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# geometry

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In triangle PQR, let M be the midpoint of QR, let N be the midpoint of PR, and let O be the intersection of QN and RM, as shown. If QN perp PR, QN = 12, and PR = 14, then find the area of triangle PQR.

Mar 5, 2023

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Since QN is perpendicular to PR, we know that triangle PQR is a right triangle with right angle at Q. Therefore, we can use the Pythagorean theorem to find the length of PQ as:

PQ^2 = PR^2 - QR^2 = 14^2 - (2MN)^2

Since M is the midpoint of QR, we have MN = NR = QR/2. Therefore:

PQ^2 = 14^2 - (QR^2/2)^2

Also, since N is the midpoint of PR, we have PN = NR = PR/2 = 7. Therefore, using the Pythagorean theorem in triangle QPN, we have:

QN^2 = PQ^2 + PN^2 = PQ^2 + 7^2

Substituting the expression for PQ^2 from above, we have:

QN^2 = 14^2 - (QR^2/2)^2 + 7^2

Since O is the intersection of QN and RM, we know that triangle QOR is similar to triangle QPN. Therefore, we have:

OR / QN = PN / PQ

Substituting the values OR = RM = QR/2 and PN = 7, we obtain:

QR/2 / QN = 7 / PQ

Solving for PQ, we obtain:

PQ = 14QN / QR

Substituting this expression for PQ into the equation for QN^2 above, we obtain:

QN^2 = 14^2 - (QR^2/2)^2 + 7^2 = (14QR)^2 / (QR^2 + 4)

Multiplying both sides by QR^2 + 4, we obtain:

QN^2 * (QR^2 + 4) = 14^2 QR^2

Expanding and simplifying, we obtain a quadratic equation in QR^2:

QR^4 - 14^2 QR^2 + 4*QN^2 = 0

Substituting the given values QN = 12, PR = 14, and NR = 7, we obtain:

QR^4 - 14^2 QR^2 + 4*12^2 = 0

Solving this quadratic equation, we obtain:

QR^2 = 49 +- 35

Since QR is a length, we must take the positive square root, so:

QR^2 = 84

Substituting this value into the expression for PQ^2 above, we obtain:

PQ^2 = 14^2 - (QR^2/2)^2 = 14^2 - 42 = 112

Therefore, the area of triangle PQR is:

(1/2) PQ * PR = (1/2) * 14 * sqrt(112) = 14sqrt(7)