Quick Revision: Area and Perimeter Video Lecture | Mathematics for Class 5
Quick Revision: Area and Perimeter Video Lecture | Mathematics for Class 5
Video Lecture and Questions for Quick Revision: Area and Perimeter Video Lecture | Mathematics for Class 5 - Class 5 full syllabus preparation | Free video for Class 5 exam to prepare for Mathematics for Class 5.
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Area and Perimeter Word Problems
Area and Perimeter Word Problems
onlinemath4all.com
2017年3月21日
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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