Question

Anyone please help me

Answer 1

Answer:

3

Step-by-step explanation:

these are 2 similar triangles (all 3 pairs of corresponding angles are equal in their pair, and the lengths of corresponding sides have the same ratio in every pair of corresponding sides and other lengths).

and they are also right-angled triangles.

so, what else do we know about them ?

PQ = 18

RT = 6

ST = QT + 9

therefore

QT = ST - 9

and

QR = QT + RT = ST - 9 + 6 = ST - 3

due to the principles of cumulative triangles

PQ/ST = QR/RT = 18/ST = (ST - 3)/6

108/ST = (ST - 3)

108 = (ST - 3)ST = ST² - 3ST

ST² - 3ST - 108 = 0

we can the write

x² - 3x - 108 = 0

and the solution for such a squared equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = -3

c = -108

x = (3 ± sqrt(9 - 4×1×-108))/2 = (3 ± sqrt(441))/2 =

= (3 ± 21)/2

x1 = (3+21)/2 = 24/2 = 12

x2 = (3-21)/2 = -18/2 = -9

a negative solution for real lengths is not valid, so we know

x = ST = 12

so,

QR = ST - 3 = 12 - 3 = 9

and then

QT = QR - RT = 9 - 6 = 3