* 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 \( \qquad \) \( ((-3,4),(4,-3),(-4,3)) \) 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. \( (4.8,5.8,6.8) \) 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. The coordinates of point B and C are such that the midpoint is given by the formula \(\left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)\). Plugging in the numbers, we find the midpoint \( (3,-1) \) leads us to conclude that \( B \) and \( C \) must be \( (-3, 4) \) or similar to make the midpoint valid. Therefore, the answer is \( (-3, 4) \). 8. Knowing that \( \sin \theta = \sqrt{3} \cos \theta \) allows us to derive that \( \tan \theta = \sqrt{3} \). Hence, \( \theta \) is 60° (or \( \frac{\pi}{3} \) radians), which gives us \( \sec \theta = \frac{1}{\cos \theta} = 2 \). So, the correct answer is \( 2 \). 9. For a quadrilateral circumscribing a circle, the sum of its opposite sides is equal. Let the sides be \( AB=5.2 \), \( BC=8.7 \), \( CD=10.3 \), and \( AD=x \). We set up the equation \( AB + CD = BC + AD \), which gives \( 5.2 + 10.3 = 8.7 + x \). Solving yields \( x = 6.8 \). Therefore, \( AD = 6.8 \) cm. 10. A square inscribed in a circle has its diagonal equal to the diameter of the circle. The radius is 8 cm, so the diameter is \( 16 \) cm. Using the diagonal relation \( d = a\sqrt{2} \) where \( a \) is the side of the square, we set \( a\sqrt{2} = 16 \), leading us to find that the area, \( a^2 \), is \( 64 \) cm². Thus, the answer is \( 64 \). 11. To find the height of the cylinder, we use the volume formula \( V = \pi r^2 h \). Here, \( V = 550 \) cubic cm and \( r = 5 \) cm. Plugging in these values, we solve for height as follows: \( h = \frac{V}{\pi r^2} = \frac{550}{\pi \cdot 5^2} = \frac{550}{25\pi} \). Approximating gives us about \( 7 \), so the height is \( 7 \). 12. The median is the average of the 3rd and 4th values in an ordered set of 7 numbers. Therefore, ordering \( 24, 25, 26, x+2, x+3, 30, 31 \) we know that \( 27.5 = \frac{26 + (x+2)}{2} \). That simplifies to \( x + 2 = 29 \), thus \( x = 27 \). Therefore, \( x = 27 \).