ML Aggarwal Solutions for Class 9 Maths Chapter 11 Mid Point Theorem is an important resource for students of the ICSE board. The solutions help students obtain a clear idea about the sort of questions that would be asked in the exam and the method of answering them. The experts at BYJU’S provide ML Aggarwal Solutions for Class 9 Maths Chapter 11 Mid Point Theorem PDF to help students with their exam preparation.
Chapter 11 has problems on triangles which can be solved by using the Mid Point Theorem. The PDF of ML Aggarwal Solutions can be referred to by students to clear their doubts during the self-study process. The main aim of providing solutions chapter-wise is to help students with their exam preparation based on the current syllabus of the ICSE board.
ML Aggarwal Solutions for Class 9 Maths Chapter 11: Mid Point Theorem
Access ML Aggarwal Solutions for Class 9 Maths Chapter 11: Mid Point Theorem
Exercise 11
1. (a) In the figure (1) given below, D, E and F are mid-points of the sides BC, CA and AB respectively of Δ 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 = 720, 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.
Solution:
(a) (i) It is given that
AB = 6 cm, BC = 4.8 cm and CA = 5.6 cm
To find: The perimeter of trapezium FBCA
It is given that
F is the mid-point of AB
We know that
BF = ½ AB = ½ × 6 = 3 cm ……. (1)
It is given that
E is the mid-point of AC
We know that
CE = ½ AC = ½ × 5.6 = 2.8 cm ……. (2)
Here F and E are the mid-point of AB and CA
FE || BC
We know that
FE = ½ BC = ½ × 4.8 = 2.4 cm …… (3)
Here
Perimeter of trapezium FBCE = BF + BC + CE + EF
Now substituting the value from all the equations
= 3 + 4.8 + 2.8 + 2.4
= 13 cm
Therefore, the perimeter of trapezium FBCE is 13 cm.
(ii) D, E and F are the midpoints of sides BC, CA and AB of Δ ABC
Here EF || BC
EF = ½ BC = ½ × 4.8 = 2.4 cm
DE = ½ AB = ½ × 6 = 3 cm
FD = ½ AC = ½ × 5.6 = 2.8 cm
We know that
Perimeter of Δ DEF = DE + EF + FD
Substituting the values
= 3 + 2.4 + 2.8
= 8.2 cm
(b) It is given that
D and E are the mid-point of sides AB and AC
BC = 5.6 cm and ∠B = 720
To find: (i) DE (ii) ∠ADE
We know that
In Δ ABC
D and E is the mid-point of the sides AB and AC
Using mid-point theorem
DE || BC
(i) DE = ½ BC = ½ × 5.6 = 2.8 cm
(ii) ∠ADE = ∠B are corresponding angles
It is given that
∠B = 720 and BC || DE
∠ADE = 720
(c) It is given that
D and E are the midpoints of AB and BC respectively
DF || BC and AF = 2.6 cm
To find: (i) BEF is a parallelogram
(ii) Calculate the value of AC
Proof:
(i) In Δ ABC
D is the midpoint of AB and DF || BC
F is the midpoint of AC ….. (1)
F and E are the midpoints of AC and BC
EF || AB ….. (2)
Here DF || BC
DF || BE ….. (3)
Using equation (2)
EF || AB
EF || DB ….. (4)
Using equation (3) and (4)
DBEF is a parallelogram
(ii) F is the midpoint of AC
So we get
AC = 2 × AF = 2 × 2.6 = 5.2 cm
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.
Solution:
It is given that
In Δ ABC
D, E and F are the mid-points of AB, BC and CA
Now join DE, EF and FD
To find:
Δ ADF ≅ Δ DBE ≅ Δ ECF ≅ Δ DEF
To prove:
In Δ ABC
D and E are the mid-points of AB and BC
DE || AC or FC
Similarly DF || EC
DECF is a parallelogram
We know that
Diagonal FE divides the parallelogram DECF in two congruent triangles DEF and CEF
Δ DEF ≅ Δ ECF …… (1)
Here we can prove that
Δ DBE ≅ Δ DEF …. (2)
Δ DEF ≅ Δ ADF ……. (3)
Using equation (1), (2) and (3)
Δ ADF ≅ Δ DBE ≅ Δ ECF ≅ Δ DEF
3. If D, E and F are mid-points of sides AB, BC and CA respectively of an isosceles triangle ABC, prove that Δ DEF is also isosceles.
Solution:
It is given that
ABC is an isosceles triangle in which AB = AC
D, E and F are the midpoints of the sides BC, CA and AB
Now D, E and F are joined
To find:
Δ DEF is an isosceles triangle
Proof:
D and E are the midpoints of BC and AC
Here DE || AB and DE = ½ AB ….. (1)
D and F are the midpoints of BC and AB
Here DF || AC and DF = ½ AC ….. (2)
It is given that
AB = BC and DE = DF
Hence, Δ DEF is an isosceles triangle.
4. The diagonals AC and BD of a parallelogram ABCD intersect at O. If P is the midpoint of AD, prove that
(i) PQ || AB
(ii) PO = ½ CD.
Solution:
It is given that
ABCD is a parallelogram in which diagonals AC and BD intersect each other
At the point O, P is the midpoint of AD
Join OP
To find: (i) PQ || AB (ii) PQ = ½ CD
Proof:
(i) In parallelogram diagonals bisect each other
BO = OD
Here O is the mid-point of BD
In Δ ABD
P and O is the midpoint of AD and BD
PO || AB and PO = ½ AB ….. (1)
Hence, it is proved that PO || AB.
(ii) ABCD is a parallelogram
AB = CD ……. (2)
Using both (1) and (2)
PO = ½ CD
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 = ½ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Solution:
It is given that
In quadrilateral ABCD
P, Q, R and S are the mid-points of sides AB, BC, CD and DA
AC is the diagonal
To find:
(i) SR || AC and SR = ½ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram
Proof:
(i) In Δ ADC
S and R are the mid-points of AD and DC
SR || AC and SR = ½ AC….. (1) Using the mid-point theorem
(ii) In Δ ABC
P and Q are the midpoints of AB and BC
PQ || AC and PQ = ½ AC ….. (2)
Using equation (1) and (2)
PQ = SR and PQ || SR
(iii) PQ = SR and PQ || SR
Hence, PQRS is a parallelogram.
6. Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square.
Solution:
It is given that
A square ABCD in which E, F, G and H are mid-points of AB, BC, CD and DA
Join EF, FG, GH and HE.
To find:
EFGH is a square
Construct AC and BD
Proof:
In Δ ACD
G and H are the mid-points of CD and AC
GH || AC and GH = ½ AC ….. (1)
In Δ ABC, E and F are the mid-points of AB and BC
EF || AC and EF = ½ AC …… (2)
Using both the equations
EF || GH and EF = GH = ½ AC ….. (3)
In the same way we can prove that
EF || GH and EH = GF = ½ BD
We know that the diagonals of square are equal
AC = BD
By dividing both sides by 2
½ AC = ½ BD ….. (4)
Using equation (3) and (4)
EF = GH = EH = GF …. (5)
Therefore, EFGH is a parallelogram
In Δ GOH and Δ GOF
OH = OF as the diagonals of parallelogram bisect each other
OG = OG is common
Using equation (5)
GH = GF
Δ GOH ≅ Δ GOF (SSS axiom of congruency)
∠GOH = ∠GOF (c.p.c.t)
We know that
∠GOH + ∠GOF = 1800 as it is a linear pair
∠GOH + ∠GOH = 1800
So we get
2 ∠GOH = 1800
∠GOH = 1800/2 = 900
So the diagonals of a parallelogram ABCD bisect and perpendicular to each other
Hence, it is proved that EFGH is a square.
7. In the adjoining figure, AD and BE are medians of Δ ABC. If DF || BE, prove that CF = ¼ AC.
Solution:
It is given that
AD and BE are the medians of Δ ABC
Construct DF || BE
To find:
CF = ¼ AC
Proof:
In Δ BCE
D is the mid-point of BC and DF || BE
F is the mid-point of EC
CF = ½ EC …… (1)
E is the mid-point of AC
EC = ½ AC ….. (2)
Using both the equations
CF = ½ EC = ½ (½ AC)
So we get
CF = ¼ AC
Hence, it is proved.
8. 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) Δ HEB = Δ HCF
(ii) GEHF is a parallelogram.
Solution:
It is given that
ABCD is a parallelogram
E and F are the mid-points of sides AB and CD
To prove:
(i) Δ HEB = Δ HCF
(ii) GEHF is a parallelogram
Proof:
(i) We know that
ABCD is a parallelogram
FC || BE
∠CEB = ∠FCE are alternate angles
∠HEB = ∠FCH ….. (1)
∠EBF = ∠CFB are alternate angles
∠EBH = ∠CFM …… (2)
Here E and F are mid-points of AB and CD
BE = ½ AB ….. (3)
CF = ½ CD …… (4)
We know that
ABCD is a parallelogram
AB = CD
Now dividing both sides by ½
½ AB = ½ CD
Using equations (3) and (4)
BE = CF ….. (5)
In Δ HEB and Δ HCF
∠HEB = ∠FCH using equation (1)
∠EBH = ∠CFH using equation (2)
BE = CF using equation (5)
So we get
Δ HEB ≅ Δ HCF (ASA axiom of congruency)
Hence, it is proved.
(ii) It is given that
E and F are the mid-points of AB and CD
AB = CD
So we get
AE = CF
Here AE || CF
AE = CF and AE || CF
So AECF is a parallelogram.
G and H are the mid-points of AF and CE
GF || EH ……. (6)
In the same way we can prove that GFHE is a parallelogram
So G and H are the points on the line DE and BF
GE || HF …….. (7)
Using equations (6) and (7) GEHF is a parallelogram.
Hence, it is proved.
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.
Solution:
It is given that
ABC is an isosceles triangle with AB = AC
D, E and F are mid-points of the sides BC, AB and AC
To find:
AD is perpendicular to EF and is bisected by it.
Proof:
In Δ ABD and Δ ACD
ABC is an isosceles triangle
∠ABD = ∠ACD
Here D is the mid-point of BC
BD = BD
It is given that AB = AC
Δ ABD ≅ Δ ACD (SAS axiom of congruency)
∠ADB = ∠AOC (c. p. c. t)
We know that
∠ADB + ∠AOC = 1800 is a linear pair
∠ADB + ∠ADB = 1800
By further calculation
2 ∠ADB = 1800
So we get
∠ADB = 180/2 = 900
So AD is perpendicular to BC ….. (1)
D and E are the mid-points of BC and AB
DE || AF …… (2)
D and F are the mid-points of BC and AC
EF || AD …. (3)
Using equation (2) and (3)
AEDF is a parallelogram.
Here the diagonals of a parallelogram bisect each other
AD and EF bisect each other
Using equations (1) and (3)
EF || BC
So AD is perpendicular to EF
Hence, it is proved.
10. (a) In the quadrilateral (1) given below, AB || DC, E and F are the mid-points of AD and BD respectively. Prove that:
(i) G is mid-point of BC
(ii) EG = ½ (AB + DC)
(b) In the quadrilateral (2) given below, AB || DC || EG. If E is mid-point of AD prove that:
(i) G is the mid-point of BC
(ii) 2EG = AB + CD
(c) In the quadrilateral (3) given below, AB || DC.
E and F are mid-point of non-parallel sides AD and BC respectively. Calculate:
(i) EF if AB = 6 cm and DC = 4 cm.
(ii) AB if DC = 8 cm and EF = 9 cm.
Solution:
(a) It is given that
AB || DC, E and F are mid-points of AD and BD
To prove:
(i) G is the mid-point of BC
(ii) EG = ½ (AB + DC)
Proof:
In Δ ABD
F is the mid-point of BD
DF = BF
E is the mid-point of AD
EF || AB and EF = ½ AB ….. (1)
It is given that AB || CD
EG || CD
F is the mid-point of BD
FG || DC
G is the mid-point of BC
FG = ½ DC …….. (2)
By adding both the equations
EF + FG = ½ AB + ½ DC
Taking ½ as common
EG = ½ (AB + DC)
Therefore, it is proved.
(b) It is given that
Quadrilateral ABCD in which AB || DC || EG
E is the mid-point of AD
To prove:
(i) G is the mid-point of BC
(ii) 2EG = AB + CD
Proof:
AB || DC
EG || AB
So we get
EG || DC
In Δ DAB,
E is the mid-point of BD and EF = ½ AB ….. (1)
In Δ BCD,
F is the mid-point of BD and FG || DC
FG = ½ CD …… (2)
By adding both the equations
EF + FG = ½ AB + ½ CD
Taking out the common terms
EG = ½ (AB + CD)
Hence, it is proved.
(c) It is given that
A quadrilateral in which AB || DC
E and F are the mid-points of non-parallel sides AD and BC
To prove:
(i) EF if AB = 6 cm and DC = 4 cm.
(ii) AB if DC = 8 cm and EF = 9 cm.
Proof:
We know that
The length of line segment joining the mid-points of two non-parallel sides is half the sum of the lengths of the parallel sides
E and F are the mid-points of AD and BC
EF = ½ (AB + CD) …… (1)
(i) AB = 6 cm and DC = 4 cm
Substituting in equation (1)
EF = ½ (6 + 4)
By further calculation
EF = ½ × 10 = 5 cm
(ii) DC = 8 cm and EF = 9 cm
Substituting in equation (1)
EF = ½ (AB + DC)
By further calculation
9 = ½ (AB + 8)
18 = AB + 8
So we get
18 – 8 = AB
AB = 10 cm
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 = 900
(ii) The line drawn through G and parallel to FE bisects DA.
Solution:
(a) It is given that
A quadrilateral ABCD in which AD = C
P, Q, R and S are mid-points of AB, BD, CD and AC
To prove:
PQRS is a rhombus
Proof:
In Δ ABD
P and Q are midpoints of AB and BD
PQ || AD and PQ = ½ AB …. (1)
In Δ BCD,
R and Q are midpoints of DC and BD
RQ || BC and RQ = ½ BC …… (2)
P and S are mid-points of AB and AC
PS || BC and PS = ½ BC ……. (3)
AD = BC
Using all the equations
PS || RQ and PQ = PS = RQ
Here PS || RQ and PS = RQ
PQRS is a parallelogram
PQ = RS = PS = RQ
PQRS is a rhombus
Therefore, it is proved.
(ii) It is given that
ABCD is a kite in which BC = CD, AB = AD, E, F, G are mid-points of CD, BC and AB
To prove:
(i) ∠EFG = 900
(ii) The line drawn through G and parallel to FE bisects DA
Construction:
Join AC and BD
Construct GH through G parallel to FE
Proof:
(i) We know that
Diagonals of a kite interest at right angles
∠MON = 900 ….. (1)
In Δ BCD
E and F are mid-points of CD and BC
EF || DB and EF = ½ DB …… (2)
EF || DB
MF || ON
Here
∠MON + ∠MFN = 1800
900 + ∠MFN = 1800
By further calculation
∠MFN = 180 – 90 = 900
So ∠EFG = 900
Hence, it is proved.
(ii) In Δ ABD
G is the mid-point of AB and HG || DB
Using equation (2)
EF || DB and EF || HG
HG || DB
Here H is the mid-point of DA
Therefore, the line drawn through G and parallel to FE bisects DA.
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
Solution:
It is given that
The straight lines l, m and n are parallel to each other
G is the mid-point of CD
To find:
(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
Proof:
(i) In Δ ACD,
G is the mid-point of CD
BG || AD as m || n
Here B is the mid-point of AC and BG = ½ AD
So we get
BG = ½ × 6 = 3 cm
(ii) In Δ CDF
G is the mid-point of CD
GE || CF as m || l
Here E is the mid-point of DF and GE = ½ CF
So we get
CF = 2GE
CF = 2 × 2.3 = 4.6 cm
(iii) From (i)
B is the mid-point of AC
AB = BC
We know that
BC = 2.4 cm
So AB = 2.4 cm
(iv) From (ii)
E is the mid-point of FD
ED = ½ FD
We know that
FD = 4.4 cm
ED = ½ × 4.4 = 2.2 cm
Chapter Test
1. ABCD is a rhombus with P, Q and R as midpoints of AB, BC and CD respectively. Prove that PQ ⊥QR.
Solution:
It is given that
ABCD is a rhombus with P, Q and R as mid-points of AB, BC and CD
To prove:
PQ ⊥QR
Construction: Join AC and BD
Proof:
Diagonals of a rhombus intersect at a right angle
∠MON = 900 ….. (1)
In Δ BCD
Q and R are mid-points of BC and CD.
RQ || DB and RQ = ½ DB ….. (2)
Here
RQ || DB
MQ || ON
We know that
∠MQN + ∠MON = 1800
Substituting the values
∠MQN + 900 = 1800
∠MQN = 180 – 90 = 900
So NQ ⊥MQ or PQ ⊥QR
Hence, it is proved.
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.
Solution:
It is given that
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
To prove:
PQRS is a rectangle
Proof:
We know that
P and Q are the mid-points of AB and BC
PQ || AC and PQ = ½ AC ….. (1)
S and R are mid-points of AD and DC
SR || AC and SR = ½ AC ….. (2)
Using both the equations
PQ || SR and PQ = SR
So PQRS is a parallelogram
AC and BD intersect at right angles
SP || BD and BD ⊥AC
So SP ⊥ AC, i.e. SP ⊥SR
∠RSP = 900
∠RSP = ∠SRQ = ∠RQS = ∠SPQ = 900
Hence, PQRS is a rectangle.
3. If D, E, F are mid-points of the sides BC, CA and AB respectively of a Δ ABC, prove that AD and FE bisect each other.
Solution:
It is given that
D, E, F are mid-points of sides BC, CA and AB of a Δ ABC
To prove:
AD and FE bisect each other
Construction:
Join ED and FD
Proof:
We know that
D and E are the midpoints of BC and AB
DE || AC and DE || AF ….. (1)
D and F are the midpoints of BC and AC
DF || AB and DF || AE …. (2)
Using both the equations
ADEF is a parallelogram
Here the diagonals of a parallelogram bisect each other
AD and EF bisect each other.
Therefore, it is proved.
4. In Δ 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.
Solution:
It is given that
In Δ ABC
D and E are the midpoints of sides AB and AC
DE is joined from 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 Δ ABC
B and E are the mid-points of AB and AC
Here DE || BC and DE = ½ BC
So EF || AB
DEFB is a parallelogram
DE = BF
So we get
DE = ½ BC = ½ × 9 = 4.5 cm
EF = ½ AB = ½ × 8 = 4 cm
We know that
Perimeter of BDEF = 2 (DE + EF)
Substituting the values
= 2 (4.5 + 4)
= 2 × 8.5
= 17 cm
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:
It is given that
ABCD is a parallelogram
E is the mid-point of AD
DL || EB meets AB produced at F
To prove:
EB = LF
B is the mid-point of AF
Proof:
We know that
BC || AD and BE || LD
BEDL is a parallelogram
BE = LD and BL = AE
Here E is the mid-point of AD
L is the mid-point of BC
In Δ FAD
E is the mid-point of AD and BE || LD at FLD
So B is the mid-point of AF
Here
EB = ½ FD = LF
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 = ½ AC.
Solution:
It is given that
ABCD is a parallelogram
P and Q are mid-points of CD and BC
To prove: CR = ¼ AC
Construction: Join AC and BD
Proof:
In parallelogram ABCD
Diagonals AC and BD bisect each other at O
AO = OC or OC = ½ AC ….. (1)
In Δ BCD
P and Q are midpoints of CD and BC
PQ || BD
In Δ BCO
Q is the mid-point of BC and PQ || OB
Here is the mid-point of CO
So we get
CR = ½ OC = ½ (½ BC)
CR = ¼ BC
Hence, it is proved.
