Question
A triangle \( K L M \) is similar to triangle \( P Q R \cdot K L=5 \mathrm{~cm}, L M=40 \mathrm{cn} \) \( \angle K L M=46^{\circ} \). Castruct triangle \( P Q R \) sach that \( Q R=6 \mathrm{~cm} \)
Ask by Cummings Long. in Kenya
Mar 19,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
To construct triangle \( PQR \) similar to triangle \( KLM \):
1. **Determine the Scale Factor:**
- Given \( KL = 5\,\text{cm} \) and \( QR = 6\,\text{cm} \), the scale factor \( k = \frac{6}{5} = 1.2 \).
2. **Find Corresponding Sides:**
- \( PQ = k \times KL = 1.2 \times 5 = 6\,\text{cm} \).
- \( PR = k \times KM \), where \( KM \) is calculated using the Law of Cosines:
\[
KM = \sqrt{5^2 + 40^2 - 2 \times 5 \times 40 \times \cos46^\circ} \approx 36.73\,\text{cm}
\]
Thus, \( PR = 1.2 \times 36.73 \approx 44.08\,\text{cm} \).
3. **Construct Triangle \( PQR \):**
- Draw base \( QR = 6\,\text{cm} \).
- At point \( Q \), construct a \( 46^\circ \) angle.
- From point \( Q \), measure \( PQ = 6\,\text{cm} \) along the angle.
- From point \( P \), draw an arc with radius \( PR \approx 44.08\,\text{cm} \) to locate point \( R \).
4. **Final Triangle:**
- Connect points \( P \), \( Q \), and \( R \) to form triangle \( PQR \) with sides \( PQ = 6\,\text{cm} \), \( QR = 6\,\text{cm} \), and \( PR \approx 44.08\,\text{cm} \), and angle \( \angle PQR = 46^\circ \).
Solution
1. Establish the correspondence between the triangles. Assume that
\[
K \leftrightarrow P,\quad L \leftrightarrow Q,\quad M \leftrightarrow R.
\]
Then side \(KL\) corresponds to side \(PQ\) and side \(LM\) corresponds to side \(QR\).
2. Given that
\[
KL = 5\,\text{cm},\quad LM = 40\,\text{cm},\quad \angle KLM = 46^\circ,
\]
and that in triangle \(PQR\) we have
\[
QR = 6\,\text{cm},
\]
determine the similarity (scale) factor from triangle \(KLM\) to triangle \(PQR\).
3. Since \(LM \leftrightarrow QR\), the scale factor is
\[
k = \frac{QR}{LM} = \frac{6}{40} = \frac{3}{20} = 0.15.
\]
4. Determine the other corresponding sides in triangle \(PQR\):
a. The side corresponding to \(KL\) is
\[
PQ = k \cdot KL = 0.15 \times 5 = 0.75\,\text{cm}.
\]
b. The third side of triangle \(KLM\) (side \(KM\)) can be found using the Law of Cosines. In triangle \(KLM\):
\[
KM^2 = KL^2 + LM^2 - 2\cdot KL\cdot LM\cos \angle KLM.
\]
Substitute the values:
\[
KM^2 = 5^2 + 40^2 - 2 \cdot 5 \cdot 40\cos 46^\circ = 25 + 1600 - 400\cos 46^\circ.
\]
Thus,
\[
KM = \sqrt{1625 - 400\cos 46^\circ}\,.
]
c. The side corresponding to \(KM\) in triangle \(PQR\) is
\[
PR = k\cdot KM = 0.15\sqrt{1625 - 400\cos 46^\circ}\,.
]
An approximate numerical evaluation gives:
- First, evaluate \(400\cos46^\circ\). Using \(\cos46^\circ\approx 0.6947\):
\[
400\cos46^\circ\approx 277.88\,.
\]
- Then,
\[
KM \approx \sqrt{1625 - 277.88} = \sqrt{1347.12} \approx 36.73\,\text{cm}\,.
\]
- Finally,
\[
PR \approx 0.15 \times 36.73 \approx 5.51\,\text{cm}\,.
\]
5. Summary of the sides in triangle \(PQR\):
- \(PQ \approx 0.75\,\text{cm}\)
- \(QR = 6\,\text{cm}\)
- \(PR \approx 5.51\,\text{cm}\)
and the corresponding angle \(\angle PQR = 46^\circ\).
6. Construction Steps for Triangle \(PQR\):
a. Draw the base \(QR = 6\,\text{cm}\).
b. At point \(Q\), construct an angle of \(46^\circ\).
c. On the ray forming the \(46^\circ\) angle from \(Q\), mark point \(P\) such that \(PQ = 0.75\,\text{cm}\).
d. With a compass, using point \(P\) as center, mark an arc with radius \(PR \approx 5.51\,\text{cm}\).
e. The intersection of this arc with the ray from \(Q\) (or with an arc centered at \(R\) if drawn from \(R\) with radius \(PR\)) locates point \(P\) precisely completing the triangle.
Thus, triangle \(PQR\) constructed with \(QR=6\,\text{cm}\), the angles and proportional sides corresponding to triangle \(KLM\) is obtained.
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Simplify this solution Bonus Knowledge
To construct triangle \( PQR \) similar to triangle \( KLM \), you need to establish a scale factor based on the lengths of corresponding sides. Since \( KL = 5 \, \text{cm} \) corresponds with \( QR = 6 \, \text{cm} \), the scale factor between triangles is \( \frac{QR}{KL} = \frac{6}{5} = 1.2 \). Consequently, the length \( LM \) in triangle \( PQR \) should be \( LM \cdot \text{scale factor} = 40 \cdot 1.2 = 48 \, \text{cm} \). To actually construct \( PQR \), start by drawing a line segment \( QR = 6 \, \text{cm} \). Then, using a protractor, construct an angle of \( 46^\circ \) at point \( Q \). From this point, mark a length of \( 48 \, \text{cm} \) along the angle to locate point \( P \). Finally, connect points \( P \) and \( R \) to complete triangle \( PQR \). Enjoy your triangle construction!
