ABCD is a trapezium in which \( \overline{\mathrm{AD}} / / \overline{\mathrm{BC}}, \mathrm{X} \) is the midpoint of \( \overline{\mathrm{AB}}, \mathrm{Y} \) is the midpoint of \( \overline{\mathrm{DC}} \), if \( \mathrm{XY}=7 \mathrm{~cm} ., \mathrm{BC}=10 \mathrm{~cm} \). and the area of the trapezium \( =35 \mathrm{~cm}^{2} \) Find the length of \( \overline{\mathrm{AD}} \) and the perpendicular distance between \( \overline{\mathrm{AD}} \) and \( \overline{\mathrm{BC}} \)

To find the lengths of \( \overline{\mathrm{AD}} \) and the perpendicular distance \( h \) between the parallel sides \( \overline{\mathrm{AD}} \) and \( \overline{\mathrm{BC}} \) in trapezium \( ABCD \) where \( \overline{\mathrm{AD}} \parallel \overline{\mathrm{BC}} \), we can use some properties of trapezoids and the given information. 1. **Finding the length of \( \overline{\mathrm{AD}} \) (let's denote it as \( a \))**: We know that the length of the segment connecting the midpoints of the two non-parallel sides of a trapezoid is given by the average of the lengths of the two parallel sides. Therefore, we have: \[ XY = \frac{a + BC}{2} \] Plugging in the values we know: \[ 7 = \frac{a + 10}{2} \] Multiplying both sides by 2 gives: \[ 14 = a + 10 \] Solving for \( a \) yields: \[ a = 14 - 10 = 4 \text{ cm} \] 2. **Finding the height \( h \)**: The area \( A \) of a trapezium can be calculated using the formula: \[ A = \frac{1}{2} \times (a + b) \times h \] Here \( a = 4 \text{ cm} \), \( b = 10 \text{ cm} \), and \( A = 35 \text{ cm}^2 \). Thus, \[ 35 = \frac{1}{2} \times (4 + 10) \times h \] Simplifying gives: \[ 35 = \frac{1}{2} \times 14 \times h \] \[ 35 = 7h \] Dividing both sides by 7 results in: \[ h = 5 \text{ cm} \] In conclusion, the length of \( \overline{\mathrm{AD}} \) is \( 4 \text{ cm} \) and the perpendicular distance \( h \) between \( \overline{\mathrm{AD}} \) and \( \overline{\mathrm{BC}} \) is \( 5 \text{ cm} \).