P is any point on base BC of ΔABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If ar(ΔABC) =12cm² then find area of ΔEPC.
Answers
Given : P is any point on base BC of ΔABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E and ar(ΔABC) = 12cm².
To find : Area of ΔEPC.
Proof :
We have, ar(ΔABC) = 12cm² and D is the mid-point of BC.
So, AD is the median of ∆ABC,
ar (∆ABD) = ar (∆ADC) = ½ × ar (∆ABC)
ar (∆ABD) = ar( = ½ × 12
ar (∆ABD) = 6 cm² …………. (1)
We know that,area of triangle between the same parallel and on the same base have equal areas :
ar (∆APD) = ar (∆APE)
ar (∆AMP) + ar (∆PDM) = ar (∆AMP) + ar (∆AME)
ar (∆PDM) = ar(∆AME) ………….(2)
ME is the median of ∆ ADC,
ar (∆ADC) = ar (MECD) + ar (∆AME)
ar (∆ADC) = ar (MECD) + ar (∆PDM)
[From eq (2)]
ar(∆ADC) = ar (ΔEPC)
6 cm² = ar (ΔEPC)
[From eq (1)]
Hence, ar (ΔEPC) is 6 cm².
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Answer:
Step-by-step explanation:
We have, ar(ΔABC) = 12cm² and D is the mid-point of BC.
So, AD is the median of ∆ABC,
ar (∆ABD) = ar (∆ADC) = ½ × ar (∆ABC)
ar (∆ABD) = ar( = ½ × 12
ar (∆ABD) = 6 cm² …………. (1)
We know that,area of triangle between the same parallel and on the same base have equal areas :
ar (∆APD) = ar (∆APE)
ar (∆AMP) + ar (∆PDM) = ar (∆AMP) + ar (∆AME)
ar (∆PDM) = ar(∆AME) ………….(2)
ME is the median of ∆ ADC,
ar (∆ADC) = ar (MECD) + ar (∆AME)
ar (∆ADC) = ar (MECD) + ar (∆PDM)
[From eq (2)]
ar(∆ADC) = ar (ΔEPC)
6 cm² = ar (ΔEPC)