# NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.2 – Part 2 (For CBSE, ICSE, IAS, NET, NRA 2022)

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- Mid-point

In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it

Q-3 ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:

- Give, ABCD is a ABCD is a rectangle and PQRS is a rhombus quadrilateral . P, Q, R
- And S is mid-points of the sides AB, BC, CD and DA respectively.

In

- P and Q are the mid-points of AB and BC respectively
- Thus, and (Mid-point theorem) … equation (1)

In ,

- S, R are the mid-points of AD and DC respectively
- and (Mid-point theorem) … equation (2)

From equation (1) and (2)

- So and

As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.

- and (Opposite sides of parallelogram) … equation (3)

Now, In ,

- Q and R are mid points of side BC and CD respectively.
- Thus, and (Mid-point theorem) … equation (4)
- (Diagonals of a rectangle are equal) … equation (5)

From equations (1) , (2) , (3) , (4) and (5) ,

- So, PQRS is a rhombus.

Q-4 ABCD is a trapezium in which , BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig.) . Show that F is the mid-point of BC.

Solution:

- Give, ABCD is a trapezium in which ,
- BD is a diagonal and E is the mid-point of AD.
- A line is drawn through E parallel to AB intersecting BC at F

Proof

- F is the mid-point of BC.
- BD intersected EF at G.
- In ,
- E is the mid-point of AD and also
- Thus, G is the mid-point of BD (Converse of midpoint theorem)

Now, In ΔBDC,

- G is the mid-point of BD and also GF || AB || DC.
- Thus, F is the mid-point of BC (Converse of midpoint theorem)