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Sagot :
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 [tex]\overline {KL}[/tex] = M; The midpoint of [tex]\overline {LJ}[/tex] = N; The midpoint of [tex]\overline {JK}[/tex] = O
The midpoint of [tex]\overline {NO}[/tex] = P; The midpoint of [tex]\overline {OM}[/tex] = Q; The midpoint of [tex]\overline {MN}[/tex] = 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;
[tex]\overline {QR}[/tex] = (1/2) × [tex]\overline {NO}[/tex]
[tex]\overline {PR}[/tex] = (1/2) × [tex]\overline {OM}[/tex]
[tex]\overline {PQ}[/tex] = (1/2) × [tex]\overline {MN}[/tex]
Given that [tex]\overline {QR}[/tex] is parallel to [tex]\overline {ON}[/tex], and [tex]\overline {PR}[/tex] 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:
https://brainly.com/question/15227899
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