Answer:
So, this triangle PQR can be broken into two right triangles, PNQ and PNR, with legs PQ = 39, PN =15, and QN = ? and PR = 17, PN = 15, and NR =? respectively.
Let's solve for what is easier first:
Since we know that 5-15-17 is a Pythagorean triplet, we can infer that NR is 5....like I said earlier, it is a right triangle, so this guess holds true.
Here comes the interesting part:
Now, we have one part of QR, which is QN.
The other part can be solved by using the Pythagorean theorem.
It is (39^2-15^2)^(1/2)..which gives you 36, the square root of 1296, which happens to be the difference between the squares of 15 and 39.
SO, QR = QN + NR
5+36 = 41
QR = 41.
Hope this helps!
Answer : The value of side QR is, 44 in.
Step-by-step explanation :
As we know that an altitude of a triangle is a segment from a vertex to the line containing its opposite side, and is perpendicular to that line.
First we have to determine the side QN.
Using Pythagoras theorem in ΔPNQ :
[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]
[tex](PQ)^2=(PN)^2+(QN)^2[/tex]
Given:
Side PQ = 39
Side PN = 15
Now put all the values in the above expression, we get the value of side QN.
[tex](39)^2=(15)^2+(QN)^2[/tex]
[tex]QN=\sqrt{(39)^2-(15)^2}[/tex]
[tex]QN=36[/tex]
Now we have to determine the side RN.
Using Pythagoras theorem in ΔPNR :
[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]
[tex](PR)^2=(PN)^2+(RN)^2[/tex]
Given:
Side PR = 17
Side PN = 15
Now put all the values in the above expression, we get the value of side RN.
[tex](17)^2=(15)^2+(RN)^2[/tex]
[tex]RN=\sqrt{(17)^2-(15)^2}[/tex]
[tex]RN=8[/tex]
As,
Side QR = Side QN + Side RN
Side QR = 36 + 8
Side QR = 44
Thus, the value of side QR is, 44 in.
