An advertisement board is in the form of an isosceles triangle with perimeter 40m and each of the equal sides are 15m. (a) Find the area of the board to the nearest square meter. (3 marks)

Katika pembetatu pacha, pande mbili ni sawa. Tuseme pande hizo ni $a = 15 \text{ m}$ na $b = 15 \text{ m}$.
Mzunguko $P = a + b + c$, ambapo $c$ ni upande wa tatu.

$$40 \text{ m} = 15 \text{ m} + 15 \text{ m} + c$$ $$40 \text{ m} = 30 \text{ m} + c$$ $$c = 40 \text{ m} - 30 \text{ m}$$ $$c = 10 \text{ m}$$

Hivyo, urefu wa upande wa tatu (base) ni $10 \text{ m}$.

Katika pembetatu pacha, kimo kinachoshuka kutoka kwenye ncha ya juu hadi kwenye base hugawanya base katika sehemu mbili sawa. Hii inaunda pembetatu mbili za pembe mraba.
Hypotenuse ya pembetatu ya pembe mraba ni $15 \text{ m}$ (upande sawa wa pembetatu pacha).
Upande mmoja wa pembetatu ya pembe mraba ni nusu ya base, yaani $\frac{10 \text{ m}}{2} = 5 \text{ m}$.
Tumia kanuni ya Pythagoras (h^2 + \text{base_half}^2 = \text{hypotenuse}^2):

$$h^2 + (5 \text{ m})^2 = (15 \text{ m})^2$$ $$h^2 + 25 \text{ m}^2 = 225 \text{ m}^2$$ $$h^2 = 225 \text{ m}^2 - 25 \text{ m}^2$$ $$h^2 = 200 \text{ m}^2$$ $$h = \sqrt{200} \text{ m}$$ $$h = 10\sqrt{2} \text{ m}$$ $$h \approx 10 \times 1.41421 \text{ m}$$ $$h \approx 14.1421 \text{ m}$$

Eneo la pembetatu $A = \frac{1}{2} \times \text{base} \times \text{height}$.

$$A = \frac{1}{2} \times 10 \text{ m} \times 10\sqrt{2} \text{ m}$$ $$A = 5 \text{ m} \times 10\sqrt{2} \text{ m}$$ $$A = 50\sqrt{2} \text{ m}^2$$ $$A \approx 50 \times 1.41421 \text{ m}^2$$ $$A \approx 70.7105 \text{ m}^2$$

$$A \approx 71 \text{ m}^2$$

Eneo la bodi ni $\text{71 m}^2$.