1. **Problem:** Find the area of the cement path around a rectangular field of length 5 m and width 8 m with a 2 m wide cement path around it. Step 1: Find the total length and width including the path. Length including path = $5 + 2 \times 2 = 5 + 4 = 9$ m Width including path = $8 + 2 \times 2 = 8 + 4 = 12$ m Step 2: Find the total area including the path. Total area = $9 \times 12 = 108$ m$^2$ Step 3: Find the area of the rectangular field. Field area = $5 \times 8 = 40$ m$^2$ Step 4: Find the area of the cement path. Cement path area = Total area - Field area = $108 - 40 = 68$ m$^2$ 2. **Problem:** In a parallelogram with sides 12 cm and 9 cm, find the distance between the longer sides if the distance between shorter sides is 8 cm. Step 1: Let the distance between longer sides be $h$. Area can be found two ways: Area = Base $\times$ Height Using shorter sides as base: Area = $9 \times 8 = 72$ cm$^2$ Step 2: Using longer side as base, area = $12 \times h$ Step 3: Equate both area expressions: $$12h = 72$$ $$h = \frac{72}{12} = 6$$ cm 3. **Problem:** Given parallelogram ABCD with sides $AB=20$ cm, $BC=13$ cm and diagonal $AC=21$ cm, find the area. Step 1: Use the formula for area using two sides and the included angle $\theta$: $$Area = AB \times BC \times \sin(\theta)$$ Step 2: Find $\theta$ using Law of Cosines in $\triangle ABC$: $$AC^2 = AB^2 + BC^2 - 2\times AB \times BC \times \cos(\theta)$$ $$21^2 = 20^2 + 13^2 - 2 \times 20 \times 13 \times \cos(\theta)$$ $$441 = 400 + 169 - 520 \cos(\theta)$$ $$441 = 569 - 520 \cos(\theta)$$ $$520 \cos(\theta) = 569 - 441 = 128$$ $$\cos(\theta) = \frac{128}{520} = 0.24615$$ Step 3: Calculate $\sin(\theta)$: $$\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - (0.24615)^2} = \sqrt{1 - 0.0606} = \sqrt{0.9394} = 0.9693$$ Step 4: Find area: $$Area = 20 \times 13 \times 0.9693 = 252.02$$ cm$^2$ 4. **Problem:** Find the surface area of a sphere with radius 28 cm. Formula for surface area of sphere: $$SA = 4\pi r^2$$ Calculate: $$SA = 4 \pi (28)^2 = 4 \pi \times 784 = 3136\pi$$ cm$^2$ 5. **Problem:** Find the radius of a sphere if surface area is 61.6 m$^2$. Using formula: $$61.6 = 4 \pi r^2$$ $$r^2 = \frac{61.6}{4 \pi} = \frac{61.6}{12.5664} = 4.9$$ $$r = \sqrt{4.9} = 2.213$$ m 6. **Problem:** Find how many liters of oil a spherical tank of radius 4 m can contain. Volume of sphere: $$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi 64 = \frac{256}{3} \pi = 268.08$$ m$^3$ Convert cubic meters to liters (1 m$^3$ = 1000 liters): $$268.08 \times 1000 = 268080$$ liters 7. **Problem:** Surface area of a sphere is $5760\pi$. Find its volume. Given: $$4 \pi r^2 = 5760 \pi$$ Divide both sides by $\pi$: $$4 r^2 = 5760$$ $$r^2 = 1440$$ $$r = \sqrt{1440} = 37.95$$ Volume: $$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (37.95)^3 = \frac{4}{3} \pi 54616.7 = 72822.2 \pi$$ 8. **Problem:** Find slant height of a cone with radius 3 m and height 5 m. Slant height $l$: $$l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} = 5.83$$ m 9. **Problem:** Radius of cone base is 16 m and slant height 15 m; find total surface area and volume in cm. Convert meters to cm: 16 m = 1600 cm, 15 m = 1500 cm Surface area: $$SA = \pi r (r + l) = \pi \times 1600 (1600 + 1500) = \pi \times 1600 \times 3100 = 4,960,000 \pi$$ cm$^2$ Volume: $$V = \frac{1}{3} \pi r^2 h$$ Find height $h$ using: $$l^2 = r^2 + h^2$$ $$1500^2 = 1600^2 + h^2$$ $$2,250,000 = 2,560,000 + h^2$$ $$h^2 = 2,250,000 - 2,560,000 = -310,000$$ (not possible, indicates correction needed: slant height must be longer than radius) Correcting units: User likely meant r=16 m = 1600 cm, l=15 m=1500 cm, so $l < r$ contradicts cone geometry. Assuming input error, keeping values as given: $$h = \sqrt{l^2 - r^2} = \sqrt{1500^2 - 1600^2} = \sqrt{2,250,000 - 2,560,000} = \sqrt{-310,000}$$ (No real solution) Suggest input error. Assuming r=15 m and l=16 m: $$h = \sqrt{16^2 - 15^2} = \sqrt{256 - 225} = \sqrt{31} = 5.57$$ m = 557 cm Now volume: $$V = \frac{1}{3} \pi (1500)^2 (557) = \frac{1}{3} \pi \times 2,250,000 \times 557 = 417,750,000 \pi$$ cm$^3$ Final answers: Area of cement path = 68 m$^2$ Distance between longer sides = 6 cm Area of parallelogram ABCD = 252.02 cm$^2$ Surface area of sphere = $3136\pi$ cm$^2$ Radius of sphere with area 61.6 m$^2$ = 2.213 m Capacity of spherical tank = 268080 liters Volume of sphere with area $5760\pi$ = $72822.2\pi$ m$^3$ Slant height of cone = 5.83 m Cone surface area = $4,960,000\pi$ cm$^2$ Cone volume = $417,750,000\pi$ cm$^3$ (assuming corrected dimensions)