Question

What are the values of x and y?

Answer 1

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:
$\sqrt{\frac{18496}{225}}=\frac{\sqrt{18496}}{\sqrt{225}}=\frac{136}{15}=y$

Answer 2

Answer:  

D. $x=\frac{64}{15}, y=\frac{136}{15}$

Step-by-step 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,

$17^2+x^2+8^2=(15+x)^2$  

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

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

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

$353=225+30x$  

$353-225=225-225+30x$  

$128=30x$  

$\frac{128}{30}=\frac{30x}{50}$

$\frac{64*2}{15*2}=x$  

$x=\frac{64}{15}$

Upon substituting $x=\frac{64}{15}$ in equation (2) we will get,

$(\frac{64}{15})^2+8^2=y^2$  

$\frac{64^2}{15^2}+64=y^2$    

$\frac{4096}{225}+\frac{14400}{225}=y^2$  

$\frac{4096+14400}{225}=y^2$  

$\frac{18496}{225}=y^2$

Taking square root of both sides we will get,

$\sqrt{\frac{18496}{225}}=y$

$\frac{136}{15}=y$

Therefore, the value of x is $\frac{64}{15}$, value y is $\frac{136}{15}$ and option D is the correct choice.