 Hi and welcome to the session. I am Neha and today I am going to help you with the following question. The question says, diagram of the adjacent picture frame has outer dimensions equal to 24 cm by 28 cm and inner dimensions 16 cm by 20 cm. Find the area of each section of the frame if the width of each section is same. So here we have a picture frame. So let's start its solution. First of all for our convenience let us mark the points on this picture frame that is A, B, C, D, E, F, G. These are the points. Now according to the question we need to find the area of the sections A, B, C, D, 2nd, D, C, F, E, 3rd, E, F, G, H and 4th A, V, G, H. Now as it is a rectangular picture frame and the width of each section is same so that means all the four sections are trapeziums. So we need to find the area of trapeziums. So first of all let us recall what is the area of trapezium? Area of trapezium is equal to 1 by 2 into sum of parallel sides into where height is the perpendicular distance between these two parallel lines. Now first of all let us find out the area of the trapezium A, B, C, D. So for trapezium A, B, C, D, B, C is parallel to A, D. So the length of B, C is 16 cm that is equal to G, F. So let us write B, C equal to 16 cm and A, D is equal to H, E that is 24 cm. Now we need to find the width of this trapezium that is the perpendicular distance between A, D and B, C. Let us name it as C, X. Let us name it as C, X and the width of the trapezium G, H, E, F as as you can see that C, X plus C, F plus F, Y is equal to 28 cm. So let us write C, X plus C, F plus F, Y equal to 28 cm. Let us give it that the width of all the sections is same that means C, X is equal to F, Y. So we will replace F, Y by C, X. Also C, F is equal to 20 cm. So we will substitute its value and we will get C, X plus 20, X equal to 28 cm. This implies X is equal to 28 minus 20 cm. X is equal to 8 cm. C, X is equal to 8 upon 2 cm is equal to C, X that is the perpendicular distance between the parallel lines A, D and B, C and that is equal to 4 cm. Let us substitute the values in the formula of area of trapezium of the area of trapezium 1 by 2 into that is A, D plus into height that is C, N. Now this is equal to 1 by 2 into A, D is equal to 24 cm plus B, C is equal to 16 cm, cm. Now here 2 and 4 will get cancelled by the common factor 2 and we will get A, B, C, D and the trapezium EFGH are of same dimensions that means the area of both of these trapeziums is therefore area of trapezium EFGH is equal to area of trapezium A, B, C, D that is equal to 80 cm square BGH and the trapezium CDEF. So let us find out the area of the trapezium ABGH first. So for trapezium EGH is parallel to BG and the length of AH is equal to 28 cm distance between the BG trapezium. Now in question we are given that the width of each section is the width of the section AHGB is same as the width of the section ABCD that is equal to 4 cm. I will use in the formula of area of trapezium find out the area of trapezium so this will be equal to 1 by 2 into sum of parallel sides that is AH plus BG into height that is B. So this will be equal to 1 by 2 into AH that is 28 cm plus BG that is 20 cm into BP that is 4 cm. Now 2 and 4 will get cancelled by the common factor 2 and we will get the dimensions of the trapezium CDEF is same as the dimensions of the trapezium ABGH. So this implies these trapeziums is of trapezium CDEF is equal to area of trapezium ABGH that is equal to 96 cm square. So our final answer for this question is area of the section ABCD that is 80 cm square section ABGH that is 96 cm square area of the section HGFE that is 80 cm square the area of the section FEDC that is 96 cm square. So with this we have finished this question hope you must have understood the question goodbye and take care.