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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 12: cho hình lăng trụ tam giác ABC.A’B’C’ ,trên cạnh B’B lấy M sao cho BM’=1/3.BB’ .biết V ABC.A’B’C’ =1,tính thể tích ABCMC’

Toán Lớp 12: cho hình lăng trụ tam giác ABC.A’B’C’ ,trên cạnh B’B lấy M sao cho BM’=1/3.BB’ .biết V ABC.A’B’C’ =1,tính thể tích ABCMC’

Comments ( 2 )

  1. Giải đáp: $\dfrac59$
    Lời giải và giải thích chi tiết:
    Ta có:
    $B’M=\dfrac13BB’$
    $\to S_{B’C’M}=\dfrac13S_{C’B’B}=\dfrac13\cdot\dfrac12S_{BCC’B’}=\dfrac16S_{BCC’B’}$
    $\to S_{BCC’M}=S_{BCC’B’}-S_{B’C’M}=\dfrac56S_{BCC’B’}$
    $\to V_{ABCMC’}=\dfrac56V_{ABCC’B’}$
    Ta có:
    $V_{AA’B’C’}=\dfrac13V_{ABCA’B’C’}$
    $\to V_{ABCB’C’}=V_{ABCA’B’C’}-V_{AA’B’C’}=V_{ABCA’B’C’}-\dfrac13V_{ABCA’B’C’}=\dfrac23V_{ABCA’B’C’}$
    $\to V_{ABCC’B’}=\dfrac23$
    $\to V_{ABCMC’}=\dfrac56\cdot \dfrac23=\dfrac{5}{9}$

    toan-lop-12-cho-hinh-lang-tru-tam-giac-abc-a-b-c-tren-canh-b-b-lay-m-sao-cho-bm-1-3-bb-biet-v-ab

  2. Có: $V_{AA’B’C’}=\dfrac{1}{3}V_{ABC.A’B’C’}=\dfrac{1}{3}$ 
    $\to V_{A.BCC’B’}=1-\dfrac{1}{3}=\dfrac{2}{3}$
    Mặt khác $\dfrac{S_{MB’C’}}{S_{BCC’B’}}=\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{1}{6}$
    $\to \dfrac{S_{MB’C’}}{S_{BCC’B’}}=\dfrac{1}{6}$
    $\to \dfrac{V_{ABCC’M}}{V_{ABCC’B’}}=\dfrac{S_{BCC’M}}{S_{BCC’B’}}=\dfrac{5}{6}$ 
    Vậy $V_{ABCC’M}=\dfrac{5}{6}.\dfrac{2}{3}=\dfrac{5}{9}$

    toan-lop-12-cho-hinh-lang-tru-tam-giac-abc-a-b-c-tren-canh-b-b-lay-m-sao-cho-bm-1-3-bb-biet-v-ab

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222-9+11+12:2*14+14 = ? ( )

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