the base BC of the ΔABC is divided at D so that BD =⅓ BC .prove that ar(ΔABD) = ½ ar (ΔADC)
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let's draw perpendicular from A on BC
LET foot of perpendicular be S
Area,of ABD = 1/2 × AS × BD
= 1/2 AS × X
area of ADC = 1/2 ×AS × DC = 1/2 × AS× 2X
SO area of ADC = 2 AREA OF ABD
area of ABD = 1/2. AREA OF ADC
LET foot of perpendicular be S
Area,of ABD = 1/2 × AS × BD
= 1/2 AS × X
area of ADC = 1/2 ×AS × DC = 1/2 × AS× 2X
SO area of ADC = 2 AREA OF ABD
area of ABD = 1/2. AREA OF ADC
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