Question

Question for Only:- ❏ Moderators ❏ Brainly Stars ❏ Best users ✰ABCD is a trapezium ab parallel to DC BFEC is the sector of circle with centre c.AB= BC is equal to 7 cm ,DEis equal to 4 cm find the area of shaded region​

Answer 1

Answer:

28,9 cm²

Step-by-step explanation:

BC and EB, being the radius of the circle, are equal.  

        ∴ EB = BC = 7 cm     [given]

Draw ⊥ from B on EC  (height of the trapezium).

  Using trigonometry,

                   height = BC.sin60°

                   height = 7 * (√3/2)

                               = 3.5√5 cm

∴ Area of trapezium:

               ⇒ 1/2 * (sum of || sides) * h

               ⇒ 1/2 * (AB + DC) * h

               ⇒ 1/2 * (AB + DE + EC) * h

               ⇒ 1/2 * (7 + 4 + 7)*3.5√3

               ⇒ 31.5√3  cm²

∴ Area of circle(sector):

              ⇒ πr²(θ/360)  

              ⇒ π(BC)² (60/360)

              ⇒ (22/7) (7)² (1/6)

              ⇒ 25.6 cm²

Observing,

  Shaded area = area of trapezium - area of circle

 ⇒ Shaded area = 31.5√3 - 25.6 cm²

                     = (31.5 * 1.73) - 25.6 cm²

                     = 28.9 cm²

Answer 2

Required Question:-

In the figure,ABCD is a trapezium with AB||CD and ∠BCD = 60°. If BFEC is a sector of a circle with

centre C and AB = BC 7cm and DE = 4cm,then find the area of the shaded figure.

Given:-

• Trapezium ABCD with AB||CD,
• ∠BCD = 60°,
• AB = BC = 7cm, and
• DE = 4cm.

To Find:-

• Area of shaded region.

Solution:-

Draw BL⊥CD.

BC = CE = 7cm (Radii of the sector)

So,DC = CE + DE = 7 + 4 = 11cm.

In right △ BLC,

$Sin60° = \frac{BL}{BC} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: (∵sin\theta = \frac{P}{H} )$

$= > \frac{ \sqrt{3} }{2} = \frac{BL}{7}$

$\therefore BL = \frac{7 \sqrt{3} }{2} = \frac{7 \times 1.73}{2} = 6.05cm$

Area of trapezium ABCD

$\: \: \: \: \: \: \: = \frac{1}{2} \times ( AB + CD) \times BL$

$= \frac{1}{2} \times (7 + 11) \times 6.05$

$= 9 \times 6.05 = 54.45cm {}^{2}$

Area of sector BCEFB

$\: \: \: \: \: \: \: = \frac{\theta}{360°} \times\pi r {}^{2}$

$= \frac{60°}{360°} \times \pi \times 7 {}^{2}$

$= \frac{1}{6} \times \frac{22}{7} \times 49$

$= 25.66cm {}^{2}$

Hence,Area of the shaded region

= Area of trapezium – Area of sector

$= (54.45 – 25.66)cm {}^{2}$

Answer:- = 28.79cm².