Maharashtra Board Class 7 Chapter 1 Set 6 Geometrical Constructions Solutions

Question 1. Write the names of pairs of congruent line segments. (Use a divider to find them.)

ℹ️ चित्र व्याख्या (Diagram Explanation): एक केंद्रीय बिंदु 'G' से विभिन्न दिशाओं में फैली हुई रेखाएँ 'N', 'M', 'D', 'E', 'C', 'R', 'B' बिंदुओं को जोड़ती हैं, जिससे कई रेखाखंड (जैसे GN, GM, GD, GE, GC, GR, GB) बनते हैं। यह आकृति रेखाखंडों की सर्वांगसमता की पहचान करने में मदद करती है।
(i) ___
(ii) ___
(iii) ___
(iv) ___
Answer:
(i) seg BG ≅ seg CG
(ii) seg NG ≅ seg MG ≅ seg EG ≅ seg RG
In simple words: This diagram shows line segments originating from a central point G. Congruent segments are those that have equal lengths, which can be verified using a divider. In this case, BG and CG are of equal length, and NG, MG, EG, RG are also of equal length.

🎯 Exam Tip: When identifying congruent line segments, pay close attention to the visual symmetry and use a divider or ruler for precise measurement if allowed.

Question 2. On the line below, the distance between any two adjoining points shown on it is equal. Hence, fill in the blanks.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सीधी रेखा पर 'Q', 'P', 'Z', 'Y', 'X', 'W', 'A', 'B', 'C' बिंदु समान दूरी पर क्रमिक रूप से अंकित हैं। यह रेखाखंडों की लंबाई और उनकी सर्वांगसमता का विश्लेषण करने के लिए उपयोग की जाती है।
(i) seg AB ≅ seg ___
(ii) seg AP ≅ seg ___
(iii) seg AC ≅ seg ___
(iv) seg ___ ≅ seg BY
(v) seg ___ ≅ seg YQ
(vi) seg BW ≅ seg ___
Answer:
(i) BC
(ii) QW
(iii) QZ
(iv) AZ
(v) AY
(vi) AC
In simple words: Since the distance between any two adjacent points on the line is equal, we can find congruent segments by counting the number of unit distances between their endpoints. For example, AB is 1 unit, so BC (1 unit) is congruent to it. AP is 3 units, and QW is also 3 units.

🎯 Exam Tip: When points are equally spaced on a line, congruent segments can be identified by comparing the number of intervals between their endpoints. Accuracy in counting is key.

Note: The above problem has many solutions. Students may write solutions other than the ones given.

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 6 Intext Questions And Activities

Question 1. Try to draw triangles with the following data. Can you draw these triangles. If not, look for the reason why you could not draw so. (Textbook pg. no. 7)
(i) ΔΑΒC in which m∠A = 85°, m∠B = 115°, l(AB) = 5cm.
Solution: m∠A + m∠B = 85° + 115° = 200° > 180° But the sum of the measures of the angles of a triangle is 180° Hence, ΔABC cannot be drawn.
(ii) ΔPQR in which l(QR) = 2cm, l(PQ) = 4cm, l(PR) = 2cm.
Solution: l(QR) + l(PR) = 2 cm + 2cm = 4 cm = l(PQ) But in a triangle, the sum of the length of any two sides of a triangle is always greater than the length of the third side. Hence, ΔPQR cannot be drawn.
In simple words: A triangle cannot be formed if the sum of two angles is greater than 180 degrees (as in part i) or if the sum of the lengths of two sides is not greater than the length of the third side (as in part ii). These are fundamental triangle properties.

🎯 Exam Tip: Remember the two key conditions for triangle formation: (1) The sum of all angles must be exactly 180°, and (2) The sum of the lengths of any two sides must be greater than the length of the third side.

Question 2. Draw ΔΑΒC such that l(BC) = 8 cm, l(CA) = 6 cm, m∠ABC = 40°. Draw a ray to make an angle of 40° with the base BC, l(BC) = 8 cm. We have to obtain point 'A' on the ray. With 'C' as the centre, draw an arc of radius 6 cm to do so. What do we observe? The arc intersects the ray in two different points. Thus, we get two triangles of two different shapes having the given measures. (Textbook pg. no. 7)
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है जहाँ आधार BC 8 सेमी है और कोण B 40° है। बिंदु C से 6 सेमी त्रिज्या का एक चाप खींचा गया है, जो 40° के कोण वाली किरण को दो बिंदुओं पर काटता है, जिससे दो संभावित त्रिभुज बनते हैं।

Here ∠B is an acute angle. ∠C can be an acute angle or an obtuse angle. Hence we get two triangles of two different shapes.
In simple words: When given two sides and a non-included angle (SSA condition), if the given angle is acute and the side opposite to it is shorter than the adjacent side but longer than the altitude, it's possible to construct two different triangles. This is known as the ambiguous case of SSA.

🎯 Exam Tip: Be aware of the ambiguous case in triangle construction (SSA). If the given angle is acute and the side opposite is shorter than the adjacent side (but not too short), two distinct triangles can often be formed.

Question 3. Can a triangle be drawn if the three angles are given, but not any side? How many such triangles can be drawn? (Textbook pg. no. 7)
Solution: Yes a triangle can be drawn. Since the length of side is not given, any length of side can be selected and then triangle can be constructed. We will get different triangles for different length of sides.
In simple words: Yes, infinitely many triangles can be drawn if only the three angles are given. All these triangles will be similar to each other, meaning they have the same shape but can have different sizes.

🎯 Exam Tip: Understanding similarity is crucial here. If only angles are given, the shape is determined, but the size is not, leading to an infinite family of similar triangles.

Question 4. Using the ruler, measure the lengths of seg AB and seg PQ. Are they of same length? Trace the seg AB on a sheet of transparent paper. Now place this new segment on PQ verify that if point A is placed on point P, then B falls on Q. (Textbook pg. no. 7)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो अलग-अलग रेखाखंडों, 'AB' और 'PQ' को दर्शाता है। छात्रों को इन रेखाखंडों की लंबाई को मापने और उनकी सर्वांगसमता की जांच करने के लिए कहा गया है।
l(AB) = ___
l(PQ) = ___
Solution: l(AB) = 4 cm l(PQ) = 4 cm Since the length of two segments is the same, if placed on one another, they will coincide.
In simple words: By measuring with a ruler, both line segments AB and PQ are found to be 4 cm long. Because they have the same length, they are congruent and will perfectly overlap if one is placed on top of the other.

🎯 Exam Tip: Congruent line segments have equal lengths. Practical measurement using a ruler or tracing paper is a good way to verify congruence.

Question 5. From the shape shown below, write the names of the pairs of congruent line segments. (Textbook pg. no. 8)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक आयताकार प्रिज्म (घनाभ) को दर्शाता है जिसके शीर्ष A, B, C, D, E, F, G, H से चिह्नित हैं। यह विभिन्न किनारों (रेखाखंडों) के बीच सर्वांगसमता की पहचान करने में मदद करता है।
(i) seg AB ≅ seg DC
(ii) seg AE ≅ seg BH
(iii) seg EF ≅ seg ___
(iv) seg DF ≅ seg ___
Solution:
(iii) seg EF ≅ seg AD ≅ seg BC ≅ seg HG
(iv) seg DF ≅ seg CG ≅ seg AE ≅ seg BH
In simple words: In a rectangular prism, opposite edges are congruent (equal in length). For example, EF is congruent to AD, BC, and HG because they are parallel and form the edges of a rectangle. Similarly, diagonals of congruent faces are also congruent.

🎯 Exam Tip: For 3D shapes like cuboids, remember that all corresponding edges and diagonals of congruent faces will also be congruent. Visualizing the parallel and equal length segments helps.

Question 6. Take a rectangular paper. Place two opposite sides upon each Other. What do you observe? (Textbook pg. no. 7)
Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक आयताकार कागज को दर्शाता है जिसे उसकी लंबाई के अनुदिश बीच से मोड़ा गया है। यह क्रिया कागज के विपरीत भुजाओं के संरेखण और सर्वांगसमता को प्रदर्शित करती है। Opposite sides of the rectangular paper coincide and hence are congruent.
In simple words: When you fold a rectangular paper so that its opposite sides meet, they perfectly overlap. This observation demonstrates that the opposite sides of a rectangle are equal in length, or congruent.

🎯 Exam Tip: This simple activity is a practical way to understand congruence in everyday objects. Congruent objects or segments perfectly match when superimposed.