20 pts in △abc, point d∈ ab with ad: db=2: 3, point m is the midpoint of cd , and aabc=30 in2. find aacd, aamc, admb, and acmb.

First, draw your diagram. Make sure all of the specifications are there.

AD : DB = 2:3

Area of ΔABC = 30 in²

This means that the sides that make up AB (AD and DB) are 2:3. The total length of side AB = 5. The altitude of ΔABC is CD, and that is 0.5 (5 × x), and if you solve that, x = 12. So CD = 12.

Since point M is the midpoint, it divides CD into 2 equal lengths, DM and CM, both of which equal 6 (12 ÷ 2 = 6)

Now:

ΔDMB = 0.5 (6 × 3) = 0.5 (18) = 9 in²

ΔDMA = 0.5 (6 × 2) = 0.5 (12) = 6 in²

ΔACD = 5/2 = 30/x, 5x = 60, x = 12 in² (Main Problem 2: If 2 triangles have congruent altitudes, then the ratio of their areas is equal to the ratio of the sides to which those altitudes are drawn.)

ΔAMC = 12 in² - 6 in² = 6 in² (Area of ΔACD - Area of ΔDMA = Area of ΔAMC)

ΔBDC = 5/3 = 30/x, 5x = 90, x = 18 in² (Main Problem 2: If 2 triangles have congruent altitudes, then the ratio of their areas is equal to the ratio of the sides to which those altitudes are drawn.)

ΔBMC = 18 in² - 9 in² = 9 in² (Area of ΔBDC - Area of ΔDMA = Area of ΔBMC)

So the answers are:

Area of ΔACD = 12 in²

Area of ΔAMC = 6 in²

Area of ΔDMB = 9 in²

Area of ΔCMB= 9 in²

Hope this helped!

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3