im very confused on this math question. geometry. can anyone help me solve it?????

Question

the question is: the medians to the legs of a right triangle are the square root of 41 and 2square root 26. Find the hypotenuse.

i hav noooooooo clue how to do that!!!!!!!

also can someone explain how to solve squared equations with roots???? like for example:

(6squ.root7 + 8squ.root.10)squared.

also, how do you figure out simple things like 3squ.rt.6 squared??????

anyone?????????

Answer 1

The median is the distance from the midpoint of the side to the opposite vertex. Let 2x and 2y be the lengths of the legs of the original right triangle. The medians are hypotenuses of two new right tiangles with legs x and 2y for one and 2x and y for the other (draw the 3 triangles). Then by pythagorean theorem applied to these two triangles:

x^2 + (2y)^2 = 41
(2x)^2 + y^2 = 104

solving
y^2 = 4
y = 2
x^2 = 25
x =5

Thus the legs are 4 and 10, the hypotenuse is given by
h^2 = 4^2 + 10^2 = 116
h = sqrt(116) = 10.77033

Answer 2

Well for your triangle question all you have to do is square your two legs and then add them together, then take the square root of that. In other words:

(sqrt(41))^2 + (2sqrt(26))^2 = hyp^2

As for solving squared equations with roots, here's how you do it.

(6sqrt(7) + 8sqrt(10)*(6sqrt(7) + 8sqrt(10) is the same as (6sqrt(7) + 8sqrt(10)^2

So just FOIL that out.

36*7 (remember sqrt(7)*sqrt(7) is just 7) + 48sqrt(70) (remember to times the numbers inside the roots) + 48sqrt(70) + 64*10

This simplifies to:

252 + 96sqrt(70) + 640

892 + 96sqrt(70)

sqrt(70) can't be simplified any further so that's your final answer for that part

Solving (3sqrt(6))^2 is quite straight forward. Just square the part outside the square root and square the square root part.

This means you'll get 9sqrt(36). Since sqrt(36) is 6, you can plug that in. Your final answer will therefor be 54.

If you are ever squaring a sqrt by itself you can just chop off the root sign.

Hope this helps.

Answer 3

You can solve squared equations with roots easily. for example, the problem

(6sqrt7 + 8sqrt10) ^2

can be solved by simply FOILing as usual (treat the square roots like a variable.)

to square square roots:

(3sqrt6)^2 = 3^2 * (sqrt6)^2 = 9*6 = 54 (property of exponents)

The geometry question can be solved thus:

The meridian bisects the legs of the original right triangle. Name each piece of one leg x (the leg is 2x in length) and the other leg y (the leg is 2y in length).
Draw one meridian (it doesn't matter which one). You can see it creates a right triangle with that meridian as the hypotenuse. I will pick sqrt 41 as my first meridian. It crosses the leg with the y pieces. So the equation for that right triangle is:

4x^2 + y^2 = 41 {Pythagorean Theorem}

Now, you can't solve that equation, so you must use the other meridian and the right triangle it forms. It crosses the leg with the x pieces. That right triangle has the equation of:

4y^2 + x^2 = 104

Now use systems of equations to figure out your x and your y. Then remember that the original triangle had two legs of 2x and 2y, and the hypotenuse. Use the Pythagorean theorem to find the hypotenuse of the original triangle.

Good luck!

Answer 4

if you draw a neat diagram, the answer will be clear.
if A, B, C are the vertices, AC being hypotenuse,
and D and E the mid-points of side AB and side BC:

then from right triangle ABE,
AB^2 + BE^2 = AE^2
AB^2 + BC^2/4 = AE^2 = 41

then from right triangle DBC,
DB^2 + BC^2 = DC^2
AB^2/4 + BC^2 = DC^2 = 4*26

Add the equations and solve for the expression (AB^2 + BC^2) which will give you hypotenuse.

Answer 5

the exterior element of a sphere A = 4 x pi x r^2, and it relatively is an common "rearrange and plug in" question. you have been given A (= 450 ) and can rearrange the formulation to detect r^2 A/(4 pi) = r^2, and discover the sq. root, giving the radius in inches. WHY purely the theory sq. root? anticipate r^2 = a hundred (it relatively is an occasion, no longer an answer!) r = +10 or r= -10 What could be your opinion of a globe with radius -10? Nuff suggested?