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Grade 10 Mathematics Learners’ Book

INNODEMS

Subsection 2.1.1 Similarity

Activity 2.1.1.

Work in pairs
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(a)

Draw triangle \(ABC\) with the following side lengths as shown in the figure below:
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\(AB = 7\) cm, \(AC = 6\)cm, and \(BC = 5\)cm.
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(b)

Label the angles in triangle \(ABC\) as follows:
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\(\angle ABC = 50^\circ\text{,}\) \(\angle BAC = 60^\circ\) ,\(\angle BCA = 70^\circ\)
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(c)

Draw triangle \(PQR\) with the following side lengths as shown in the figure below:
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\(PQ = 21\) cm, \(PR = 18\)cm, and \(QR = 15\)cm.
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(d)

Label the angles in triangle \(PQR\) as follows:
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\(\angle PQR = 50^\circ\text{,}\) \(\angle QRP = 70^\circ\) ,\(\angle QPR = 60^\circ\)
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(e)

Find the ratio of corresponding sides:
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  • \(QR\) to \(BC\) \((QR/BC)\text{.}\)
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  • \(PQ\) to \(AB\) \((PQ/AB)\text{.}\)
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  • \(PR\) to \(AC\) \((PR/AC)\text{.}\)
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(f)

What do you notice about the ratios of corresponding sides above?
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(g)

What do you observe between \(\angle ABC \) and \(\angle PQR \text{,}\) \(\angle BCA \) and \(\angle QRP \) ,\(\angle BAC \) and \(\angle QPR \)
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(h)

What do you observe about the two triangles based on their corresponding sides and angles?
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(i)

What is the relationship between triangle \(ABC\) and triangle \(PQR\text{?}\)
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(j)

Discuss your findings and share your conclusions with the class.
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\(\textbf{Key Takeaway }\)
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  • Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio.
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  • Similarity refers to a relationship between two shapes or figures where one can be transformed into the other through scaling (enlargement or reduction), without changing its shape. The two shapes are similar if they have the same shape but may differ in size.
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Example 2.1.1.

In the figure below, the triangles \(PQR\) and \(ABC\) are similar. Calculate the lengths marked with letters \(x\) and \(y\text{.}\)
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Solution.
\(AB\) corresponds to \(PQ\text{,}\) \(BC\) corresponds to \(QR\text{,}\) and \(AC\) corresponds to \(PR\text{.}\)
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\begin{align*} \text{Therefore,} \amp \frac{AB}{PQ}= \frac{AC}{PR}= \frac {BC}{QR} \\ =\amp \frac{6\, cm}{y}= \frac{9\, cm}{12\, cm}= \frac {x}{14\, cm} \\ \frac{6\, cm}{y}=\amp \frac{9\, cm}{12\, cm}\\ y \times 9\, cm= \amp 6\, cm \times 12\, cm \\ y= \amp \frac{72\, cm^2}{9\, cm} \\ y =\amp 8\, cm \\ \text{Therefore PQ}= \amp 8 cm\\ \frac{9\, cm}{12\, cm}=\amp \frac{x}{14\, cm}\\ x \times 12\, cm= \amp 9\, cm \times 14\, cm \\ x= \amp \frac{126\, cm^2}{12\, cm} \\ x =\amp 10.5\, cm \\ \text{Therefore BC}= \amp 10.5\, cm \end{align*}
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Example 2.1.2.

Given that triangles \(XYZ\) and \(PQR\) in figure 2.1.3 are similar, Find the size of \(\angle\, QPR\text{,}\) \(\angle\, PQR\) and the length of line \(PR\)
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Figure 2.1.3.
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Solution.
Since the two triangles are equal, then their corresponding angles must be equal
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\(\angle\, ZYX\) corresponds to \(\angle\, QPR\)
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Since \(\angle\, ZYX=72^\circ\text{,}\) then
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\(\angle\, QPR= 72^\circ \)
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\(\angle\, YZX\) corresponds to \(\angle\, PQR\)
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Since \(\angle\, YZX=61^\circ\text{,}\) then
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\(\angle\, PQR= 61^\circ \)
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Now to find the length of \(PQ\text{,}\) We use the concept of similarity
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\begin{align*} \frac{YZ}{PQ} =\amp\frac{XY}{PR} \\ \frac{16}{40}= \amp \frac{12}{PR}\\ PR \times 16= \amp 40 \times 12\\ PR= \amp \frac{480}{16}\\ PR= \amp 30\, cm \end{align*}
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Checkpoint 2.1.4. Using Similar Triangles to Find Unknown Side Lengths.

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Checkpoint 2.1.5. Using Similarity to Determine Unknown Lengths and Scale Factor.

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Exercises Exercises

1.

In the triangles below, Determine which triangles are similar by comparing their corresponding sides:
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2.

Given that triangle \(ABC\) is similar to triangle \(DEF\text{,}\) as shown in the diagram below,
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Figure 2.1.6.
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(a). Determine the measure of angle \(\theta\)
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(b). Calculate the value of \(x\)
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4.

Triangle \(ABE\) is similar to triangle \(ACD\text{,}\) as shown in the figure below, Given that \(DC=24\, cm\) , \(AE=6\, cm\)\(ED=12\, cm\text{,}\) determine the length of \(BE\text{.}\)
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Figure 2.1.8.
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