The Pythagorean's Theorem for our situation would look like this:

So let's call the short leg s, the long leg l and the hypotenuse h. It appears that all our measurements are based on the measurement of the short leg. The long leg is 4 more than twice the short leg, so that expression is l=2s+4; the hypotenuse measure is 6 more than twice the short leg, so that expression is h=2s+6. And the short leg is just s. Now we can rewrite our formula accordingly:

And of course we have to expand. Doing that will leave us with

Combining like terms we have

Our job now is to get everything on one side of the equals sign and solve for s

That is now a second degree polynomial, a quadratic to be exact, and it can be factored several different ways. The easiest is to figure what 2 numbers add to be -8 and multiply to be -20. Those numbers would be 10 and -2. Since we are figuring out the length of the sides, AND we know that the two things in math that will never EVER be negative are time and distance/length, -2 is not an option. That means that the short side, s, measures 10. The longer side, 2s+4, measures 2(10)+4 which is 24, and the hypotenuse, 2s+6, measures 2(10)+6 which is 26. So there you go!
Answer:
Step-by-step explanation:
Sum of angle in a triangle = 180°
<A+<B+<C = 180
Given <B = 50°
Substituting into the formula
<A+50+<C = 180
<A+<C = 180-50
<A+<C = 130°
Since the ∆ABC is an acute triangle, the angles <A and <C must be angles less than 90° since acute angles are angles less than 90°
The possible values of <A and <C that will be acute and give a sum of 130° are;
∠A= 58° and ∠C= 72°
∠A= 80° and ∠C= 50°
∠A= 60° and ∠C= 70°
You can see that all the Angles are less than 90° and their sum is 130°
Answer:
B. 42-10
Step-by-step explanation:
It is the amount that Sid has minus what he ows
Answer:
12 1/6
It wants me to add more but there's nothing to add
1. AB 3-2/-1-1
-1/2
BC 2+1/1+3
3/4
AC 3+1/-1+3
4/2=2
It is a right triangle
2. JK -1-2/4+3
-3/7
LM -2+5/-5-2
-3/7
KL -1+5/4-2
<u>4/2=2</u>
JM 2+2/-3+5
<u>4/2=2
</u><u />The quadrilateral is a parallelogram