1. **Problem Statement:** We have a children's park consisting of a quadrilateral ABCD and a semicircle AED. Given side lengths AB = 24 m, BC = 7 m, CD = 25 m, and the semicircle with diameter AE. 2. **Find the arc length AED:** - The arc length of a semicircle is half the circumference of a full circle. - Formula for circumference of a circle: $$C = 2\pi r$$ - Since AED is a semicircle, arc length $$= \pi r$$ - Diameter AE is equal to AB = 24 m (since AE lies on the same line as AB and the semicircle is on the left side). - Radius $$r = \frac{AE}{2} = \frac{24}{2} = 12$$ m - Therefore, arc length $$= \pi \times 12 = 12\pi$$ m 3. **Find the total distance walked around the park three times:** - The total distance for one round is the perimeter of quadrilateral ABCD plus the arc length AED. - Quadrilateral sides: AB = 24 m, BC = 7 m, CD = 25 m, DA = AE = 24 m (since AE is diameter and corresponds to side DA) - Perimeter of ABCD $$= AB + BC + CD + DA = 24 + 7 + 25 + 24 = 80$$ m - Total distance for one round $$= 80 + 12\pi$$ m - For three rounds: $$3 \times (80 + 12\pi) = 240 + 36\pi$$ m 4. **Find the area of the children’s park:** - Area of quadrilateral ABCD plus area of semicircle AED. - Area of semicircle $$= \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (12)^2 = 72\pi$$ m² - To find area of ABCD, split into rectangle ABFE and triangle FCD (assuming F is foot of perpendicular from C to AB extended). - Since AB = 24 m and BC = 7 m, and right angles at A and B, ABCD is a trapezoid or can be approximated as rectangle plus triangle. - Area of rectangle ABFE $$= AB \times BC = 24 \times 7 = 168$$ m² - Area of triangle FCD (right triangle with base CD = 25 m and height BC = 7 m) $$= \frac{1}{2} \times 25 \times 7 = 87.5$$ m² - Total area of ABCD $$= 168 + 87.5 = 255.5$$ m² - Total area of park $$= 255.5 + 72\pi$$ m² 5. **Change quadrilateral ABCD to rectangle AGFD without changing area:** - The new rectangle AGFD must have the same area as ABCD, i.e., 255.5 m². - Let length AG = 25 m (same as CD), then width GF $$= \frac{255.5}{25} = 10.22$$ m - Draw rectangle AGFD with length 25 m and width 10.22 m on the diagram. **Final answers:** - i) Arc length AED $$= 12\pi$$ m - ii) Total distance for three rounds $$= 240 + 36\pi$$ m - iii) Area of park $$= 255.5 + 72\pi$$ m² - iv) Rectangle AGFD with dimensions 25 m by 10.22 m has the same area as ABCD.