Find the value of x in the isosceles triangle shown

x = 8√5

Step-by-step explanation:

The large triangle is divided into two smaller ones.  The right one has one leg of length 8, one leg of length x/2, and hypotenuse of length √80.

Appling the Pythagorean Theorem, we can find x:

8² + (x/2)² = (√80)²,

or:                              x^2

64 + x^2/4 = 80, or   = 80

                                    4

Then x^2 = 320 = 64*5.  Taking the square root of both sides, we get:

x = 8√5

there still no table

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !