Jharkhand Board JAC Class 9 Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

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Question 1.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig). AC is a diagonal. Show that:

(i) SR || AC and SR = \(\frac {1}{2}\)AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Answer:

(i) In ΔDAC,

R is the mid-point of DC and S is the mid-point of DA.

Thus by mid-point theorem, SR || AC and SR = \(\frac {1}{2}\)AC

(ii) In ΔBAC,

P is the mid-point of AB and Q is the mid-point of BC.

Thus by mid-point theorem, PQ || AC

and PQ = \(\frac {1}{2}\) AC

Also, SR = \(\frac {1}{2}\) AC

Thus, PQ = SR

(iii) SR || AC (Proved)

and, PQ || AC (Proved)

⇒ SR || PQ (Lines parallel to same line are parallel to each other)

Also, PQ = SR(Proved)

Thus, PQRS is a parallelogram.

(A quadrilateral in which a pair of sides is equal and parallel is a parallelogram).

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.

Answer:

Given: ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

To Prove: PQRS is a rectangle.

Construction: AC and BD are joined.

Proof: P and Q are mid-points of AB and BC respectively

⇒ PQ || AC and PQ = \(\frac{1}{2}\) AC …(i)

(By mid-point theorem)

Also, S are R are mid-points of AD and CD respectively, so, by mid-point theorem,

SR || AC and SR = \(\frac{1}{2}\)AC …(ii)

From (i) and (ii)

∴ PQ || SR and PQ = SR

⇒ PQRS is a parallelogram

As ABCD is a rhombus

AB = BC (sides of rhombus)

⇒ \(\frac{1}{2}\) AB = \(\frac{1}{2}\) BC

⇒ BP = BQ (As P and Q are mid-points of AB and BC respectively)

So, in ΔBPQ

∠BPQ = ∠BQP …(i)

(Angles opposite to equal sides are equal)

Also, BC = CD (Sides of rhombus)

⇒ \(\frac{1}{2}\) BC = \(\frac{1}{2}\) CD

⇒ CQ = CR (As Q and R are midpoint of BC and CD respectively)

So, in ACQR

∠CQR = ∠CRQ …(ii) (Angle opposites to equal sides are equal)

AB || CD (Opposite sides of a parallelogram)

⇒ ∠B + ∠C =180 …(iii)

(sum of angles on same sides of transversal is 180°)

In ΔBPQ, ∠BPQ + ∠BQP + ∠B = 180° (Angles sum property)

In ΔCQR, ∠CQR + ∠CRQ + ∠C = 180° (Angle sum property)

On adding both the equations, we get

∠BPQ + ∠BQP + ∠B + ∠CQR + ∠CRQ + ∠C = 360°

⇒ 180° + ∠BPQ + ∠BQP + ∠CQR + ∠CRQ = 360° [From (ii)]

⇒ 180° + 2 ∠BQP + 2 ∠CQR = 360° [From (i) and (ii)]

⇒ 2 (∠BQP + ∠CQR) = 360° – 180° = 180°

⇒ ∠BQP + ∠CQR = \(\frac{180°}{2}\) = 90° …(iv)

Now, ∠BQP + ∠CQR + ∠PQR = 180° (sum of angles on a straight line is 180°)

90° + ∠PQR = 180° [From (iv)]

⇒ ∠PQR = 180° – 90° = 90°

So, PQRS is a parallelogram in which one angle is 90°

⇒ PQRS is a rectangle.

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.

Answer:

Given: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

To Prove PQRS is a rhombus.

Construction: AC and BD are joined.

Proof: In ∆ABC

P and Q are the mid-points of AB and BC respectively

Thus, PQ || AC and PQ = \(\frac{1}{2}\) AC (Mid-point theorem) …(i)

In ∆ADC, SR AC and SR = \(\frac{1}{2}\) AC (Mid-point theorem) …(ii)

From (i) and (ii)

∴ PQ = SR and PQ || SR

⇒ PQRS is a parallelogram.

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

AB = CD (opposite sides of rectangle)

⇒ AB = CD

⇒ BP = CR(As P and R are midpoints of AB and CD respectively)

Consider ∆PBQ and ∆RCQ

BP = CR (Proved)

BQ = CQ [As Q is a mid-point of BC]

∠B = ∠C = 90° (As each angle of a rectangle ABCD is of 90°)

⇒ ΔPBQ ≅ ΔRCQ by SAS congruence criterion.

⇒ PQ = QR (CPCT)

Also, PQ = RS and PS = QR (As opposite sides of parallelogram are equal)

∴ PQ = QR = RS = PS

So, PQRS is a parallelogram in which all sides are equal.

⇒ PQRS is a rhombus.

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 (see Fig). Show that F is the mid-point of BC.

Answer:

Given: ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.

To prove: F is the mid-point of BC.

Proof: BD intersects EF at G.

In ∆BAD,

E is the mid-point ofAD andalsoEG || AB.

Thus, G is the mid-point of BD (Converse of mid-point theorem)

Now, EG || AB

⇒ EF || AB

⇒ GF || AB

Also, AB || CD

∴ CD || GF (Lines parallel to same line are parallel to each other)

In ABCD, G is a mid-point of BD and CD || GF

⇒ F is a mid-point of BC (By converse of mid-point theorem)

Now, in ∆BDC,

G is the mid-point of BD and also GF || AB || DC.

Thus, F is the mid-point of BC (Converse of mid-point theorem)

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Question 5.

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig). Show that the line segments AF and EC trisect the diagonal BD.

Answer:

Given: ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.

To prove: AF and EC trisect the diagonal BD.

Proof: ABCD is a parallelogram.

Therefore, AB || CD, also, AE || FC

Now, AB = CD (Opposite sides of parallelogram ABCD)

⇒ \(\frac{1}{2}\) AB = \(\frac{1}{2}\) CD

⇒ AE = FC (E and F are midpoints of sides AB and CD)

AECF is a parallelogram

(AE and CF are parallel and equal to each other)

AF || EC (Opposite sides of a parallelogram)

Now, in ADQC,

⇒ F is mid-point of side DC and FP || CQ (as AF || EC).

P is the mid-point of DQ (Converse of mid-point theorem)

⇒ DP = PQ …………(i)

Similarly, in AAPB,

E is mid point of side AB and EQ || AP (as AF || EC).

Q is the mid-point of PB (Converse of mid-point theorem)

⇒ PQ = QB ………….. (ii)

From equations (i) and (i),

DP = PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD.

Question 6.

Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Answer:

Let ABCD be a quadrilateral and p, Q, R and S are the mid-points of AB, BC, CD and DA respectively.

Now, In ∆ACD, R and S are the mid-points of CD and DA respectively.

Thus SR || AC and SR = \(\frac{1}{2}\) AC (by mid-point theorem)

Also, in DABC, P and Q are mid-point of AB and BC respectively

⇒ PQ || AC and PQ = \(\frac{1}{2}\) AC (by mid-point theorem)

So, PQ = SR and PQ || SR.

Thus, PQRS is parallelogram, [since one pair of opposite sides is equal and parallel]

PR and QS are the diagonals of the parallelogram PQRS.

So, they will bisect each other. (As diagonals of a parallelogram bisect each other.)

Question 7.

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

(i) D is the mid-point of AC

(ii) MD ⊥ AC

(iii) CM = MA = \(\frac {1}{2}\)AB

Answer:

(i) In ∆ACB

M is the mid-point of AB and MD|| BC

Thus, D is the mid-point of AC (Converse of mid-point theorem)

(ii) ∠ACB = ∠ADM (Corresponding angles)

As ∠ACB = 90°

Thus, ∠ADM = 90° i.e. MD ⊥ AC

(iii) In ∆AMD and ∆CMD,

AD = CD (D is the mid-point of side AC)

∠ADM = ∠CDM (Each 90°)

DM = DM (common)

Thus, ΔADM ≅ ΔCDM by SAS congruence criterion.

AM = CM by CPCT

also, AM = \(\frac{1}{2}\) AB (M is mid point of AB)

Hence, CM = MA = \(\frac{1}{2}\) AB