P is any point on base BC of ΔABC and D is the mid-point of BC. DE is drawn parallel toPA to meet AC at E. If ar (ΔABC) = 12 cm2, then find area of ΔEPC.
Given: Area (ABC) = 12 cm2, D is midpoint of BC and AP is parallel to ED. We need to find area of the triangle EPC.
Since, AP||ED, and we know that the area of triangles between the same parallel and on the same base are equal. So,
Area (APE) = Area (APD)
⇒ Area (APM) + Area (AME) = Area (APM) + Area (PMD)
⇒ Area (AME) = Area (PMD) …… (1)
Since, median divide triangles into two equal parts. So,
Area (ADC) = `1/2` Area (ABC) = `12/2` = 6 cm2
⇒ Area (ADC) = Area (MDCE) + Area (AME)
⇒Area (ADC) = Area (MDCE) + Area (PMD) (from equation (1))
⇒ Area (ADC) = Area (PEC)
Area (PEC) = 6 cm2.