in a right triangle one leg measures 4 inches and the other leg measures 6 inches?

Question

what is the lenght of the hypotenuse

Answer 1

Keep in mind that the square of the hypotenuse is equal to the sum of the squares of the legs.

So the hypotenuse is equal to the square root of 4 squared plus 6 squared.
c^2 = a^2 + b^2 if that looks familiar (c is the hypotenuse).

So, in this case:
c^2 = 4^2 + 6^2
c^2 = 16 + 36
c^2 = 52
c = SQRT(52)

Answer 2

As per Pythagoras Theorem,
In a right angled triangle, the square of the hypotenuse is equal to sum of square of the remaining two sides.

Since you have got the remaining two sides it is easy to find the hypotenuse.

hypotenuse square = 4 square + 6 square
or that is, hypotenuse square = 16 + 36 = 52
Hence hypotenuse = square root of 52 = 7.21 inches approx.

Answer 3

Using the Pythagorean Theorem of a^2 + b^2 = c^2,
4^2 + 6^2 = c^2
16 + 36 = c^2
c^2 = 52
c = square root of 52
The length of the hypotenuse is the square root of 52.

Answer 4

c=the hypotenuse
so 4^2+6^2=c^2
16+36=c^2
52=c^2 so
c=52 square rooted so its c=7.21 inches

Answer 5

4^2 + 6^2 = c^2
16 + 36 = c^2
52 = c^2
c = sqrt(52)
c = sqrt(4 * 13)
c = 2sqrt(13)

ANS : about 7.2111 inches

Answer 6

4 squared (16) + 6 squared (36)= 52

The square root of 52 = 7.21 inches.

Answer 7

properly to sparkling up those you should use the Pythagorean theorem, it really is a squared + b squared = c squared. a and b continually signify the legs in a suitable triangle and c represents the hypotenuse. save in options the Pythagorean theorem in uncomplicated words works in proper triangles. on your first difficulty a might want to = 3 centimeters and b might want to = 6 centimeters. So merely sparkling up out, 3 squared + 6 squared = c squared. 3 squared = 9 and six squared = 36, upload them at the same time, 40 5 = c squared, so now to end the project you're taking the sq. root of each area to cancel out the squared so your answer is sq. root of 40 5 = c. Heres the actually complications set up: a squared + b squared = c squared 3 squared + 6 squared = c squared 9 + 36 = c squared 40 5 = c squared sq. root of 40 5 =c on your 2d difficulty a might want to = 15 and c might want to = 20. So merely sparkling up out, 15 squared + b squared = 20 squared. 15 squared = 225 and 20 squared = four hundred. you at present have 225 + b squared = four hundred. Subtract 225 from each and each area, now having b squared = one hundred seventy 5. sq. root each and each area to cancel out the squared, leaving b = sq. root of one hundred seventy 5 Heres the actually complications set up: a squared + b squared = c squared 15 squared + b squared = 20 squared 225 + b squared = four hundred b squared = one hundred seventy 5 b = sq. root of one hundred seventy 5 on your very last difficulty a might want to = 4 and b might want to = 6. sparkling up out, 4 squared + 6 squared = c squared. 4 squared = 16 and six squared = 36, so for this reason you difficulty is, fifty 2 = c squared. sq. root each and each area to cancel out the squared, leaving sq. root of fifty 2 = c squared. Heres the actually complications set up: a squared + b squared = c squared 4 squared + 6 squared = c squared 16 + 36 = c squared fifty 2 = c squared sq. root of fifty 2 = c

Answer 8

16+36=h^2
h=sqrt(52)
= 7.2111 in.

Answer 9

rev kev has the right answer

Answer 10

Is your name Ilene?