Ex 12.2 Class 9 Maths Ganita Manjari Solutions | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 12 Exercise 12.2 Properties of Parallelograms, Rhombuses, Rectangles, and Angle Bisectors. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough covers true/false diagnostic properties of special quadrilaterals, formal geometric proofs for angle bisectors and rhombuses, converse property evaluations, and quadrilateral angle bisector constructions entirely clear.
🔑 Key Concepts & Properties:
- Parallelogram & Rectangle: A parallelogram with at least one right angle (or equal diagonals) is a rectangle.
- Rhombus & Angle Bisectors: A parallelogram with a diagonal bisecting an interior angle is a rhombus, and adjacent angle bisectors in a parallelogram always intersect at right angles (\(90^\circ\)).
NCERT Class 9 Maths Ganita Manjari Chapter 12 Exercise 12.2 Solutions
Question 1
True or false?
(i) A parallelogram with a right angle is a rectangle.
(ii) A rhombus with perpendicular diagonals is a square.
(iii) If the diagonals of a parallelogram are equal, then it is a rectangle.
Solution:
* (i) True. By definition, a parallelogram has opposite sides parallel and equal. If one interior angle is a right angle (\(90^\circ\)), consecutive interior angles sum to \(180^\circ\), making all four interior angles right angles, which defines a rectangle.
* (ii) False. Every rhombus already has perpendicular diagonals by definition! Therefore, having perpendicular diagonals does not restrict a general rhombus to be a square (a square also requires equal adjacent angles or equal adjacent side ratios, which are already guaranteed for all rhombuses, but perpendicular diagonals alone are universal to all rhombuses).
* (iii) True. In a parallelogram, equal diagonals lead to congruent triangles (\(\triangle ABC \cong \triangle DCB\)), which forces adjacent sides to meet at right angles, proving the parallelogram is a rectangle.
Question 2
The diagonal \(AC\) of a parallelogram \(ABCD\) bisects \(\angle A\). Show that it also bisects \(\angle C\) and that \(ABCD\) is a rhombus.
Solution:
Given: \(ABCD\) is a parallelogram, and diagonal \(AC\) bisects \(\angle A\) (\(\angle DAC = \angle BAC\)).
Proof Steps:
1. Let \(\angle DAC = \angle BAC = \theta\).
2. Since \(ABCD\) is a parallelogram, opposite sides are parallel: \(AB \parallel DC\) and \(AD \parallel BC\).
3. Using transversal \(AC\) with parallel lines \(AB \parallel DC\), alternate interior angles are equal: \(\angle BCA = \angle BAC = \theta\).
4. Using transversal \(AC\) with parallel lines \(AD \parallel BC\), alternate interior angles are equal: \(\angle DCA = \angle DAC = \theta\).
5. Therefore, \(\angle DCA = \angle BCA = \theta\), which proves that diagonal \(AC\) also bisects \(\angle C\).
6. In \(\triangle ABC\), since \(\angle BAC = \angle BCA = \theta\), sides opposite to equal angles are equal: \(AB = BC\).
7. Since \(ABCD\) is a parallelogram, opposite sides are equal (\(AB = CD\) and \(AD = BC\)). Thus, all four sides are equal (\(AB = BC = CD = DA\)), proving that \(ABCD\) is a rhombus.
Question 3
The following questions examine converses of true properties. Answer them with Yes or No. If your answer is No, what extra condition can you add so that the answer becomes Yes?
(i) If the diagonals of a quadrilateral \(ABCD\) bisect its angles, must \(ABCD\) be a rhombus?
(ii) If the diagonals of a quadrilateral \(ABCD\) bisect each other at right angles, must \(ABCD\) be a rhombus?
(iii) If the diagonals of a quadrilateral \(ABCD\) are of equal length, must \(ABCD\) be a rectangle?
Solution:
* (i) No. A kite can have its diagonals bisecting the vertex angles without being a rhombus or parallelogram.
Extra condition: If it is known to be a parallelogram whose diagonals bisect its interior angles, then it must be a rhombus.
* (ii) Yes. A quadrilateral whose diagonals bisect each other is a parallelogram, and if those diagonals intersect at right angles (\(90^\circ\)), it is a rhombus.
* (iii) No. An isosceles trapezoid has equal diagonals but is not a rectangle.
Extra condition: If the quadrilateral is a parallelogram whose diagonals are equal in length, then it must be a rectangle.
Question 4
Let \(ABCD\) be a parallelogram with \(AB \neq BC\). Show that the pairwise intersection points of the four angle bisectors form the vertices of a rectangle (see Fig. 12.11). Why did we assume \(AB \neq BC\)?
Solution:
Proof Steps:
1. In parallelogram \(ABCD\), consecutive interior angles (\(\angle A\) and \(\angle B\)) sum to \(180^\circ\).
2. Their angle bisectors meet at point \(P\). In \(\triangle APB\), the sum of half-angles is \(\frac{1}{2}\angle A + \frac{1}{2}\angle B = \frac{1}{2}(180^\circ) = 90^\circ\). Thus, \(\angle APB = 90^\circ\), meaning the opposite interior angle at \(P\) in quadrilateral \(PQRS\) is also \(90^\circ\).
3. Similarly, all four interior angles of the intersection quadrilateral \(PQRS\) are right angles (\(90^\circ\)), proving that \(PQRS\) is a rectangle.
Why did we assume \(AB \neq BC\)?
If \(AB = BC\), the parallelogram is a rhombus. In a rhombus, the angle bisectors of all four vertices coincide with the diagonals. Because the diagonals of a rhombus intersect each other at a single point (or degenerate into overlapping lines depending on vertex alignments), the region \(PQRS\) collapses into a single line segment or point rather than forming a non-degenerate rectangle with positive area.
Frequently Asked Questions (FAQs)
Q1: What is the difference between a rhombus and a rectangle regarding diagonals?
A: A rhombus has perpendicular diagonals that bisect the vertex angles, whereas a rectangle has equal diagonals that bisect each other into equal segments.
Q2: Why do adjacent angle bisectors in a parallelogram always form a rectangle?
A: Because consecutive angles of a parallelogram sum to \(180^\circ\), their bisectors intersect at \(90^\circ\), creating right angles at all four vertices of the inner quadrilateral.
Quick Navigation:
Last updated according to the latest NCERT Ganita Manjari syllabus.