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Ad is a Median of a δAbc.P is Any Point on Ad. Show that the Area of δAbp is Equal to the Area of δAcp. - Mathematics

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प्रश्न

AD is a median of a ΔABC.P is any point on AD. Show that the area of ΔABP is equal to the area of ΔACP.

बेरीज

उत्तर


AD is the median of ΔABC, so, it will divide ΔABC into two triangles of equal areas.
Therefore, Area(ΔABD) = area(ΔACD) ...(1)
Now PD is the median of ΔPBC.
Therefore, Area(ΔPBD) = area(ΔPCD) ...(2)
Subtract equation (2) from equation (1), we have
Area(ΔABD) - area(ΔPBD) = Area(ΔACD) - Area(ΔPCD)
Area(ΔABP) = area(ΔACP).

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पाठ 21: Areas Theorems on Parallelograms - Exercise 21.1

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फ्रँक Mathematics [English] Class 9 ICSE
पाठ 21 Areas Theorems on Parallelograms
Exercise 21.1 | Q 17

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