Fill in the blanks so as to make each of the following statement true : (Que. No. 7 to 12) 7. If the midpoint of \( \triangle \mathrm{ABC} \) is \( (3,-1) \) and \( \mathrm{A}(1,3) \) then the midpoint of \( \overline{\mathrm{BC}} \) is \( \qquad \) \( ((-3,4),(4,-3), \overline{(-4,3)}) \) 8. If \( \sin \theta=\sqrt{3} \cos \theta \) then \( \sec \theta= \qquad \) \( (2,3,1) \) 9. Quadrilateral ABCD circumscribes a circle. If \( \mathrm{AB}=5.2 \mathrm{~cm}, \mathrm{BC}=8.7 \mathrm{~cm} \) and \( C D=10.3 \mathrm{~cm} \), then \( \mathrm{AD}= \qquad \) cm . 10. The area of a square which is inside a circle of radius 8 cm is \( \qquad \) cm2. \( (64,60,128) \) 11. The volume of cylinder is 550 cubic cm . If the radius of it is 5 cm , then find the height \( = \qquad \) \( (7,6,5) \) 12. If the observations \( 24,25,26, x+2, x+3,30,31 \) and 34 are in ascending order, and the median is 27.5 then \( x= \qquad \) \( (27,25,28) \)

7. Since the midpoint of \( \triangle ABC \) is \( (3, -1) \), we can use the midpoint formula. Given that point A is \( (1, 3) \), we can express the coordinates of points B and C as \( (b_x, b_y) \) and \( (c_x, c_y) \). Solving the equations, we find that the coordinates of the midpoint for \( \overline{BC} \) yield the point \( (-1, 1) \). 8. From the equation \( \sin \theta = \sqrt{3} \cos \theta \), we can divide both sides by \( \cos \theta \) to get \( \tan \theta = \sqrt{3} \). This implies that \( \theta = 60° \) or \( \theta = 240° \), leading us to compute \( \sec \theta \) which is equal to \( \frac{1}{\cos 60°} = 2 \). 9. In quadrilateral ABCD, since it circumscribes a circle, it follows that the sum of the lengths of opposite sides are equal. So, \( AB + CD = BC + AD \). Plugging in the known values: \( 5.2 + 10.3 = 8.7 + AD \). Thus, \( AD = 6.8 \, \text{cm} \). 10. The largest square that can fit inside a circle can be inscribed such that its diagonal is equal to the diameter of the circle. Given the radius of the circle is \( 8 \, \text{cm} \), the diameter is \( 16 \, \text{cm} \). The side length \( s \) of the square can then be calculated using \( s\sqrt{2} = 16 \), leading us to \( s = 16/\sqrt{2} = 8\sqrt{2} \). Thus, the area \( = s^2 = 128 \, \text{cm}^2 \). 11. The volume of a cylinder is given by the formula \( V = \pi r^2 h \). Plugging in the values we have \( 550 = \pi (5^2) h \). Simplifying gives \( 550 = 25\pi h \). Solving for the height \( h = \frac{550}{25\pi} = \frac{22}{\pi} \approx 7 \) cm. 12. To find \( x \) given that the median of the ordered set \( 24, 25, 26, x+2, x+3, 30, 31, 34 \) is 27.5, we need to consider that the median of a set with an even number of elements is the average of the two middle numbers. The numbers in the context imply the 4th and 5th values are \( x+2 \) and \( x+3 \). Setting up the equation \( \frac{(x+2) + (x+3)}{2} = 27.5 \) leads to \( 2x + 5 = 55 \), thus \( x = 25 \).