Question

What are the values of x and y?

Answer 1

D.
`x=(64)/(15), y=(136)/(15)`

Explanation:

We have been given an image of two similar triangle. We are asked to find the values of x and y for our given triangles.

We will use Pythagoras theorem to solve for x and y.

In triangle ABC:

`17^2+y^2=(15+x)^2...(1)`

In triangle BCD:

`x^2+8^2=y^2...(2)`

Upon substituting equation (1) in equation (1) we will get,

`17^2+x^2+8^2=(15+x)^2`

`289+x^2+64=225+30x+x^2`

`353+x^2=225+30x+x^2`

`353+x^2-x^2=225+30x+x^2-x^2`

`353=225+30x`

`353-225=225-225+30x`

`128=30x`

`(128)/(30)=(30x)/(50)`

`(64*2)/(15*2)=x`

`x=(64)/(15)`

Upon substituting
`x=(64)/(15)` in equation (2) we will get,

`((64)/(15))^2+8^2=y^2`

`(64^2)/(15^2)+64=y^2`

`(4096)/(225)+(14400)/(225)=y^2`

`(4096+14400)/(225)=y^2`

`(18496)/(225)=y^2`

Taking square root of both sides we will get,

`\sqrt{(18496)/(225)}=y`

`(136)/(15)=y`

Therefore, the value of x is
`(64)/(15)`, value y is
`(136)/(15)` and option D is the correct choice.

Answer 2

The correct answer is:

x=64/15, y=136/15.

Explanation:

We will use the Pythagorean theorem to solve this.

For the smaller triangle, the legs of the triangle are x and 8, and the hypotenuse is 8. In the Pythagorean theorem, this gives us:
x²+8²=y²

Simplifying, we have:
x²+64=y².

For the large triangle, the legs of the triangle are y and 17, and the hypotenuse is 15+x. In the Pythagorean theorem, this gives us:
y²+17²=(15+x)².

Simplifying the left hand side we have:
y²+289 = (15+x)².

In order to simplify the right hand side, we write (15+x)² as the product of two binomials:
y²+289=(15+x)(15+x).

Multiplying the right hand side, we have:
y²+289=15*15+15*x+x*15+x*x
y²+289=225+15x+15x+x²
y²+289=225+30x+x².

We can now substitute the value from the smaller triangle in place of y²:
x²+64+289=225+30x+x².

Combining like terms, we have:
x²+353=225+30x+x².

Subtracting x² from both sides, we have:
x²+353-x²=225+30x+x²-x²
353=225+30x.

Subtract 225 from both sides:
353-225=225+30x-225
128=30x.

Divide both sides by 30:
128/30=30x/30
128/30 = x.

Both the numerator and denominator of this fraction are divisible by 2:
(128/2)/(30/2)=64/15=x.

Now we use this to find y; in our Pythagorean theorem equation from earlier, we have:
x²+64=y².

Using our value for x, we have:
(64/15)²+64=y².

Squaring the fraction, we have:
4096/225+64=y².

In order to write 64 as a fraction using 225 as the denominator, we multiply:
(64*225)/225=14400/225.

This gives us:
4096/225+14400/225=y²

Combining like terms,
18496/225=y².

Taking the square root of both sides, we have:

`\sqrt{(18496)/(225)}=(√(18496))/(√(225))=(136)/(15)=y`