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.
+941 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(QN2+PR2)=sqrt(122+142)=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:
h2=PQ2−RO2=142−(19√2)2
So,
h=sqrt(142−(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.
