Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 5 Set 5.3 Quadrilaterals Solutions

Question 1. Diagonals of a rectangle ABCD intersect at point O. If AC = 8 cm, then find BO and if ∠CAD = 35°, then find ∠ACB.
Answer: ℹ️ चित्र व्याख्या (Diagram Explanation): एक आयत ABCD दिखाया गया है जिसके विकर्ण AC और BD बिंदु O पर प्रतिच्छेद करते हैं। कोण CAD 35 डिग्री के रूप में चिह्नित है। A शीर्ष-बाएँ, B नीचे-बाएँ, C नीचे-दाएँ और D शीर्ष-दाएँ कोने हैं।
i. AC = 8 cm ...(i) [Given]
ABCD is a rectangle [Given]
∴ BD = AC [Diagonals of a rectangle are congruent]
∴ BD = 8 cm [From (i)]
BO = \(\frac{1}{2}\) BD [Diagonals of a rectangle bisect each other]
∴ BO = \(\frac{1}{2} \times 8\)
∴ BO = 4 cm
ii. side AD || side BC and seg AC is their transversal. [Opposite sides of a rectangle are parallel]
∴ ∠ACB = ∠CAD [Alternate angles]
∠ACB = 35° [ ∴∠CAD = 35°]
∴ BO = 4 cm, ∠ACB = 35°
In simple words: In a rectangle, diagonals are equal and bisect each other, so BO is half of AC. Parallel sides and transversal create alternate angles, making ∠ACB equal to ∠CAD.

🎯 Exam Tip: Remember the properties of a rectangle, especially that its diagonals are congruent and bisect each other, and opposite sides are parallel, forming alternate interior angles with a transversal.

Question 2. In a rhombus PQRS, if PQ = 7.5 cm, then find QR. If ∠QPS = 75°, then find the measures of ∠PQR and ∠SRQ.
Answer: ℹ️ चित्र व्याख्या (Diagram Explanation): एक समचतुर्भुज PQRS दिखाया गया है। P नीचे-बाएँ, Q नीचे-दाएँ, R ऊपर-दाएँ और S ऊपर-बाएँ कोने हैं। भुजा PQ की लंबाई 7.5 सेमी है और कोण QPS 75 डिग्री चिह्नित है। सभी भुजाएँ समान लंबाई की हैं।
i. PQ = 7.5 cm [Given]
PQRS is a rhombus. [Given]
∴ QR = PQ [Sides of a rhombus are congruent]
∴ QR = 7.5 cm
ii. ∠QPS = 75° [Given]
\(\angle\text{QPS} + \angle\text{PQR} = 180^\circ\) [Adjacent angles of a rhombus are supplementary]
∴ \(75^\circ + \angle\text{PQR} = 180^\circ\)
∴ \(\angle\text{PQR} = 180^\circ - 75^\circ\)
∴ \(\angle\text{PQR} = 105^\circ\)
iii. ∠SRQ = ∠QPS [Opposite angles of a rhombus]
∴ ∠SRQ = 75°
∴ QR = 7.5 cm, ∠PQR = 105°, ∠SRQ = 75°
In simple words: In a rhombus, all sides are equal, so QR is the same as PQ. Adjacent angles are supplementary (add up to 180°), and opposite angles are equal.

🎯 Exam Tip: For rhombuses, ensure you recall that all sides are equal, adjacent angles sum to 180°, and opposite angles are congruent. These properties are key for finding missing side lengths and angles.

Question 3. Diagonals of a square IJKL intersects at point M. Find the measures of ∠IMJ, ∠JIK and ∠LJK.
Answer: ℹ️ चित्र व्याख्या (Diagram Explanation): एक वर्ग IJKL दिखाया गया है। I शीर्ष-बाएँ, J नीचे-बाएँ, K नीचे-दाएँ और L शीर्ष-दाएँ कोने हैं। विकर्ण IK और JL बिंदु M पर प्रतिच्छेद करते हैं, जो एक समकोण पर हैं।
IJKL is a square. [Given]
∴ seg IK ⊥ seg JL [Diagonals of a square are perpendicular to each other]
∠IMJ = 90°
∠JIL = 90° ....... (i) [Angle of a square]
ii. \(\angle\text{JIK} = \frac{1}{2} \angle\text{JIL}\) [Diagonals of a square bisect the opposite angles]
\(\angle\text{JIK} = \frac{1}{2} (90^\circ)\) [From (i)]
∴ ∠JIK = 45°
∠IJK = 90° (ii) [Angle of a square]
iii. \(\angle\text{LJK} = \frac{1}{2} \angle\text{IJK}\) [Diagonals of a square bisect the opposite angles]
\(\angle\text{LJK} = \frac{1}{2} (90^\circ)\) [From (ii)]
∴ LJK = 45°
∴ ∠IMJ = 90°, ∠JIK = 45°, ∠LJK = 45°
In simple words: In a square, diagonals are perpendicular bisectors of each other and bisect the angles. So, the intersection angle is 90°, and the angles formed by a diagonal are half of the square's 90° corner angles.

🎯 Exam Tip: Remember that diagonals of a square are perpendicular bisectors of each other and bisect the vertex angles. This means all intersection angles are 90°, and diagonal-formed angles are 45°.

Question 4. Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of rhombus and its Perimeter.
Answer: ℹ️ चित्र व्याख्या (Diagram Explanation): एक समचतुर्भुज ABCD दिखाया गया है जिसके विकर्ण AC और BD बिंदु O पर प्रतिच्छेद करते हैं। विकर्ण लंबवत रूप से प्रतिच्छेद करते हैं। A शीर्ष-बाएँ, B नीचे-बाएँ, C नीचे-दाएँ और D शीर्ष-दाएँ कोने हैं।
i. Let ABCD be the rhombus.
AC = 20 cm, BD = 21 cm
\(\text{AQ} = \frac{1}{2} \text{AC}\) [Diagonals of a rhombus bisect each other]
= \(\frac{1}{2} \times 20 = 10 \text{ cm}\) (i)
Also, \(\text{BO} = \frac{1}{2} \text{BD}\) [Diagonals of a rhombus bisect each other]
= \(\frac{1}{2} \times 21 = \frac{21}{2} \text{ cm}\) (ii)
ii. In ΔAOB, ∠AOB = 90° [Diagonals of a rhombus are perpendicular to each other]
∴ \(\text{AB}^2 = \text{AO}^2 + \text{BO}^2\) [Pythagoras theorem]
= \( (10)^2 + \left(\frac{21}{2}\right)^2 \) [From (i) and (ii)]
= \(100 + \frac{441}{4}\)
= \(\frac{400+441}{4}\)
∴ \(\text{AB}^2 = \frac{841}{4}\)
\(\text{AB} = \sqrt{\frac{841}{4}}\) [Taking square root of both sides]
= \(\frac{29}{2} = 14.5 \text{ cm}\)
iii. Perimeter of ABCD
= \(4 \times \text{AB} = 4 \times 14.5 = 58 \text{ cm}\)
∴ The side and perimeter of the rhombus are 14.5 cm and 58 cm respectively.
In simple words: Diagonals of a rhombus bisect each other perpendicularly. Using the Pythagorean theorem on the right-angled triangle formed by half-diagonals and a side, we can find the side length. The perimeter is then four times the side length.

🎯 Exam Tip: When given diagonals of a rhombus, use the fact that they bisect each other at right angles. This creates four right-angled triangles where the side of the rhombus is the hypotenuse, allowing you to apply the Pythagorean theorem.

Question 5. State with reasons whether the following statements are 'true' or 'false'.
i. Every parallelogram is a rhombus.
ii. Every rhombus is a rectangle,
iii. Every rectangle is a parallelogram.
iv. Every square is a rectangle,
v. Every square is a rhombus.
vi. Every parallelogram is a rectangle.
Answer: 
i. False.
All the sides of a rhombus are congruent, while the opposite sides of a parallelogram are congruent.
ii. False.
All the angles of a rectangle are congruent, while the opposite angles of a rhombus are congruent.
iii. True.
The opposite sides of a parallelogram are parallel and congruent. Also, its opposite angles are congruent.
The opposite sides of a rectangle are parallel and congruent. Also, all its angles are congruent.
iv. True.
The opposite sides of a rectangle are parallel and congruent. Also, all its angles are congruent.
All the sides of a square are parallel and congruent. Also, all its angles are congruent.
v. True.
All the sides of a rhombus are congruent. Also, its diagonals are perpendicular bisectors of each other.
All the sides of a square are congruent. Also, its diagonals are perpendicular bisectors of each other.
vi. False.
All the angles of a rectangle are congruent, while the opposite angles of a parallelogram are congruent.
In simple words: This question tests the hierarchical relationships between quadrilaterals. A figure inherits properties from the categories above it. For example, a square is both a rectangle and a rhombus because it possesses all their defining properties.

🎯 Exam Tip: Understand the properties that define each quadrilateral (parallelogram, rhombus, rectangle, square) and how they relate to each other. A Venn diagram can be helpful for visualizing these relationships and determining true/false statements efficiently.