Chapter 8: Quadrilaterals (Exercise 8.2)

Chapter 8: Quadrilaterals (Exercise 8.2)

Question 1:
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that:
1. SR ∥ AC and SR = ½ AC
2. PQ = SR
3. PQRS is a parallelogram.

Solution:
1. In ΔADC:
* S is the midpoint of AD and R is the midpoint of CD.
* By the Mid-point Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it:
SR ∥ AC and SR = ½ AC --- (Equation 1)

2. In ΔABC:
* P is the midpoint of AB and Q is the midpoint of BC.
* By Mid-point Theorem:
PQ ∥ AC and PQ = ½ AC --- (Equation 2)

* From Equation 1 and Equation 2:
PQ = SR

3. To prove PQRS is a parallelogram:
* From Equation 1 and Equation 2, PQ ∥ AC and SR ∥ AC ⇒ PQ ∥ SR.
* Since one pair of opposite sides (PQ and SR) is equal and parallel, PQRS is a parallelogram.
(Hence proved.)

Question 2:
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Solution:
* Join diagonals AC and BD of rhombus ABCD.
* Diagonals of a rhombus intersect at 90°.

* In ΔABC, P and Q are midpoints of AB and BC ⇒ PQ ∥ AC and PQ = ½ AC.
* In ΔADC, S and R are midpoints of AD and CD ⇒ SR ∥ AC and SR = ½ AC.
* Therefore, PQ ∥ SR and PQ = SR, making PQRS a parallelogram.

* Since PQ ∥ AC and PS ∥ BD, and diagonals AC ⊥ BD, the adjacent sides PQ ⊥ PS.
* Thus, ∠SPQ = 90°.
* Since a parallelogram with one right angle is a rectangle, PQRS is a rectangle.
(Hence proved.)

Question 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:
* Join diagonal AC.
* In ΔABC, P and Q are midpoints of AB and BC ⇒ PQ = ½ AC.
* In ΔADC, S and R are midpoints of AD and CD ⇒ SR = ½ AC.
* Join diagonal BD.
* In ΔABD, P and S are midpoints of AB and AD ⇒ PS = ½ BD.

* Since ABCD is a rectangle, its diagonals are equal: AC = BD.
* Therefore, PQ = QR = RS = SP = ½ AC.
* Since all four sides are equal, PQRS is a rhombus.
(Hence proved.)

Question 4:
ABCD is a trapezium in which AB ∥ DC, 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. Show that F is the mid-point of BC.

Solution:
* Let line EF intersect diagonal BD at point G.

1. In ΔDAB:
* E is the midpoint of AD and EG ∥ AB (since EF ∥ AB).
* By the Converse of Mid-point Theorem, a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.
* Therefore, G is the mid-point of BD.

2. In ΔBCD:
* G is the midpoint of BD and GF ∥ DC (since AB ∥ DC and EF ∥ AB).
* By the Converse of Mid-point Theorem, line GF bisects side BC.
* Therefore, F is the mid-point of BC.
(Hence proved.)