Ex 12.3 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 12.3 Midpoint Theorems, Varignon Parallelograms, and Triangle Congruence Properties. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough covers midpoint segment theorems, triangle partitioning, rhombus and square Varignon properties, and geometric reconstructions entirely clear.
🔑 Key Concepts & Theorems:
- Midpoint Theorem: The line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it.
- Varignon's Theorem: The quadrilateral formed by joining the midpoints of the consecutive sides of any quadrilateral is always a parallelogram.
NCERT Class 9 Maths Ganita Manjari Chapter 12 Exercise 12.3 Solutions
Question 1
If P, Q, R are the midpoints of sides AB, AC, BC respectively of \(\triangle ABC\):
(i) Show that \(\triangle PQR\) is congruent to \(\triangle QPA\) and to two other triangles which you should identify.
(ii) Suppose someone erases \(\triangle ABC\), leaving only \(\triangle PQR\) on the paper. Can you reconstruct \(\triangle ABC\) from \(\triangle PQR\)?
Solution:
(i) Congruent Triangles:
By the midpoint theorem, segments \(PQ, QR, RP\) divide \(\triangle ABC\) into four congruent small triangles. Thus, \(\triangle PQR\) is congruent to:
* \(\triangle QPA\)
* \(\triangle RBP\)
* \(\triangle RCQ\)
(All are congruent by SSS or SAS congruence since corresponding sides equal half the main triangle's sides).
(ii) Reconstructing \(\triangle ABC\):
Yes. Through each vertex of \(\triangle PQR\) (i.e., through \(P, Q, R\)), we can construct lines parallel to the opposite sides of \(\triangle PQR\). The intersection of these three parallel lines forms the outer vertices \(A, B, C\), successfully reconstructing \(\triangle ABC\).
Question 2
In \(\triangle ABC\), let M and N be midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD.
Solution:
Proof Steps:
1. By the Midpoint Theorem, since \(M\) and \(N\) are midpoints of \(AB\) and \(AC\), the segment \(MN \parallel BC\).
2. Let \(MN\) intersect \(AD\) at point \(X\).
3. In \(\triangle ABD\), since \(M\) is the midpoint of \(AB\) and \(MX \parallel BD\) (as \(MN \parallel BC\)), by the converse of the midpoint theorem, \(X\) is the midpoint of \(AD\).
4. Therefore, \(MN\) bisects \(AD\).
Question 3
In a quadrilateral \(ABCD\), suppose \(AB \parallel DC\) and \(AB \neq CD\). Suppose G and H are the midpoints of AC and BD respectively. Prove that \(GH \parallel AB\). (Why did we assume \(AB \neq CD\)?
Solution:
Proof Steps:
1. Consider trapezoid \(ABCD\) with \(AB \parallel DC\).
2. Using vector properties or midline construction through diagonals \(AC\) and \(BD\), segment \(GH\) connects the midpoints of the diagonals.
3. The line segment joining the midpoints of the diagonals of a trapezoid is parallel to the parallel bases and its length equals half the difference of the bases (\(GH = \frac{|AB - CD|}{2}\)).
4. Thus, \(GH \parallel AB\).
Why did we assume \(AB \neq CD\)?
If \(AB = CD\) and \(AB \parallel DC\), the quadrilateral \(ABCD\) becomes a parallelogram. In a parallelogram, the midpoints of the two diagonals \(AC\) and \(BD\) coincide at the exact same geometric center (the intersection point of the diagonals), making \(G\) and \(H\) the same point. Thus, \(GH\) would degenerate into a zero-length point rather than forming a distinct parallel segment of positive length.
Question 4
Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively.
(i) Show that PR and QS bisect each other.
(ii) Show that if \(AC = BD\), then PR and QS are perpendicular. Is the converse true?
Solution:
(i) Bisecting each other:
By Varignon's Theorem, joining the midpoints \(P, Q, R, S\) forms a parallelogram \(PQRS\). Since \(PQRS\) is a parallelogram, its diagonals \(PR\) and \(QS\) bisect each other.
(ii) Perpendicularity and Converse:
* If \(AC = BD\), the adjacent sides of parallelogram \(PQRS\) are equal (\(PS = PQ\), since each equals half of diagonals \(AC\) and \(BD\)). A parallelogram with equal adjacent sides is a rhombus, and the diagonals of a rhombus are perpendicular. Thus, \(PR \perp QS\).
* Converse: Yes, the converse is true. If \(PR\) and \(QS\) are perpendicular, \(PQRS\) is a rhombus, meaning its adjacent sides are equal, which implies that the diagonals of the original quadrilateral are equal (\(AC = BD\)).
Question 5
Suppose PQRS is the Varignon parallelogram of ABCD.
(i) Copy only PQRS on another paper. Show how you will recreate a congruent copy \(A'B'C'D'\) of ABCD from PQRS.
(ii) Justify why the quadrilateral \(A'B'C'D'\) constructed is congruent to ABCD.
(iii) Show that if PQRS is a square, then AC and BD are perpendicular and equal. Prove the converse.
Solution:
(i) & (ii) Reconstruction and Justification:
* Choose any arbitrary point \(S'\) (corresponding to \(S\)), then construct vector paths using the property that midpoints divide outer segments equally. By applying triangle congruence (such as \(\triangle SDR \cong \triangle SD'R\)), we can position vertices \(A', B', C', D'\) such that \(P, Q, R, S\) act as their exact midpoints.
(iii) Square Varignon Parallelogram:
* If \(PQRS\) is a square, its diagonals are equal in length and perpendicular to each other. Since the diagonals of Varignon parallelogram \(PQRS\) correspond to half the lengths and orientations of the diagonals \(AC\) and \(BD\) of \(ABCD\), equal and perpendicular diagonals in \(PQRS\) imply that \(AC\) and \(BD\) are also equal and perpendicular in \(ABCD\).
* Converse: If \(AC\) and \(BD\) are equal and perpendicular, then \(PQRS\) is a square.
Frequently Asked Questions (FAQs)
Q1: What is Varignon's Theorem?
A: Varignon's Theorem states that the quadrilateral formed by joining the midpoints of the consecutive sides of any quadrilateral is always a parallelogram.
Q2: How do diagonals of the outer quadrilateral relate to the Varignon parallelogram?
A: The sides of the Varignon parallelogram are parallel to the diagonals of the outer quadrilateral and have lengths equal to half of those diagonals.
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