Points M, N, and O are the midpoints of sides KL, LJ, and JK, respectively, of triangle JKL. Points P, Q, and R are the midpoints of NO, OM, and MN, respectively. If the area of triangle PQR is 12, then what is the area of triangle JQR?

 Apr 13, 2024

We can solve this problem by recognizing that subdividing a triangle into smaller similar triangles reduces the area by a factor based on the ratio of side lengths.


Similarity of Triangles:


Since M, N, and O are midpoints of sides, segments NO, OM, and MN are parallel to sides KL, LJ, and JK respectively (corresponding sides theorem).


Due to alternate interior angles, angles in triangles JKL, JMN, JNO, etc. are congruent (alternate interior angles theorem).


Therefore, triangles JKL, JMN, JNO, etc. are similar by Angle-Angle (AA) Similarity.


Ratio of Side Lengths:


Since M, N, and O are midpoints, segments NO, OM, MN, PN, QM, and PR are all half the length of their corresponding sides in the larger triangle (JKL).


Area Ratio of Similar Triangles:


The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding side lengths.


Let k be the scaling factor between triangles JKL and PQR (i.e., the ratio of side lengths between corresponding sides).


The area of triangle PQR is 12.


Relating Areas of JKL and PQR:


Area(PQR) / Area(JKL) = k^2 (ratio of areas of similar triangles)


Substitute the given value:


12 / Area(JKL) = k^2


Relating Side Lengths:


Since PN, QM, and PR are half the length of their corresponding sides in JKL, k = 1/2 (ratio of side lengths).


Finding Area(JKL):


Substitute k = 1/2 in the equation from step 4:


12 / Area(JKL) = (1/2)^2


Multiply both sides by Area(JKL):


12 = Area(JKL) / 4


Solve for Area(JKL):


Area(JKL) = 12 * 4 = 48


Therefore, the area of triangle JKL is 48 square units.

 Apr 13, 2024

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