Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 5 Set 5.1 Quadrilaterals Solutions

Question 1. Diagonals of a parallelogram WXYZ intersect each other at point O. If \( \angle XYZ = 135^\circ \), then measure of \( \angle XWZ \) and \( \angle YZW \)? If l(OY) = 5 cm, then l(WY) = ?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज WXYZ को दर्शाता है, जिसमें उसके विकर्ण WY और XZ बिंदु O पर प्रतिच्छेद करते हैं। कोण XYZ 135 डिग्री के रूप में चिह्नित है, और भुजाओं और कोणों को समझने के लिए W, X, Y, Z शीर्षों को लेबल किया गया है।
i. \( \angle XYZ = 135^\circ \)
WXYZ is a parallelogram.
\( \angle XWZ = \angle XYZ \)
\( \therefore \angle XWZ = 135^\circ \) .....(i)
ii. \( \angle YZW + \angle XYZ = 180^\circ \) [Adjacent angles of a parallelogram are supplementary]
\( \therefore \angle YZW + 135^\circ = 180^\circ \) [From (i)]
\( \therefore \angle YZW = 180^\circ - 135^\circ \)
\( \therefore \angle YZW = 45^\circ \)
iii. l(OY) = 5 cm [Given]
l(OY) = \( \frac{1}{2} \) l(WY) [Diagonals of a parallelogram bisect each other]
\( \therefore \) l(WY) = 2 x l(OY)
= 2 x 5
\( \therefore \) l(WY) = 10 cm
\( \therefore \angle XWZ = 135^\circ \), \( \angle YZW = 45^\circ \), l(WY) = 10 cm
In simple words: In a parallelogram, opposite angles are equal and adjacent angles are supplementary. Diagonals bisect each other. Using these properties, we find the unknown angles and the length of the diagonal.

🎯 Exam Tip: Remember the key properties of parallelograms: opposite angles are equal, adjacent angles sum to 180°, and diagonals bisect each other. These are fundamental for solving such problems.

Question 2. In a parallelogram ABCD, if \( \angle A = (3x + 12)^\circ \), \( \angle B = (2x - 32)^\circ \), then liptl the value of x and the measures of \( \angle C \) and \( \angle D \).
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज ABCD को दर्शाता है। कोण A को (3x + 12) डिग्री और कोण B को (2x - 32) डिग्री के रूप में लेबल किया गया है। शीर्षों A, B, C, D को स्पष्ट रूप से दर्शाया गया है, और भुजाएं समानांतर हैं।
ABCD is a parallelogram. [Given]
\( \therefore \angle A + \angle B = 180^\circ \) [Adjacent angles of a parallelogram are supplementary]
\( \therefore (3x + 12)^\circ + (2x - 32)^\circ = 180^\circ \)
\( \therefore 3x + 12 + 2x - 32 = 180 \)
\( \therefore 5x - 20 = 180 \)
\( \therefore 5x = 180 + 20 \)
\( \therefore 5x = 200 \)
\( \therefore x = \frac{200}{5} \)
\( \therefore x = 40 \)
ii. \( \angle A = (3x + 12)^\circ \)
\( = [3(40) + 12]^\circ \)
\( = (120 + 12)^\circ = 132^\circ \)
\( \angle B = (2x - 32)^\circ \)
\( = [2(40) - 32]^\circ \)
\( = (80 - 32)^\circ = 48^\circ \)
\( \therefore \angle C = \angle A = 132^\circ \)
\( \angle D = \angle B = 48^\circ \) [Opposite angles of a parallelogram]
\( \therefore \) The value of x is 40, and the measures of \( \angle C \) and \( \angle D \) are \( 132^\circ \) and \( 48^\circ \) respectively.
In simple words: By using the property that adjacent angles of a parallelogram are supplementary, we can set up an equation to find the value of 'x'. Once 'x' is known, we can calculate the measures of all angles using the property that opposite angles are equal.

🎯 Exam Tip: When dealing with angles in a parallelogram, always remember that adjacent angles add up to 180 degrees, and opposite angles are equal. This allows you to find unknown variables and angle measures efficiently.

Question 3. Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज ABCD को दर्शाता है। भुजा AD को 'x' cm और भुजा AB को 'x + 25' cm के रूप में लेबल किया गया है। शीर्षों को A, B, C, D नाम दिया गया है, और विपरीत भुजाओं को बराबर लंबाई के साथ दिखाया गया है।
i. Let □ABCD be the parallelogram and the length of AD be x cm.
One side is greater than the other by 25 cm.
\( \therefore \) AB = x + 25 cm
AD = BC = x cm
AB = DC = (x + 25) cm [Opposite angles of a parallelogram]
ii. Perimeter of □ABCD = 150 cm [Given]
\( \therefore \) AB + BC + DC + AD = 150
\( \therefore \) (x + 25) + x + (x + 25) + x = 150
\( \therefore \) 4x + 50 = 150
\( \therefore \) 4x = 150 - 50
\( \therefore \) 4x = 100
\( \therefore x = \frac{100}{4} \)
\( \therefore \) x = 25
iii. AD = BC = x = 25 cm
AB = DC = x + 25 = 25 + 25 = 50 cm
\( \therefore \) The lengths of the sides of the parallelogram are 25 cm, 50 cm, 25 cm and 50 cm.
In simple words: We define the sides of the parallelogram in terms of 'x' based on the given relationship. Then, using the perimeter formula for a parallelogram (2 * (length + width)), we solve for 'x' to find the lengths of all sides.

🎯 Exam Tip: For perimeter problems involving parallelograms, remember that opposite sides are equal. Define the side lengths using a variable and the given conditions, then apply the perimeter formula: Perimeter = 2 * (side1 + side2).

Question 4. If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all angles of the parallelogram.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज ABCD को दर्शाता है। कोण A को x डिग्री और कोण B को 2x डिग्री के रूप में लेबल किया गया है। शीर्षों को A, B, C, D नाम दिया गया है, जो समांतर चतुर्भुज के कोणों और भुजाओं को समझने में मदद करता है।
i. Let □ABCD be the parallelogram.
The ratio of measures of two adjacent angles of a parallelogram is 1 : 2.
Let the common multiple be x.
\( \therefore \angle A = x^\circ \) and \( \angle B = 2x^\circ \)
\( \angle A + \angle B = 180^\circ \) [Adjacent angles of a parallelogram are supplementary]
\( \therefore x + 2x = 180 \)
\( \therefore 3x = 180 \)
\( \therefore x = \frac{180}{3} \)
\( \therefore x = 60 \)
ii. \( \angle A = x^\circ = 60^\circ \)
\( \angle B = 2x^\circ = 2 \times 60^\circ = 120^\circ \)
\( \angle A = \angle C = 60^\circ \)
\( \angle B = \angle D = 120^\circ \) [Opposite angles of a parallelogram]
\( \therefore \) The measures of the angles of the parallelogram are \( 60^\circ, 120^\circ, 60^\circ \) and \( 120^\circ \).
In simple words: Since adjacent angles of a parallelogram are supplementary, we can represent them using the given ratio and form an equation. Solving this equation gives us the value of the common multiple, which then allows us to find all four angles, remembering that opposite angles are equal.

🎯 Exam Tip: When given a ratio of adjacent angles in a parallelogram, let the angles be \( x \) and \( 2x \). Apply the supplementary angle property (\( x + 2x = 180^\circ \)) to find \( x \), then use the opposite angles property to find all angles.

Question 5. Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO show that ABCD is a rhombus.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज ABCD को दर्शाता है, जिसके विकर्ण AC और BD बिंदु O पर प्रतिच्छेद करते हैं। त्रिभुज AOB को दर्शाया गया है, जिसकी भुजाएँ AO, BO और AB हैं। यह विकर्णों के प्रतिच्छेदन और भुजाओं के बीच संबंध को समझने में मदद करता है।
To prove: ABCD is a rhombus.
Solution:
Proof:
AO = 5, BO = 12, AB = 13 [Given]
\( AO^2 + BO^2 = 5^2 + 12^2 \)
\( = 25 + 144 \)
\( \therefore AO^2 + BO^2 = 169 \) .....(i)
\( AB^2 = 13^2 = 169 \) ....(ii)
\( \therefore AB^2 = AO^2 + BO^2 \) [From (i) and (ii)]
\( \therefore \triangle AOB \) is a right-angled triangle. [Converse of Pythagoras theorem]
\( \therefore \angle AOB = 90^\circ \)
\( \therefore \) seg AC \( \perp \) seg BD .....(iii) [A-O-C]
\( \therefore \) In parallelogram ABCD,
\( \therefore \) seg AC \( \perp \) seg BD [From (iii)]
\( \therefore \) ABCD is a rhombus. [A parallelogram is a rhombus perpendicular to each other]
In simple words: We use the given side lengths and the Pythagorean theorem to show that the diagonals of the parallelogram intersect at a right angle. A parallelogram with perpendicular diagonals is a rhombus.

🎯 Exam Tip: To prove a parallelogram is a rhombus, a common method is to show that its diagonals are perpendicular. This can often be done using the converse of the Pythagorean theorem if side lengths are given.

Question 6. In the adjoining figure, PQRS and □ABCR are two parallelograms. If \( \angle P = 110^\circ \), then find the measures of all the angles of ABCR.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो संलग्न समांतर चतुर्भुजों, PQRS और ABCR को दर्शाता है, जो भुजा CR साझा करते हैं। कोण P को 110 डिग्री के रूप में लेबल किया गया है। शीर्षों P, Q, R, S और A, B, C को दर्शाया गया है, जिससे कोणों और उनके संबंधों को समझने में मदद मिलती है।
PQRS is a parallelogram. [Given]
\( \therefore \angle R = \angle P \) [Opposite angles of a parallelogram]
\( \therefore \angle R = 110^\circ \) .....(iii)
ABCR is a parallelogram. [Given]
\( \therefore \angle A + \angle R = 180^\circ \) [Adjacent angles of a parallelogram are supplementary]
\( \therefore \angle A + 110^\circ = 180^\circ \) [From (i)]
\( \therefore \angle A = 180^\circ - 110^\circ \)
\( \therefore \angle A = 70^\circ \)
\( \therefore \angle C = \angle A = 70^\circ \)
\( \therefore \angle B = \angle R = 110^\circ \) [Opposite angles of a parallelogram]
\( \therefore \angle A = 70^\circ, \angle B = 110^\circ, \)
\( \therefore \angle C = 70^\circ, \angle R = 110^\circ \)
In simple words: We first find angle R in parallelogram PQRS using the property of opposite angles. Then, using this angle R as an adjacent angle in parallelogram ABCR, we find angle A using the supplementary property. Finally, we determine the remaining angles of ABCR using the opposite angles property.

🎯 Exam Tip: When dealing with multiple parallelograms sharing a side, use the properties of one parallelogram to find an angle, then transfer that information to the adjacent parallelogram to solve for its angles. Clearly state which parallelogram's properties you are using at each step.

Question 7. In the adjoining figure, ABCD is a parallelogram. Point E is on the ray AB such that BE = AB, then prove that line ED bisects seg BC at point F.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज ABCD को दर्शाता है। किरण AB पर एक बिंदु E इस प्रकार है कि BE = AB। एक रेखाखंड ED खींचा गया है जो भुजा BC को बिंदु F पर काटता है। शीर्षों और रेखाखंडों को चिह्नित किया गया है ताकि ज्यामितीय प्रमाण को समझने में मदद मिल सके।
Given: ABCD is a parallelogram.
BE = AB
To prove: Line ED bisects seg BC at point F i.e. FC = FB
Solution:
Proof:
ABCD is a parallelogram. [Given]
\( \therefore \) seg AB = seg DC .......(i) [Opposite angles of a parallelogram]
seg AB = seg BE ........(ii) [Given]
seg DC = seg BE ........(iii) [From (i) and (ii)]
side DC || side AB [Opposite sides of a parallelogram]
i.e. side DC || seg AE and seg DE is their transversal. [A-B-E]
\( \therefore \angle CDE \cong \angle AED \)
\( \therefore \angle CDF \cong \angle BEF \) .....(iv) [D-F-E, A-B-E]
In \( \triangle DFC \) and \( \triangle EFB \),
seg DC = seg EB [From (iii)]
\( \angle CDF = \angle BEF \) [From (iv)]
\( \angle DFC = \angle EFB \) [Vertically opposite angles]
\( \therefore \triangle DFC \cong \triangle EFB \) [SAA test]
\( \therefore \) FC \( \cong \) FB [c.s.c.t]
\( \therefore \) Line ED bisects seg BC at point F.
In simple words: We prove that two triangles, DFC and EFB, are congruent using the SAA (Side-Angle-Angle) test. This congruence is established by identifying equal sides (DC = BE) and equal angles (alternate interior angles and vertically opposite angles), which then allows us to conclude that FC = FB, meaning ED bisects BC.

🎯 Exam Tip: For proving bisection, congruence of triangles is often the key. Identify corresponding equal sides and angles (e.g., opposite sides of parallelogram, alternate interior angles, vertically opposite angles) to satisfy a congruence test (SAA, ASA, SAS, SSS).

Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Practice Set 5.1 Intext Questions And Activities

Question 1. Write the following pairs considering ABCD. (Textbook pg. no 57)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक सामान्य चतुर्भुज ABCD को दर्शाता है जिसके शीर्ष A, B, C, D हैं। इस चित्र का उपयोग आसन्न भुजाओं, आसन्न कोणों, विपरीत भुजाओं और विपरीत कोणों के युग्मों की पहचान करने के लिए एक संदर्भ के रूप में किया जाता है।
Pairs of adjacent sides:
(i) AB, AD
(ii) AD, DC
(iii) DC, BC
(iv) BC, AB
Pairs of adjacent angles:
(i) \( \angle A, \angle B \)
(ii) \( \angle C, \angle D \)
(iii) \( \angle B, \angle C \)
(iv) \( \angle D, \angle A \)
Pairs of opposite sides:
(i) AB, DC
(ii) AD, BC
Pairs of opposite angles:
(i) \( \angle A, \angle C \)
(ii) \( \angle B, \angle D \)
In simple words: This exercise requires identifying and listing the pairs of adjacent sides, adjacent angles, opposite sides, and opposite angles in a general quadrilateral ABCD based on their definitions.

🎯 Exam Tip: Understand the definitions of adjacent (sharing a vertex or side) and opposite (not sharing a vertex or side) for both sides and angles in any quadrilateral. This is a basic but essential concept.

Question 2. Complete the following tree diagram. (Textbook pg. no 57)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृक्ष आरेख (tree diagram) है जो चतुर्भुजों के वर्गीकरण को दर्शाता है। यह "I am a quadrilateral" से शुरू होता है और फिर विभिन्न प्रकार के चतुर्भुजों - समांतर चतुर्भुज (Parallelogram), समलंब चतुर्भुज (Trapezium), आयत (Rectangle), समचतुर्भुज (Rhombus) और वर्ग (Square) - में विभाजित होता है, साथ ही उनके गुणों का भी वर्णन करता है।
I am a quadrilateral
My both pairs of
opposite sides are
parallel
Parallelogram
My properties
i. Opposite sides are
congruent.
ii. Opposite angles
are congruent.
iii. Diagonals bisect
each other.
My all angles are
right angles
Rectangle
My properties
i. Opposite sides
are congruent.
ii. Diagonals are
congruent and
bisect
other.
each
My all sides are
equal in length
Rhombus
My properties
i. Opposite sides
are congruent.
ii. Diagonals are
perpendicular
bisectors
each other.
of
My all angles are
equal and all sides
are equal
Square
My properties
i. Diagonals are
congruent and
are
perpendicular
bisectors
each other.
of
My only one pair of opposite sides is parallel
Trapezium
In simple words: This tree diagram classifies quadrilaterals based on their properties, branching from a general quadrilateral to more specific types like parallelograms, trapeziums, rectangles, rhombuses, and squares, each with unique characteristics related to their sides, angles, and diagonals.

🎯 Exam Tip: Thoroughly memorize the definitions and properties of different types of quadrilaterals (parallelogram, rectangle, rhombus, square, trapezium). Understanding how they relate to each other (e.g., a square is a rectangle and a rhombus) is crucial.

Question 3. In the above theorem, to prove \( \angle DAB = \angle BCD \), is any change in the construction needed? If so, how will you write the proof making the change? (Textbook pg. no. 60)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक चतुर्भुज ABCD को दर्शाता है जिसके अंदर एक विकर्ण BD खींचा गया है। यह विकर्ण चतुर्भुज को दो त्रिभुजों, ABD और BCD में विभाजित करता है, जो कोणों DAB और BCD के बीच संबंधों को सिद्ध करने के लिए महत्वपूर्ण है।
Solution:
Yes
Construction: Draw diagonal BD.
Proof:
side AB || side CD and diagonal BD is their transversal. [Given]
\( \therefore \angle ABD \cong \angle CDB \) ........(i) [Alternate angles]
side BC || side AD and diagonal BD is their transversal. [Given]
\( \therefore \angle ADB \cong \angle CBD \) ........(ii) [Alternate angles]
In \( \triangle DAB \) and \( \triangle BCD \),
\( \angle ABD \cong \angle CDB \) [From (i)]
seg BD = seg DB [Common side]
\( \therefore \angle ADB \cong \angle CBD \) [From (ii)]
\( \therefore \triangle DAB \cong \triangle BCD \) [ASA test]
\( \therefore \angle DAB \cong \angle BCD \) [c.a.c.t.]
Note: \( \angle DAB \) s \( \angle BCD \) can be proved using the same construction as in the above theorem.
\( \angle BAC \cong \angle DCA \) .....(i)
\( \angle DAC \cong \angle BCA \) ......(ii)
\( \therefore \angle BAC + \angle DAC = \angle DCA + \angle BCA \) [Adding (i) and (ii)]
\( \therefore \angle DAB = \angle BCD \) [Angle addition property]
In simple words: To prove that opposite angles \( \angle DAB \) and \( \angle BCD \) are equal, we need to draw a diagonal (BD). This allows us to prove the congruence of the two triangles formed (\( \triangle DAB \) and \( \triangle BCD \)) using the ASA test, which then implies the equality of the corresponding angles.

🎯 Exam Tip: When proving properties of a parallelogram, drawing diagonals is a common construction technique. For proving angle equality, look for alternate interior angles formed by parallel lines and transversals, and then use triangle congruence tests.