Q.6, Ex 9.2 Perimeter and area Class 7 Maths |Ex 9.2 Q.no6 Perimeter and area Class 7 Maths new book
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Q.6, Ex 9.2 Perimeter and area Class 7 Maths |Ex 9.2 Q.no6 Perimeter and area Class 7 Maths new book
Q.6, Ex 9.2 Perimeter and area Class 7 Maths |Ex 9.2 Q.no6 Perimeter and area Class 7 Maths new book
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7th maths | perimeter and area | Lesson - 12 | Part - 1 | Maharashtra Board
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Word Problems and Operations #1 SOLUTIONS: 1. The perimeter of a rectangle is given by the formula: Perimeter = 2(length width) In this case, the perimeter is given as 50 cm, so we can set up the equation: 50 cm = 2((2w - 2 cm) w) Simplifying the equation, we have: 50 cm = 2(3w - 2 cm) 50 cm = 6w - 4 cm 54 cm = 6w Dividing both sides of the equation by 6, we get: w = 9 cm Therefore, the width of the rectangle is 9 cm. 2. Let's denote the shortest side of the triangle as "x" cm. According to the
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Word Problems and Operations #1 SOLUTIONS: 1. The perimeter of a rectangle is given by the formula: Perimeter = 2(length width) In this case, the perimeter is given as 50 cm, so we can set up the equation: 50 cm = 2((2w - 2 cm) w) Simplifying the equation, we have: 50 cm = 2(3w - 2 cm) 50 cm = 6w - 4 cm 54 cm = 6w Dividing both sides of the equation by 6, we get: w = 9 cm Therefore, the width of the rectangle is 9 cm. 2. Let's denote the shortest side of the triangle as "x" cm. According to the
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