 Hello friends welcome to the session I am Alka, let's discuss the given question. The vertices of a triangle A, B, C, R, A, 4, 6, V, 1, 5 and C, 7, 2. A line is drawn to intersect sides A, B and A, C, R, B and E respectively such that A, B upon A, B equal to A, C upon A, E upon A, C equal to 1 upon 4. Calculate the area of triangle A, B, E and compare it with the area of triangle A, B, C. Here is the given figure A, B, C with coordinates of A, B and C and we see that the line A, B and A, C are intersected at the point D and E respectively such that A, B upon A, B equal to A, E upon A, C equal to 1 by 4. Now let's begin with the solution. Here we are given A, D upon A, B equal to 1 by 4. This implies A, D upon D, B equal to 1 by 3 and we are also given A, E upon A, C equal to 1 by 4. This implies A, E upon E, C equal to 1 by 3. Now we find the coordinates of the point D and E. So to find the coordinate of D and E we will apply the section formula and we know that the section formula is m1 x2 plus m2 x1 upon m1 plus m2 and m1 y2 plus m2 y1 upon m1 plus m2. Now we are going to use this formula to find the coordinates of D and E. Therefore, coordinates B are 1 into 1 plus 3 into 4 upon 1 plus 3 and 1 into 5 plus 3 into 6 upon 1 plus 3. This is equal to 13 upon 4 and 23 upon 4. So the coordinates of D are 13 upon 4, 23 upon 4. Now coordinates of E are 1 into 7 plus 3 into 4 upon 1 plus 3 and 1 into 2 plus 3 into 6 upon 1 plus 3. This is equal to 19 upon 4, 5. So 19 upon 4 and 5 are the coordinates of E. Now we are having the coordinates of the point D and E. Therefore, now we will find the area of the triangle A, D, E and A, B, C. And we know that area of whose vertices x1 by 1, x2 by 2 and x3 by 3 is given by 1 by 2 into x1 by 2 minus y3 plus x2 y3 minus y2 x2 y3 minus y1 plus x3 y1 minus y2. So this is the formula to find the area of the triangle. Therefore, area of triangle A, D, E is 1 by 2 into 4, 23 upon 4 minus 5 plus 13 upon 4 into 5 minus 6 plus 19 upon 4 into 6 minus 23 upon 4. This is equal to 1 by 2 into 3 minus 13 by 4 plus 19 by 16. What this can also be written as 1 by 2. Now on taking LCM, we get 48 minus 52 plus 19. This is equal to 5 of 15 upon 32 square units. So this is the area of the triangle A, D, E. Now we will find the area of triangle A, B, C. So area of triangle A, B, C is 1 by 2 into 4 into 5 minus 2 plus 1 into 2 minus 6 plus 7 into 6 minus 5. So this is equal to 1 by 2 into 12 minus 4 plus 7. This is equal to 15 by 2 square units. So this is the area of the triangle A, B, C. Now we have to compare the area of the triangle A, D, E to the area of the triangle A, B, C. So it is 50 area of triangle A, D, E to area of triangle A, B, C is 15 upon 32 upon 15 upon 2. So this is equal to 1 is to 16. Therefore the area of the triangle A, D, E is 15 upon 32 square units and the ratio of the triangle A, D, E to the area of the triangle A, B, C is 1 is to 16. So hope you understood the solution and enjoyed the session. Goodbye and take care.