Points $M$, $N$, and $O$ are the midpoints of sides $\overline{KL}$, $\overline{LJ}$, and $\overline{JK}$, respectively, of triangle $JKL$. Points $P$, $Q$, and $R$ are the midpoints of $\overline{NO}$, $\overline{OM}$, and $\overline{MN}$, respectively. If the area of triangle $PQR$ is $21$, then what is the area of triangle $LPQ$?

The midpoint theorem states that the line joining the mid points of two sides of a triangle is parallel to the third and facing side and equal to half of the length of the third side

Based on the midpoint theorem, the area of triangle ΔLPQ is 63 square units

The reason the value of the area of triangle ΔLPQ as given above is correct is as follows:

The given parameters;

The midpoint of = M; The midpoint of = N; The midpoint of = O

The midpoint of = P; The midpoint of = Q; The midpoint of = R

The area of triangle ΔPQR = 21

The required parameter:

Calculate the area of triangle ΔLPQ

Method:

The definition of midpoint, area ratio, and area of a triangle formula can be used to find the area of triangle ΔLPQ

Solution:

According to the midpoint theorem, we have;

= (1/2) ×

= (1/2) ×

= (1/2) ×

Given that is parallel to , and is parallel to, we have;

∠MON = ∠PRQ

Similarly, we have, ∠MNO = ∠PQR

Therefore, ΔPQR is similar to triangle ΔMON, which is also similar to ΔJKS

The area of triangle ΔPQR = 21, by area ratio = (Side ratio)², we have;

The sides of ΔMON = 2 × The side length of ΔPQR

The area of triangle ΔMON = 2² × The area of ΔPQR

∴ The area of triangle ΔMON = 4 × 21

Similarly the area of ΔJKS = 4 × 4 × 21

PQ = JK/4

The area of LPQ = (1/2) × PQ × h

h = (3/4×JL) × sin(x°)

∴ The area of LPQ = (1/2)×JK/4×(3/4×JL) × sin(x°)

However; (1/2)×JK×JL× sin(x°) = Area of ΔJKS = 4 × 4 × 21

Therefore;

The area of ΔLPQ = (Area of ΔJKS)/4×(3/4) = (4 × 4 × 21)/4×(3/4) = 63

The area of triangle ΔLPQ = 63 square units

Learn more about the midpoint theorem here:

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8