ML Aggarwal Solution Class 9 Chapter 11 Mid Point Theorem Exercise 11

 Exercise 11

Question 1

(a) In the figure (1) given below, D, E and F are mid-points of the sides BC, CA and AB respectively of ∆ ABC. If AB = 6 cm, BC = 4.8 cm and CA= 5.6 cm, find the perimeter of (i) the trapezium FBCE (ii) the triangle DEF.

(b) In the figure (2) given below, D and E are mid-points of the sides AB and AC respectively. If BC=5.6 cm and∠B = 72°, compute (i) DE (ii)∠ADE.

(c) In the figure (3) given below, D and E are mid-points of AB, BC respectively and DF || BC. Prove that DBEF is a parallelogram. Calculate AC if AF = 2.6 cm.

Sol :

(a) (i) 

Given : AB = 6 cm, BC = 4.8 cm, and CA = 5.6 cm

Required : The perimeter of trapezium FBCA.

∵F is the mid-point of AB (given)
∴BF$=\frac{1}{2}$AB$=\frac{1}{2}\times 6$ cm

=3 cm..(1)

∵E is the mid-point of AC (given)
∴CE$=\frac{1}{2}$AC
$=\frac{1}{2}\times 5.6$ cm=2.8cm...(2)

Now F and E are the mid point of the AB and CA 

∴FE||BC and FE$=\frac{1}{2}\times \text{BC}$
⇒FE$=\frac{1}{2}\times 4.8$ cm=2.4 cm...(3)

∴Perimeter of trapezium FBCE
[Substituting the value from (1), (2) and (3)]

=BF+BC+CE+EF

=3 cm+4.8 cm+2.8 cm+2.4 cm=13 cm

Hence , perimeter of trapezium FBCE=13 cm

(ii) ∵D,E and F are the mid-points of the sides BC, CA and AB of ΔABC respectively

∴EF||BC and EF$=\frac{1}{7}$ BC
⇒EF$=\frac{1}{2}\times 4.8$=2.4 cm

Similarly,

$\mathrm{DE}=\frac{1}{2} \mathrm{AB}=\frac{1}{2} \times 6 \mathrm{~cm}$=3 cm

and FD$=\frac{1}{2} \mathrm{AC}=\frac{1}{2} \times 5.6 \mathrm{~cm}$=2.8 cm

∴Perimeter of ΔDEF

=DE+EF+FD

=3 cm+2.4 cm+2.8 cm

=8.2 cm


(b) Given : D and E are mid-point of the sides AB and AC respectively. BC=5.6 cm and ∠B=72°
Required : (i) DE (ii) ∠ADE

Sol :

In ΔABC

∵D and E are mid-point of the sides AB and AC respectively.

∴By mid point theorem

DE||BC and DE$=\frac{1}{2}$ BC
(i) DE$=\frac{1}{2}$BC $=\frac{1}{2} \times 5.6 \mathrm{~cm}$ =2.8 cm

(ii) ∠ADE=∠B (corresponding angles)
∴ ∠ADE=72° [∵BC||DE]
[∠B=72° (given)]

(c) Given : D and E are mid points of AB and , BC respectively and DF||BC,

AF=2.6 cm

To prove : (i) BEF is a parallelogram

(ii) To calculate the value of AC

Proof : (i) In ΔABC
∵D id the mid point of AB and DF||BC

∴F is the mid point of AC..(1)

Now , F and E are mid points of AC and BC respectively

∴EF||AB..(2)

Now, DF||BC

⇒DF||BE ...(3)

∵EF||AB [From (2)]

⇒EF||DB..(4)

From (3) and (4), DBEF is a parallelogram

(ii) ∵F is mid point of AC

∴AC=2×AF=2×2.6 cm

=5.2 cm

Question 2

Prove that the four triangles formed by joining in pairs the mid-points of the sides c of a triangle are congruent to each other.

Sol :

Given: In ∆ABC, D, E and r,

F are mid-points of AB, BC and CA respectively. Join DE, EF and FD.

To prove : 
⇒∆ADF≅∆DBE≅∆ECF≅∆DEF

Proof : In ∆ABC, D and E are mid point of AB and BC respectively

∴DE||AC or FC
Similarly, DF||EC
∴DECF is a parallelogram

∴Diagonal FE divides the parallelogram DECF in two congruent triangle DEF and CEF

∴∆DEF≅∆FCF...(1)

Similarly we can prove that,
∆DBE≅∆DEF...(2)
and ∆DEF≅∆ADF ...(3)

From (1), (2) and (3)
∆ADF≅∆DBE≅∆ECF≅∆DEF

(Q.E.D)

Question 3

If D, E and F are mid-points of sides AB, BC and CA respectively of an isosceles triangle ABC, prove that ∆DEF is also F, isosceles.

Sol :

Given : ABC is an isosceles triangle in which AB = AC

D, E and F are mid point of the sides BC , CA and AB respectively D,E ,F are joined.

To prove : ΔDEF is an isosceles triangle

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

∴DE||AB and DE$=\frac{1}{2} \mathrm{AB}$ ...(1)

Again , D and F are the mid-points of BC and AB respectively

∴DF||AC and DF$\mathrm{DF}=\frac{1}{2} \mathrm{AC}$...(2)

∵AB=BC (given)

∴DE=DF

∴ΔDEF is an isosceles triangle
(Q.E.D)

Question 4

The diagonals AC and BD of a parallelogram ABCD intersect at O. If P is the mid-point of AD, prove that

(i) PQ || AB

(ii) PO=12CD.

Sol :

(i) Given : 

ABCD is a parallelogram in which diagonals AC and BD intersect each other. At point O, P is the mid-point of AD. Join OP.

To prove : (i) PQ||AB

(ii) $P Q=\frac{1}{2} C D$

Proof : We know that in parallelogram diagonals bisect each other

∴BO=OD

i.e. O is the mid point of BD

Now in ΔABD

P and O is the mid-point of AD and BD respectively 

∴PO||AB and $\mathrm{PO}=\frac{1}{2} \mathrm{AB}$ ..(1)

i.e. PO||AB  [proved (i) part]

(ii) Now 

∵ABCD is a parallelogram

∴AD=CD..(2)

From (1) and (2)

$\mathrm{PO}=\frac{1}{2} \mathrm{CD}$

(Q.E.D)

Question 5

In the adjoining figure, ABCD is a quadrilateral in which P, Q, R and S are mid-points of AB, BC, CD and DA respectively. AC is its diagonal. Show that

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

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Sol :

Given : In quadrilateral ABCD

P,Q,R and S are the mid points of sides AB, BC, CD and DA respectively AC is the diagonal.

To prove :

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

(ii)  PQ=SR

(iii) PQRS is a parallelogram

Proof : 

(i) In ΔADC

S and R are the mid points of AD and DC

∴SR||AC and SR$=\frac{1}{2} \mathrm{AC}$ ..(i)

(Mid points theorem)

(ii) Similarly in ΔABC,

P and Q are mid points of AB and BC

PQ||AC and PQ$=\frac{1}{2} \mathrm{AC}$..(ii)

From (i) and (ii)

PQ=SR and PQ||SR

(iii) ∵PQ=SR and PQ||SR

∴PQRS is a parallelogram

Question 6

Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square,

Sol :

Given : A square ABCD in which E,F,G and H are mid points of AB, BC ,CD and DA respectively. Join EF,FG,GH and HE

To prove : 

EFGH is a square

Construction : Join AC and BD

Proof : In ΔACD, G and H are mid points of CD and AC respectively

∴GH||AC and GH$=\frac{1}{2} \mathrm{AC}$...(1)

Now, in ΔABC, E and F are mid points of AB and BC respectively 

∴EF||AC and EF$=\frac{1}{2} A C$...(2)

From (1) and (2)

EF||GH and EF=GH$=\frac{1}{2} \mathrm{AC}$..(3)

Similarly , we can prove that

EF||GH and EH=GF$=\frac{1}{2}$BD

But AC=BD  (∵Diagonals of square are equal)

Dividing both sides by 2,

$\frac{1}{2} \mathrm{AC}=\frac{1}{2} \mathrm{BD}$...(4)

From (3) and (4)

⇒EF=GH=EH=GF...(5)

∴EFGH is a parallelogram

Now, in ΔGOH and ΔGOF 

⇒OH=OF

(Diagonals of parallelogram bisects each other)

⇒OG=OG (common)
⇒GH=GH [From (5)]

∴ΔGOH≅ΔGOF

[By S.S.S axiom of congruency]

∴∠GOH=∠GOF (c.p.c.t)

Now, ∠GOH=∠GOF=180° (Linear pair)

or ∠GOH+∠GOH=180°

or 2∠GOH

∴∠GOH$=\frac{180^{\circ}}{2}=90^{\circ}$

∴Diagonals of parallelogram ABCD bisects and perpendicular to each other

∴EFGH is a square

(Q.E.D)

Question 7

In the adjoining figure, AD and BE are medians of ∆ABC. If DF U BE, prove that CF $=\frac{1}{4} \mathrm{AC}$

Sol :

Given : In the given figure 

AD an BE are the medians of ΔABC

⇒DF||BE is drawn

To prove : 

$\mathrm{CF}=\frac{1}{4} \mathrm{AC}$

Proof :

In ΔBCE

∴D is the mid point of BC and DF||BE

∴F is the mid points of EC

⇒$\mathrm{CF}=\frac{1}{2} \mathrm{EC}$ ...(i)

∵E is the mid point of AC

∴$\mathrm{EC}=\frac{1}{2} \mathrm{AC}$...(ii)

From (i) and (ii)

⇒$\mathrm{CF}=\frac{1}{2} \mathrm{EC}=\frac{1}{2}\left(\frac{1}{2} \mathrm{AC}\right)$

⇒$=\frac{1}{4} \mathrm{AC}$

Question 8

(a) In the figure (1) given below, ABCD is a parallelogram. E and F are mid-points of the sides AB and CO respectively. The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively. Prove that

(i) ∆HEB = ∆HCF

(ii) GEHF is a parallelogram.\

(b) In the diagram (2) given below, ABCD is a parallelogram. E is mid-point of CD and P is a point on AC such that PC = 14 AC. EP produced meets BC at F. Prove that

(i) F is mid-point of BC (ii) 2EF = BD

Sol :

Given : ABCD is a parallelogram. E and F are mid point of the side AB and CD respectively

To prove : 

(i) ∆HEB ≅ ∆HCF

(ii) GEHF is a parallelogram 

Proof : (i) ABCD is a parallelogram

FC||BE

∴∠CEB=∠FCE  (alternate angles)

⇒∠HEB=∠FCH ..(1)

Also ∠EBF=∠CFB (alternate angles)

⇒∠EBH=∠CFM ..(2)

E and F are mid points if AB and CD respectively

∴$\mathrm{BE}=\frac{1}{2} \mathrm{AB}$..(3)

and $\mathrm{CF}=\frac{1}{2} \mathrm{CD}$...(4)

But ABCD is a parallelogram

∴AB=CD

$\frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{CD}$ (Dividing both sides by $\frac{1}{2}$)

⇒BE=CF [From (3) and (4)]

Now , in ΔHEB and ΔHCF

⇒∠HEB=∠FCH  [From (1)]

⇒∠EBH=∠CFH  [From (2)]

⇒BE=CF  [From (5)]

∴ ∆HEB ≅ ∆HCF

(by A.S.A axiom of congruency)

[(i) part is proved]

(ii) Since E and F are mid points of AB and CD

∴AE=CF  [∵AB=CD]

Now AE||CF (given)

∴AE=CF and AE||CF

∴AECF is a ||gm

Now, G and H points in the AF and CE respectively

∴GF||EH...(6)

Similarly we can prove that GFHE is a ||gm

Now point G and H on the line DE and BF respectively.

∴GE||HF...(7)

From (6) and (7)

GEHF is a parallelogram (Q.E.D)

(b) Given : ABCD is a parallelogram in which E is the mid point of DC and P is a point on AC such that $\mathrm{PC}=\frac{1}{4} \mathrm{AC}$ EP produced meets BC at F

To prove : (i) F is the mid point of BC

(ii) 2EF=BD

Construction : Join BD to intersects AC at O

Proof : Diagonals of parallelogram bisects each other

∴AO=CO

But $\mathrm{CP}=\frac{1}{4} \mathrm{AC}$ (given)

∴ $\mathrm{CP}=\frac{1}{4}(2 \mathrm{CO})$

⇒$\mathrm{CP}=\frac{1}{2} \mathrm{CO}$

i.e. P is mid point of CO

∴In COD, E and P are mid points of DC and CO

∴EP||DO

i.e. EF||DO

Further, in ΔCBD, E is mid point of DC and EF||BD

∴F is the mid point of BC and EF$=\frac{1}{2} \mathrm{BD}$

i.e. 2EF=BD  (Q.E.D)

Question 9

ABC is an isosceles triangle with AB = AC. D, E and F are mid-points of the sides BC, AB and AC respectively. Prove that the line segment AD is perpendicular to EF and is bisected by it.

Sol :

Given : ABC is an isosceles triangle with AB=AC, D , E and F are mid points of the sides BC , AB and AC respectively

To prove : AD⟂EF and AD bisects the EF

Proof : In ΔABD and ΔACD

⇒∠ABD=∠ACD (ABC is an isosceles triangle)

⇒BD=BC (given D is mid point of BC)

⇒AB=AC (given)

∴ΔABD≅ΔACD

(by S.A.S axiom of congruency)

∴∠ADB=∠AOC (c.p.c.t)

Also, ∠ADB+∠AOC=180°  (Linear pair)

⇒∠ADB+∠ADB=180° (by above)

⇒2∠ADB$=\frac{180^{\circ}}{2}$

⇒∠ADB=90°

∴AD⟂BC..(1)

Now D and E are mid points of BC and AB (given)

∴DE||AF ...(2)

Again D and F are mid points of BC and AC 

∴EF||AD..(3)

From (2) and (3)

AEDF is a parallelogram 

∵Diagonals of a parallelogram bisect each other

∴AD and EF bisects each other

From (1) and (3)

⇒AD⊥EF  (EF||BC)

(Q.E.D)

Question 10

(a) In the quadrilateral (1) given below, AB || DC, E and F are mid-points of AD and BD respectively. Prove that:

(i) G is mid point of BC

(ii) EG$=\frac{1}{2}$(AB+DC)

(b) In the quadrilateral (2) given below , AB||DC||EG. If E is mid point of AD prove that

(i) EF if AB=6 cm and DC=4 cm

(ii) AB if DC=8 cm and EF=9 cm

Sol :

(a) Given : AB||DC, E and F are mid points of AD and BD respectively

To prove : 

(i) G is mid point of BC

(ii) EG$=\frac{1}{2}$(AB+DC)

Proof : 

In ΔABD , DF=BF  (∵F is mid point of BD)

Also, E is the mid point of AD  (given)

∵EF||AB and EF$=\frac{1}{2} \mathrm{AB}$...(1)

⇒EG||CD  [AB||CD (given)]

Now F is mid point of BD and FG||DC

∴G is mid point of BC

⇒FG$=\frac{1}{2}$ DC ...(2)

Adding (1) and (2), we get

⇒EF+FG$=\frac{1}{2}$AB+$=\frac{1}{2}$DC

⇒EG$=\frac{1}{2}$(AB+DC)

Hence, the result

(b) Given : Quadrilateral ABCD in which AB||DC||EG. E is mid point of AD

To prove : (i) G is mid point of BC

(ii) 2EG=AB+CD

Proof : ∵AB||DC (given)

and EG||AB (given)

⇒EG||DC

In ΔDAB,

E is mid point of AD and EG||AB  (given)

∴F is the mid point of BD and EF$=\frac{1}{2} \mathrm{AB}$...(1)

In ΔBCD,

F is mid point of BD and FG||DC

⇒FG$=\frac{1}{2} \mathrm{CD}$ ..(2)

Adding (1) and (2)

⇒$\mathrm{EF}+\mathrm{FG}=\frac{1}{2} \mathrm{AB} \frac{1}{2} \mathrm{CD}$

⇒$\mathrm{EG}=\frac{1}{2}(\mathrm{AB}+\mathrm{CD})$

(Q.E.D)

(c) Given : A quadrilateral in which AB||DC, E and F are mid points of non parallel sides AD and BC respectively.

Required : (i) EF if AB=6 cm and DC=4 cm 

(ii) AB if DC=8 cm and EF=9 cm

Now , the length of line segment joining the mid point of two non-parallel sides is half the sum of the lengths of the parallel sides

∵E and F are mid points of AD and BC respectively

∴EF$=\frac{1}{2}$(AB+CD)..(1)

(i) AB=6 cm and DC=4 cm

Putting these in (1), we get

EF$=\frac{1}{2}(6+4)=\frac{1}{2} \times 10$=5 cm

(ii) DC=8 cm and EF=9 cm

Putting these in (1), we get

⇒$\mathrm{EF}=\frac{1}{2}(\mathrm{AB}+\mathrm{DC})$

⇒$9=\frac{1}{2}(\mathrm{AB}+8)$

⇒18=AB+8

⇒18-8=AB

∴AB=10 cm

Question 11

(a) In the quadrilateral (1) given below, AD = BC, P, Q, R and S are mid-points of AB, BD, CD and AC respectively. Prove that PQRS is a rhombus.

(b) In the figure (2) given below, ABCD is a kite in which BC = CD, AB = AD, E, F, G are mid-points of CD, BC and AB respectively. Prove that:

(i) ∠EFG = 90

(ii) The line drawn through G and parallel to FE bisects DA.

Sol :

(a) Given : A quadrilateral ABCD in which AD=BC. P,Q, R and S are mid point of AB, BD , CD and AC respectively

To prove : PQRS is a rhombus

Proof : In ΔABD, P and Q are mid points of AB and BD respectively

∴PQ||AD and PQ (given)

$=\frac{1}{2} \mathrm{AB}$..(1)

Again in ΔBCD, R and Q are mid points of DC and BD respectively (given)

∴RQ||BC and $\mathrm{RQ}=\frac{1}{2} \mathrm{BC}$...(2)

Also, P and S are mid points of AB and AC respectively. (given)

PS||BC and $\mathrm{PS}=\frac{1}{2} \mathrm{BC}$..(3)

∵AD=BC (given)

∴From (1), (2) and (3)

PS||RQ and PQ=PS=RQ

Now, PS||RQ and PS=RQ

⇒PQRS is parallelogram

Further PQ=RS=PS=RQ

⇒PQRS is a rhombus

Hence, the result 

(b) Given : ABCD is a kite in which BC=CD , AB=AD , E , F, G are mid points of CD, BC and AB respectively

To prove : (i) ∠EFG=90°

(ii) The line drawn through

G and parallel to FE bisects DA

Construction : Join AC and BD and Draw GH through G parallel to FE

Proof : (i) Diagonals of a kite intersect at right angle

∴∠MON=90°..(1)

In ΔBCD

E and F are mid points of CD and BC respectively

∴EF||DB and EF$=\frac{1}{2}$DB..(2)

∴EF||DB 

⇒MF||ON

∴∠MON+∠MFN=180°

⇒∠MFN=180°-90°

⇒∠MFN=90°

⇒∠EFG=90° (proved)

(ii) In ΔABD

⇒G is mid point of AB and HG||DB

[From (2), EF||DB and EF||HG (given)]

⇒HG||DB

∴H is mid point of DA

Hence , the line drawn through G and parallel to FE bisects DA

(Q.E.D)

Question 12

In the adjoining figure, the lines l, m and n are parallel to each other, and G is mid-point of CD. Calculate:

(i) BG if AD = 6 cm

(ii) CF if GE = 2.3 cm

(iii) AB if BC = 2.4 cm

(iv) ED if FD = 4.4 cm.

Sol :

Given : The straight line l,m and n are parallel to each other . G is the mid point of CD

To calculate : (i) BG if AD=6cm

(ii) CF if GE=2.3 cm

(iii) AB if BC=2.4 cm

(iv) ED if FD=4.4 cm

(i) In ΔACD,

⇒G is the mid point of CD (given)

and BG||AD (m||n given)

∴B is the mid point of AC and BG$=\frac{1}{2} \mathrm{AD}$

⇒$B G=\frac{1}{2} \times 6 \mathrm{~cm}$=3 cm

(ii) In ΔCDF

G is the mid point of CD and GE||CF (m||l given)

∴E is mid point of DF and GE$=\frac{1}{2}$CF

⇒CF=2GE=2×2.3 cm=4.6cm

(iii) Since B is mid point of AC (In part (i))

∴AB=BC 

⇒AB=2.4 cm  (BC=2.4 cm (given))

(iv) Since E is mid point of FD (In part (ii))

⇒ED$=\frac{1}{2}$FD

∴ED$=\frac{1}{2}\times 4.4 cm$  (FD=4.4 cm (given))

⇒ED=2.2 cm