The RD Sharma Solutions for Class 7 Chapter 16 are designed to help students clearly understand the concepts of congruence in geometry. This chapter explains how two shapes or figures can be exactly the same in size and shape, and how we can prove their congruence using different rules. By using RD Sharma Solutions for Class 7 Chapter 16 Congruence, students can practise step-by-step methods for solving problems and gain confidence in applying these concepts to various questions in their exams.
In Class 7 Chapter 16 RD Sharma Solutions, each question is solved in a simple and easy-to-understand way. The solutions follow the same logical steps taught in the textbook, so students can easily relate to what they learn in class. Whether it is understanding the basic meaning of congruence, learning about corresponding sides and angles, or applying rules like SSS, SAS, ASA, and RHS, the Class 7 Chapter 16 Congruence RD Sharma Solutions give a complete learning experience. One of the main advantages of studying from RD Sharma Solutions for Class 7 Chapter 16 is that they not only provide answers but also explain why a particular step is taken. This helps students develop problem-solving skills and makes them ready for competitive exams in the future. The RD Sharma Solutions for Class 7 Chapter 16 Congruence are also useful for revision before tests, as they cover all important questions and examples from the chapter.
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By regularly practising with Class 7 Chapter 16 RD Sharma Solutions, students can improve their accuracy, speed, and understanding of the topic. These solutions act as a perfect guide for mastering the concept of congruence and scoring better in mathematics exams.
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You can easily download RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence PDF from here to study anytime and anywhere. This PDF contains all the solutions from the chapter in a clear and simple format, so you can revise quickly without confusion. The RD Sharma Solutions for Class 7 Chapter 16 Congruence in PDF form are well-organized and follow the textbook pattern, making it easy to find and practise any question. Whether you want to prepare for your exams, complete homework, or revise before a test, the Class 7 Chapter 16 Congruence RD Sharma Solutions PDF is a handy resource you can keep on your phone, tablet, or computer for instant access.
Access Answers to RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence
Q1. Which of the following statements are true or false?
(i) All squares are congruent.
Answer: False
Detailed Explanation: A square must have all sides equal and all angles 90 degrees, but different squares can have different side lengths. If side lengths are different, their sizes are different, so the squares are not congruent.
(ii) If two squares have equal areas, they are congruent.
Answer: True
Detailed Explanation: For a square, area depends only on the side length. If two squares have the same area, their sides are equal. Equal sides make the squares equal in size and shape, so they are congruent.
(iii) If two rectangles have equal areas, they are congruent.
Answer: False
Detailed Explanation: Two rectangles can have the same area but different lengths and widths. For example, 2 × 12 and 3 × 8 both have area 24, but their side lengths are not the same, so the rectangles are not congruent.
(iv) If two triangles have equal areas, they are congruent.
Answer: False
Detailed Explanation: Triangles can share the same area yet have different side lengths and angles. Since size and shape can differ, equal area alone does not guarantee congruence.
Q2. Triangles ABC and PQR are isosceles with AB = AC and PQ = PR. It is also given that AB = PQ and BC = QR. Are the triangles congruent? If ∠B = 50°, what is the value of the corresponding angle in ΔPQR?
Answer: Yes, the triangles are congruent. The corresponding angle is 50°.
Detailed Explanation: From AB = AC and PQ = PR, we get AC = PR because AB matches PQ. Also, BC = QR is given. Thus, the three pairs of sides match: AB = PQ, AC = PR, and BC = QR. So ΔABC ≅ ΔPQR by SSS. In this matching, vertex B corresponds to vertex Q. Therefore, if ∠B = 50°, then ∠Q = 50°.
Q3. Triangles ABC and DBC are isosceles on the same base BC. Points A and D lie on the same side of BC. Are triangles ADB and ADC congruent? If ∠BAC = 40° and ∠BDC = 100°, find ∠ADB.
Answer: Yes, ΔADB and ΔADC are congruent. The value of ∠ADB is 50°.
Detailed Explanation: Since ΔABC is isosceles with base BC, AB = AC. Since ΔDBC is isosceles with base BC, DB = DC. Segment AD is common to both triangles. So in ΔADB and ΔADC, we have AB = AC, DB = DC, and AD = AD. Hence ΔADB ≅ ΔADC by SSS. At point D, angle ∠BDC is 100°. Line AD lies inside angle ∠BDC, so ∠BDA and ∠ADC together make 100°. Because the two triangles are congruent, ∠BDA = ∠ADC. Therefore, each is 50°, and ∠ADB = 50°.
Q4. Lines AC and BD intersect at O and bisect each other. Triangles AOC and BOD are formed by joining AC and BD. State three matching parts. Are the triangles congruent? Which rule is used?
Answer: Yes, the triangles are congruent by SAS.
Detailed Explanation: Since AC and BD bisect each other at O, we have AO = OC and BO = OD. Also, ∠AOC and ∠BOD are vertically opposite angles, so they are equal. Thus, two sides and the included angle match: AO = OC, BO = OD, and ∠AOC = ∠BOD. Therefore, ΔAOC ≅ ΔBOD by SAS.
Q5. ΔABC is isosceles with AB = AC. Line AD bisects ∠A and meets BC at D.
(i) Are triangles ADB and ADC congruent?
Answer: Yes, by SAS.
(ii) State the matching parts used.
Answer: AB = AC, ∠BAD = ∠CAD because AD is the angle bisector, and AD is common.
(iii) Is BD = DC?
Answer: Yes. They are corresponding parts of congruent triangles, so BD = DC.
Detailed Explanation: With AB = AC, AD common, and ∠BAD = ∠CAD, the two triangles ΔADB and ΔADC satisfy SAS. Hence they are congruent and the matching parts such as BD and DC are equal.
Q6. ΔABC is isosceles with AB = AC. AD is perpendicular to BC.
(i) Are triangles ABD and ACD congruent?
Answer: Yes, by RHS.
(ii) State the matching parts used.
Answer: AB = AC as the hypotenuses, ∠ADB = ∠ADC = 90°, and AD is common.
(iii) Is BD = DC?
Answer: Yes. By corresponding parts of congruent triangles, BD = DC.
Detailed Explanation: Right angles at D and equal hypotenuses with a common side AD satisfy the RHS condition. So ΔABD ≅ ΔACD, which gives BD = DC and other equal parts.
Q7. ΔABC is isosceles with AB = AC and AD ⟂ BC. Are triangles ABD and ACD congruent? Which side and angle are equal in both?
Answer: Yes, ΔABD and ΔACD are congruent by RHS.
Detailed Explanation: AB = AC are the hypotenuses, AD is common, and ∠ADB = ∠ADC = 90°. Therefore, ΔABD ≅ ΔACD by RHS. From this, BD = DC and the angles at B and C near D are equal as corresponding parts of congruent triangles.