Mensuration
- If the sides of an equilateral triangle are increased by 20%, 30% and 50% respectively to form a new triangle, the increase in the perimeter of the equilateral triangle is
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Let the side of equilateral triangle be x units.
∴ Perimeter = 3x units.
After increase, Perimeter = 1.2x + 1.3x +1.5x = 4x units
Increase = 4 x – 3x = x units∴ % Increase = x × 100 = 100 33 = 1 % 3x 3 3 Correct Option: B
Let the side of equilateral triangle be x units.
∴ Perimeter = 3x units.
After increase, Perimeter = 1.2x + 1.3x +1.5x = 4x units
Increase = 4 x – 3x = x units∴ % Increase = x × 100 = 100 33 = 1 % 3x 3 3
- A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope stretched and describes 88 metres when it has traced out 72° at the centre, the length of the rope is (Take π = 22/7)
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θ = 72° → 88m∵ 360° → 88 × 360 = 440 m 72
⇒ 2pr = 440r = 440 . 7 2 22
∴ r = 70m = OA which is the length of the rope.Correct Option: A
θ = 72° → 88m∵ 360° → 88 × 360 = 440 m 72
⇒ 2pr = 440r = 440 . 7 2 22
∴ r = 70m = OA which is the length of the rope.
- Three circles of radii 3.5 cm, 4.5 cm and 5.5 cm touch each other externally. Then the perimeter of the triangle formed by joining the centres of the circles, in cm, is
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AE = AD = 3.5 cm
BE = BF = 4.5 cm
CD = CF = 5.5 cm
∴ Perimeter of triangle = AB + BC + CA
= AE + EB + BF + FC + CD+ DA
= 2(AE + BE + CD) = 2(3.5 + 4.5 + 5.5)
= 2 × 13.5 = 27 cmCorrect Option: A
AE = AD = 3.5 cm
BE = BF = 4.5 cm
CD = CF = 5.5 cm
∴ Perimeter of triangle = AB + BC + CA
= AE + EB + BF + FC + CD+ DA
= 2(AE + BE + CD) = 2(3.5 + 4.5 + 5.5)
= 2 × 13.5 = 27 cm
- ABCD is a parallelogram in which diagonals AC and BD intersect at O. If E, F, G and H are the mid points of AO, DO, CO and BO respectively, then the ratio of the perimeter of the quadrilateral EFGH to the perimeter of parallelogram ABCD is
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In ∆OAB,
Mid-point of OA = E
Mid-point of OB = H
∴ EH || AB and HE = 1/2 AB
Similarly, HG = 1/2 BC,
FG = 1/2 CD and EF = 1/2 AD
∴ EH + HG + FG + EF = 1/2 (AB + BC + CD + AD)
⇒ Perimeter of EFGH = 1/2 × Perimeter of ABCD
∴ Required ratio = 1: 2Correct Option: C
In ∆OAB,
Mid-point of OA = E
Mid-point of OB = H
∴ EH || AB and HE = 1/2 AB
Similarly, HG = 1/2 BC,
FG = 1/2 CD and EF = 1/2 AD
∴ EH + HG + FG + EF = 1/2 (AB + BC + CD + AD)
⇒ Perimeter of EFGH = 1/2 × Perimeter of ABCD
∴ Required ratio = 1: 2
- A circular wire of diameter 112 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 9 : 7. The smaller side of the rectangle is
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Using Rule 14,
Circumference of circular shape = π × diameter= 22 × 112 = 352 cm 7
= length of wire
∴ Perimeter of rectangle = 2 (length + breadth)
⇒ 2 (l + b) = 352⇒ l + b = 352 = 176 2 ∴ Smaller side of rectangle = 7 × 176 = 77 16 Correct Option: A
Using Rule 14,
Circumference of circular shape = π × diameter= 22 × 112 = 352 cm 7
= length of wire
∴ Perimeter of rectangle = 2 (length + breadth)
⇒ 2 (l + b) = 352⇒ l + b = 352 = 176 2 ∴ Smaller side of rectangle = 7 × 176 = 77 16
