NCERT Solutions Class 8 Maths Chapter 11 Exercise 11.1 Mensuration
NCERT solutions for class 8 maths chapter 11 exercise 11.1 mensuration is geared towards helping kids revise concepts that they learned in previous classes. These include finding the area and perimeter of rectangles, triangles, and circles. The questions are in a word problem format where students might have to compare the areas of two shapes, find the cost of fencing a rectangular figure, or solve problems based on a combinatorial configuration. NCERT solutions class 8 maths chapter 11 exercise 11.1 has a total of 5 sums that are of medium level of difficulty.
The point to remember while solving these questions is that kids must be well-versed with the formulas of such shapes. In order to move on to higher-order problems that can be found in this exercise students must have a strong foundational understanding of these figures. The pdf of class 8 maths NCERT solutions chapter 11 exercise 11.1 mensuration is linked below and can be used to recall certain concepts that will be put to further use in this lesson.
☛ Download NCERT Solutions Class 8 Maths Chapter 11 Exercise 11.1
Exercise 11.1 Class 8 Chapter 11
More Exercises in Class 8 Maths Chapter 11
- NCERT Solutions Class 8 Maths Chapter 11 Ex 11.2
- NCERT Solutions Class 8 Maths Chapter 11 Ex 11.3
- NCERT Solutions Class 8 Maths Chapter 11 Ex 11.4
NCERT Solutions Class 8 Maths Chapter 11 Exercise 11.1 Tips
NCERT solutions class 8 maths chapter 11 exercise 11.1 mensuration sees the use of the following formulas:
- Area of a rectangle = a × b, where b is the breadth and a is the length. By multiplying these two dimensions we can find the complete area of a rectangle.
- Area of a square = a × a, where a is the length of each side. As all 4 sides are of equal measure hence, we take its square.
- Area of a triangle = 1/2 * b * h, where b is the base and h is the height. We use Heron’s formula and the Pythagoras theorem to find this formula.
- Area of a parallelogram = b × h, where b is the base and h is the corresponding perpendicular.
- Area of a circle = πr 2, where π is a constant given by either 22/7 or 3.14 and r is the radius of that circle.
NCERT solutions class 8 maths chapter 11 exercise 11.1 aids students in recalling and applying the above-written formulas to complex questions. It is a good test for them. In case kids experience roadblocks they can revisit these concepts detailed in previous chapters.
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NCERT Video Solutions for Class 8 Maths Chapter 11 Exercise 11.1
|Video Solutions for Class 8 Maths Chapter 11 Exercise 11.1|
|Exercise 11.1 Question 1||Exercise 11.1 Question 4|
|Exercise 11.1 Question 2||Exercise 11.1 Question 5|
|Exercise 11.1 Question 3|