Mensuration
- The sides of a triangle having area 7776 sq. cm are in the ratio 3 : 4 : 5. The perimeter of the triangle is
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Using Rule 1,
Let the sides of triangle be 3x, 4x and 5x units.
Here, (3x)² + (4x)² = (5x)²
Hence, it is a right angled triangle.
Area of ∆ABC = 1 × AB × BC 2 ⇒ 1 × 3x × 4x = 7776 2
⇒ 6x² = 7776⇒ x² = 7776 1296 6
⇒ x = 1296 = 36 cm.
∴ Perimeter of triangle = 3x + 4x + 5x = 12x = 12 × 36 = 432 cm.Correct Option: A
Using Rule 1,
Let the sides of triangle be 3x, 4x and 5x units.
Here, (3x)² + (4x)² = (5x)²
Hence, it is a right angled triangle.Area of ∆ABC = 1 × AB × BC 2 ⇒ 1 × 3x × 4x = 7776 2
⇒ 6x² = 7776⇒ x² = 7776 1296 6
⇒ x = 1296 = 36 cm.
∴ Perimeter of triangle = 3x + 4x + 5x = 12x = 12 × 36 = 432 cm.
- The diameter of each wheel of a car is 70 cm. If each wheel rotates 400 times per minute, then the speed of the car (in km/hr) is (Take π= 22/7)
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Using Rule 7,
Circumference of the wheel of car = π × d= 22 × 70 = 220 cm 7
= Distance covered in one rotation
∴ Distance covered by car in 1 minute = (400 × 220) cm.
∴ Distance covered by car in 1 hour = (400 × 220 × 60) cm.= 
400 × 220 × 60 
km. 1000 × 100
= 52.8 km.
∴ Speed of car = 52.8 kmphCorrect Option: C
Using Rule 7,
Circumference of the wheel of car = π × d= 22 × 70 = 220 cm 7
= Distance covered in one rotation
∴ Distance covered by car in 1 minute = (400 × 220) cm.
∴ Distance covered by car in 1 hour = (400 × 220 × 60) cm.= 400 × 220 × 60 km. 1000 × 100
= 52.8 km.
∴ Speed of car = 52.8 kmph
- Quadrilateral ABCD is circumscribed about a circle. If the lengths of AB, BC and CD are 7 cm, 8.5 cm, and 9.2 cm respectively, then the length (in cm) of DA is
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Since tangents drawn from an exterior point to a circle are equal in length.
∴ AP = AS
BP = BQ
CR = CQ
DR = DS
On adding all these,
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = BC + DA
⇒ 7 + 9.2 = 8.5 + DA
⇒ 16.2 = 8.5 + DA
⇒ DA = 16.2 – 8.5 = 7.7 cm.Correct Option: A
Since tangents drawn from an exterior point to a circle are equal in length.
∴ AP = AS
BP = BQ
CR = CQ
DR = DS
On adding all these,
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = BC + DA
⇒ 7 + 9.2 = 8.5 + DA
⇒ 16.2 = 8.5 + DA
⇒ DA = 16.2 – 8.5 = 7.7 cm.
- The perimeter of a rhombus is 60 cm and one of its diagonal is 24 cm. The area of the rhombus is
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Using Rule 12,
Side of rhombus = Perimeter = 60 = 15 cm 4 4
d1 = AC = 24 cm.
OC = 12 cm. CD = 15 cm.
∠COD = 90°
∴ In ∆ OCD,
OD = √CD² - OC²
= √15² - 12²
= √225 - 144 = √81
= 9 cm.
∴ d2 = BD = 2 × 9 = 18 cm.= 1 d1d2 = 1 24 × 18 = 216 sq. cm. 2 2 Correct Option: B
Using Rule 12,
Side of rhombus = Perimeter = 60 = 15 cm 4 4
d1 = AC = 24 cm.
OC = 12 cm. CD = 15 cm.
∠COD = 90°
∴ In ∆ OCD,
OD = √CD² - OC²
= √15² - 12²
= √225 - 144 = √81
= 9 cm.
∴ d2 = BD = 2 × 9 = 18 cm.= 1 d1d2 = 1 24 × 18 = 216 sq. cm. 2 2
- The ratio of circumference and diameter of a circle is 22 : 7. If the circumference be
then the radius of the circle is :1 1 m 7
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Using Rule 14,
Circumference = 22 Diameter 7 ⇒ 11 7 = 22 2r 7 ⇒ 11 = 22 14r 7
⇒ 14r × 22 = 11 × 7⇒ r = 11 × 7 = 1 metre 14 × 22 4 Correct Option: C
Using Rule 14,
Circumference = 22 Diameter 7 ⇒ 11 7 = 22 2r 7 ⇒ 11 = 22 14r 7
⇒ 14r × 22 = 11 × 7⇒ r = 11 × 7 = 1 metre 14 × 22 4
