1. **(A) Area of triangle using base and altitude:** The area of a triangle is given by the formula: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude}$$ Given base $= 6$ and altitude, multiply by $\frac{1}{3}$: $$\text{Area} = \frac{1}{2} \times 6 \times \frac{1}{3} = \frac{1}{2} \times 2 = 1$$ 2. **(B) Area of rectangle (length × width):** Area = length $\times$ width (no values given, formula only) 3. **(C) Formula for area of triangle with variables:** $$\text{Area} = \frac{1}{2} \times x_0 \times 6 \times x = 3x x_0$$ (interpreted multiplication) 4. **(D) Multiplication for square:** Side $\times$ side gives the area of a square. 5. **Types of triangles:** - Scalene: مختلف الاطلاع (all sides different) - Right Angled: قاعدہ الزاویہ (has a right angle) - Equilateral: ساماں الاطلاع (all sides equal) - Isosceles: تادلی الزاویہ (two sides equal) 6. **Adjoint of matrix** Given matrix: $$A=\begin{bmatrix}4 & -2 \\ -1 & 6\end{bmatrix}$$ Steps to find adjoint (transpose of cofactor matrix): Cofactors: $$C_{11} = 6, C_{12} = 1, C_{21} = 2, C_{22} = 4$$ Adjoint (transpose): $$\text{adj}(A) = \begin{bmatrix}6 & 2 \\ 1 & 4\end{bmatrix}$$ 7. **Calculate** $$(\frac{9}{16})^{-\frac{1}{2}}$$ Rewrite: $$= \left(\frac{9}{16}\right)^{-\frac{1}{2}} = \left(\frac{9}{16}\right)^{\frac{-1}{2}} = \frac{1}{\left(\frac{9}{16}\right)^{\frac{1}{2}}}$$ Square root: $$\sqrt{\frac{9}{16}} = \frac{3}{4}$$ Therefore: $$\frac{1}{\frac{3}{4}} = \frac{4}{3}$$ 8. **Correct answer for** $$(\frac{9}{16})^{-\frac{1}{2}}$$ is $\frac{4}{3}$. 9. **Logarithm of any number to itself as base:** $$\log_a a = 1$$ (the logarithm of a number to its own base is always 1) 10. **Product of** $(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$: Apply difference of squares: $$=(\sqrt{x})^2 - (\sqrt{y})^2 = x - y$$ 11. **Factors of** $a^4 - 4b^4$: Rewrite: $$a^4 - (2b^2)^2 = (a^2)^2 - (2b^2)^2$$ Apply difference of squares: $$(a^2 - 2b^2)(a^2 + 2b^2)$$ 12. **LCM of** $15x^2$, $45xy$, and $30xyz$: Prime factors: - $15x^2 = 3 \times 5 \times x^2$ - $45xy = 3^2 \times 5 \times x \times y$ - $30xyz = 2 \times 3 \times 5 \times x \times y \times z$ LCM takes highest powers: - $3^2$, $5$, $2$, $x^2$, $y$, $z$ So: $$LCM = 2 \times 3^2 \times 5 \times x^2 \times y \times z = 90 x^2 y z$$ 13. **Solution of inequality** $-2 < x < \frac{3}{2}$: $x=\frac{3}{2}$ (three halves) satisfies this as it is the upper bound. 14. **Ordered pair satisfying** $y = 2x$: Check options: - (2,1): $1 \stackrel{?}{=} 2 \times 2 = 4$ no - (1,2): $2 = 2 \times 1 = 2$ yes - (0,1): $1 = 2 \times 0 = 0$ no - (2,2): $2 = 2 \times 2 = 4$ no Answer: (1,2) 15. **Distance between** $(0,0)$ and $(1,1)$: Use distance formula: $$\sqrt{(1-0)^2 + (1-0)^2} = \sqrt{1 + 1} = \sqrt{2}$$ 16. **Each angle of an equilateral triangle:** $$60^\circ$$ 17. **Diagonals of a parallelogram:** Diagonals ______ each other. Answer: bisect 18. **Bisectors of angles of a triangle are:** Answer: concurrent 19. **Comparison of two alike quantities is called:** Answer: proportion 20. **For matrix options given:** Original matrix: $$\begin{bmatrix}4 & -2 \\ -1 & 6\end{bmatrix}$$ Adjoint: $$\begin{bmatrix}6 & 2 \\ 1 & 4\end{bmatrix}$$ 21. **Summary answers:** - $(\frac{9}{16})^{-\frac{1}{2}} = \frac{4}{3}$ - Logarithm base itself = 1 - $(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = x - y$ - Factors of $a^4 - 4b^4$ are $(a^2 - 2b^2)(a^2 + 2b^2)$ - LCM is $90 x^2 y z$ - Solution of $-2 < x < \frac{3}{2}$ is $\frac{3}{2}$ - Point satisfies $y=2x$ is $(1,2)$ - Distance (0,0) and (1,1) is $\sqrt{2}$ - Equilateral triangle angles: $60^\circ$ - Diagonals of parallelogram bisect each other - Angle bisectors of triangle are concurrent - Comparison of quantities is proportion Final matrix adjoint: $$\text{adj} \begin{bmatrix}4 & -2 \\ -1 & 6\end{bmatrix} = \begin{bmatrix}6 & 2 \\ 1 & 4\end{bmatrix}$$