Mensuration
- The diameter of two circles are the side of a square and the diagonal of the square. The ratio of the area of the smaller circle and the larger circle is
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Using Rule 10 and 14,
Let Side of square = x units Diagonal of square = √2x units
then Radius of smaller circle = x/2 units Radius of larger circle= √2x = x units 2 √2
∴ Required ratio of areas= π x² : πx² 4 2
= 2 : 4 = 1 : 2Correct Option: A
Using Rule 10 and 14,
Let Side of square = x units Diagonal of square = √2x units
then Radius of smaller circle = x/2 units Radius of larger circle= √2x = x units 2 √2
∴ Required ratio of areas= π x² : πx² 4 2
= 2 : 4 = 1 : 2
- Three circles of equal radius ‘a’ cm touch each other. The area of the shaded region is :
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Using Rule 6 and 17,
Let AB = BC = CA = 2a cm.
∠BAC = ∠ACB = ∠ABC = 60°Area of ∆ABC = √3 × (side)² 4 = √3 × 4a² 4
= √3a² sq.cm.Area of three sectors = 3 × 60 × π × a² 360 = πa² × sq.cm. 2 Area of the shaded region = √3a² - π × sq.cm. 2 = 
2√3 - π 
a² sq.cm. 2 Correct Option: D
Using Rule 6 and 17,
Let AB = BC = CA = 2a cm.
∠BAC = ∠ACB = ∠ABC = 60°Area of ∆ABC = √3 × (side)² 4 = √3 × 4a² 4
= √3a² sq.cm.Area of three sectors = 3 × 60 × π × a² 360 = πa² × sq.cm. 2 Area of the shaded region = √3a² - π × sq.cm. 2 = 2√3 - π a² sq.cm. 2
- The radii of two circles are 10 cm and 24 cm. The radius of a circle whose area is the sum of the area of these two circles is
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Using Rule 14,
Let the required radius = r cm, then
πr² = πr1² + πr2²
→ r² = r1² + r2²
= 10² + 24²
= 100 + 576 = 676
⇒ r = 676 = 26 cmCorrect Option: D
Using Rule 14,
Let the required radius = r cm, then
πr² = πr1² + πr2²
→ r² = r1² + r2²
= 10² + 24²
= 100 + 576 = 676
⇒ r = 676 = 26 cm
- The area of the greatest circle inscribed inside a square of side 21 cm is (Take π = 22 7 )
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Using Rule 14,
The diameter of the greatest circle inscribed inside a square will be equal to the side of square i.e., 21 cm.
∴ Radius of the circle = 21/2
∴ Area of the circle = π × (radius)²= 22 × 21 × 21 = 693 cm² 7 2 2 2
= 346.5 cm².Correct Option: C
Using Rule 14,
The diameter of the greatest circle inscribed inside a square will be equal to the side of square i.e., 21 cm.
∴ Radius of the circle = 21/2
∴ Area of the circle = π × (radius)²= 22 × 21 × 21 = 693 cm² 7 2 2 2
= 346.5 cm².
- The area of the greatest circle, which can be inscribed in a square whose perimeter is 120 cm, is :
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Using Rule 14,
Side of the square = 120 = 30 cm. 4
Clearly, diameter of the greatest circle = Side of the square = 30 cm∴ Radius = 30 = 15 cm. 2
Required area = π × (radius)²= 22 × (15)² cm² 2 Correct Option: A
Using Rule 14,
Side of the square = 120 = 30 cm. 4
Clearly, diameter of the greatest circle = Side of the square = 30 cm∴ Radius = 30 = 15 cm. 2
Required area = π × (radius)²= 22 × (15)² cm² 2
