In the figure PS = 3 SQ=6 QR=5 PT=x TR=y give any two pairs of values of x and y such that line ST|| side QR
Answers
Given : PS = 3, SQ=6 QR = 5, PT = x & TR = y
To Find : the pair of value of x &y such that ST II side QR.
Solution:
Thales theorem ( BPT)
if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
ST II QR
=> PS/SQ = PT / TR
=> 3 / 6 = x/y
=> 1/2 = x/ y
=> y = 2x
PR = PT + TR = x + 2x = 3x
in ΔPQR
PQ = PS + SQ = 3 + 6 = 9
QR = 5
PR = 3x
Sum of two sides of a triangle is > third side
=> 5 + 3x > 9
=> 3x > 4
also
5 + 9 > 3x
=> 14 > 3x
=> 4 < 3x < 14
=> x = 2 , 3 , 4 ( considering integral values only )
y = 4 , 6 , 8
the pair of value of x & y = ( 2 , 4) , ( 3 , 6) and ( 4, 8)
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Answer:
Step-by-step explanation:
It is given that,
ST // QR
PS = 3
SQ = 6
QR = 5
PT = x
TR = y
We know that → if a line is parallel to one side of the triangle intersects the other two sides, then it divides the other two sides proportionally.
Based on the above theorem, we get
∴ on substituting the given values
⇒ 3/6 = x/y
⇒ 1/2 = x/y
⇒ y = 2x
In Δ PQR, we have
PQ = PS + SQ = 3 + 6 = 9
and
PR = PT + TR = x + y = x + 2x = 3x
QR = 5
Also, we know that → the sum of any two sides of a triangle is greater than the 3rd side.
So, based on this, we get
PQ + QR > PR
⇒ 9 + 5 > 3x
⇒ 14 > 3x . . . . Equation 1
and
PR + QR > PQ
⇒ 3x + 5 > 9
⇒ 3x > 4 . . . . Equation 2
From equation 1 and equation 2, we get
4 < 3x < 14
Now,
The possible pair of values (considering integral values only) of x and y can be:
If x = 2 then y = 4 ⇒ value pair → (2, 4)
If x = 3 then y = 6 ⇒ value pair → (3, 6)
If x = 4 then y = 8 ⇒ value pair → (4, 8)
Thus, any two pair of values of x and y that satisfies the line ST // side QR will be (2, 4) & (3, 6).