Class 9 Mathematics
Chapter 8: Quadrilaterals
Exercise 8.2 – Stepwise Solutions
Important Concepts Used
1. Mid-point theorem: The line joining midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
2. Properties of parallelogram: Opposite sides are equal and parallel.
3. Diagonals of parallelogram bisect each other.
4. Properties of rhombus and rectangle.
Q1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively. AC is a diagonal. Show that:
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram
(i) Proof:
In ΔADC, S and R are midpoints of AD and DC.
By midpoint theorem:
SR || AC and SR = 1/2 AC
(ii) Proof:
In ΔABC, P and Q are midpoints of AB and BC.
So, PQ || AC and PQ = 1/2 AC
Thus,
PQ = SR
(iii) Proof:
PQ || AC and SR || AC ⇒ PQ || SR
PQ = SR
Hence, one pair of opposite sides equal and parallel.
PQRS is a parallelogram
Q2. ABCD is a rhombus and P, Q, R, S are midpoints of sides. Show that PQRS is a rectangle.
From Q1, PQRS is a parallelogram.
In rhombus, diagonals are perpendicular.
Thus, sides of PQRS are parallel to diagonals of rhombus.
Hence adjacent sides are perpendicular.
PQRS is a rectangle
Q3. ABCD is a rectangle and P, Q, R, S are midpoints. Show that PQRS is a rhombus.
From midpoint theorem:
PQ = QR = RS = SP
Thus all sides equal.
PQRS is a rhombus
Q4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is midpoint of AD. A line through E parallel to AB meets BC at F. Show that F is midpoint of BC.
Given: E is midpoint of AD
EF || AB
In ΔABD:
E is midpoint of AD
EF || AB
So, by midpoint theorem, F is midpoint of BC
Hence proved
Q5. In parallelogram ABCD, E and F are midpoints of AB and CD. Show that AF and EC trisect diagonal BD.
Let AF and EC intersect BD at P and Q.
Using midpoint theorem and properties of parallelogram, triangles formed are congruent.
Thus,
BP = PQ = QD
Hence AF and EC trisect BD
Q6. ABC is a right triangle right-angled at C. M is midpoint of hypotenuse AB. A line through M parallel to BC meets AC at D. Show that:
(i) D is midpoint of AC
(ii) MD ⟂ AC
(iii) CM = MA = 1/2 AB
(i) Proof:
M is midpoint of AB and MD || BC
By midpoint theorem:
D is midpoint of AC
(ii) Proof:
BC ⟂ AC
MD || BC
Thus, MD ⟂ AC
(iii) Proof:
In right triangle, midpoint of hypotenuse is equidistant from vertices.
So,
CM = MA = MB = 1/2 AB