Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 5 Set 5 Quadrilaterals Solutions

Question 1. Choose the correct alternative answer and fill in the blanks.
(i) If all pairs of adjacent sides of a quadrilateral are congruent, then it is called —-
(A) rectangle
(B) parallelogram
(C) trapezium
(D) rhombus
Answer: (D) rhombus
In simple words: A rhombus is defined by all its adjacent sides being congruent, which in turn means all four sides are congruent. Rectangles, parallelograms, and trapeziums do not necessarily have this property.

🎯 Exam Tip: Understanding the basic definitions and properties of different quadrilaterals is crucial for accurately identifying them in questions.

Question 1.
(ii) If the diagonal of a square is \(12\sqrt{2}\) cm, then the perimeter of square is —-
(A) 24 cm
(B) \(24\sqrt{2}\) cm
(C) 48 cm
(D) \(48\sqrt{2}\) cm
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वर्ग ABCD दिखाया गया है। विकर्ण AC की लंबाई \(12\sqrt{2}\) सेमी है। बिंदु B पर एक समकोण दर्शाया गया है, जिससे यह त्रिभुज ABC को समकोण त्रिभुज बनाता है।
Answer: (C) 48 cm In ΔABC,
AC\(^2\) = AB\(^2\) + BC\(^2\)
\(\therefore\) \((12\sqrt{2})^2\) = AB\(^2\) + AB\(^2\)
\(\therefore\) AB\(^2\) = \(\frac{12^2 \times 2}{2}\) = 12\(^2\)
\(\therefore\) AB = 12 cm
\(\therefore\) Perimeter of \(\Box\)ABCD = 4 x 12 = 48 cm
In simple words: For a square, the diagonal length is side times \(\sqrt{2}\). Given the diagonal is \(12\sqrt{2}\) cm, the side length is 12 cm. The perimeter of a square is 4 times its side, so 4 * 12 cm = 48 cm.

🎯 Exam Tip: Remember the relationship between the side and diagonal of a square (diagonal = side \(\times \sqrt{2}\)) as it's a common shortcut in geometry problems. Pythagoras theorem is key here.

Question 1.
(iii) If opposite angles of a rhombus are \((2x)^\circ\) and \((3x – 40)^\circ\), then the value of x is —-
(A) 100°
(B) 80°
(C) 160°
(D) 40°
Answer: (D) 40° \(2x = 3x – 40\) ... [Opposite angles of a rhombus are equal]
\(x = 40^\circ\)
In simple words: In a rhombus, opposite angles are equal. So, set the two given angle expressions equal to each other and solve for x.

🎯 Exam Tip: Knowing the properties of a rhombus, specifically that opposite angles are congruent, is essential for solving this type of problem quickly.

Question 2. Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक आयत ABCD दर्शाया गया है। भुजा AB 7 सेमी है और भुजा BC 24 सेमी है। बिंदु B पर एक समकोण दर्शाया गया है, जिससे त्रिभुज ABC एक समकोण त्रिभुज बनाता है, और AC विकर्ण है।
Solution: Let ABCD be the rectangle.
AB = 7 cm, BC = 24 cm
In ΔABC, \(\angle\)B = 90° [Angle of a rectangle]
AC\(^2\) = AB\(^2\) + BC\(^2\) [Pythagoras theorem]
= \(7^2 + 24^2\)
= 49 + 576
= 625
AC = \(\sqrt{625}\) [Taking square root of both sides]
= 25 cm
\(\therefore\) The length of the diagonal of the rectangle is 25 cm.
In simple words: The diagonal of a rectangle forms a right-angled triangle with its two adjacent sides. Using the Pythagorean theorem, the square of the diagonal is the sum of the squares of the adjacent sides. Calculating \(7^2 + 24^2\) gives 625, and its square root is 25 cm.

🎯 Exam Tip: Recognize that the diagonal of a rectangle acts as the hypotenuse of a right-angled triangle formed by its sides, allowing for direct application of the Pythagorean theorem.

Question 3. If diagonal of a square is 13 cm, then find its side.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वर्ग PQRS दर्शाया गया है। विकर्ण PR की लंबाई 13 सेमी है। बिंदु Q पर एक समकोण दर्शाया गया है, जिससे त्रिभुज PQR एक समकोण त्रिभुज बनाता है। भुजा PQ और QR को 'x' के रूप में लेबल किया गया है।
Solution: Let PQRS be the square of side x cm.
\(\therefore\) PQ = QR = x cm .....(i) [Sides of a square]
\(\therefore\) In ΔPQR, \(\angle\)Q = 90° [Angle of a square]
\(\therefore\) PR\(^2\) = PQ\(^2\) + QR\(^2\) [Pythagoras theorem]
\(\therefore\) \(13^2\) = \(x^2 + x^2\) [From (i)]
\(\therefore\) 169 = \(2x^2\)
\(\therefore\) \(x^2 = \frac{169}{2}\)
\(\therefore\) \(x = \sqrt{\frac{169}{2}}\) [Taking square root of both sides]
\(\therefore\) \(x = \frac{13}{\sqrt{2}}\)
\(= \frac{13}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\) [Multiplying the numerator and denominator by \(\sqrt{2}\)]
\(= \frac{13\sqrt{2}}{2} = 6.5\sqrt{2}\) cm
The length of the side of the square is \(6.5\sqrt{2}\) cm.
In simple words: In a square, the diagonal is \(\sqrt{2}\) times the side length. So, if the diagonal is 13 cm, the side is \(13/\sqrt{2}\) cm. Rationalizing the denominator gives \((13\sqrt{2})/2\), which is \(6.5\sqrt{2}\) cm.

🎯 Exam Tip: Remember the formula that relates the diagonal (d) and side (a) of a square: \(d = a\sqrt{2}\). This directly helps in solving such problems. Rationalizing denominators is also an important skill.

Question 4. Ratio of two adjacent sides of a parallelogram is 3: 4, and its perimeter is 112 cm. Find the length of its each side.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समांतर चतुर्भुज STUV दर्शाया गया है। भुजा ST और UV को '3x' के रूप में लेबल किया गया है, जबकि भुजा TU और SV को '4x' के रूप में लेबल किया गया है। यह समांतर चतुर्भुज की विपरीत भुजाओं की समान लंबाई को दर्शाता है।
Solution: Let STUV be the parallelogram.
Ratio of two adjacent sides of a parallelogram is 3 : 4.
Let the common multiple be x.
ST = 3x cm and TU = 4x cm
\(\therefore\) ST = UV = 3x cm
TU = SV = 4x cm .....(i) [Opposite sides of a parallelogram]
Perimeter of \(\Box\)STUV = 112 [Given]
\(\therefore\) ST + TU + UV + SV = 112
\(\therefore\) \(3x + 4x + 3x + 4x = 112\) [From (i)]
\(\therefore\) \(14x = 112\)
\(\therefore\) \(x = \frac{112}{14}\)
\(\therefore\) \(x = 8\)
\(\therefore\) ST = UV = \(3x = 3 \times 8 = 24\) cm
\(\therefore\) TU = SV = \(4x = 4 \times 8 = 32\) cm [From (i)]
\(\therefore\) The lengths of the sides of the parallelogram are 24 cm, 32 cm, 24 cm and 32 cm.
In simple words: Represent the adjacent sides of the parallelogram as 3x and 4x based on their ratio. Since opposite sides are equal, the perimeter is \(2 \times (3x + 4x)\). Set this equal to 112 cm, solve for x, and then find the actual lengths of the sides.

🎯 Exam Tip: When given ratios of sides, always introduce a common multiple (like 'x') to represent the actual side lengths. Remember that the perimeter of a parallelogram is \(2 \times (\text{sum of adjacent sides})\).

Question 5. Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समचतुर्भुज PQRS दर्शाया गया है। विकर्ण PR और QS बिंदु T पर प्रतिच्छेद करते हैं। विकर्णों की लंबाई PR = 20 सेमी और QS = 48 सेमी है। बिंदु T पर विकर्णों के बीच एक समकोण दर्शाया गया है, जिससे त्रिभुज PQT एक समकोण त्रिभुज बनाता है।
Solution: \(\Box\)PQRS is a rhombus. [Given]
PR = 20 cm and QS = 48 cm [Given]
\(\therefore\) PT = \(\frac{1}{2}\) PR [Diagonals of a rhombus bisect each other]
\(= \frac{1}{2} \times 20 = 10\) cm
Also, QT = \(\frac{1}{2}\) QS [Diagonals of a rhombus bisect each other]
\(= \frac{1}{2} \times 48 = 24\) cm
In ΔPQT, \(\angle\)PTQ = 90° [Diagonals of a rhombus are perpendicular to each other]
\(\therefore\) PQ\(^2\) = PT\(^2\) + QT\(^2\) [Pythagoras- theorem]
= \(10^2 + 24^2\)
= 100 + 576
\(\therefore\) PQ\(^2\) = 676
\(\therefore\) PQ = \(\sqrt{676}\) [Taking square root of both sides]
= 26 cm
\(\therefore\) The length of side PQ is 26 cm.
In simple words: The diagonals of a rhombus bisect each other at right angles. So, half of each diagonal forms the legs of a right-angled triangle, and the side of the rhombus is the hypotenuse. Calculate half the diagonals (10 cm and 24 cm) and use Pythagoras theorem to find the side length (26 cm).

🎯 Exam Tip: Key properties of a rhombus – diagonals bisect each other at right angles – are crucial for solving problems involving side lengths and diagonals. Recognize the right-angled triangle formed at the intersection of diagonals.

Question 6. Diagonals of a rectangle PQRS are intersecting in point M. If \(\angle\)QMR = 50°, then find the measure of \(\angle\)MPS.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक आयत PQRS दर्शाया गया है। विकर्ण PR और QS बिंदु M पर प्रतिच्छेद करते हैं। \(\angle\)QMR 50° के रूप में चिन्हित है।
Solution: \(\Box\)PQRS is a rectangle.
\(\therefore\) PM = \(\frac{1}{2}\) PR ...(i)
MS = \(\frac{1}{2}\) QS ...(ii) [Diagonals of a rectangle bisect each other]
Also, PR = QS .....(iii) [Diagonals of a rectangle are congruent]
\(\therefore\) PM = MS ....(iv) [From (i), (ii) and (iii)]
In ΔPMS,
PM = MS [From (iv)]
\(\therefore\) \(\angle\)MSP = \(\angle\)MPS = x° .....(v) [Isosceles triangle theorem]
\(\angle\)PMS = \(\angle\)QMR = 50° .....(vi) [Vertically opposite angles]
In ΔMPS,
\(\angle\)PMS + \(\angle\)MPS + \(\angle\)MSP = 180° [Sum of the measures of the angles of a triangle is 180°]
\(\therefore\) 50° +\(x\) + \(x\) = 180° [From (v) and (vi)]
\(\therefore\) \(50^\circ + 2x = 180^\circ\)
\(\therefore\) \(2x = 180^\circ - 50^\circ\)
\(\therefore\) \(2x = 130^\circ\)
\(\therefore\) \(x = \frac{130}{2} = 65^\circ\)
\(\therefore\) \(\angle\)MPS = 65° [From (v)]
In simple words: In a rectangle, diagonals bisect each other and are equal, making PM = MS. This forms an isosceles triangle PMS. \(\angle\)PMS is vertically opposite to \(\angle\)QMR, so \(\angle\)PMS = 50°. In \(\triangle\)PMS, the sum of angles is 180°. Since \(\angle\)MSP = \(\angle\)MPS, we have \(50^\circ + 2 \times \angle\)MPS = 180°, which gives \(\angle\)MPS = 65°.

🎯 Exam Tip: Recall that the diagonals of a rectangle are equal and bisect each other. This property often leads to isosceles triangles, which can be used with angle sum property of triangles and vertically opposite angles to find unknown angles.

Question 7. In the adjoining figure, if seg AB || seg PQ, seg AB = seg PQ, seg AC || seg PR, seg AC = seg PR, then prove that seg BC || seg QR and seg BC = seg QR.
ℹ️ चित्र व्याख्या (Diagram Explanation): दो समांतर चतुर्भुजों के साथ एक ज्यामितीय आकृति दिखाई गई है। एक खंड AB है जो PQ के समानांतर और समान है। एक खंड AC है जो PR के समानांतर और समान है। बिंदु A, B, C एक त्रिभुज बनाते हैं, और बिंदु P, Q, R एक और त्रिभुज बनाते हैं। आकृति का उद्देश्य ABQP और ACRP समांतर चतुर्भुजों को प्रदर्शित करना है।
Solution: Given: seg AB || seg PQ, seg AB = seg PQ,
seg AC || seg PR, seg AC = seg PR
To prove: seg BC || seg QR, seg BC = seg QR
Proof:
Consider \(\Box\)ABQP,
seg AB || seg PQ [Given]
seg AB = seg PQ [Given]
\(\therefore\) \(\Box\)ABQP is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
\(\therefore\) segAP || segBQ .....(i)
\(\therefore\) seg AP = seg BQ .....(ii) [Opposite sides of a parallelogram]
Consider \(\Box\)ACRP,
seg AC || seg PR [Given]
seg AC = seg PR [Given]
\(\therefore\) \(\Box\)ACRP is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
\(\therefore\) seg AP || seg CR ...(iii)
\(\therefore\) seg AP = seg CR .......(iv) [Opposite sides of a parallelogram]
Consider \(\Box\)BCRQ,
seg BQ || seg CR
seg BQ = seg CR
\(\therefore\) \(\Box\)BCRQ is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
\(\therefore\) seg BC || seg QR
\(\therefore\) seg BC = seg QR [Opposite sides of a parallelogram]
In simple words: By showing that quadrilaterals ABQP and ACRP are parallelograms (due to given parallel and congruent sides), we establish that AP is parallel and equal to BQ and CR respectively. This implies BQ is parallel and equal to CR, which makes BCRQ a parallelogram. Therefore, BC must be parallel and equal to QR.

🎯 Exam Tip: This proof relies on the definition of a parallelogram (one pair of opposite sides parallel and congruent). Systematically apply this definition to construct the required parallelogram BCRQ and prove the final statements.

Question 8. In the adjoining figure, \(\Box\)ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that PQ || AB and PQ = \(\frac{1}{2}\) (AB + DC).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समलंब ABCD दर्शाया गया है जहाँ AB, DC के समानांतर है। P भुजा AD का मध्यबिंदु है और Q भुजा BC का मध्यबिंदु है। एक रेखाखंड PQ मध्यबिंदुओं को जोड़ता है।
Solution: Given: \(\Box\)ABCD is a trapezium.
To prove:
Construction: Join points A and Q. Extend seg AQ and let it meet produced DC at R.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समलंब ABCD को दर्शाया गया है जहाँ AB, DC के समानांतर है। P भुजा AD का मध्यबिंदु है और Q भुजा BC का मध्यबिंदु है। AQ को बढ़ाया गया है ताकि यह DC के विस्तारित भाग को बिंदु R पर मिले।
Proof:
seg AB || seg DC [Given]
and seg BC is their transversal.
\(\therefore\) \(\angle\)ABC \(\cong\) \(\angle\)RCB [Alternate angles]
\(\therefore\) \(\angle\)ABQ = \(\angle\)RCQ ....(i) [B-Q-C]
In ΔABQ and ΔRCQ,
\(\angle\)ABQ = \(\angle\)RCQ [From (i)]
ℹ️ चित्र व्याख्या (Diagram Explanation): त्रिभुज ABQ और RCQ को अलग से दर्शाया गया है। बिंदु B, Q, C एक रेखा में हैं। \(\angle\)ABQ और \(\angle\)RCQ को समान दर्शाया गया है, साथ ही BQ और CQ को समान दर्शाया गया है (Q मध्यबिंदु है), और \(\angle\)BQA और \(\angle\)CQR को ऊर्ध्वाधर विपरीत कोणों के रूप में दर्शाया गया है।
seg BQ = seg CQ [Q is the midpoint of seg BC]
\(\angle\)BQA = \(\angle\)CQR [Vertically opposite angles]
\(\therefore\) ΔABQ \(\cong\) ΔRCQ [ASA test]
seg AB = seg CR ...(ii) [c. s. c. t.]
seg AQ = seg RQ [c. s. c. t.]
\(\therefore\) Q is the midpoint of seg AR. ....(iii)
In ΔADR,
Points P and Q are the midpoints of seg AD and seg AR respectively. [Given and from (iii)]
ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज ADR दर्शाया गया है। P भुजा AD का मध्यबिंदु है और Q भुजा AR का मध्यबिंदु है। रेखाखंड PQ मध्यबिंदुओं को जोड़ता है और DR के समानांतर दर्शाया गया है।
\(\therefore\) seg PQ || seg DR [Midpoint theorem]
i.e. seg PQ || seg DC ........(iv) [D-C-R]
But, seg AB || seg DC .......(v) [Given]
\(\therefore\) seg PQ || seg AB [From (iv) and (v)]
In ΔADR,
PQ = \(\frac{1}{2}\) DR [Midpoint theorem]
\(= \frac{1}{2}\) (DC + CR) [D-C-R]
\(= \frac{1}{2}\) (DC + AB) [From (ii)]
PQ = \(\frac{1}{2}\) (AB+ DC)
In simple words: To prove the midpoint theorem for a trapezium, extend AQ to meet DC (extended) at R. This creates congruent triangles ABQ and RCQ, proving Q is the midpoint of AR and AB = CR. Then, in \(\triangle\)ADR, PQ connects midpoints P and Q, so by the midpoint theorem, PQ is parallel to DR (and thus AB and DC) and PQ is half of DR. Substituting DR = DC + CR = DC + AB completes the proof.

🎯 Exam Tip: This is a standard proof using construction and the midpoint theorem. Clearly state the construction steps, use congruence criteria (ASA) and corresponding parts of congruent triangles (c.s.c.t.) correctly. Ensure logical flow and refer to properties like alternate angles.

Question 9. In the adjoining figure, \(\Box\)ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonals AC and DB respectively, then prove that MN || AB.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समलंब ABCD दर्शाया गया है जहाँ AB, DC के समानांतर है। विकर्ण AC और DB को दर्शाया गया है। M विकर्ण AC का मध्यबिंदु है और N विकर्ण DB का मध्यबिंदु है। रेखाखंड MN मध्यबिंदुओं को जोड़ता है।
Solution: Given: ABCD is a trapezium. AB || DC.
Points M and N are midpoints of diagonals AC and DB respectively.
To prove: MN || AB
Construction: Join D and M. Extend seg DM to meet seg AB at point E such that A-E-B.
Proof:
seg AB || seg DC and seg AC is their transversal. [Given]
\(\therefore\) \(\angle\)CAB = \(\angle\)ACD [Alternate angles]
\(\therefore\) \(\angle\)MAE \(\cong\) \(\angle\)MCD ....(i) [C-M-A, A-E-B]
In ΔAME and ΔCMD,
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समलंब ABCD को दर्शाया गया है। विकर्ण AC और DB खींचे गए हैं, M और N उनके मध्यबिंदु हैं। DM को E तक बढ़ाया गया है ताकि यह AB को E पर मिले। त्रिभुज AME और CMD को दर्शाया गया है, जिनमें \(\angle\)AME और \(\angle\)CMD ऊर्ध्वाधर विपरीत कोण हैं, और \(\angle\)MAE और \(\angle\)MCD समान हैं।
\(\angle\)AME = \(\angle\)CMD [Vertically opposite angles]
seg AM = seg CM [M is the midpoint of seg AC]
\(\angle\)MAE = \(\angle\)MCD [From (i)]
\(\therefore\) ΔAME \(\cong\) ΔCMD [ASA test]
\(\therefore\) seg ME = seg MD [c.s.c.t]
\(\therefore\) Point M is the midpoint of seg DE. ...(ii)
In ΔDEB,
ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज DEB दर्शाया गया है। M भुजा DE का मध्यबिंदु है और N भुजा DB का मध्यबिंदु है। रेखाखंड MN मध्यबिंदुओं को जोड़ता है और EB के समानांतर दर्शाया गया है।
Points M and N are the midpoints of seg DE and seg DB respectively. [Given and from (ii)]
\(\therefore\) seg MN || seg EB [Midpoint theorem]
\(\therefore\) seg MN || seg AB [A-E-B]
In simple words: To prove MN || AB in a trapezium, construct a line DM extended to meet AB at E. By proving \(\triangle\)AME \(\cong\) \(\triangle\)CMD using ASA test, we establish that M is the midpoint of DE. Since N is the midpoint of DB, by the midpoint theorem in \(\triangle\)DEB, MN || EB. As EB is part of AB, MN || AB is proven.

🎯 Exam Tip: This proof involves strategic construction and application of congruence criteria (ASA) and the midpoint theorem. Carefully identifying corresponding angles (alternate, vertically opposite) and sides is critical for proving triangle congruence and subsequently applying the midpoint theorem.

Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Problem Set 5 Intext Questions And Activities

Question 1. Draw five parallelograms by taking various measures of lengths and angles.
Answer: [Self-exercise for students to draw various parallelograms]
In simple words: A parallelogram is a quadrilateral with two pairs of parallel sides. Students should draw different shapes that fit this definition, varying side lengths and angles.

🎯 Exam Tip: Practicing drawing different types of quadrilaterals helps in visualizing their properties and understanding their definitions better, which is crucial for geometry concepts.

Question 2. Draw a parallelogram PQRS. Draw diagonals PR and QS. Denote the intersection of diagonals by letter O. Compare the two parts of each diagonal with a divider. What do you find? (Textbook page no. 60)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समांतर चतुर्भुज PQRS दर्शाया गया है। इसके विकर्ण PR और QS बिंदु O पर प्रतिच्छेद करते हैं। यह दर्शाया गया है कि O, PR और QS दोनों का मध्यबिंदु है, जिससे OP=OR और OQ=OS होता है।
Answer: seg OP = seg OR, and seg OQ = seg OS
Thus we can conclude that, point O divides the diagonals PR and QS in two equal parts.
In simple words: When you draw a parallelogram and its diagonals, you will observe that the point where they cross divides each diagonal into two equal segments. This shows that the diagonals bisect each other.

🎯 Exam Tip: This activity reinforces a fundamental property of parallelograms: their diagonals bisect each other. Visual verification helps in remembering theoretical concepts for exams.

Question 3. To verify the different properties of quadrilaterals.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक लकड़ी के तख्ते पर पेंच और धागे का उपयोग करके विभिन्न चतुर्भुजों के निर्माण को दर्शाया गया है। इसमें पेंचों को ग्रिड पैटर्न में लगाया गया है, और धागे का उपयोग करके आयत, वर्ग, समांतर चतुर्भुज, आदि जैसे चतुर्भुज बनाए जा सकते हैं, जिससे उनकी भुजाओं और कोणों के गुणों का पता चलता है।
Answer: Material: A piece of plywood measuring about 15 cm x 10 cm, 15 thin screws, twine, scissor.
Note: On the plywood sheet, fix five screws in a horizontal row keeping a distance of 2 cm between any two adjacent screws. Similarly make two more rows of screws exactly below the first one. Take care that the vertical distance between any two adjacent screws is also 2 cm.
With the help of the screws, make different types of quadrilaterals of twine. Verify the properties of sides and angles of the quadrilaterals. (Textbook page no. 75)
In simple words: This activity uses screws and thread on a board to physically construct and explore different quadrilaterals. By changing how the thread is strung around the screws, students can create shapes like squares, rectangles, or parallelograms and directly observe their side and angle properties.

🎯 Exam Tip: Practical activities like this help build an intuitive understanding of geometric properties. While not directly tested in exams, a strong conceptual base can improve problem-solving skills in geometry questions.