Mensuration
- A square and an equilateral triangle are drawn on the same base. The ratio of their area is
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Using Rule 6 and 10,
As a square and an equilateral triangle are drawn on the same base, side of triangle and square will be the same. Let the side be x units.∴ Area of square Area of triangle = x² √3 x² 4 = 4 ⇒ 4 : √3 √3 Correct Option: D
Using Rule 6 and 10,
As a square and an equilateral triangle are drawn on the same base, side of triangle and square will be the same. Let the side be x units.∴ Area of square Area of triangle = x² √3 x² 4 = 4 ⇒ 4 : √3 √3
- A wire, when bent in the form of a square, encloses a region having area 121 cm2. If the same wire is bent into the form of a circle, then the area of the circle
is (Take π = 22/7)
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Using Rule 10 and 14,
Side of the square = √121 = 11cm
∴ Length of the wire = 4 × side = 4 × 11 = 44 cm
Now the wire is bent into the form of a circle.
If the radius of the circle be r cm, then,
∴ 2πr = 44⇒ r = 44 = 44 × 7 = 7 cm 2π 2 × 22
⇒ Area of the circle = πr²= 22 × 7 × 7 = 154cm² 7 Correct Option: C
Using Rule 10 and 14,
Side of the square = √121 = 11cm
∴ Length of the wire = 4 × side = 4 × 11 = 44 cm
Now the wire is bent into the form of a circle.
If the radius of the circle be r cm, then,
∴ 2πr = 44⇒ r = 44 = 44 × 7 = 7 cm 2π 2 × 22
⇒ Area of the circle = πr²= 22 × 7 × 7 = 154cm² 7
- A copper wire is bent in the form of an equilateral triangle and has area 121√3 cm². If the same wire is bent into the form of a circle, the area (in cm²) enclosed by the wire is (Take π = 22/7)
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Using Rule 6 and 14,
Area of the equilateral triangle= √3 side² 4 ⇒ 121√3 = √3 side² 4 ∴ side² = 121√3 × 4 = 121 × 4 √3
∴ Side = √121 × 4
= 11 × 2 = 22 cm
× Total length of wire= 3 × 22 = 66 cm
Let the radius of the circle be r cm, then 2πr = 66⇒ 2 × 22 × r = 66 7 ⇒ r = 66 × 7 = 21 cm 2 × 22 2
∴ Area of the circle = πr²= 22 × 21 × 21 7 2 2
= 346.5 cm²Correct Option: C
Using Rule 6 and 14,
Area of the equilateral triangle= √3 side² 4 ⇒ 121√3 = √3 side² 4 ∴ side² = 121√3 × 4 = 121 × 4 √3
∴ Side = √121 × 4
= 11 × 2 = 22 cm
× Total length of wire= 3 × 22 = 66 cm
Let the radius of the circle be r cm, then 2πr = 66⇒ 2 × 22 × r = 66 7 ⇒ r = 66 × 7 = 21 cm 2 × 22 2
∴ Area of the circle = πr²= 22 × 21 × 21 7 2 2
= 346.5 cm²
- A copper wire is bent in the shape of a square of area 81cm2. If the same wire is bent in the form of a semicircle, the radius (in cm) of the semicircle is (Take π = 22/7)
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Using Rule 10 and 14,
Side of a square = √81 = 9 cm
∴ Length of the wire = 4 × 9 = 36 cm.
∴ Perimeter of semi-circle = (π + 2)r where r = radius⇒ 
22 + 2 
r = 36 cm 7 ⇒ 36 r = 36 7 ⇒ r = 36 × 7 = 7 cm 36 Correct Option: D
Using Rule 10 and 14,
Side of a square = √81 = 9 cm
∴ Length of the wire = 4 × 9 = 36 cm.
∴ Perimeter of semi-circle = (π + 2)r where r = radius⇒ 22 + 2 r = 36 cm 7 ⇒ 36 r = 36 7 ⇒ r = 36 × 7 = 7 cm 36
- At each corner of a triangular field of sides 26 m, 28 m and 30 m, a cow is tethered by a rope of length 7 m. The area (in m²) ungrazed by the cows is
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Using Rule 17,
Area grazed by all cows = 180° πr² = πr² 360° 2 = 1 × 22 × 7 × 7 = 77 sq.metre 2 7
Semi-perimeter of triangular fieldS = 26 + 28 + 30 = 42 metres 2
∴ Area of the field = √s(s - a)(s - b)(s - c)
= √42(42 - 26)(42 - 28)(42 - 30)
= √42 × 16 × 14 × 12 = 336 sq.metre
∴ Area ungrazed by the cows = 336 – 77 = 259 sq.metreCorrect Option: B
Using Rule 17,
Area grazed by all cows = 180° πr² = πr² 360° 2 = 1 × 22 × 7 × 7 = 77 sq.metre 2 7
Semi-perimeter of triangular fieldS = 26 + 28 + 30 = 42 metres 2
∴ Area of the field = √s(s - a)(s - b)(s - c)
= √42(42 - 26)(42 - 28)(42 - 30)
= √42 × 16 × 14 × 12 = 336 sq.metre
∴ Area ungrazed by the cows = 336 – 77 = 259 sq.metre
