 30°
  8. Next, taking Cand H as centres and with the radius more than \(\frac { 1 }{ 2 }\)CH, draw
    arcs to intersect each other, say at l.
    Draw the ray OI. This ray OI is the bisector of the ∠GOE,
    i. e., ∠GOI = ∠EOI = \(\frac { 1 }{ 2 }\) ∠GOE = \(\frac { 1 }{ 2 }\) (30°) = 15°
    Thus, ∠IOA = ∠IOE + ∠EOA
    =15°+ 60° = 75°
    On measuring the ∠IOA by protractor, we find that ∠IOA = 15°
    Thus, the construction is verified.

(ii) Steps of construction
NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 10

  1. Taking O as centre and some radius, draw an arc of a circle which intersects OA say at a point B.
  2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
  3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at a point D.
  4. Draw the ray Of passing through C. Then, ∠EOA = 60°.
  5. Draw the ray OF passing through D. Then, ∠FOE = 60°.
  6. Next, taking Cand Das centres and with the radius more than \(\frac { 1 }{ 2 }\) CD, draw arcs to intersect each other, say at G.
  7. Draw the ray OG intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the ∠FOE, i.e.,
    NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 11
  8. Next, taking H and D as centres and with the radius more than \(\frac { 1 }{ 2 }\) HD, draw arcs to intersect each other, say at l.
  9. Draw the ray Ol. This ray Ol is the bisector of the ∠FOG, i.e.,
    NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 12
    Thus, ∠lOA = ∠IOG + ∠GOA = 15° + 90° = 105°. On measuring the ∠lOA by protractor, we find that ∠FOA = 105°.
    Thus, the construction is verified.

(iii) Steps of construction
NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 13

  1. Produce AO to A’ to form ray OA’.
  2. Taking O as centre and some radius, draw an arc of a circle which intersects OA at a point B and OA’ at a point B’.
  3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.
  4. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
  5. Draw the ray OE passing through C, then ∠EOA = 60°.
  6. Draw the ray OF passing through D, then ∠FOE = 60°.
  7. Next, taking C and D as centres and with the radius more than \(\frac { 1 }{ 2 }\) CD, draw arcs to intersect each other, say at G.
  8. Draw the ray OGintersecting the arc drawn in step 1 at H. This ray OG is the bisector of the ∠FOE i,e.,
    NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 14
  9. Next, taking B’ and H as centres and with the radius more than \(\frac { 1 }{ 2 }\) B’H, drawn arcs to intersect each other, say at l.
  10. Draw the ray Ol. This ray Ol is the bisector of the ∠B’OG i.e.,
    NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 15
    On measuring the ∠IOA by protractor, we find that ∠lOA = 135°.
    Thus, the construction is verified.

Question 5.
Construct an equilateral triangle, given its side and justify the construction.
Solution:
Steps of construction
NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 img 16

  1. Take a ray AX with initial point A From AX, cut off AB = 4 cm.
  2. Taking A as centre and radius (= 4 cm), draw an arc of a circle, which intersects AX, say at a point B.
  3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
  4. Draw the ray AE passing through C.
  5. Next, taking B as centre and radius (= 4 cm), draw an arc of a circle, which intersects AX, say at a point A
  6. Taking A as centre and with the same radius as in step 5, draw an arc intersecting the previously drawn arc, say at a point C.
    Draw the ray BF passing through C.
    Then, ∆ ABC is the required triangle with gives side 4 cm.

Justification
AB = BC (By construction)
AB = AC (By construction)
∴ AB = BC = CA
∴ ∆ ABC is an equilateral triangle.
∴ The construction is justified.

We hope the NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.1, drop a comment below and we will get back to you at the earliest.