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# Another triangle problem with no explanation last time I asked

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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?

Apr 22, 2020

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The area of triangle LPQ is 189.

Apr 22, 2020
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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?

$$\begin{array}{|rclcrcl|} \hline \text{area}\ [JKL] &=& A \\ \text{area}\ [MNO] &=& \frac14 A \\ \text{area}\ [PQR] &=& \frac14 *[MNO] \\ &=& \frac14 \cdot \frac14 A \\ [PQR] &=& \frac{1}{16} A \quad | \quad [PQR]= 21 \\ 21 &=& \frac{1}{16} A \\ 21*16 &=& A \\ \mathbf{A} &=& \mathbf{16\cdot 21} \\\\ \text{area}\ [LPQ] &=& \frac{JK}{4} \cdot \frac{h}{2} \quad & | & \sin(J) &=& \frac{h}{ \frac{JL}{2} +\frac{JL}{4} } \\ & & & | & \sin(J) &=& \frac{h}{ \frac34 * JL } \\ & & & | & h &=& \frac34 *JL \sin(J) \\ \triangle LPQ &=& \frac{JK}{4} \cdot \frac{ \frac34 JL \sin(J) }{2} \\ &=& \frac{3}{32} \cdot JK\cdot JL\cdot \sin(J) \quad & | & JK\cdot JL\cdot \sin(J) &=& 2A \\ &=& \frac{3}{32} \cdot 2A \\ &=& \frac{6}{32} A \quad & | & A &=& 16\cdot 21 \\ &=& \frac{6}{32} \cdot 16\cdot 21 \\ &=& \frac{6}{2} \cdot 21 \\ &=& 3 \cdot 21 \\ \mathbf{\text{area}\ [LPQ]} &=& \mathbf{63} \\ \hline \end{array}$$

Apr 22, 2020
edited by heureka  Apr 22, 2020
edited by heureka  Apr 22, 2020