Prove that the area of triangle is half of the product of its Base and corresponding height.
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CP parallel to BA (By Construction)
AP parallel to BC (By Construction)
Therefore BCPA is a parallelogram.
Hence AC is a diagonal of parallelogram BCPA.
Therefore, area of triangle = ½ ar(IIgm BCPA) = ½ × BC × AD (As BC is the base and AD is the altitude of IIgm BCPA).
AP parallel to BC (By Construction)
Therefore BCPA is a parallelogram.
Hence AC is a diagonal of parallelogram BCPA.
Therefore, area of triangle = ½ ar(IIgm BCPA) = ½ × BC × AD (As BC is the base and AD is the altitude of IIgm BCPA).
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mira71:
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Answer:
follow the explanation part
Step-by-step explanation:
Let base and height of a parallelogram be a and b respectively. We know that the area A of a parallelogram is A = bh.
Now draw a diagonal to form two congruent triangles within the parallelogram. Now shade one of the triangles, which represents the area t of the triangle. We see that this triangle has base b and height h as well. If we shade the other triangle with base b and height h, we get the total area of the parallelogram.
Since there are two triangles with equal area t, we have
area of triangle * 2 = area of parallelogram
2t = bh or t = 1/2*bh
We see that the area of a triangle is one half the base times the height.
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