Question

ABCD is a rombus with area 32root3cm²
find the area of shaded region or major section
O is center of circle ​

Answer 1

Gɪᴠᴇɴ :-

• Area of Rhombus = 32√3cm².
• AO , CO sides & Diagonal BO are radius of given circle .

Tᴏ Fɪɴᴅ :-

• Area of shaded Region ?

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• Rhombus diagonals bisect each other at 90°.
• Area of Rhombus = (1/2) * Diagonal₁ * Diagonal₂ .
• Area of circle = π * (radius)².

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume that, radius of circle is r cm.

So,

→ BO = AO = CO = r cm.

Now, Let the diagonals OB & AC intersect at D.

Than,

→ OD = DB = r/2

→ AD = CD .

So, In right ΔODA, By Pythagoras theorem, we get :-

→ OA² = OD² + AD²

→ r² = (r/2)² + AD²

→ AD² = r² - (r/2)²

→ AD² = 3r²/4

→ AD = (√3r/2)

Than,

→ AC = 2AD = √3r

Therefore,

→ Area of Rhombus = 1/2 * D₁ * D₂ = 1/2 * OD * AC

→ 32√3 = 1/2 * r * √3r

→ 32 = (1/2) * r²

→ r² = 64

→ r = 8cm.

Hence,

→ Area of given Circle = πr² = 3.14 * 64 = 200.96cm.²

And,

→ Area of Rhombus = 32√3 = 32 * 1.73 = 55.36cm².

So,

→ Shaded Area = Area of circle - Area of Rhombus

→ Shaded Area = 200.96 - 55.36 = 145.6cm². (Ans.)

Answer 2

$\bigstar$ Answer :

• We get shaded region by subtracting total circle area from rhombus area.

Let us assume that radius of the circle is " r " cm

Let the diagonals DB & AC intersect at O.

⇒ OD = OB = r/2

& OA = OC

Now in Δ AOD , by using Pythagoras Theorem ,

AD² = OA² + OD²

⇒ r² = (OA)² + (r/2)²

⇒ r² - (r/2)² = (OA)²

⇒ 3r²/4 = (OA)²

⇒ OA = √3 r/2

AC = 2 (OA)= √3 r

So , AC = √3 r , BD = r

So , we find the two diagonals of rhombus .

• Area of Rhombus = 1/2*d1*d2

⇒ Area of rhombus = 32√3

⇒ 1/2*√3 r*r = 32√3

⇒  r² = 64

⇒ r = 8 cm

So ,

• Area of circle = πr²

⇒ Area of circle = 64π cm² ... (1)

Area of Rhombus = 32√3 cm² ... (2)

• Area of shaded region = Area of circle - Area of Rhombus

⇒ Area of shaded region = 64π - 32√3

⇒ Area of shaded region = 200.96 - 55.36

⇒ Area of shaded region = 145.6 cm²