PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 8 Quadrilaterals Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2

Question 1.
ABCD is a quadrilateral in which F Q, R and S are midpoints of the sides AB, BC, CD and DA respectively (see the given figure 1). AC is a diagonal. Show that:
(i) SR || AC and SR = \(\frac{1}{2}\) AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 1
Answer:
In ∆ DAC, S and R are the midpoints of DA and DC respectively.
Through C draw a line parallel to AD which intersects line SR at T.
In ∆ DRS and ∆ CRT
∠ DRS = ∠ CRT (Vertically opposite angles)
∠ RSD = ∠ RTC (Alternate angles formed by transversal ST of DS || TC)
DR = CR (R is the midpoint of DC.)
∴ ∆ DRS ≅ ∆ CRT (AAS rule)
∴ DS = CT and SR = RT (CPCT)
As S is the midpoint of DA, we have DS = SA.
∴ SA = CT
And, by construction, SA || CT.
∴ Quadrilateral SACT is a parallelogram.
∴ ST || AC
∴ SR || AC ………… (1)
Now, SR = RT gives SR = \(\frac{1}{2}\)ST
In parallelogram SACT, ST = AC.
∴ SR = \(\frac{1}{2}\)AC ……………. (2)
Taking (1) and (2) together,
SR || AC and SR = \(\frac{1}{2}\)AC ….. Result (1)
Similarly, in ∆ ABC, P and Q are the midpoints of AB and BC respectively. ,
∴ PQ || AC and PQ = \(\frac{1}{2}\)AC
Now, SR = \(\frac{1}{2}\)AC and PQ = \(\frac{1}{2}\)AC
∴ PQ = SR …… Result (ii)
Similarly, SR || AC and PQ || AC.
∴ PQ || SR
Thus, in quadrilateral PQRS, PQ = SR and PQ || SR.
Hence, by theorem 8.8, PQRS is a parallelogram. … Result (iii)

PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Question 2.
ABCD is a rhombus and F Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Answer:
PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 2
ABCD is a rhombus and F Q, R and S are the midpoints of sides AB, BC, CD and DA respectively.
∴ In ∆ ABC, PQ || AC and PQ = \(\frac{1}{2}\)AC.
∴ In ∆ ADC, SR || AC and SR = \(\frac{1}{2}\)AC.
Hence, in quadrilateral PQRS, PQ || SR and PQ = SR.
∴ Quadrilateral PQRS is a parallelogram.
Now, since ABCD is a rhombus, AC and BD bisect each other at right angles at M.
∴ ∠ AMB = 90°
Now, AC || PQ and MN is their transversal.
∴ ∠ AMN + ∠ MNP = 180° (Interior angles on the same side of transversal)
∴ ∠ AMB + ∠MNP = 180°
∴ 90° + ∠ MNP = 180°
∴ ∠ MNP = 90°
In ∆ ABD, P and S are the midpoints of AB and AD respectively.
∴ PS || BD and NP is their transversal.
∴ ∠ DNP + ∠ NPS = 180°
∴ ∠ MNP + ∠ NPS =180°
∴ 90° + ∠ NPS = 180°
∴ ∠ NPS = 90°
∴ ∠ SPQ = 90°
Thus, in parallelogram PQRS, one angle ∠P is a right angle.
Hence, quadrilateral PQRS is a rectangle.

PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Question 3.
ABCD is a rectangle and P, Q, R and S are midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Answer:
PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 3
Since ABCD is a rectangle, its diagonals are equal.
∴ AC = BD
∴ \(\frac{1}{2}\)AC = \(\frac{1}{2}\)BD
In ∆ ABC, P and Q are the midpoints of AB and BC respectively.
∴ PQ = \(\frac{1}{2}\)AC
Similarly, in ∆ ADC, SR = \(\frac{1}{2}\)AC; in ∆ ABD, SP = \(\frac{1}{2}\) BD and in ∆ BCD, QR = \(\frac{1}{2}\) BD.
Now, PQ = SR = \(\frac{1}{2}\)AC, SP = QR = \(\frac{1}{2}\)BD and \(\frac{1}{2}\)AC = \(\frac{1}{2}\)BD
Hence, in quadrilateral PQRS,
PQ = QR = RS = SP
Thus, all the sides of quadrilateral PQRS are equal.
Hence, quadrilateral PQRS is a rhombus.

PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Question 4.
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the midpoint of BC.
PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 4
Answer:
Suppose line EF drawn through E and parallel to AB intersects BD at M.
EF || AB and AB || DC
∴ EF || DC
Trapezium ABCD is divided into two triangles, ∆ ABD and ∆ BCD, by diagonal BD.
In ∆ ABD, E is the midpoint of AD and a line through E and parallel to AB intersects BD at M.
Hence, by theorem 8.10, M is the midpoint of BD.
Now, in ∆ BCD, M is the midpoint of BD and a line through M and parallel to CD intersects BC at F.
Hence, by theorem 8.10, F is the midpoint of BC.
Note: The following result about the length of EF can also be derived:
EF = \(\frac{1}{2}\)(AB + CD)
Moreover, if X and Y are the midpoints of the diagonals of above trapezium ABCD, then XY = \(\frac{1}{2}\)|AB – CD|.

PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Question 5.
In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.
PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 5
Answer:
E and F are the midpoints of AB and CD respectively.
∴ AE = \(\frac{1}{2}\)AB and CF = \(\frac{1}{2}\)CD
In parallelogram ABCD, AB = CD and AB || CD.
∴ AE = CF and AE || CF
Hence, quadrilateral AECF is a parallelogram.
∴ AF || EC
∴ AP || EQ
In ∆ ABP E is the midpoint of AB and EQ || AR
∴ Q is the midpoint of PB. (Theorem 8.10)
∴PQ = QB …………… (1)
Similarly, in ∆ DQC, F is the midpoint of DC and FP || CQ.
∴ P is the midpoint of DQ. (Theorem 8.10)
∴ DP = PQ …………….. (2)
From (1) and (2), DP = PQ = QB.
Moreover, DP + PQ + QB = BD.
Thus, AF and EC trisect the diagonal BD.

PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Question 6.
Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.
Answer:
PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 6
In quadrilateral ABCD, P Q, R and S are the midpoints of sides AB, BC, CD and DA respectively.
In ∆ ABC, P and Q are the midpoints of AB and BC respectively.
∴ PQ || AC and PQ = \(\frac{1}{2}\)AC …………….. (1)
In ∆ ADC, S and R are the midpoints of DA and DC respectively.
∴ SR || AC and SR = \(\frac{1}{2}\)AC ……………… (2)
From (1) and (2),
PQ = SR and PQ || SR.
Thus, in quadrilateral PQRS, sides in one pair of opposite sides are equal and parallel. Hence, quadrilateral PQRS is a parallelogram. The diagonals of a parallelogram bisect each other. [Theorem 8.6]
∴ PR and SQ bisect each other.
Thus, the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.

PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Question 7.
ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that:
(i) D is the midpoint of AC.
(ii) MD ⊥ AC
(iii) CM = MA = \(\frac{1}{2}\)AB
Answer:
PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 7
In ∆ ABC, ∠ C is a right angle and M is the midpoint of hypotenuse AB. A line through M and parallel to BC intersects AC at D.
Hence, by theorem 8.10, DM bisects AC.
∴ D is the midpoint of AC. ….. Result (i)
In ∆ ABC, ∠ C is a right angle.
∴ ∠ C = 90°
Now, BC || DM and DC is their transversal.
∴ ∠ MDC + ∠ DCB = 180° (Interior angles on the same side of transversal)
∴ ∠ MDC + 90° = 180°
∴ ∠ MDC = 90°
Thus, MD is perpendicular to AC.
∴ MD ⊥ AC …… Result (ii)
Now, in ∆ ADM and ∆ CDM,
AD = CD (D is the midpoint of AC)
∠ ADM = ∠ CDM (Right angles)
DM = DM (Common)
∴ ∆ ADM ≅ ∆ CDM (SAS rule)
∴ AM = CM (CPCT) ……………. (1)
Now, M is the midpoint of AB.
∴ AM = \(\frac{1}{2}\)AB …… (2)
< Prom (1) and (2),
CM = MA = \(\frac{1}{2}\)AB …… Result (iii)

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