Question

the triangles are similar. solve for x

Answer 1

Answer:

x= 6.25

Step-by-step explanation:

The hypotenuse of triangle JKL is taken as 96, in the similar way the hypotenuse of triangle JTU is 36. So x can be found by first finding the value of JU, using the Pythagoras theorem.

$a^{2} = b^{2} + c^{2}$

In this shape the formula will be:

$TU^{2} = JT^{2} + JU^{2}$

Substitute the given values in the shape into the formula.

TU = 34

JT = 27

TU = Lets take 'y'

$34^{2} = 27^{2} + y^{2}$

1156 = 729 + $y^{2}$

$y^{2}$ = 1156 - 729

$y^{2}$ = 427

y = $\sqrt{427}$ = 20.7 ( rounded to 1 decimal place) = 21 (rounded to whole number).

So the original value of JU is '21', but we have to find the value of 'x'. So the expression '-4 + 4x' is equal to 21. This can be written as:

-4 + 4x = 21

4x = 21 + 4

4x = 25

∴ x = $\frac{25}{4}$ = 6.25

x= 6.25