A school playground features a seating area shaped like a rhombus ABCD. The perimeter of the rhombus is 40 metres and the length of one of its diagonals (AC) is 16 metres. Calculate the total area of the seating space.

Rombasi ina pande nne zenye urefu sawa. Mzunguko (perimeter) wa rombasi hutolewa na fomula $P = 4s$, ambapo $s$ ni urefu wa upande mmoja.
Tunapewa mzunguko $P = 40$ mita.

$$4s = 40$$

Gawanya pande zote mbili kwa 4:

$$s = \frac{40}{4}$$ $$s = 10 \text{ m}$$

Hivyo, urefu wa kila upande wa rombasi ni mita 10.

Tunajua kwamba diagonals za rombasi hugawana kila mmoja kwa pembe za kulia. Hii inaunda pembetatu nne za kulia ndani ya rombasi.
Urefu wa diagonal moja (AC) ni mita 16. Hivyo, nusu ya diagonal hii ni $AO = \frac{16}{2} = 8$ mita.
Upande wa rombasi ($s$) ni hypotenuse ya pembetatu hizi za kulia.
Hebu tuchukue pembetatu ya kulia iliyoundwa na nusu ya diagonals na upande mmoja wa rombasi.
Tuna $s = 10$ m (hypotenuse) na $AO = 8$ m (moja ya miguu). Hebu $BO$ iwe mguu mwingine.
Kwa theoremu ya Pythagoras: $AO^2 + BO^2 = s^2$.

$$8^2 + BO^2 = 10^2$$ $$64 + BO^2 = 100$$

Toa 64 kutoka pande zote mbili:

$$BO^2 = 100 - 64$$ $$BO^2 = 36$$

Chukua mzizi wa mraba wa pande zote mbili:

$$BO = \sqrt{36}$$ $$BO = 6 \text{ m}$$

Diagonal ya pili ($d_2$) ni mara mbili ya $BO$.

$$d_2 = 2 \times BO$$ $$d_2 = 2 \times 6 \text{ m}$$ $$d_2 = 12 \text{ m}$$

Eneo la rombasi ($A$) hutolewa na fomula $A = \frac{1}{2} d_1 d_2$, ambapo $d_1$ na $d_2$ ni urefu wa diagonals.
Tuna $d_1 = 16$ m na $d_2 = 12$ m.

$$A = \frac{1}{2} (16 \text{ m})(12 \text{ m})$$ $$A = 8 \times 12 \text{ m}^2$$ $$A = 96 \text{ m}^2$$

Eneo lote la nafasi ya kuketi ni:

$$96 \text{ m}^2$$