Question

In the given figure, A D = B C = 5 c m , A B = 4 c m , C D = 10 c m . What is the length of CF if B is the midpoint of ED and EF ⊥ DF?

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Answer 1

Answer:

(i) It is given that

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

To find: The perimeter of trapezium FBCA

ML Aggarwal Solutions for Class 9 Maths Chapter 11 Image 2

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

ML Aggarwal Solutions for Class 9 Maths Chapter 11 Image 3

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

Answer 2

Therefore the length of CF is 3.856 cm.

Given:

AD = BC = 5cm

AB = 4 cm

CD = 10 cm

B is the midpoint of ED and EF ⊥ DF

To Find:

The length of CF.

Solution:

The given question can be simply solved in the following way.

Let a perpendicular be dropped on to CD from point B and the point where it touches CD named as X.

Similarly  a perpendicular be dropped on to CD from point A and the point where it touches CD named as Y.

Due to symmetry ( AD = BC = 5cm ) in the given figure,

⇒ DY + YX + XC = CD

⇒ DY + AB + XC = CD     ( ∵ AB = XY )

⇒ 2XC + AB = CD            ( ∵ DY = XC due to symmetry )

⇒ 2XC + 4 = 10

⇒ 2XC = 6 ⇒  XC = 3cm

Now in ΔCBX,

By pythagoras theorem, BC² = XC² + BX²

⇒ 5² = 3² + BX²

⇒ BX = 4 cm

Now in ΔDBX,

By pythagoras theorem,

⇒ BD² = BX² + DX²

⇒ BD² = 4² + 7²     ( ∵ DX = CD - XC = 10 - 3 )

⇒ BD² = 16 + 49 = 65

⇒ BD = √65 = 8.06 ≅ 8 cm

As 'B' is the midpoint of ED, DB = BE = 8 cm

In ΔDBX and ΔDEF,

According to Proportionality rule ,

⇒ BX / BD = EF / DE

⇒ 4 / 8 = EF / 16       ( ∵ DE = DB + BE = 8 + 8 = 16 cm )

⇒ EF = 8

Now in ΔDEF,

By pythagoras theorem,

⇒ DE² = EF² + DF²

⇒ 16² = 8² + DF²

⇒ DF² = 192

⇒ DF = √192 ⇒ DF = 13.856 cm

Now CF = DF - DC = 13.856 - 10 = 3.856

Therefore the length of CF is 3.856 cm.

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