* Fill in the blanks so as to make each of the following statement true : (Que. No. 7 to 12) 7. If the midpoint of $$ riangle \mathrm{ABC} $$ 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 $$ \sin heta=\sqrt{3} \cos heta $$ then $$ \sec heta= \qquad $$ $$ (2,3,1) $$ 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 $$ x= \qquad $$ $$ (27,25,28) $$

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* Fill in the blanks so as to make each of the following statement true : (Que. No. 7 to 12) 7. If the midpoint of $$ 	riangle \mathrm{ABC} $$ 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 $$ \sin 	heta=\sqrt{3} \cos 	heta $$ then $$ \sec 	heta= \qquad $$ $$ (2,3,1) $$ 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 $$ x= \qquad $$ $$ (27,25,28) $$

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**7. Midpoint of $$ \overline{\mathrm{BC}} $$**

Given:
- Centroid (midpoint of the triangle) = $$ (3, -1) $$
- Vertex $$ \mathrm{A} = (1, 3) $$

The centroid formula for a triangle $$ \triangle ABC $$ is:
$$
\left( \frac{A_x + B_x + C_x}{3}, \frac{A_y + B_y + C_y}{3} \right) = (3, -1)
$$
Plugging in the values:
$$
\frac{1 + B_x + C_x}{3} = 3 \implies B_x + C_x = 8
$$
$$
\frac{3 + B_y + C_y}{3} = -1 \implies B_y + C_y = -6
$$
The midpoint of $$ \overline{BC} $$ is:
$$
\left( \frac{B_x + C_x}{2}, \frac{B_y + C_y}{2} \right) = \left( \frac{8}{2}, \frac{-6}{2} \right) = (4, -3)
$$
**Answer:** $$ (4, -3) $$

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**8. Value of $$ \sec \theta $$**

Given:
$$
\sin \theta = \sqrt{3} \cos \theta
$$
Divide both sides by $$ \cos \theta $$:
$$
\tan \theta = \sqrt{3} \implies \theta = 60^\circ
$$
$$
\sec \theta = \frac{1}{\cos 60^\circ} = 2
$$
**Answer:** $$ 2 $$

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**9. Length of $$ \mathrm{AD} $$ in Quadrilateral ABCD**

Given:
- $$ \mathrm{AB} = 5.2 \, \text{cm} $$
- $$ \mathrm{BC} = 8.7 \, \text{cm} $$
- $$ \mathrm{CD} = 10.3 \, \text{cm} $$

For a tangential quadrilateral:
$$
\mathrm{AB} + \mathrm{CD} = \mathrm{BC} + \mathrm{AD}
$$
Plugging in the values:
$$
5.2 + 10.3 = 8.7 + \mathrm{AD} \implies 15.5 = 8.7 + \mathrm{AD} \implies \mathrm{AD} = 6.8 \, \text{cm}
$$
**Answer:** $$ 6.8 \, \text{cm} $$

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**10. Area of the Square Inside the Circle**

Given:
- Radius of the circle = $$ 8 \, \text{cm} $$

The diagonal of the square inscribed in the circle is equal to the diameter of the circle:
$$
\text{Diagonal} = 2 \times 8 = 16 \, \text{cm}
$$
Side length of the square:
$$
s = \frac{16}{\sqrt{2}} = 8\sqrt{2} \, \text{cm}
$$
Area of the square:
$$
\text{Area} = s^2 = (8\sqrt{2})^2 = 128 \, \text{cm}^2
$$
**Answer:** $$ 128 \, \text{cm}^2 $$

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**11. Height of the Cylinder**

Given:
- Volume $$ V = 550 \, \text{cm}^3 $$
- Radius $$ r = 5 \, \text{cm} $$

Volume of a cylinder:
$$
V = \pi r^2 h \implies 550 = \pi \times 25 \times h \implies h = \frac{550}{25\pi} \approx 7 \, \text{cm}
$$
**Answer:** $$ 7 $$

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**12. Value of $$ x $$**

Given observations in ascending order: $$ 24, 25, 26, x+2, x+3, 30, 31, 34 $$

Total number of observations = 8, so the median is the average of the 4th and 5th terms:
$$
\text{Median} = \frac{(x+2) + (x+3)}{2} = 27.5
$$
Solving:
$$
2x + 5 = 55 \implies 2x = 50 \implies x = 25
$$
**Answer:** $$ 25 $$

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**Summary of Answers:**
7. $$ (4, -3) $$  
8. $$ 2 $$  
9. $$ 6.8 \, \text{cm} $$  
10. $$ 128 \, \text{cm}^2 $$  
11. $$ 7 $$  
12. $$ 25 $$
7. $$ (4, -3) $$ 8. $$ 2 $$ 9. $$ 6.8 \, \text{cm} $$ 10. $$ 128 \, \text{cm}^2 $$ 11. $$ 7 $$ 12. $$ 25 $$

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