Mensuration
- One side of a square is increased by 30%. To maintain the same area, the other side will have to be decreased by
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Using Rule 10,
If the required percentage be x, then30 - x - 30x = 0 100
⇒ 300 – 10x – 3x = 0Percentage Effect = x + y + xy % 100
⇒ 13x = 300⇒ x = 300 = 23 1 % 13 13 Correct Option: A
Using Rule 10,
If the required percentage be x, then30 - x - 30x = 0 100
⇒ 300 – 10x – 3x = 0Percentage Effect = x + y + xy % 100
⇒ 13x = 300⇒ x = 300 = 23 1 % 13 13
- The length and breadth of a rectangle are doubled. Percentage increase in area is
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Using Rule 10,
Percentage increase in area = 100 + 100 + 100 × 100 % = 300% 100 Correct Option: C
Using Rule 10,
Percentage increase in area = 100 + 100 + 100 × 100 % = 300% 100
- If the area of a triangle with base 12 cm is equal to the area of a square with side 12 cm, the altitude of the triangle will be
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Using Rule 1 and 10,
Area of square = (12)² = 144 cm²Area of triangle = 1 × base × height 2 = 1 × 12 × height 2 ⇒ 1 × 12 × height = 144 2 ⇒ Height = 144 × 2 = 24cm. 12 Correct Option: B
Using Rule 1 and 10,
Area of square = (12)² = 144 cm²Area of triangle = 1 × base × height 2 = 1 × 12 × height 2 ⇒ 1 × 12 × height = 144 2 ⇒ Height = 144 × 2 = 24cm. 12
- The ratio of the areas of the circumcircle and the incircle of an equilateral triangle is
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For the equilateral triangle of side a,
In radius = a 2√3 Circum-radius = a √3 ∴ Required ratio = a ² : π a ² √3 2√3 = 1 : 1 = 4 : 1 3 12 Correct Option: B
For the equilateral triangle of side a,
In radius = a 2√3 Circum-radius = a √3 ∴ Required ratio = a ² : π a ² √3 2√3 = 1 : 1 = 4 : 1 3 12
- Area of the incircle of an equilateral triangle with side 6 cm is
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DB = DC = 3cm.
AD = √AB² - BD² = √6² - 3²
= √36 - 9 = √27 = 3√3 cm.∴ OD = In-radius = 1 × 3√3 = √3 cm. √3
∴ Area of the in-circle = πr²
= π × √3 × √3 = 3π sq.cm.Correct Option: D
DB = DC = 3cm.
AD = √AB² - BD² = √6² - 3²
= √36 - 9 = √27 = 3√3 cm.∴ OD = In-radius = 1 × 3√3 = √3 cm. √3
∴ Area of the in-circle = πr²
= π × √3 × √3 = 3π sq.cm.