ML Aggarwal Solution Class 9 Chapter 11 Mid Point Theorem TEST

  TEST

Question 1

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

Sol :
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 and BD
Proof : Diagonals of rhombus intersects at right angle






∴∠MON=90°

In ΔBCD

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.

Sol :









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 intersects at right angles.

∴SP||BD and BD⊥AC

∴SP⊥SR

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.

Sol :
Given : D , E, F are mid points of the sides BC, CA and AB respectively of a ΔABC

To prove : AD and FE bisects each other

Construction : Join ED and FD









Proof : D and E are mid points of BC and AC 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 parallelogram 

∵Diagonals of a parallelogram bisect each other

∴AD and EF bisects 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.

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








To prove :

(i) BDEI is a parallelogram

(ii) Find the perimeter of BDEF

Proof : In ΔABC

∵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$=\frac{1}{2}\times 9$=4.5 cm

EF$=\frac{1}{2}$AB$=\frac{1}{2}\times 8$=4 cm

∴Perimeter of BDEF=2(DE+EF)

=2(4.5+4)

=8.5×2=17


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.








Sol :

Given In the 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=LD and BL=AE
∵E is mid point of AD
∴L is mid point of BC

In ΔFAD

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 $=\frac{1}{2}$AC






Sol :

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 parallelogram ABCD, diagonals AC and BD bisects each other at O

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

In ΔBCD

P and Q are mid points of CD and BC 

∴PQ||BD

∵In ΔBCO

Q is mid point of BC and PQ||OB

∴R is mid point of CO

∴$CR=\frac{1}{2}OC$

$=\frac{1}{2}\left(\frac{1}{2}BC\right)$

∴$CR=\frac{1}{4}BC$

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