answer:**Problem 11.1**. In the following figure, (ABCD) is a square of length (10text{ cm}). (AEB, FED), and (FBC) are straight lines. The area of (triangle AED) is larger than that of (triangle FEB) by (10text{ cm}^2). If the area of (triangle DFB) is (P text{ cm}^2), find the value of (P). **Step-by-step solution:** 1. Let the area of (triangle BDE) be (x). 2. Consider the given condition ( text{Area of } triangle AED - text{Area of } triangle FEB = 10 text{ cm}^2 ). # Method 1: 1. Express the areas with known dimensions: [ text{Area of } triangle ABD = frac{1}{2} cdot 10 cdot 10 = 50 text{ cm}^2 ] Use the given condition: [ 50 - text{Area of } triangle BDF = 10 ] 2. Solve for the area of (triangle BDF): [ text{Area of } triangle BDF = 50 - 10 = 40 text{ cm}^2 ] # Method 2: Detailed steps: 1. Using the given area difference: [ text{Area of } triangle AED - text{Area of } triangle FEB = 10 ] 2. Express area forms to find the unknown triangle's area: [ text{Area of } triangle AED + text{Area of } triangle AEF - text{Area of } triangle FEB - text{Area of } triangle AEF = 10 ] Simplify: [ text{Area of } triangle ADF - text{Area of } triangle AFB = 10 ] 3. Recall from square properties: [ text{Area of } triangle ABD = frac{1}{2} cdot 10 cdot 10 = 50 text{ cm}^2 ] And rearrange: [ 50 - text{Area of } triangle DFB = 10 ] Therefore: [ text{Area of } triangle DFB = 50 - 10 = 40 text{ cm}^2 ] **Conclusion:** [ boxed{40} ] **Problem 11.2**. Workman A needs 90 days to finish a task independently while workman B needs (Q) days for the same task. If they only need (P) days to finish the task when working together, find the value of (Q). **Step-by-step solution:** 1. Work rate formula: [ frac{1}{90} + frac{1}{Q} = frac{1}{P} ] 2. Given that total work together is completed in 40 days, set (P = 40): [ frac{1}{90} + frac{1}{Q} = frac{1}{40} ] 3. Rearrange to solve for (Q): [ frac{1}{Q} = frac{1}{40} - frac{1}{90} = frac{9}{360} - frac{4}{360} = frac{5}{360} = frac{1}{72} ] Hence: [ Q = 72 ] **Conclusion:** [ boxed{72} ] **Problem 11.3**. In the following figure, (AB parallel CD), the area of trapezium (ABCD) is (R text{ cm}^2). Given that the areas of (triangle ABE) and (triangle CDE) are (Q text{ cm}^2) and (4Q text{ cm}^2) respectively, find the value of (R). **Step-by-step solution:** 1. Identify similarity and ratios: [ triangle ABE sim triangle CDE quad (equiangular) ] which gives: [ frac{Q}{4Q} = left( frac{AB}{CD} right)^2 implies frac{AB}{CD} = frac{1}{2} ] 2. Use the ratios of sides in similar triangles: [ AE : EC = BE : ED = 1 : 2 ] 3. Calculate areas involved: [ S_{triangle AEB} = Q, quad S_{triangle CDE} = 4Q ] and heights remain consistent. Compare other triangles: [ S_{triangle AED} = 2Q, quad S_{triangle BEC} = 2Q ] 4. Add up to find total trapezium area: [ S_{ABCD} = Q + 4Q + 2Q + 2Q = 9Q ] Given (R = 648 text{ cm}^2): [ 9Q = 648 implies Q = 72 ] **Conclusion:** [ boxed{648} ] **Problem 11.4**. In the following figure, (O) is the centre of the circle, (HJ) and (IK) are diameters and (angle HKI = S^circ). Given that (angle H KI + angle HOI + angle HJI = frac{1}{4} R^circ), find the value of (S). **Step-by-step solution:** 1. Given: [ S + 2S + S = frac{1}{4} times 648^circ ] 2. Combine like terms: [ 4S = 162^circ ] Solve for (S): [ S = frac{162}{4} = 40.5 ] **Conclusion:** [ boxed{40.5} ]

answer:The coordinates of point P are given as P(-6,-9). When we refer to the coordinates of a point with respect to the x-axis, we are essentially reflecting the point over the x-axis. This reflection changes the sign of the y-coordinate while keeping the x-coordinate the same. Therefore, the transformation can be detailed as follows: - Original coordinates: (-6, -9) - Reflect over the x-axis: The x-coordinate remains -6, and the y-coordinate changes sign from -9 to 9. Thus, the coordinates of point P with respect to the x-axis are boxed{(-6,9)}.

answer:From begin{vmatrix} a & b c & d end{vmatrix} = 7, it follows that ad - bc = 7. By the properties of determinants: [ begin{vmatrix} 3a & 3b 3c & 3d end{vmatrix} = (3a)(3d) - (3b)(3c) = 9(ad - bc) = 9 times 7 ] Thus, we have: [ begin{vmatrix} 3a & 3b 3c & 3d end{vmatrix} = 63 ] So, the answer is boxed{63}.

answer:Consider the function y=f(x)=x^2-2x. This can be rewritten in vertex form by completing the square: [ y = (x-1)^2 - 1 ] Now let's analyze the nature of this function over the given interval [-1, 3]. We can see that: - The function has a vertex at (1, -1). - Since the coefficient of the x^2 term is positive, the parabola opens upwards. - Therefore, the function is decreasing on the interval [-1, 1] and increasing on the interval [1, 3]. At the vertex, x = 1, the function attains its minimum value: [ f(1) = (1-1)^2 - 1 = 0 - 1 = -1 ] At the endpoints of the interval: - When x = -1, f(-1) = (-1-1)^2 - 1 = 4 - 1 = 3. - When x = 3, f(3) = (3-1)^2 - 1 = 4 - 1 = 3. Both f(-1) and f(3) give us the maximum value of the function on the interval. Therefore, the range of the function y = x^2 - 2x for -1 leq x leq 3 is [-1, 3]. Hence, the correct answer is: [ boxed{B: [-1, 3]} ]