 Hello and how are you all today? My name is Priyanka and let us discuss the question it says find the perimeters of first of all triangle ABE then the rectangle BCDE in this figure whose perimeter is greater. Now this is the figure which is given to us in this question. Now before proceeding on with the solution we should be well aware that we can find out the perimeter of the triangle let's say this is a triangle of length A, B and C respectively. So we can find out the perimeter of triangle having length ABC by adding the length of all the sides and we can find out the perimeter of a rectangle by the formula that is to length plus width where L is the length and B is the width and the knowledge of these two formulas is the key idea for this question. Now let's proceed with the solution keeping the key idea in our mind. Now first of all we need to find out the perimeter of triangle ABE. We have the sides of triangle ABE as 5 by 2, 2, 3 by 4 and 3, 3 by 5. So we have the perimeter of triangle ABE as 5 by 2 plus 2, 3 by 4 plus 3, 3 by 5 and the answer will be in centimeter. Now let's solve the bracket. We have 5 by 2 plus this can be written as 11 by 4 plus we have 18 by 5. Now to add these unlike fractions we need to first take the LCM. So the LCM of 2, 4 and 5 is 20. So we have 5 by 2 multiplied by 10 the numerator and the denominator and in the numerator we have 50 plus it will get multiplied by 5. So we have plus 55 plus we have 18 into 4 that is 72. Now we have the sum as 177 upon 20 centimeter which can be written as 2617 centimeter. Right? So sorry it will be 8 over here. So we have the perimeter of triangle ABE as 177 by 20 centimeter or A17 by 20 centimeter. This completes the first part. Now let's proceed on with the second part. In the second part we need to find out the perimeter of the rectangle that is BCD. Now we have the length of this rectangle as CD which is equal to BE that is 2, 3 by 4 centimeter and we have the breadth that is B as 7 by 6 centimeter. So what we need to do is we need to apply the formula that we learned in our key idea that is perimeter of a rectangle is 2 bracket length plus breadth. So we have 2 bracket 2, 3 by 4 plus 7 by 6 that is equal to 2. Now here we have 8 sorry 11 by 4 plus 7 by 6. We have converted this mixed fraction into an improper fraction. Now we need to add these two unlike fractions by taking their LCM. Their LCM is 12. So we know that 4 into 3 is 12. So we will multiply the numerator that is 11 also by 3 to get 33 plus 6 into 2 is 12. So 7 into 2 is 14. So we have 2 into 33 plus 14 gives us 47 upon 12. So we have 2 into 13 plus 14. Now I am simplifying we have 47 by 6 centimeter as our perimeter for the rectangle which can also be written as over here we can have 6 you know that 7 into 6 is 42. So here we are left with 5. Now thirdly we need to answer whose perimeter is greater. We have the perimeter of the triangle as 8 17 by 20 and we have the perimeter of the rectangle as 7 5 by 6. So we need to compare. So let's name it as or let's compare 8 17 by 20 to 7 5 by 6. Let's convert it into improper fraction. We have it as 1 77 by 20 and here we have it as 47 by 6. Now for comparing we need to convert these unlike fractions into light fractions for that we need to take the LCM of 20 and 6 that is 60. So now we need to convert these two fractions into an equivalent fraction having the denominator as 60. So that means 20 into 3 we know is 60. So 1 77 into 3 is 5 31 and 6 sorry here we had 7. So we have 6 into 10 gives us 60. So we have 47 into 10 as 4 70. So clearly we can see that 5 31 upon 60 is greater than 4 70 upon 60 or we can write that 8 17 by 20 is greater than 7 5 by 6 and in words we can say that therefore perimeter of triangle E B E is greater. Right. So this completes the session. Hope you understood it well and enjoyed it too. Have a nice day.