G_Similarity_8_Lesson 8

Introduction

What is similarity? Looking around yourself, you will see many objects which have same shape but different sizes. For examples, leaves of a tree have almost the same shape but different sizes. Similarly, photographs of different sizes, tree of different sizes, book of different size, circle of different sized. But most of these objects are of same shapes. All those objects which have the same shape but not necessarily the same size are called similar objects.

Three similar triangle. 🌴 🌴 🌴

Three similar book. 📕 📕 📕

Three similar rectangle. 🏼 🏼 🏼

Three similar circle. 𖧋 𖧋 𖧋

Three similar sphere. 🌐 🌐 🌐

What is similar figures?

1. Line-segments are similar.
──── ──── ────
2. Circles are similar.
𖧋 𖧋 𖧋
3. Equilateral triangles are similar.
🔺 🔺 🔺
4. Squares are similar.
🔶 🔶 🔶

Any two polygons, with corresponding angles equal and corresponding sides proportional, are similar. Thus, two polygons are similar, if they satisfiy if both of the following conditions:

1. Corresponding angles are equal.
2. The corresponding sides are proportional.

Even if one of the conditions does not hold, the polygons are not similar as in the case of a rectangle and square given below. Here all the corresponding angles are equal but the corresponding sides are not proportional.

Similar Triangles: Corresponding Angles and Proportional Sides

Two triangles \( \triangle ABC \) and \( \triangle DEF \) are said to be similar, denoted as \[ \triangle ABC \sim \triangle DEF, \] if and only if the following two conditions hold:

1. Corresponding angles are equal
   \[ \angle A = \angle D, \quad \angle B = \angle E, \quad \angle C = \angle F. \]
2. Corresponding sides are proportional
   \[ \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}. \]

This definition follows directly from the AA (Angle-Angle) Similarity Criterion: if two angles of one triangle are respectively equal to two angles of another triangle, the triangles are similar. Consequently, the third pair of angles must also be equal (since the sum of interior angles in any triangle is \(180^\circ\)), and the ratios of corresponding side lengths are equal.

Solved Exercise

1. Given that \( BC \parallel DE \), show that \( \triangle ABC \) and \( \triangle ADE \) are similar.[2HA]
 
   From the figure, in \( \triangle ABC \) and \( \triangle ADE \), we can write that

1. \(\angle BAC = \angle DAE\) (common angle)
2. \(\angle ABC = \angle ADE\) (corresponding angles, transversal AB)
3. \(\angle ACB = \angle AED\) (corresponding angles, transversal AC)
Since pair of corresponding angles equal , therefore, by AAA similarity,
\(\triangle ABC \sim \triangle ADE\)
2. Given that \( AB \parallel CD \), prove that \( \triangle AOB \) and \( \triangle COD \) are similar.[2HA]
From the figure, in \( \triangle AOB \) and \( \triangle COD \), we can write that

1. \(\angle AOB = \angle COD\) (vertically opposite angles)
2. \(\angle OAB = \angle OCD\) (alternate interior angles, transversal AD, \(AB \parallel CD\))
3. \(\angle OBA = \angle ODC\) (alternate interior angles, transversal BC, \(AB \parallel CD\))
Since three of corresponding angles are equal, therefore, by AAA similarity,
\(\triangle AOB \sim \triangle COD\)
3. Given that \( BC \parallel PQ \), show that \( \triangle ABC \sim \triangle APQ \).[2HA]
From the figure, in \( \triangle ABC \) and \( \triangle APQ \), we can write that

1. \(\angle BAC = \angle PAQ\) (common angle)
2. \(\angle ABC = \angle APQ\) (corresponding angles, transversal AB, \(BC \parallel PQ\))
3. \(\angle ACB = \angle AQP\) (corresponding angles, transversal AC, \(BC \parallel PQ\))
Since three corresponding angles are equal, therefore, by AA similarity,
\(\triangle ABC \sim \triangle APQ\)
4. Given that \( AB \parallel CD \), prove that \( \triangle AOB \sim \triangle COD \).[2HA]
From the figure, in \( \triangle AOB \) and \( \triangle COD \), we can write that

1. \(\angle AOB = \angle COD\) (vertically opposite angles)
2. \(\angle OAB = \angle OCD\) (alternate interior angles, transversal AD, \(AB \parallel CD\))
3. \(\angle OBA = \angle ODC\) (alternate interior angles, transversal BC, \(AB \parallel CD\))
Since two pairs of corresponding angles are equal, therefore, by AA similarity,
\(\triangle AOB \sim \triangle COD\)
5. In the given figure, \( AB \parallel CD \), \( BO = 6 \, \text{cm} \), \( AB = 8 \, \text{cm} \), and \( CD = 4 \, \text{cm} \). Find the value of \( x \).[2U]
From the figure, \( \triangle ABO \sim \triangle CDO \), so by Proportionality Theorem we have
\(\frac{BO}{OC} = \frac{AB}{CD}\)
or\(\frac{6}{x} = \frac{8}{4}\)
or\(\frac{6}{x} = 2\)
or\(x = \frac{6}{2} = 3\)
or\(x = 3\)cm
6. In the given figure, \( \triangle PQR \) and \( \triangle XYR \) are similar. Find the measure of \( YR \).[2U]
From the figure, \( \triangle PQR \sim \triangle XYR \), so by Proportionality Theorem we have
\(\frac{PQ}{XY} = \frac{QR}{YR}\)
or\(\frac{25}{10} = \frac{15}{YR}\)
or\(\frac{5}{2} = \frac{15}{YR}\)
or\(YR = \frac{15 \times 2}{5} = 6\)
or\(YR = 6\)cm
7. In the given figure, if \( \triangle PQR \sim \triangle PTR \), \( PQ = 6 \, \text{cm} \), and \( PR = 8 \, \text{cm} \), find the measure of \( TR \).[2U]
Here
\(QR=\sqrt{6^2+8^2}=10\)
From the figure, \( \triangle PQR \sim \triangle PTR \), so by Proportionality Theorem we have
\(\frac{QR}{PR}=\frac{PQ}{PT} = \frac{PR}{TR}\)
or\(\frac{10}{8} = \frac{8}{TR}\)
or\(TR= \frac{64}{10}\)
or\(TR= 6.4\)cm

8. If \( \triangle ABC \sim \triangle PQR \) in the given figure, find the values of \( x \) and \( y \). [2U]
Here
\(\triangle ABC \sim \triangle PQR\)
\(\frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}\)
or\(\frac{8}{y} = \frac{10}{15} = \frac{x}{9}\)
or\(\frac{8}{y} = \frac{2}{3} = \frac{x}{9}\)
or\(\frac{8}{y} = \frac{2}{3}\)
or\(y = 8 \times \frac{3}{2} = 12\)
and\(\frac{x}{9} = \frac{2}{3}\)
or\(x = 9 \times \frac{2}{3} = 6\)
or\(x = 6\) cm, \(y = 12\) cm
9. In the given figure, if \( \triangle ADO \sim \triangle BCO \), find the length of \( BC \).
Here
\(\triangle ADO \sim \triangle BCO\)
\(\frac{AO}{BO} = \frac{AD}{BC}\)
or\(\frac{8}{4} = \frac{10}{BC}\)
or\(BC=5\)
10. If \( \triangle PQR \) and \( \triangle PST \) are similar, and \( PQ = 10 \, \text{cm} \), \( ST = 12 \, \text{cm} \), and \( SP = 4 \, \text{cm} \), find the length of \( RQ \).
Here
\(\triangle PQR \sim \triangle PST\)
\(\frac{PQ}{PS} = \frac{PR}{PT} = \frac{RQ}{ST}\)
or\(\frac{10}{4} = \frac{RQ}{12}\)
or\(RQ = 12 \times \frac{10}{4}\)
or\(RQ = 30\) cm
11. In the adjoining figure, if \( \triangle LMN \sim \triangle LEF \), then find the value of \( EF \).
    Also, write the corresponding angles.
Here
\(\triangle LMN \sim \triangle LEF\)
0r\(\frac{LM}{LE} = \frac{LN}{LF} = \frac{MN}{EF}\)
or\(\frac{LN}{LF} = \frac{MN}{EF}\)
or\(\frac{10}{4} = \frac{5}{EF}\)
or\(EF = \frac{5 \times 4}{10} \)
or\(EF = 2\) cm
12. In the adjoining figure, by which axiom are \( \triangle ABO \) and \( \triangle COD \) similar? Also, write the corresponding angles.
Here
\(AB \parallel CD\)
\(\angle OAB = \angle OCD\) (alternate interior angles)
\(\angle OBA = \angle ODC\) (alternate interior angles)
\(\angle AOB = \angle COD\) (vertically opposite angles)
Since three corresponding angles are equal,
\(\triangle ABO \sim \triangle COD\) by AAA axiom

The corresponding angles are
\(\angle OAB = \angle OCD\)
\(\angle OBA = \angle ODC\)
\(\angle AOB = \angle COD\)
13. In the given figure, if \( \triangle AOB \) and \( \triangle COD \) are similar, find the values of \( x \) and \( y \).
Here
\(\triangle AOB \sim \triangle COD\)
\(\frac{AO}{DO} = \frac{BO}{CO} = \frac{AB}{CD}\)
or\(\frac{y}{4} = \frac{3}{x} = \frac{3.5}{7}\)
or\(x=7\)cm and \(y=2\)cm
14. In right-angled \( \triangle PQR \), \( \angle P = 90^\circ \). From vertex \( P \), perpendicular \( PS \) is drawn on base \( QR \). Prove that \( \triangle PQR \sim \triangle SQP \).
From the figure, in \( \triangle PQR \) and \( \triangle SQP \), we can write that

1. \(\angle PQR = \angle PQS\) (common angle)
2. \(\angle PRQ = \angle SPQ=x\)
3. \(\angle QPR = \angle PSQ=90\)
Since three orresponding angles are equal, therefore, by AAA similarity,
\(\triangle PQR \sim \triangle SQP\)
15. In the adjoining figure, \( \angle Y = \angle Z \), \( XY = 20 \, \text{cm} \), \( AY = 15.5 \, \text{cm} \), and \( XZ = 15 \, \text{cm} \).
    (i) Prove that: \( \triangle XAZ \sim \triangle XBY \)
    (ii) Find the measure of \( XB \).
(i) Proof:
\(\angle Y = \angle Z\) (given)
\(\angle XBY=\angle XAZ\) (given 90)
\(\angle YXB=\angle ZXA \) (given 90)
Thus, by AAA similarity,
\(\triangle XAZ \sim \triangle XBY\)

(ii) Find \( XB \):
\(\triangle XAZ \sim \triangle XBY\)
Thus,
\(\frac{XA}{XB} = \frac{XZ}{XY}\)
or\(\frac{15.5}{XB} = \frac{15}{20}\)
or\(XB = 15.5 \times \frac{4}{3}\)
or\(XB = \frac{62}{3}\) cm
16. In the given figure, if \( AB \parallel CD \), prove that \( \triangle CDO \sim \triangle OAB \).
From the figure, in \( \triangle CDO \) and \( \triangle OAB \), we can write that

1. \(\angle COD = \angle AOB\) (vertically opposite angles)
2. \(\angle OCD = \angle OBA\) (alternate interior angles, transversal AD, \( AB \parallel CD \))
3. \(\angle ODC = \angle OAB\) (alternate interior angles, transversal BC, \( AB \parallel CD \))
Since three corresponding angles are equal, therefore, by AAA similarity,
\(\triangle CDO \sim \triangle OAB\)
17. In the given right-angled triangle \( PQR \), \( \angle P = 90^\circ \) and \( MN \perp QR \). Prove that \( \triangle PQR \) and \( \triangle MQN \) are similar.
From the figure, in \( \triangle PQR \) and \( \triangle MQN \), we can write that

1. \(\angle PQR = \angle MQN\) (common)
2. \(\angle QPR = \angleMNQ\) (both 90)
Since two pairs of corresponding angles are equal, therefore, by AA similarity,
\(\triangle PQR \sim \triangle MQN\)
18. In the given figure, if \( \angle QPS = \angle PRQ = 30^\circ \), then prove that \( \triangle PQS \sim \triangle PQR \). Also, find the length of \( QS \).
From the figure, in \( \triangle PQS \) and \( \triangle PQR \), we can write that

1. \(\angle QPS = \angle PRQ = 30^\circ\) (given)
2. \(\angle PQS = \angle PQR\) (common angle at Q)
Since two pairs of corresponding angles are equal, therefore, by AA similarity,
\(\triangle PQS \sim \triangle PQR\)

Now, find \( QS \):
\(\triangle PQS \sim \triangle PQR\)
\(\frac{PQ}{QS} = \frac{QR}{PQ}\)
or\(\frac{8}{QS} = \frac{16}{8}\)
or\(QS = 4\) cm