How many different triangles can you make if you are given these two lengths for sides?

Infinitely many triangles.

Step-by-step explanation:

Given the lengths of two sides are 8 inches and 10 inches.

Let's assume third side = x inches.

Using the Triangle Inequalities given as follows:-

1. a+b > c,

2. b+c > a,

3. c+a > b.

Using the lengths given in the problem, we can write:-

1. x+8 > 10 ⇔ x > 10-8 ⇔ x > 2.

2. x+10 > 8 ⇔ x > 8-10 ⇔ x > -2.

3. 8+10 > x ⇔ x < 18.

So, the solution set is 2 < x < 18. It means third side can take any value in interval (2, 18).

Hence, there are infinitely many triangles.

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11 -1> 1 -1+ a

-12 > a