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Tafuta eneo la mcheduara ufuatao uwapo umefungwa upande mmoja. (Tumia x = 22/7)

Mathematics

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9. Tafuta eneo la msambamba ufuatao:
(Pata eneo la trapeza ifuatayo, kwa kudhani ni trapeza pacha (isosceles trapezoid) kwani umbo linaonyesha trapeza na si msambamba (parallelogram) kama ilivyoelezwa.)

  • Hatua ya 1: Tambua vipimo vya trapeza. Trapeza ina urefu wa msingi wa juu a=24 Sma = 24 \text{ Sm} na msingi wa chini b=42 Smb = 42 \text{ Sm}. Upande mmoja mteremko ni c=30 Smc = 30 \text{ Sm}. Kwa trapeza pacha, tofauti ya nusu ya misingi inatoa urefu wa sehemu ya msingi wa chini inayounda pembetatu ya kulia na urefu hh.

    x=b−a2=42 Sm−24 Sm2=18 Sm2=9 Smx = \frac{b - a}{2} = \frac{42 \text{ Sm} - 24 \text{ Sm}}{2} = \frac{18 \text{ Sm}}{2} = 9 \text{ Sm}
  • Hatua ya 2: Tafuta urefu (hh) wa trapeza kwa kutumia kanuni ya Pythagoras. Katika pembetatu ya kulia, xx ni upande mmoja, hh ni upande mwingine, na cc ni hipotenusi.

    h2+x2=c2h^2 + x^2 = c^2 h2+(9 Sm)2=(30 Sm)2h^2 + (9 \text{ Sm})^2 = (30 \text{ Sm})^2 h2+81 Sm2=900 Sm2h^2 + 81 \text{ Sm}^2 = 900 \text{ Sm}^2 h2=900 Sm2−81 Sm2h^2 = 900 \text{ Sm}^2 - 81 \text{ Sm}^2 h2=819 Sm2h^2 = 819 \text{ Sm}^2 h=819 Smh = \sqrt{819} \text{ Sm}
  • Hatua ya 3: Kokotoa eneo la trapeza. Eneo la trapeza hutolewa na fomula: A=12(a+b)hA = \frac{1}{2}(a+b)h.

    A=12(24 Sm+42 Sm)819 SmA = \frac{1}{2}(24 \text{ Sm} + 42 \text{ Sm})\sqrt{819} \text{ Sm} A=12(66 Sm)819 SmA = \frac{1}{2}(66 \text{ Sm})\sqrt{819} \text{ Sm} A=33819 Sm2A = 33 \sqrt{819} \text{ Sm}^2 A≈33×28.618 Sm2A \approx 33 \times 28.618 \text{ Sm}^2 A≈944.39 Sm2A \approx 944.39 \text{ Sm}^2

Eneo la trapeza ni:

944.39 Sm2\boxed{\text{944.39 Sm}^2}

10. Tafuta eneo la mcheduara ufuatao iwapo umefungwa upande mmoja. (Tumia π=22/7\pi = 22/7)
(Pata eneo la umbo lifuatalo. Umbo hili lina mstatili katikati na nusu-duara mbili kwenye ncha zake.)

  • Hatua ya 1: Tambua vipimo vya umbo. Urefu jumla wa umbo ni 100 Sm100 \text{ Sm}. Kipenyo cha nusu-duara ni 14 Sm14 \text{ Sm}. Hivyo, rediasi ya nusu-duara ni r=14 Sm2=7 Smr = \frac{14 \text{ Sm}}{2} = 7 \text{ Sm}.

  • Hatua ya 2: Kokotoa vipimo vya mstatili. Upana wa mstatili ni sawa na kipenyo cha duara, yaani 14 Sm14 \text{ Sm}. Urefu wa mstatili ni urefu jumla ukiondoa rediasi mbili (moja kila upande): Urefu wa mstatili =100 Sm−(7 Sm+7 Sm)=100 Sm−14 Sm=86 Sm= 100 \text{ Sm} - (7 \text{ Sm} + 7 \text{ Sm}) = 100 \text{ Sm} - 14 \text{ Sm} = 86 \text{ Sm}.

  • Hatua ya 3: Kokotoa eneo la mstatili. Eneo la mstatili = urefu ×\times upana

    Amstatili=86 Sm×14 Sm=1204 Sm2A_{\text{mstatili}} = 86 \text{ Sm} \times 14 \text{ Sm} = 1204 \text{ Sm}^2
  • Hatua ya 4: Kokotoa eneo la duara kamili (kutoka nusu-duara mbili). Nusu-duara mbili huunda duara kamili. Eneo la duara = πr2\pi r^2

    Aduara=227×(7 Sm)2A_{\text{duara}} = \frac{22}{7} \times (7 \text{ Sm})^2 Aduara=227×49 Sm2A_{\text{duara}} = \frac{22}{7} \times 49 \text{ Sm}^2 Aduara=22×7 Sm2=154 Sm2A_{\text{duara}} = 22 \times 7 \text{ Sm}^2 = 154 \text{ Sm}^2
  • Hatua ya 5: Kokotoa eneo jumla la umbo. Eneo jumla = Eneo la mstatili + Eneo la duara

    Ajumla=1204 Sm2+154 Sm2=1358 Sm2A_{\text{jumla}} = 1204 \text{ Sm}^2 + 154 \text{ Sm}^2 = 1358 \text{ Sm}^2

Eneo la umbo ni:

1358 Sm2\boxed{\text{1358 Sm}^2}
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