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.
+938 Since QN is perpendicular to PR, we know that triangle QNO is a right triangle. Using the Pythagorean theorem, we can find the length of QO:
\(QO = sqrt(QN^2 + PR^2) = sqrt(12^2 + 14^2) = sqrt(288) = 12√2\)
Now, we can find the length of RO by adding the length of QO and NO:
\(RO = QO + NO = 12√2 + 7 = 19√2\)
Since triangle PQR is isosceles with midpoints M and N, we know that PM = PR/2 = 7 and PQ = PR = 14.
Let's call the height of triangle PQR "h". Then, using the area formula for triangle PQR, we get:
\(Area = (1/2)bh = (1/2)(14)(h)\)
Using the Pythagorean theorem, we can find h in terms of RO:
\(h^2 = PQ^2 - RO^2 = 14^2 - (19√2)^2\)
So,
\(h = sqrt(14^2 - (19√2)^2) = sqrt(196 - 368) = sqrt(-172)\)
Since the square of a real number is always non-negative, we know that this value of h is not possible. Therefore, triangle PQR does not exist.
