Am is a median in △abc (m∈ bc ). a line drawn through point m intersects ab at its midpoint p. find areas of △apc and △pmc, if area of apm=35m^2.

Area of ΔAPC = 70 square meter and Area of △PMC= 35 square meter

Step-by-step explanation:

Here, AM is a median in △ABC (M∈ BC ). A line drawn through point M intersects AB at its midpoint P.

Since, we know that median divides a triangle into equal areas.

Therefore, By the diagram we can say that If Area of ΔAPM=35 .

Therefore Area of ΔBPM = 35 ( Because MP is the median of triangle MAB.)

⇒ Area of ΔPMC = 35 ( Because MP is the median of triangle PBC.)

Now, Area of ΔAPC= Area of ΔPBC ( CP is the median of ΔABC)

But area of ΔPBC= Area of ΔBPM+ area of ΔPMC= 35+35=70 square unit.

⇒ Area of  Δ APC= 70 square unit.

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step-by-step explanation:

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,