This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
10.1 Write down AH : HG. Step 1: Identify parallel lines and similar triangles. Since DECF is a parallelogram, DE FC. As F lies on BC, this implies DE BC. Because DE BC, ADE is similar to ABC ( ADE ABC). Step 2: Use the ratio of sides from the parallelogram and given information. From the parallelogram, DE = FC. We are given BF : FC = 2 : 3. This means FC = (3)/(2+3) BC = (3)/(5) BC. Therefore, DE = (3)/(5) BC. Step 3: Use the similarity ratio for heights. For similar triangles ADE and ABC, the ratio of their corresponding heights is equal to the ratio of their corresponding bases: AHAG = DEBC Substitute DE = (3)/(5) BC: AHAG = (3)/(5) BCBC = (3)/(5) So, AH = (3)/(5) AG. Step 4: Determine the ratio AH : HG. We know that AG = AH + HG. Therefore, HG = AG - AH = AG - (3)/(5) AG = (2)/(5) AG. The ratio AH : HG is: AH : HG = (3)/(5) AG : (2)/(5) AG = 3 : 2 The ratio is 3 : 2. 10.2 Calculate t if the area of the parallelogram is a maximum. Step 1: Express the base and height of the parallelogram in terms of t. The parallelogram is DECF. The base of the parallelogram is FC. From 10.1, FC = (3)/(5) BC. Given BC = (5-t) units. So, FC = (3)/(5)(5-t). The perpendicular height of the parallelogram is HG. From 10.1, HG = (2)/(5) AG. Given AG = t units. So, HG = (2)/(5)t. Step 2: Write the formula for the area of the parallelogram, A(t). The area of a parallelogram is given by base × perpendicular height. A(t) = FC × HG A(t) = ((3)/(5)(5-t)) ((2)/(5)t) A(t) = (6)/(25) (5t - t^2) Step 3: Differentiate the area function with respect to t. To find the maximum area, we need to find the critical points by setting the first derivative to zero. (dA)/(dt) = (d)/(dt) ((6)/(25) (5t - t^2)) (dA)/(dt) = (6)/(25) (5 - 2t) Step 4: Set the derivative to zero and solve for t. (6)/(25) (5 - 2t) = 0 5 - 2t = 0 2t = 5 t = (5)/(2) Step 5: Verify that this value of t corresponds to a maximum (optional, but good practice). The second derivative is (d^2A)/(dt^2) = (6)/(25)(-2) = -(12)/(25). Since the second derivative is negative, t = (5)/(2) corresponds to a maximum area. Also, t must be positive (AG > 0) and 5-t must be positive (BC > 0), so 0 < t < 5. Our value t = (5)/(2) is within this range. The value of t that will maximise the area of the parallelogram is (5)/(2). That's 2 down. 3 left today — send the next one.