(a + b) (a + b + c) = 41 (b + c) (a + b + c) = 38 (a + c) (a + b + c) = 25 find value of a + b + c = ?

a+b+c=7.2111

Step-by-step explanation:

Let's find the answer but first for clarity we will use the following:

k=a+b+c

Now, using the given equations we have:

(a + b)*(a + b + c) = 41 which is:

(k-c)*k=41 (eq. 1)

(b + c)*(a + b + c) = 38 which is:

(k-a)*k = 38 (eq. 2)

(a + c)*(a + b + c) = 25 which is:

(k-b)*k=25 (eq. 3)

Now, adding all the equations we have:

(k-c)*k + (k-a)*k + (k-b)*k = 41 + 38 + 25 which is:

k*(k-c+k-a+k-b)=104

k*(3k-(a+b+c))=104 but remember that k=a+b+c so:

k*(3k-k)=104

2*k^2=104

k=7.2111

a+b+c=7.2111

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)