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Calculate: (a) the perimeter of the prism, (b) The total surface area of the prism, (c) The volume of prism, (d) The angles between the planes AFED and BCEF.

Mathematics

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4 parts
a

82.1 m\text{82.1 m}

b

288.9 m2\text{288.9 m}^2

c

70.56 m3\text{70.56 m}^3

d

67.93∘\text{67.93}^\circ

Step-by-step solution

6 steps

Hapa kuna suluhisho la maswali yako kuhusu mche mrefu wa pembetatu (triangular prism):

Kutokana na mchoro na data uliyotoa:
Urefu wa msingi wa pembetatu (AB) = 7.6 m7.6 \text{ m}
Upana wa msingi wa pembetatu (BC) = 8.2 m8.2 \text{ m}
Upana wa msingi wa pembetatu (AC) = 3.2 m3.2 \text{ m}
Urefu wa mche mrefu (AD, BE, CF) = 14.7 m14.7 \text{ m}
Urefu wa pembetatu ya msingi (kutoka B hadi AC) = 3.0 m3.0 \text{ m}

Part a· 1 step

Kokotoa mzunguko wa mche mrefu.

Mzunguko wa mche mrefu ni jumla ya urefu wa kingo zake zote. Kuna kingo 9: kingo 3 za msingi wa chini, kingo 3 za msingi wa juu, na kingo 3 za wima.
Kingo za msingi wa chini: AB, BC, AC
Kingo za msingi wa juu: DE, EF, DF (zinafanana na AB, BC, AC)
Kingo za wima: AD, BE, CF (zote zina urefu wa mche mrefu)

  1. Jumlisha urefu wa kingo zote.

    Mzunguko=(AB+BC+AC)+(DE+EF+DF)+(AD+BE+CF)\text{Mzunguko} = (AB + BC + AC) + (DE + EF + DF) + (AD + BE + CF)

    Kwa kuwa DE = AB, EF = BC, DF = AC, na AD = BE = CF:

    Mzunguko=2×(AB+BC+AC)+3×AD\text{Mzunguko} = 2 \times (AB + BC + AC) + 3 \times AD Mzunguko=2×(7.6 m+8.2 m+3.2 m)+3×14.7 m\text{Mzunguko} = 2 \times (7.6 \text{ m} + 8.2 \text{ m} + 3.2 \text{ m}) + 3 \times 14.7 \text{ m} Mzunguko=2×(19.0 m)+44.1 m\text{Mzunguko} = 2 \times (19.0 \text{ m}) + 44.1 \text{ m} Mzunguko=38.0 m+44.1 m\text{Mzunguko} = 38.0 \text{ m} + 44.1 \text{ m} Mzunguko=82.1 m\text{Mzunguko} = 82.1 \text{ m}

    Mzunguko wa mche mrefu ni 82.1 m\boxed{\text{82.1 m}}.

Part b

288.9 m2\text{288.9 m}^2

Eneo lote la uso wa mche mrefu.

Eneo lote la uso ni jumla ya eneo la nyuso mbili za pembetatu (msingi na juu) na nyuso tatu za mstatili za kando.

  1. Kokotoa eneo la msingi wa pembetatu (ABC).

    Eneo la pembetatu = 12×msingi×urefu\frac{1}{2} \times \text{msingi} \times \text{urefu}
    Msingi AC = 3.2 m3.2 \text{ m}
    Urefu wa pembetatu (kutoka B hadi AC) = 3.0 m3.0 \text{ m}

    Eneo la ABC=12×3.2 m×3.0 m\text{Eneo la ABC} = \frac{1}{2} \times 3.2 \text{ m} \times 3.0 \text{ m} Eneo la ABC=1.6 m×3.0 m\text{Eneo la ABC} = 1.6 \text{ m} \times 3.0 \text{ m} Eneo la ABC=4.8 m2\text{Eneo la ABC} = 4.8 \text{ m}^2

  2. Kokotoa eneo la nyuso za mstatili za kando.

    Urefu wa mche mrefu = 14.7 m14.7 \text{ m}
    Eneo la ABED = AB ×\times AD = 7.6 m×14.7 m=111.72 m27.6 \text{ m} \times 14.7 \text{ m} = 111.72 \text{ m}^2
    Eneo la BCFE = BC ×\times BE = 8.2 m×14.7 m=120.54 m28.2 \text{ m} \times 14.7 \text{ m} = 120.54 \text{ m}^2
    Eneo la ACFD = AC ×\times AD = 3.2 m×14.7 m=47.04 m23.2 \text{ m} \times 14.7 \text{ m} = 47.04 \text{ m}^2

  3. Jumlisha maeneo yote.

    Eneo lote la uso=2×(Eneo la ABC)+Eneo la ABED+Eneo la BCFE+Eneo la ACFD\text{Eneo lote la uso} = 2 \times (\text{Eneo la ABC}) + \text{Eneo la ABED} + \text{Eneo la BCFE} + \text{Eneo la ACFD} Eneo lote la uso=2×4.8 m2+111.72 m2+120.54 m2+47.04 m2\text{Eneo lote la uso} = 2 \times 4.8 \text{ m}^2 + 111.72 \text{ m}^2 + 120.54 \text{ m}^2 + 47.04 \text{ m}^2 Eneo lote la uso=9.6 m2+111.72 m2+120.54 m2+47.04 m2\text{Eneo lote la uso} = 9.6 \text{ m}^2 + 111.72 \text{ m}^2 + 120.54 \text{ m}^2 + 47.04 \text{ m}^2 Eneo lote la uso=288.9 m2\text{Eneo lote la uso} = 288.9 \text{ m}^2

    Eneo lote la uso wa mche mrefu ni 288.9 m2\boxed{\text{288.9 m}^2}.

Part c

70.56 m3\text{70.56 m}^3

Kiasi cha mche mrefu.

Kiasi cha mche mrefu = Eneo la msingi ×\times Urefu wa mche mrefu.

  1. Kokotoa kiasi.

    Kiasi=Eneo la ABC×AD\text{Kiasi} = \text{Eneo la ABC} \times \text{AD} Kiasi=4.8 m2×14.7 m\text{Kiasi} = 4.8 \text{ m}^2 \times 14.7 \text{ m} Kiasi=70.56 m3\text{Kiasi} = 70.56 \text{ m}^3

    Kiasi cha mche mrefu ni 70.56 m3\boxed{\text{70.56 m}^3}.

Part d

67.93∘\text{67.93}^\circ

Pembe kati ya nyuso za AFED na BCEF.

Tunachukulia kuwa mche mrefu ni mche mrefu wa kulia (right prism), ambapo kingo za wima (AD, BE, CF) ziko wima kwa nyuso za msingi.
Nyuso za AFED na BCEF ni nyuso za kando za mche mrefu.
AFED inawakilisha uso ACFD.
BCEF inawakilisha uso BCFE.
Nyuso hizi mbili zinakutana kwenye kingo CF. Pembe kati ya nyuso hizi ni pembe kati ya mistari miwili, mmoja katika kila uso, ambayo iko wima kwa mstari wa makutano (CF) kwenye sehemu moja. Mistari hii ni AC na BC katika msingi wa pembetatu ABC. Kwa hivyo, pembe inayohitajika ni ∠ACB\angle ACB.

  1. Tumia Kanuni ya Kosaini (Cosine Rule) kupata ∠ACB\angle ACB katika pembetatu ABC.

    Kanuni ya Kosaini inasema: c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab \cos C
    Ambapo a=BC=8.2 ma = BC = 8.2 \text{ m}, b=AC=3.2 mb = AC = 3.2 \text{ m}, na c=AB=7.6 mc = AB = 7.6 \text{ m}.
    Tunataka kupata ∠C\angle C (yaani ∠ACB\angle ACB).

    (7.6)2=(8.2)2+(3.2)2−2(8.2)(3.2)cos⁡(∠ACB)(7.6)^2 = (8.2)^2 + (3.2)^2 - 2(8.2)(3.2) \cos(\angle ACB) 57.76=67.24+10.24−52.48cos⁡(∠ACB)57.76 = 67.24 + 10.24 - 52.48 \cos(\angle ACB) 57.76=77.48−52.48cos⁡(∠ACB)57.76 = 77.48 - 52.48 \cos(\angle ACB) 52.48cos⁡(∠ACB)=77.48−57.7652.48 \cos(\angle ACB) = 77.48 - 57.76 52.48cos⁡(∠ACB)=19.7252.48 \cos(\angle ACB) = 19.72 cos⁡(∠ACB)=19.7252.48\cos(\angle ACB) = \frac{19.72}{52.48} cos⁡(∠ACB)≈0.375762\cos(\angle ACB) \approx 0.375762 ∠ACB=arccos⁡(0.375762)\angle ACB = \arccos(0.375762) ∠ACB≈67.93∘\angle ACB \approx 67.93^\circ

    Pembe kati ya nyuso za AFED na BCEF ni 67.93∘\boxed{\text{67.93}^\circ}.

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