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| BSEB Class 7 Maths Chapter 7 सर्वांगसमता Textbook Solutions PDF: Download Bihar Board STD 7th Maths Chapter 7 सर्वांगसमता Book Answers |
Bihar Board Class 7th Maths Chapter 7 सर्वांगसमता Textbooks Solutions PDF
Bihar Board STD 7th Maths Chapter 7 सर्वांगसमता Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 7th Maths Chapter 7 सर्वांगसमता Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 7th Maths Chapter 7 सर्वांगसमता solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 7th Maths Chapter 7 सर्वांगसमता Textbooks. These Bihar Board Class 7th Maths Chapter 7 सर्वांगसमता Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.Bihar Board Class 7th Maths Chapter 7 सर्वांगसमता Books Solutions
| Board | BSEB |
| Materials | Textbook Solutions/Guide |
| Format | DOC/PDF |
| Class | 7th |
| Subject | Maths Chapter 7 सर्वांगसमता |
| Chapters | All |
| Provider | Hsslive |
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BSEB Class 7th Maths Chapter 7 सर्वांगसमता Textbooks Solutions with Answer PDF Download
Find below the list of all BSEB Class 7th Maths Chapter 7 सर्वांगसमता Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:Bihar Board Class 7 Maths सर्वांगसमता Ex 7.1
प्रश्न 1.
(i) AB = KL
CD = IJ
EF = GH
(ii) AB और KL रेखाखण्डों की माप = 3 cm.
CD और IJ रेखाखण्डों की माप = 2.4 cm.
EF और GH रेखाखण्डों की माप = 4 cm.
इनके माप बराबर हैं, इसलिए ये रेखाखण्ड सर्वांगसम हैं।
प्रश्न 2.
(i) ∠ABC = ∠JKL
∠DEF = ∠GHI
∠MNO = ∠PQR
(ii) ∠ABC = ∠JKL
∠DEF = ∠GHI
∠MNO = ∠PQR
प्रश्न 3.
दो कोण सर्वांगसम होते हैं, जब उनके माप बराबर हैं।
∠ABC = ∠DEF
∠ABC = 70°
∠DEF = 70°
प्रश्न 4.
(i) ∆XYZ = ∆ABC
भुजा – कोण
XY ↔ AC – ∠X ↔ ∠A
YZ ↔ BC – ∠Y ↔ ∠B
XZ ↔ AB – ∠Z ↔ ∠C
(ii) ∆ABC = ∆ABD
भुजा – कोण
AB ↔ AB – ∠ABC ↔ ∠A
BC ↔ AD – ∠BCA ↔ ∠D
AC ↔ DB – ∠BAC ↔ ∠B
(iii) ∆ABD = ∆CDB
भुजा – कोण
AB ↔ DC – ∠ABD ↔ ∠D
BD ↔ BD – ∠BCA ↔ ∠B
AD ↔ BC – ∠DAB ↔ ∠C
प्रश्न 5.
हाँ
प्रश्न 6.
प्रश्नानुसार,
पहले आयत की लम्बाई = 10 cm.
पहले आयत की चौड़ाई = 8 cm.
दूसरे आयत की लम्बाई = 12 cm.
दूसरे आयत की चौड़ाई = 8 cm.
दोनों आयत को सर्वांगसम करने के लिए-
पहले आयत की लम्बाई = 10 + 2 = 12
पहले आयत की लम्बाई में 2 cm. बढ़ाना होगा।
Bihar Board Class 7 Maths सर्वांगसमता Ex 7.2
प्रश्न 1.
सर्वांगसम प्रतिबंधों का प्रयोग करें-
(i) SSS – (भुजा भुजा-भुजा)
(ii) SAS – (भुजा-कोण-भुजा)
(iii) ASA – (कोण-भुजा-कोण)
(iv) RHS – (समकोण -कर्ण-भुजा)
प्रश्न 2.
कारण
(i) AC = FD क्योंकि ΔABC = ΔDEF
(ii) ∠BAC = ∠FDE क्योंकि ΔABC = ΔDEF
(iii) ∠ACB = ∠EFD क्योंकि ΔABC = ΔDEF
प्रश्न 3.
(iii) ΔABC = ΔBAD
प्रश्न 4.
ΔABC में
∠A = 40°
∠C = 35°
∠B = 180° – (35° + 40°) = 105°
AB = 2.5
ΔDEF में
∠E = 105°
∠F = 35°
∠D = 40°
DE = 2.5 cm
कोण-भुजा-कोण से
ΔABC = ΔDEF
प्रश्न 5.
ΔABD = ΔBDC
यहाँ AB = DC
AD = BC
BD = BD
भु. = भु० भु० से
ΔABC = ΔBDC
प्रश्न 6.
(i) AB = PQ
AC = PR
BC = QR
प्रश्न 7.
ΔABD और ΔACD
AB = AC
BD = CD
AD = AD
ΔPQR और ΔSOQ
OP = OS
OR = OQ
∠POR = ∠SOQ
प्रश्न 8.
(i) AC = PR.
BC = QR
∠R = ∠C
भुजा-कोण-भुजा (SAS)
(ii) 4 cm = 4 cm
3 cm = 3 cm
2 cm = 2 cm
भुजा-भुजा-भुजा से सर्वांगसम है।
(iii) ∠90° = ∠90° (समकोण Δ)
∠30° = ∠30°
5 cm = 5 cm
कोण-भुजा-कोण से सर्वांगसम है।
प्रश्न 9.
AB = PR
BC = PQ
∠B = ∠P = 90°
भुजा-कोण-भुजा से
ΔABC = ΔPQR सर्वांगसम है।
प्रश्न 10.
ΔBAC तथा ΔCDB में
AC = BD
∠ABC = ∠BCD
∠ACB = ∠DBC
कोण-भुजा-कोण से
ΔBAC = ΔCOB
(ii) ΔRPQ तथा ΔRST में
PQ = ST
QR = RT
∠PQR = ∠STR
भुजा-कोण-भुजा से
ΔRPQ = ΔRST
BSEB Textbook Solutions PDF for Class 7th
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