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Cho tam giác ABC có diện tích là 120cm².Điểm M nằm trên cạnh AC sao cho MA =1/4

cho tam giác ABC có diện tích là 120cm².Điểm M nằm trên cạnh AC sao cho MA =1/4.NốiB với M.Điểm G nằm trên cạnh MB sao cho GM =1/3 BM.AG cắt BC tại N. a)Tính diện tích tam giác ABM. b)Tính tỉ số BM/NC
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a) Ta có diện tích tam giác ABM là:
S(ABM) = 1/2 * AB * h
Với h là đường cao từ B đến AM.
Ta có: h = AM = 1/4 AC = 1/4 (AB + BC)
= 1/4 (AB + AB + BM)
= 1/4 (2AB + BM)
= 1/2 AB + 1/4 BM
= 1/2 AB + 1/4 (3GM)
= 1/2 AB + 3/4 GM
= 1/2 AB + 3/4 (1/3 BM)
= 1/2 AB + 1/4 BM
= 1/4 (2AB + BM)
= 1/4 AC
= 1/4 * 120
= 30 cm²

Vậy diện tích tam giác ABM là 30 cm².

b) Ta có tỉ số BM/NC:
Áp dụng định lí Menelaus cho tam giác ABC và đường thẳng AGN ta có:
BM/NC = (AB/AN) * (GM/GN) * (AC/BC)
= (AB/(AB + BN)) * (GM/(GM + MN)) * (AC/BC)
= (AB/(AB + BN)) * (GM/(GM + 1/3 BM)) * (AC/BC)
= (AB/(AB + BN)) * (GM/(GM + 1/3 BM)) * (AC/BC)
= (AB/(AB + BN)) * (GM/(GM + 1/3 BM)) * (AC/BC)
= (AB/(AB + BN)) * (1/(1 + 1/3)) * (AC/BC)
= (AB/(AB + BN)) * (3/4) * (AC/BC)
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2
1
Quang Huy
12/05 15:25:49
+5đ tặng

a: S ABM=1/4*S ABC=30cm2

b: Kẻ MK//AN

Xét ΔANC có MK//AN

nên CK/CN=CM/CA=3/4

=>CK=3/4CN

=>NK=1/4CN

Xét ΔBMK có GN//MK

nên BN/BK=BG/BM=2/3

=>BN=2/3BK

=>BK=3/2BN

=>NK=1/3BK

=>1/4CN=1/3BK

=>CN/BK=1/3:1/4=4/3

=>CN=4/3BK=4/3*3/2BN=4/2BN=2BN

=>BN/CN=1/2

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+4đ tặng
: S ABM=1/4*S ABC=30cm2

b: Kẻ MK//AN

Xét ΔANC có MK//AN

nên CK/CN=CM/CA=3/4

=>CK=3/4CN

=>NK=1/4CN

Xét ΔBMK có GN//MK

nên BN/BK=BG/BM=2/3

=>BN=2/3BK

=>BK=3/2BN

=>NK=1/3BK

=>1/4CN=1/3BK

=>CN/BK=1/3:1/4=4/3

=>CN=4/3BK=4/3*3/2BN=4/2BN=2BN

=>BN/CN=1/2

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