Mensuration
- If the difference between areas of the circumcircle and the incircle of an equilateral triangle is 44 cm2, then the area of the triangle is (Take π = 22/7)
-
View Hint View Answer Discuss in Forum
Let the each side of the equilateral triangle be 2x cm.
Then BD = xRadius of incircle = OD = 1 AD 3 = 1 √(2x)² - x² 3 = √3x = x cm 3 √3
Radius of circum circle = BO = √BD² + OD²= √x² + x² = 2x cm 3 √3
According to the question,π 
2x 
² - π x ² = 44 √3 √3 ⇒ 4πx² - πx² = 44 3 3
⇒ πx² = 44⇒ x² = 44 × 7 = 14 22
∴ Area of the equilateral triangle= √3 × side² = √3 ×(2x)² 4 4
= √3 x² = 14√3 Sq. cm.Correct Option: C
Let the each side of the equilateral triangle be 2x cm.
Then BD = xRadius of incircle = OD = 1 AD 3 = 1 √(2x)² - x² 3 = √3x = x cm 3 √3
Radius of circum circle = BO = √BD² + OD²= √x² + x² = 2x cm 3 √3
According to the question,π 2x ² - π x ² = 44 √3 √3 ⇒ 4πx² - πx² = 44 3 3
⇒ πx² = 44⇒ x² = 44 × 7 = 14 22
∴ Area of the equilateral triangle= √3 × side² = √3 ×(2x)² 4 4
= √3 x² = 14√3 Sq. cm.
- If the area of a circle inscribed in a square is 9π cm²,then the area of the square is
-
View Hint View Answer Discuss in Forum
Using Rule 10 and 14,
∴ Area of the circle = πr² = 9π
⇒ r² = 9
⇒ r = √9 = 3 cm
∴ Side of the square = 2r = 6 cm
∴ Area of the Square = side² = 6 × 6 = 36 cm²Correct Option: C
Using Rule 10 and 14,
∴ Area of the circle = πr² = 9π
⇒ r² = 9
⇒ r = √9 = 3 cm
∴ Side of the square = 2r = 6 cm
∴ Area of the Square = side² = 6 × 6 = 36 cm²
- The sides of a triangle are 6 cm, 8 cm and 10 cm. The area of the greatest square that can be inscribed in it, is
-
View Hint View Answer Discuss in Forum
Using Rule 1,
Here, 6² + 8² = 10²
Hence, ∆ABC is right angled BD is perpendicular to AC∴ 1 × AB × BC = 1 × AC × BD 2 2 ⇒ 1 × 6 × 8 = 1 × 10 × BD 2 2 ⇒ BD = 48 = 24 10 5
∴ BD = diagonal of square∴ Area of square = 24 × 24 2 × 5 × 5 = 576 cm² 50 Correct Option: D
Using Rule 1,
Here, 6² + 8² = 10²
Hence, ∆ABC is right angled BD is perpendicular to AC∴ 1 × AB × BC = 1 × AC × BD 2 2 ⇒ 1 × 6 × 8 = 1 × 10 × BD 2 2 ⇒ BD = 48 = 24 10 5
∴ BD = diagonal of square∴ Area of square = 24 × 24 2 × 5 × 5 = 576 cm² 50
- The length of a side of an equilateral triangle is 8 cm. The area of the region lying between the circumference and the incircle of the triangle is (Use π = 22/7)
-
View Hint View Answer Discuss in Forum
Let AD ⊥ BC
∴ BD = 4 cm and
AB = 8 cm
∴ AD = √AB² - BD² = √8² - 4²
= √64 - 16 = √48
= 4√3cm
∴ OD = radius of the in circle= 1 × 4√3 cm = 4 3 √3
∴ Area of the in circle= π 4 ² cm² = 16 π cm² √3 3
AO = radius of circum-circle= 2 × 4√3 = 8 3 √3
∴ Area of the cirum-circle= π 83 ² = 64 π cm² √3 3
∴ Area of the required region= 64 π - 16 π cm² 3 3 = 48π = 16π cm² 3 = 16 × 22 = 352 = 50 2 cm² 7 7 7 Correct Option: B
Let AD ⊥ BC
∴ BD = 4 cm and
AB = 8 cm
∴ AD = √AB² - BD² = √8² - 4²
= √64 - 16 = √48
= 4√3cm
∴ OD = radius of the in circle= 1 × 4√3 cm = 4 3 √3
∴ Area of the in circle= π 4 ² cm² = 16 π cm² √3 3
AO = radius of circum-circle= 2 × 4√3 = 8 3 √3
∴ Area of the cirum-circle= π 83 ² = 64 π cm² √3 3
∴ Area of the required region= 64 π - 16 π cm² 3 3 = 48π = 16π cm² 3 = 16 × 22 = 352 = 50 2 cm² 7 7 7
- The length of each side of an equilateral triangle is 14√3 cm. The area of the incircle (in cm²), is
-
View Hint View Answer Discuss in Forum
BD = DC = 7√3 cm
AD = √AB² - BD²
= √(14√3)² - (7√3)²
= √(14√3 + 7√3)(14√3 - 7√3)
= √21√3 × 7√3 = 21 cm∴ OD = Radius of circle = 1 × 21 = 7cm 3
∴ Area of circle = πr&sub2;= 22 × 7 × 7 = 154 sq. cm. 7 Correct Option: C
BD = DC = 7√3 cm
AD = √AB² - BD²
= √(14√3)² - (7√3)²
= √(14√3 + 7√3)(14√3 - 7√3)
= √21√3 × 7√3 = 21 cm∴ OD = Radius of circle = 1 × 21 = 7cm 3
∴ Area of circle = πr&sub2;= 22 × 7 × 7 = 154 sq. cm. 7
