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分析:設(shè)DE是△ABC平行于BC的中位線,可得當(dāng)P點(diǎn)位于△ABC內(nèi)部的線段DE上方時(shí),能使△PAB的面積大于
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,因此所求的概率等于△ADE的面積與△ABC的面積比值,根據(jù)相似三角形的性質(zhì)求出這個(gè)面積比即可.
解答:
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分別取AB、AC中點(diǎn)D、E,連接DE
∵DE是△ABC的中位線,
∴DE上一點(diǎn)到BC的距離等于A到BC距離的一半
設(shè)A到BC的距離為h,則當(dāng)動(dòng)點(diǎn)P位于線段DE上時(shí),
△PAB的面積S=
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BC•
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h=
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S
△ABC=
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S
因此,當(dāng)點(diǎn)P位于△ABC內(nèi)部,且位于線段DE上方時(shí),△PAB的面積大于
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.
∵△ADE∽△ABC,且相似比
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=
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∴S
△ADE:S
△ABC=
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由此可得△PAB的面積大于
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的概率為P=
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=
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.
故答案為:
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.
點(diǎn)評(píng):本題給出三角形ABC內(nèi)部一點(diǎn)P,求三角形PBC面積大于或等于三角形ABC面積的一半的概率,著重考查了相似三角形的性質(zhì)和幾何概型的計(jì)算等知識(shí),屬于基礎(chǔ)題.