In the triangle ABC, point M is the midpoint of AB, point D belongs to AC so that AD:DC=2:5. If the Area of ABC = 56 yd^2, find the Area of BMC, Area of AMD, and the Area of CMD.
Since M is the midpoint,
area AMC = area BMC , they have the same height and base
so BMC = (1/2)(56) = 28
now look at triangles AMD and CMD, they have the same height and their bases are in the ratio of 2 : 5
so triangle AMD = (2/7)(28) = 8
and triangle CMD = (5/7)(28) = 20
To find the areas of BMC, AMD, and CMD, we'll first find the lengths of all the sides of triangle ABC.
Let's denote the length of AB as x. Since M is the midpoint of AB, AM and MB are both equal to x/2.
Since AD:DC = 2:5, we can assign lengths to AD and DC. Let AD = 2k and DC = 5k, where k is a positive constant.
Therefore, AC = AD + DC = 2k + 5k = 7k.
Now, we can use Heron's formula to find the area of triangle ABC.
Area of triangle ABC = √[s(s-AB)(s-AC)(s-BC)]
where s is the semi-perimeter of the triangle.
In this case, s = (AB + AC + BC)/2.
Substituting the known values, we have:
s = (x + 7k + BC)/2
The formula for the area becomes:
Area of triangle ABC = √[s(s-AB)(s-AC)(s-BC)]
= √[(x + 7k + BC)(x + 7k - x/2)(x + 7k - BC/2)(7k - x/2)]
Given that the area of triangle ABC is 56 yd², we have:
√[(x + 7k + BC)(x + 7k - x/2)(x + 7k - BC/2)(7k - x/2)] = 56
Since we only need to find the ratios of the areas, we can skip solving this equation and proceed to calculating the areas of the smaller triangles.
Now, let's find the lengths of the sides of triangle BMC.
BM = x/2
BC = AC - AB = 7k - x
Using the formula for the area of a triangle (Area = 1/2 * base * height), we can find the area of triangle BMC.
Area of BMC = 1/2 * BM * BC
= 1/2 * (x/2) * (7k - x)
Next, let's find the lengths of the sides of triangle AMD.
AM = BM = x/2
AD = 2k
The area of triangle AMD can be calculated as follows:
Area of AMD = 1/2 * AM * AD
= 1/2 * (x/2) * 2k
= 1/2 * xk
Lastly, let's find the lengths of the sides of triangle CMD.
CM = AM + AC = x/2 + 7k
CD = AD + DC = 2k + 5k = 7k
The area of triangle CMD can be calculated as follows:
Area of CMD = 1/2 * CM * CD
= 1/2 * (x/2 + 7k) * 7k
Therefore, to summarize:
Area of BMC = 1/2 * (x/2) * (7k - x)
Area of AMD = 1/2 * xk
Area of CMD = 1/2 * (x/2 + 7k) * 7k
Please note that the values of x and k have not been given in the question, so we cannot calculate the exact areas. However, the formulas provided above will give you the areas in terms of x and k.
To find the areas of BMC, AMD, and CMD, we first need to determine the lengths of the sides of triangle ABC.
Since point M is the midpoint of AB, we can deduce that AM = BM.
Let's assume the length of AB is x. Since M is the midpoint, AM = BM = x/2.
For the ratio AD:DC = 2:5, let's assume AD is 2k and DC is 5k, where k is a positive constant.
Therefore, AC = AD + DC = 2k + 5k = 7k.
Now, we can determine the lengths of BC and AC using the Pythagorean theorem.
BC^2 = AB^2 - AC^2
BC^2 = (x)^2 - (7k)^2
BC = √(x^2 - 49k^2)
We know that the area of triangle ABC is given by the formula: Area = (1/2) * base * height.
The base of triangle ABC is BC, and the corresponding height is AM.
Area of ABC = (1/2) * BC * AM
56 = (1/2) * (√(x^2 - 49k^2)) * (x/2)
Now, let's solve this equation to find the value of x and k.
112 = (√(x^2 - 49k^2)) * (x/2)
Divide both sides of the equation by 2:
56 = (√(x^2 - 49k^2)) * (x)
Square both sides of the equation to eliminate the square root:
3136 = (x^2 - 49k^2) * (x^2)
Divide both sides of the equation by x^2:
3136/x^2 = x^2 - 49k^2
Rearrange the equation:
x^4 - 49k^2x^2 - 3136 = 0
Now, we can solve this quadratic equation to find the values of x and k.