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Construction engineer is using a triangular prism shaped beam in a roof structure. Calculate (a) Perimeter (b) Total surface area (c) Volume (d) Angle between the planes AEFD and BCEF

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3 parts
b

288.70 vitengo vya mraba288.70 \text{ vitengo vya mraba}

c

198.73 vitengo vya ujazo198.73 \text{ vitengo vya ujazo}

d

44.65∘44.65^\circ

Step-by-step solution

5 steps

Huu ni mchoro wa prism yenye umbo la pembetatu.
Vipimo vilivyotolewa ni:
Pande za msingi wa pembetatu: s1=5.2s_1 = 5.2, s2=5.2s_2 = 5.2, s3=7.4s_3 = 7.4.
Urefu wa prism (umbali kati ya nyuso za pembetatu): L=14.7L = 14.7.

Part a· 1 step

Kokotoa Mzingo (Perimeter)

Mzingo wa prism kwa kawaida hurejelea mzingo wa sehemu yake ya msalaba, ambayo ni msingi wa pembetatu.

  1. Tafuta mzingo wa msingi wa pembetatu.

    Mzingo wa pembetatu ni jumla ya urefu wa pande zake.

    Pmsingi=s1+s2+s3P_{msingi} = s_1 + s_2 + s_3 Pmsingi=5.2+5.2+7.4P_{msingi} = 5.2 + 5.2 + 7.4 Pmsingi=17.8P_{msingi} = 17.8

    Mzingo wa msingi ni 17.8.

Part b

288.70 vitengo vya mraba288.70 \text{ vitengo vya mraba}

Kokotoa Eneo la Jumla la Uso (Total Surface Area)

Eneo la jumla la uso wa prism ni jumla ya eneo la nyuso zote mbili za msingi na eneo la nyuso zote za mstatili.

Ajumla=2×Amsingi+Pmsingi×LA_{jumla} = 2 \times A_{msingi} + P_{msingi} \times L

  1. Tafuta eneo la msingi wa pembetatu.

    Msingi ni pembetatu pacha (isosceles triangle) yenye pande 5.2, 5.2, na 7.4.
    Tafuta urefu (hh) wa pembetatu kwa kutumia theorem ya Pythagoras. Gawanya msingi wa 7.4 kwa nusu (3.7).

    h2+3.72=5.22h^2 + 3.7^2 = 5.2^2 h2+13.69=27.04h^2 + 13.69 = 27.04 h2=27.04−13.69h^2 = 27.04 - 13.69 h2=13.35h^2 = 13.35 h=13.35≈3.65376h = \sqrt{13.35} \approx 3.65376

    Eneo la msingi:

    Amsingi=12×msingi×urefuA_{msingi} = \frac{1}{2} \times msingi \times urefu Amsingi=12×7.4×13.35A_{msingi} = \frac{1}{2} \times 7.4 \times \sqrt{13.35} Amsingi=3.7×13.35≈3.7×3.65376A_{msingi} = 3.7 \times \sqrt{13.35} \approx 3.7 \times 3.65376 Amsingi≈13.5189 vitengo vya mrabaA_{msingi} \approx 13.5189 \text{ vitengo vya mraba}

  2. Kokotoa eneo la jumla la uso.

    Ajumla=2×Amsingi+Pmsingi×LA_{jumla} = 2 \times A_{msingi} + P_{msingi} \times L Ajumla=2×13.5189+17.8×14.7A_{jumla} = 2 \times 13.5189 + 17.8 \times 14.7 Ajumla=27.0378+261.66A_{jumla} = 27.0378 + 261.66 Ajumla=288.6978A_{jumla} = 288.6978

    Eneo la jumla la uso ni takriban 288.70 vitengo vya mraba\boxed{288.70 \text{ vitengo vya mraba}}.

Part c

198.73 vitengo vya ujazo198.73 \text{ vitengo vya ujazo}

Kokotoa Kiasi (Volume)

Kiasi cha prism ni eneo la msingi wake kuzidisha kwa urefu wake.

V=Amsingi×LV = A_{msingi} \times L

  1. Kokotoa kiasi.

    Tumia eneo la msingi lililokokotolewa hapo juu na urefu wa prism.

    V=13.5189×14.7V = 13.5189 \times 14.7 V=198.72723V = 198.72723

    Kiasi ni takriban 198.73 vitengo vya ujazo\boxed{198.73 \text{ vitengo vya ujazo}}.

Part d

44.65∘44.65^\circ

Kokotoa Pembe kati ya nyuso AEFD na BCEF

Kutokana na uandishi wa kawaida wa prismu ya pembetatu, nyuso AEFD na BCEF hazitakuwa zote nyuso za nje za prismu. BCEF ni uso wa mstatili. AEFD ni ndege inayopita katikati ya prismu.
Kwa kawaida, swali kama hili huuliza pembe kati ya nyuso mbili za karibu za mstatili au kati ya uso wa mstatili na msingi.
Tuchukulie kuwa swali linamaanisha pembe kati ya nyuso mbili za karibu za mstatili, ambazo ni sawa na pembe za ndani za msingi wa pembetatu.
Hebu tuweke lebo kwenye pembetatu ya msingi kama ifuatavyo: A, B, C.
Pande: AB=5.2AB = 5.2, BC=5.2BC = 5.2, AC=7.4AC = 7.4.
Hii ni pembetatu pacha (isosceles triangle) ambapo B ni kilele.
Nyuso za mstatili ni ABED, BCFE, na ACFD.
Uso BCEF ni uso wa mstatili.
Ikiwa AEFD ilikuwa kosa la kuandika na ilikusudiwa kuwa ACFD, basi tunatafuta pembe kati ya nyuso ACFD na BCFE. Hii ni pembe ∠BCA\angle BCA (au ∠C\angle C) ya pembetatu ya msingi.

  1. Tumia Sheria ya Kosine (Law of Cosines) kupata pembe za msingi wa pembetatu.

    Kwa pembetatu yenye pande a=5.2a=5.2, b=7.4b=7.4, c=5.2c=5.2 (ambapo aa ni upande kinyume na A, bb kinyume na B, cc kinyume na C).
    Tunatafuta pembe CC (pembe kati ya pande aa na bb).

    c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab \cos C 5.22=5.22+7.42−2(5.2)(7.4)cos⁡C5.2^2 = 5.2^2 + 7.4^2 - 2(5.2)(7.4) \cos C 27.04=27.04+54.76−76.96cos⁡C27.04 = 27.04 + 54.76 - 76.96 \cos C 0=54.76−76.96cos⁡C0 = 54.76 - 76.96 \cos C 76.96cos⁡C=54.7676.96 \cos C = 54.76 cos⁡C=54.7676.96≈0.711538\cos C = \frac{54.76}{76.96} \approx 0.711538 C=arccos⁡(0.711538)C = \arccos(0.711538) C≈44.65∘C \approx 44.65^\circ

    Pembe kati ya nyuso ACFD na BCFE ni takriban 44.65∘\boxed{44.65^\circ}.

    Tafadhali kumbuka kuwa nimechukulia kuwa "AEFD" ilikuwa kosa la kuandika na ilikusudiwa kuwa "ACFD" ili swali liwe na maana katika muktadha wa prismu. Ikiwa "AEFD" ilikusudiwa kama ilivyoandikwa, basi ni ndege ya diagonal na hesabu ya pembe itakuwa ngumu zaidi na isiyo ya kawaida kwa aina hii ya swali.

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