## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 11 Mid Point Theorem Chapter Test

Question 1.

ABCD is a rhombus with P, Q and R as mid¬points of AB, BC and CD respectively. Prove that PQ ⊥ QR.

Answer:

Given : ABCD is a rhombus with P, Q and R as mid-points of AB, BC and CD respectively.

To Prove: PQ ⊥ QR

Construction : Join AC & BD.

Proof : Diagonals of rhombus intersect at right angle.

∴ ∠MON = 90°

In ∆BCD

Q and R are mid-points of BC and CD respectively.

∴ RQ || DB and RQ = \(\frac{1}{2}\) DB …..(2)

∴ RQ || DB ⇒ MQ || ON

∴ ∠MQN + ∠MON = 180°

⇒ ∠MQN + 90°= 180° ⇒ ∠MQN = 180°-90°

⇒ ∠MQN = 90° ⇒ NQ ⊥ MQ

or PQ ⊥ QR (Q.E.D.)

Question 2.

The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral formed by joining the mid-points of its adjacent sides is a rectangle.

Answer:

Given : ABCD is a quadrilateral in which diagonals AC and BD are perpendicular to each other. P, Q, R and S are mid-points of AB, BC, CD and DA respectively.

To prove: PQRS is a rectangle.

Proof : P and Q are mid-points of AB and BC (given)

∴ PQ || AC and PQ = \(\frac{1}{2}\) AC ….(1)

Again S and R are mid-points of AD and DC (given)

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

From (1) and (2)

PQ || SR and PQ = SR

∴ PQRS is a parallelogram

Further AC and BD intersect at right angles

∴ SP || BD and BD ⊥ AC.

∴ SP ⊥ AC

i.e. SP ⊥ SR

i.e. ∠RSP =90″

∴ ∠RSP = ∠SRQ – ∠RQS = ∠SPQ = 90°

∴ PQRS is a rectangle (Q.E.D.)

Question 3.

If D, E, F are mid-points of the sides BC, CA and AB respectively of a ∆ ABC, Prove that AD and FE bisect each other.

Answer:

Given : D, E, F are mid-points of the sides BC, CA and AB respectively of a ∆ ABC

To Prove: AD and FE bisect each other.

Const: Join ED and FD

Proof : D and E are mid-points of BC and AB respectively (given).

∴ DE || AC ⇒ DE || AF ….(1)

Again D and F are mid-points of BC and AC respectively (given)

∴ DF || AB ⇒ DF || AE …..(2)

From (1) and (2)

ADEF is a ||gm

∵ Diagonals of a ||gm bisect each other

∴ AD and EF bisect each other.

Hence, the result. (Q.E.D.)

Question 4.

In ∆ABC, D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 8 cm and BC = 9 cm, find the perimeter of the parallelogram BDEF.

Answer:

Given. In ∆ABC, D and E are the mid-points of sides AB and AC respectively. DE is joined and from E, EF || AB is drawn AB = 8 cm and BC = 9cm.

To prove.

(i) BDEI is a parallelogram.

(ii) Find the perimeter of BDEF

Proof: In ∆ABC,

∵ B and E are the mid-points of AB and AC respectively

∴ DE || BC and DE = \(\frac{1}{2}\)BC

∵ EF || AB

∴ DEFB is a parallelogram.

∴ DE = BF

∵ DE = \(\frac{1}{2}\) BC = \(\frac{1}{2}\) × 9 = 4·5 cm

EF = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) × 8 = 4 cm.

∴ Perimeter of BDEF = 2 (DE + EF)

=2 (4·5 + 4)

= 8·5 × 2 = 17 cm.

Question 5.

In the given figure, ABCD is a parallelogram and E is mid-point of AD. DL || EB meets AB produced at F. Prove that B is mid-point of AF and EB = LF.

Solution:

Given : In the given figure,

ABCD is a parallelogram

E is mid-point of AD

DL || EB meets AB produced at F

To prove : EB = LF

B is mid-point of AF

Proof: ∵ BC || AD and BE || LD

∴ BEDL is a parallelogram

∴ BE = LDandBL = AE

∵ E is mid-point of AD

∴ L is mid-point of BC

In ∆FAD,

E is mid-point of AD and BE || LD at FLD

∴ B is mid point of AF

∵ EB = \(\frac{1}{2}\) FD = LF

Question 6.

In the given figure, ABCD is a parallelogram. If P and Q are mid-points of sides CD and BC respectively. Show that CR = \(\frac{1}{4}\) AC.

Solution:

Given : In the figure, ABCD is a parallelogram P and Q are the mid-points of sides CD and BC respectively.

To prove : CR = \(\frac{1}{4}\) AC

Construction : Join AC and BD.

Proof: In ||gm ABCD, diagonals AC and BD bisect each other at O

AO = OC or OC = \(\frac{1}{2}\) AC …….(i)

In ∆BCD,

P and Q are mid points of CD and BC

∴ PQ || BD

∵ In ∆BCO,

Q is mid-point of BC and PQ || OB

∴ R is mid-point of CO

∴ CR = \(\frac{1}{2}\) OC = \(\frac{1}{2}\) (\(\frac{1}{2}\)BC)

∴ CR = \(\frac{1}{4}\)BC