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ML Aggarwal Solutions Class 9 Mathematics Solutions for Mid Point Theorem Exercise 11 in Chapter 11 - Mid Point Theorem
(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 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^{0}
compute (i) DE (ii) ∠ADE.
(c) In 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.
(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 = 72^{0}
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 =
72^{0}
and BC || DE
∠ADE =
72^{0}
(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
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