Calculate the surface area of a rectangular base pyramid with base dimensions 19cm and 40cm, and a slant height of 13cm.

Piramidi ina msingi wa mstatili wenye urefu ($L$) wa $19 \text{ cm}$ na upana ($W$) wa $13 \text{ cm}$.
Urefu wa piramidi ($h$) ni $10 \text{ cm}$.

Msingi ni mstatili.

Kuna jozi mbili za nyuso za pembeni zenye urefu mteremko tofauti.

• Urefu mteremko $l_1$ (kwa nyuso zenye msingi $W = 13 \text{ cm}$): Tunatumia theorem ya Pythagoras na urefu wa piramidi ($h$) na nusu ya urefu wa msingi ($L/2$). $$l_1^2 = h^2 + \left(\frac{L}{2}\right)^2$$ $$l_1^2 = (10 \text{ cm})^2 + \left(\frac{19 \text{ cm}}{2}\right)^2$$ $$l_1^2 = 100 \text{ cm}^2 + (9.5 \text{ cm})^2$$ $$l_1^2 = 100 \text{ cm}^2 + 90.25 \text{ cm}^2$$ $$l_1 = \sqrt{190.25} \text{ cm} \approx 13.7931 \text{ cm}$$
• Urefu mteremko $l_2$ (kwa nyuso zenye msingi $L = 19 \text{ cm}$): Tunatumia theorem ya Pythagoras na urefu wa piramidi ($h$) na nusu ya upana wa msingi ($W/2$). $$l_2^2 = h^2 + \left(\frac{W}{2}\right)^2$$ $$l_2^2 = (10 \text{ cm})^2 + \left(\frac{13 \text{ cm}}{2}\right)^2$$ $$l_2^2 = 100 \text{ cm}^2 + (6.5 \text{ cm})^2$$ $$l_2^2 = 100 \text{ cm}^2 + 42.25 \text{ cm}^2$$ $$l_2 = \sqrt{142.25} \text{ cm} \approx 11.9268 \text{ cm}$$

• Eneo la nyuso mbili zenye msingi $W = 13 \text{ cm}$ na urefu mteremko $l_1$: $$A_{pembeni1} = 2 \times \left(\frac{1}{2} \times W \times l_1\right) = W \times l_1$$ $$A_{pembeni1} = 13 \text{ cm} \times \sqrt{190.25} \text{ cm} \approx 13 \text{ cm} \times 13.7931 \text{ cm} \approx 179.3103 \text{ cm}^2$$
• Eneo la nyuso mbili zenye msingi $L = 19 \text{ cm}$ na urefu mteremko $l_2$: $$A_{pembeni2} = 2 \times \left(\frac{1}{2} \times L \times l_2\right) = L \times l_2$$ $$A_{pembeni2} = 19 \text{ cm} \times \sqrt{142.25} \text{ cm} \approx 19 \text{ cm} \times 11.9268 \text{ cm} \approx 226.6092 \text{ cm}^2$$

Jumla ya eneo la uso ($A_{jumla}$) ni jumla ya eneo la msingi na eneo la nyuso zote nne za pembeni.

Kuzungusha hadi sehemu mbili za desimali: