Bar Graphs – Maharashtra Board Class 6 Solutions for Mathematics (English Medium)
MathematicsGeneral ScienceMaharashtra Board Solutions
Exercise-85
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MathematicsGeneral ScienceMaharashtra Board Solutions
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MathematicsGeneral ScienceMaharashtra Board Solutions
Solution 1:
By the property of exterior angle of a triangle,
m∠XZP = m∠XYZ + m∠YXZ
∴ 90° = m∠XYZ + 55°
∴ m∠XYZ = 90° – 55°
∴ m∠XYZ = 35°
Solution 2:
By the property of exterior angle of a triangle,
m∠BCD = m∠BAC + m∠ABC
∴ 130° = m∠BAC + 60°
∴ m∠BAC = 130° – 60°
∴ m∠BAC = 70°
Solution 3:
The sum of the measures of the three angles of a triangle is 180°.
Thus, in ∆PST,
m∠P + m∠S + m∠PTS = 180°
∴ 60° + 80° + m∠PTS = 180°
∴ 140° + m∠PTS = 180°
∴ m∠PTS = 180° – 140°
∴ m∠PTS = 40°
Now, By the property of exterior angle of a triangle,
m∠PTN = m∠P + m∠S
∴ m∠PTN = 60° + 80°
∴ m∠PTN = 140°
Solution 4:
The sum of the measures of the three angles of a triangle is 180°.
Thus, in ∆RST,
m∠R + m∠S + m∠T = 180°
∴ 70° + 30° + m∠T = 180°
∴ 100° + m∠T = 180°
∴ m∠T = 180° – 100°
∴ m∠T = 80°
Solution 5:
By the property of exterior angle of a triangle,
m∠TRM = m∠N + m∠T
∴ m∠TRM = 30° + 80°
∴ m∠TRM = 110°
Solution 6:
By the property of exterior angle of a triangle,
m∠ACD = m∠A + m∠B
∴ m∠ACD = m∠A + m∠A (∵ m∠A = m∠B)
∴ m∠ACD = 2 × m∠A
∴ 140° = 2 × m∠A
∴ m∠A = 70° (dividing both sides by 2)
∴ m∠A = m∠B = 70°
Solution 1:
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
AlgebraGeometryScience and TechnologyHindi
Exercise -6.1
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Let the common multiple be x.
∴ The measures of the angles are 2x°, 3x°, 4x° and x° respectively.
By the angle sum property of a quadrilateral,
2x° + 3x° + 4x° + x° = 360°
∴ 10x° = 360°
∴ x° = 36°
m∠D = x° = 36°
m∠A = 2x° = 2 × 36° = 72°
m∠B = 3x° = 3 × 36° = 108°
m∠C = 4x° = 4 × 36° = 144°
Now, m∠A + m∠B = 72° + 108° = 180° and
m∠C + m∠D = 144° + 36° = 180°
side AD ∥ side BC (By interior angle test for parallel lines.)
Again, m∠A + m∠D = 72° + 36° ≠ 180° and
m∠B + m∠C = 108° + 144° ≠ 180°
∴ Side AB is not parallel to side CD.
∴ Only one pair of opposite sides is parallel,
∴ □ABCD is a trapezium.
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Exercise – 6.2
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Exercise – 6.3
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Exercise – 6.4
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Exercise – 6.5
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MathematicsGeneral ScienceMaharashtra Board Solutions
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Solution 2(1):
The principal remains the same but the period increases 5 times.
∴ The interest will also increase fivefold.
∴ Interest for 5 years = 720 × 5 = Rs. 3,600.
Solution 2(2):
The principal remains the same but the period increases 3 times. (2 years × 3 years = 6 years)
∴ The interest will also triple.
∴ Interest for 6 years = 3300 × 3 = Rs. 9,900.
Solution 1:
Solution 2(1):
The period and the rate of interest remain the same. The new principal Rs. 15000 is 3 times the old principal Rs. 5000.
∴ Interest also will be 3 times of Rs. 1200∴ Interest on Rs. 15000 = 1200 × 3 = Rs. 3600
Solution 2(2):
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MathematicsGeneral ScienceMaharashtra Board Solutions
Exercise 66:
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Length of the rectangle = 4 cm
Breadth of the rectangle = 2 cm
Area of a rectangle = length × breadth
= 4 × 2
= 8 sq. cm
Thus, the area of the rectangle is 8 sq. cm.
Solution 3:
Side of the square = 40 cm
Area of a square = (side)2 = (40)2 = 1600 sq. cm
Thus, the area of the square is 1600 sq. cm
Exercise 67:
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Exercise 68:
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Length of the pit, l = 2 m
Breadth of the pit, l = 2 m
Area of the pit =l × b = 2 × 2 = 4 sq. m
Length of the plot, l = 12.4 m,
Breadth of the plot, b = 10.2 m
Area of the plot = l × b = 12.4 × 10.2 = 126.48 sq. m
Now, area of the plot after the pit is dug = 126.48 − 4= 122.48 sq. m
Thus, the area of the plot after the pit is dug is 122.48 sq. m.
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Exercise 69:
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Exercise 70:
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