5. Congruency and similarities

# 5. Congruency and similarities ## Congruency and congruent plane figures ### Introduction Welcome to the lesson on Congruency and Congruent Plane Figures! In geometry, we often compare different shapes and sizes. Sometimes, two figures might look exactly alike, even if one is rotated or flipped. When figures have the *exact* same size and shape, we say they are **congruent**. Understanding congruency is fundamental in mathematics and helps us analyze geometric relationships more deeply. ### Key Concepts #### What is Congruency? Two geometric figures are said to be **congruent** if they have the same size and the same shape. This means that if you were to place one figure on top of the other, they would perfectly overlap. The symbol for congruency is \( \cong \). For example, two identical coins are congruent, two identical squares are congruent, and two identical triangles are congruent. #### Properties of Congruent Figures When two figures are congruent, their **corresponding parts** are equal. This means: * **Corresponding sides** are equal in length. * **Corresponding angles** are equal in measure. #### Congruent Plane Figures ##### Congruent Polygons Two polygons are congruent if their corresponding sides are equal in length and their corresponding angles are equal in measure. For example, if pentagon \( ABCDE \) is congruent to pentagon \( PQRST \) (written as \( ABCDE \cong PQRST \)), then: * \( AB = PQ \), \( BC = QR \), \( CD = RS \), \( DE = ST \), \( EA = TP \) * \( \angle A = \angle P \), \( \angle B = \angle Q \), \( \angle C = \angle R \), \( \angle D = \angle S \), \( \angle E = \angle T \) ##### Congruent Triangles Triangles are a special type of polygon, and their congruency is a very important concept. When two triangles are congruent, their vertices must be listed in an order that shows their corresponding parts. If \( \triangle ABC \cong \triangle DEF \), it means: * **Corresponding Vertices:** \( A \) corresponds to \( D \), \( B \) to \( E \), and \( C \) to \( F \). * **Corresponding Sides:** \( AB = DE \), \( BC = EF \), \( AC = DF \). * **Corresponding Angles:** \( \angle A = \angle D \), \( \angle B = \angle E \), \( \angle C = \angle F \). #### Criteria for Congruent Triangles We don't always need to check all three sides and all three angles to prove that two triangles are congruent. There are five specific criteria that allow us to determine congruency with fewer measurements: 1. **Side-Side-Side (SSS) Congruence Criterion:** If three sides of one triangle are equal in length to the three corresponding sides of another triangle, then the two triangles are congruent. \[ \text{If } AB = DE, BC = EF, AC = DF, \text{ then } \triangle ABC \cong \triangle DEF \] 2. **Side-Angle-Side (SAS) Congruence Criterion:** If two sides and the *included* angle (the angle between the two sides) of one triangle are equal to two corresponding sides and the included angle of another triangle, then the two triangles are congruent. \[ \text{If } AB = DE, \angle B = \angle E, BC = EF, \text{ then } \triangle ABC \cong \triangle DEF \] 3. **Angle-Side-Angle (ASA) Congruence Criterion:** If two angles and the *included* side (the side between the two angles) of one triangle are equal to two corresponding angles and the included side of another triangle, then the two triangles are congruent. \[ \text{If } \angle B = \angle E, BC = EF, \angle C = \angle F, \text{ then } \triangle ABC \cong \triangle DEF \] 4. **Angle-Angle-Side (AAS) or SAA Congruence Criterion:** If two angles and a *non-included* side of one triangle are equal to two corresponding angles and the non-included side of another triangle, then the two triangles are congruent. \[ \text{If } \angle A = \angle D, \angle B = \angle E, BC = EF, \text{ then } \triangle ABC \cong \triangle DEF \] 5. **Right-angle-Hypotenuse-Side (RHS) Congruence Criterion:** This criterion applies specifically to **right-angled triangles**. If the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and the corresponding side of another right-angled triangle, then the two triangles are congruent. \[ \text{If } \angle B = \angle E = 90^\circ, AC = DF \text{ (hypotenuses)}, BC = EF \text{ (sides)}, \text{ then } \triangle ABC \cong \triangle DEF \] ### Vocabulary * **Congruent:** Having exactly the same size and shape. * **Corresponding Sides:** Sides that occupy the same relative position in congruent figures. * **Corresponding Angles:** Angles that occupy the same relative position in congruent figures. * **Included Angle:** An angle formed by two specific sides of a polygon. * **Included Side:** A side that connects the vertices of two specific angles in a polygon. * **Hypotenuse:** The longest side of a right-angled triangle, opposite the right angle. ### Examples #### Example 1: Identifying Congruent Figures Are these figures congruent? 1. Two squares, one with side length 5 cm and another with side length 5 cm. **Yes**, they are congruent. 2. Two circles, one with radius 3 cm and another with radius 4 cm. **No**, they are not congruent (different sizes). 3. Two equilateral triangles, one with side length 6 cm and another with side length 6 cm. **Yes**, they are congruent. #### Example 2: Proving Triangle Congruence (SSS) Given \( \triangle PQR \) with sides \( PQ = 6 \text{ cm} \), \( QR = 8 \text{ cm} \), \( PR = 10 \text{ cm} \), and \( \triangle XYZ \) with sides \( XY = 6 \text{ cm} \), \( YZ = 8 \text{ cm} \), \( XZ = 10 \text{ cm} \). Prove that \( \triangle PQR \cong \triangle XYZ \). **Solution:** 1. \( PQ = XY \) (Given as 6 cm) 2. \( QR = YZ \) (Given as 8 cm) 3. \( PR = XZ \) (Given as 10 cm) Since all three corresponding sides are equal, by the SSS Congruence Criterion, \( \triangle PQR \cong \triangle XYZ \). #### Example 3: Proving Triangle Congruence (SAS) Given \( \triangle ABC \) and \( \triangle DEF \). \( AB = DE = 7 \text{ cm} \), \( \angle B = \angle E = 50^\circ \), and \( BC = EF = 9 \text{ cm} \). Prove that \( \triangle ABC \cong \triangle DEF \). **Solution:** 1. \( AB = DE \) (Given as 7 cm) 2. \( \angle B = \angle E \) (Given as \( 50^\circ \) and it is the included angle) 3. \( BC = EF \) (Given as 9 cm) Since two corresponding sides and their included angle are equal, by the SAS Congruence Criterion, \( \triangle ABC \cong \triangle DEF \). #### Example 4: Proving Triangle Congruence (ASA) Given \( \triangle GHI \) and \( \triangle JKL \). \( \angle G = \angle J = 75^\circ \), \( GH = JK = 12 \text{ cm} \), and \( \angle H = \angle K = 40^\circ \). Prove that \( \triangle GHI \cong \triangle JKL \). **Solution:** 1. \( \angle G = \angle J \) (Given as \( 75^\circ \)) 2. \( GH = JK \) (Given as 12 cm and it is the included side) 3. \( \angle H = \angle K \) (Given as \( 40^\circ \)) Since two corresponding angles and their included side are equal, by the ASA Congruence Criterion, \( \triangle GHI \cong \triangle JKL \). #### Example 5: Proving Triangle Congruence (RHS) Consider two right-angled triangles, \( \triangle MNO \) (right-angled at N) and \( \triangle PQR \) (right-angled at Q). If the hypotenuse \( MO = PR = 13 \text{ cm} \) and a side \( NO = QR = 5 \text{ cm} \). Prove that \( \triangle MNO \cong \triangle PQR \). **Solution:** 1. \( \angle N = \angle Q = 90^\circ \) (Both are right-angled triangles) 2. \( MO = PR \) (Hypotenuses are equal, given as 13 cm) 3. \( NO = QR \) (Corresponding sides are equal, given as 5 cm) Since the triangles are right-angled, their hypotenuses are equal, and one pair of corresponding sides are equal, by the RHS Congruence Criterion, \( \triangle MNO \cong \triangle PQR \). ### Exercises 1. State whether the following pairs of figures are congruent or not congruent. Give a reason for your answer. a. Two regular hexagons with side length 4 cm. b. A rectangle with length 8 cm and width 5 cm, and another rectangle with length 5 cm and width 8 cm. c. Two triangles with angles \( 30^\circ, 60^\circ, 90^\circ \). 2. Given that \( \triangle ABC \cong \triangle XYZ \): a. Which side corresponds to \( AC \)? b. Which angle corresponds to \( \angle B \)? c. If \( BC = 12 \text{ cm} \), what is the length of \( YZ \)? 3. In the figure below, identify the congruence criterion (SSS, SAS, ASA, AAS, RHS) that proves \( \triangle PQS \cong \triangle RQS \) if: a. \( PQ = RQ \), \( PS = RS \), and \( QS \) is common. b. \( PQ = RQ \), \( \angle PQS = \angle RQS \), and \( QS \) is common. c. \( \angle P = \angle R = 90^\circ \), \( PS = RS \), and \( QS \) is common. ### Assignment 1. In the figure, \( AD \) and \( BC \) bisect each other at \( E \). Prove that \( \triangle ABE \cong \triangle DCE \). *Hint: Remember what "bisect" means and look for vertically opposite angles.* 2. Given a square \( ABCD \). Points \( E \) and \( F \) are on \( AB \) and \( BC \) respectively, such that \( AE = BF \). Prove that \( \triangle ADE \cong \triangle BCF \). 3. A ladder is leaning against a vertical wall. The base of the ladder is 3 meters from the wall, and it reaches a height of 4 meters on the wall. Another ladder of the same length is leaning against another wall. Its base is 3 meters from the wall. Are the two right-angled triangles formed by the ladders, walls, and ground congruent? Justify your answer using a congruence criterion. ### Summary In this lesson, we explored the concept of **congruency**, which means having the exact same size and shape. We learned that congruent figures have equal corresponding sides and angles. We focused on congruent triangles and discovered the five key criteria to prove triangle congruency: * **SSS** (Side-Side-Side) * **SAS** (Side-Angle-Side) * **ASA** (Angle-Side-Angle) * **AAS** (Angle-Angle-Side) * **RHS** (Right-angle-Hypotenuse-Side) Mastering these criteria is essential for solving problems involving geometric proofs and understanding the relationships between different shapes in mathematics. ## Congruency in triangles (SSS, SAS, ASA, RHS) ### Introduction Welcome to this lesson on congruency in triangles! In geometry, we often compare shapes to see if they are identical. When two figures are exactly the same size and shape, we say they are congruent. In this lesson, we will focus specifically on triangles and learn about four important conditions that help us determine if two triangles are congruent without having to check every side and every angle. ### Key Concepts #### What is Congruency? Two geometric figures are **congruent** if they have the same size and shape. This means that one figure can be perfectly superimposed on the other. If two triangles are congruent, all their corresponding sides are equal in length, and all their corresponding angles are equal in measure. #### Conditions for Congruent Triangles To prove that two triangles are congruent, we don't need to show that all six corresponding parts (three sides and three angles) are equal. We only need to satisfy one of the following four conditions: ##### 1. SSS (Side-Side-Side) Congruence Criterion If the three sides of one triangle are equal in length to the three corresponding sides of another triangle, then the two triangles are congruent. * **Condition:** Side1 = Side1', Side2 = Side2', Side3 = Side3' * **Example:** If triangle ABC has sides AB, BC, CA and triangle PQR has sides PQ, QR, RP. If AB = PQ, BC = QR, and CA = RP, then \( \triangle ABC \cong \triangle PQR \). ##### 2. SAS (Side-Angle-Side) Congruence Criterion If two sides and the **included angle** (the angle between the two sides) of one triangle are equal to the corresponding two sides and included angle of another triangle, then the two triangles are congruent. * **Condition:** Side1 = Side1', Included Angle = Included Angle', Side2 = Side2' * **Example:** If in \( \triangle ABC \) and \( \triangle PQR \), we have AB = PQ, \( \angle B = \angle Q \), and BC = QR, then \( \triangle ABC \cong \triangle PQR \). ##### 3. ASA (Angle-Side-Angle) Congruence Criterion If two angles and the **included side** (the side between the two angles) of one triangle are equal to the corresponding two angles and included side of another triangle, then the two triangles are congruent. * **Condition:** Angle1 = Angle1', Included Side = Included Side', Angle2 = Angle2' * **Example:** If in \( \triangle ABC \) and \( \triangle PQR \), we have \( \angle B = \angle Q \), BC = QR, and \( \angle C = \angle R \), then \( \triangle ABC \cong \triangle PQR \). ##### 4. RHS (Right-angle-Hypotenuse-Side) Congruence Criterion This criterion applies only to **right-angled triangles**. If the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and the corresponding side of another right-angled triangle, then the two triangles are congruent. * **Condition:** Right Angle = Right Angle (always \(90^\circ\)), Hypotenuse = Hypotenuse', Side = Side' * **Example:** If in right-angled \( \triangle ABC \) (with \( \angle B = 90^\circ \)) and right-angled \( \triangle PQR \) (with \( \angle Q = 90^\circ \)), we have AC = PR (hypotenuses) and AB = PQ (one corresponding side), then \( \triangle ABC \cong \triangle PQR \). ### Vocabulary * **Congruent:** Having the same size and shape. * **Corresponding Sides:** Sides that are in the same relative position in two different figures. * **Corresponding Angles:** Angles that are in the same relative position in two different figures. * **Included Angle:** The angle formed by two specific sides of a triangle. * **Included Side:** The side that is between two specific angles of a triangle. * **Hypotenuse:** The longest side of a right-angled triangle, opposite the right angle. ### Examples #### Example 1: SSS Congruence Consider two triangles, \( \triangle XYZ \) and \( \triangle LMN \). Given: XY = LM = 5 cm, YZ = MN = 7 cm, ZX = NL = 9 cm. **Solution:** Since all three corresponding sides are equal (XY=LM, YZ=MN, ZX=NL), the triangles are congruent by the SSS criterion. Thus, \( \triangle XYZ \cong \triangle LMN \). #### Example 2: SAS Congruence Consider two triangles, \( \triangle PQR \) and \( \triangle STU \). Given: PQ = ST = 6 cm, \( \angle Q = \angle T = 70^\circ \), QR = TU = 8 cm. **Solution:** We have two sides (PQ and QR) and their included angle (\( \angle Q \)) equal to the corresponding two sides (ST and TU) and their included angle (\( \angle T \)). Therefore, the triangles are congruent by the SAS criterion. Thus, \( \triangle PQR \cong \triangle STU \). #### Example 3: ASA Congruence Consider two triangles, \( \triangle DEF \) and \( \triangle GHI \). Given: \( \angle E = \angle H = 50^\circ \), EF = HI = 10 cm, \( \angle F = \angle I = 60^\circ \). **Solution:** We have two angles (\( \angle E \) and \( \angle F \)) and their included side (EF) equal to the corresponding two angles (\( \angle H \) and \( \angle I \)) and their included side (HI). Therefore, the triangles are congruent by the ASA criterion. Thus, \( \triangle DEF \cong \triangle GHI \). #### Example 4: RHS Congruence Consider two right-angled triangles, \( \triangle JKL \) (right-angled at K) and \( \triangle MNO \) (right-angled at N). Given: JL (hypotenuse) = MO (hypotenuse) = 13 cm, KL = NO = 5 cm. **Solution:** Both triangles are right-angled, their hypotenuses are equal (JL = MO), and one pair of corresponding sides are equal (KL = NO). Therefore, the triangles are congruent by the RHS criterion. Thus, \( \triangle JKL \cong \triangle MNO \). ### Exercises 1. Given two triangles \( \triangle ABC \) and \( \triangle DEF \) with the following measurements: AB = 7 cm, BC = 9 cm, CA = 11 cm DE = 7 cm, EF = 9 cm, FD = 11 cm Are the triangles congruent? If so, state the criterion. 2. In \( \triangle PQR \) and \( \triangle STU \): PQ = 5 cm, \( \angle Q = 40^\circ \), QR = 6 cm ST = 5 cm, \( \angle T = 40^\circ \), TU = 6 cm Are the triangles congruent? If so, state the criterion. (Careful with the angle placement!) 3. In the figure below, AD and BC bisect each other at O. Prove that \( \triangle AOB \cong \triangle DOC \). *(Imagine a diagonal line AD and BC crossing at O. A, O, D are collinear, B, O, C are collinear.)* 4. \( \triangle XYZ \) is a right-angled triangle with \( \angle Y = 90^\circ \). \( \triangle RST \) is a right-angled triangle with \( \angle S = 90^\circ \). Given: XZ = RT = 15 cm and XY = RS = 12 cm. Are the triangles congruent? If so, state the criterion. ### Assignment 1. For each pair of triangles, determine if they are congruent. If they are, state the congruency criterion (SSS, SAS, ASA, or RHS) and write the congruence statement (e.g., \( \triangle ABC \cong \triangle PQR \)). a) \( \triangle ABC \): AB = 8, BC = 10, \( \angle B = 60^\circ \) \( \triangle DEF \): DE = 8, EF = 10, \( \angle E = 60^\circ \) b) \( \triangle PQR \): PQ = 12, QR = 15, RP = 18 \( \triangle STU \): ST = 12, TU = 15, US = 18 c) \( \triangle JKL \): \( \angle J = 45^\circ \), JK = 7, \( \angle K = 70^\circ \) \( \triangle MNO \): \( \angle M = 45^\circ \), MN = 7, \( \angle N = 70^\circ \) d) \( \triangle XYZ \) (right-angled at Y): XZ = 17, YZ = 8 \( \triangle ABC \) (right-angled at B): AC = 17, BC = 8 2. Prove that the diagonals of a rectangle divide it into two congruent triangles. Which congruency criterion applies? 3. Two vertical poles of equal height are standing on a horizontal ground. A rope connects the top of one pole to the bottom of the other. Show that the triangles formed by the poles, the ground, and the ropes are congruent. Which criterion is used? ### Summary In this lesson, we explored the concept of congruent triangles, which are triangles that are identical in both size and shape. We learned about four key criteria to determine if two triangles are congruent without needing to compare all six parts: * **SSS (Side-Side-Side):** All three corresponding sides are equal. * **SAS (Side-Angle-Side):** Two corresponding sides and their included angle are equal. * **ASA (Angle-Side-Angle):** Two corresponding angles and their included side are equal. * **RHS (Right-angle-Hypotenuse-Side):** For right-angled triangles, the hypotenuse and one corresponding side are equal. Understanding these criteria is fundamental for solving many problems in geometry and for proving properties of various shapes. ## Similar plane figures and the constant of proportionality ### Introduction Welcome to this lesson on similar plane figures and the constant of proportionality! In mathematics, we often encounter shapes that look alike but are different in size. These shapes are called similar figures. Understanding similarity is crucial for various applications, from architecture to map-making. In this lesson, we will explore what makes figures similar, how to identify them, and how to use the constant of proportionality to find unknown measurements. ### Key Concepts #### What are Similar Figures? Two plane figures are said to be **similar** if they have the same shape but not necessarily the same size. This means one figure is an enlargement or reduction of the other. For two figures to be similar, two conditions must be met: 1. **Corresponding angles are equal:** All angles in the same relative position in both figures must have the same measure. 2. **Corresponding sides are proportional:** The ratio of the lengths of corresponding sides must be constant. This constant ratio is known as the **constant of proportionality** or **scale factor**. #### Properties of Similar Polygons Let's consider two polygons, Polygon A and Polygon B. If they are similar: * **Angles:** If Polygon A has angles \( heta_1, heta_2, heta_3, ext{...} \) and Polygon B has corresponding angles \( heta'_1, heta'_2, heta'_3, ext{...} \), then \( heta_1 = heta'_1 \), \( heta_2 = heta'_2 \), and so on. * **Sides:** If Polygon A has side lengths \( L_1, L_2, L_3, ext{...} \) and Polygon B has corresponding side lengths \( L'_1, L'_2, L'_3, ext{...} \), then the ratio of their corresponding sides is equal to the constant of proportionality, \( k \): \[ \frac{L'_1}{L_1} = \frac{L'_2}{L_2} = \frac{L'_3}{L_3} = k \] #### The Constant of Proportionality (Scale Factor) The constant of proportionality, often denoted by \( k \), is the ratio of the length of a side in the second figure to the length of the corresponding side in the first figure. It tells us how much the figure has been enlarged or reduced. * If \( k > 1 \), the second figure is an enlargement of the first. * If \( 0 < k < 1 \), the second figure is a reduction of the first. * If \( k = 1 \), the figures are congruent (same shape and same size). #### Ratio of Areas of Similar Figures If two figures are similar with a constant of proportionality \( k \) (ratio of corresponding side lengths), then the ratio of their areas is equal to the square of the constant of proportionality, \( k^2 \). Let \( A_1 \) be the area of the first figure and \( A_2 \) be the area of the second figure. If \( L_1 \) and \( L_2 \) are corresponding side lengths, then: \[ \frac{A_2}{A_1} = \left(\frac{L_2}{L_1}\right)^2 = k^2 \] ### Vocabulary * **Similar Figures:** Figures that have the same shape but not necessarily the same size. * **Corresponding Angles:** Angles that are in the same relative position in similar figures. * **Corresponding Sides:** Sides that are in the same relative position in similar figures. * **Proportional:** Having a constant ratio. * **Constant of Proportionality (Scale Factor):** The constant ratio of the lengths of corresponding sides in similar figures. Denoted by \( k \). * **Congruent Figures:** Figures that are exactly the same in shape and size (a special case of similar figures where \( k=1 \)). ### Examples **Example 1: Identifying Similar Rectangles** Rectangle A has length 6 cm and width 3 cm. Rectangle B has length 12 cm and width 6 cm. Are they similar? If so, find the constant of proportionality. * **Check Angles:** All angles in both rectangles are 90 degrees, so corresponding angles are equal. * **Check Sides:** * Ratio of lengths: \( \frac{\text{Length of B}}{\text{Length of A}} = \frac{12 \text{ cm}}{6 \text{ cm}} = 2 \) * Ratio of widths: \( \frac{\text{Width of B}}{\text{Width of A}} = \frac{6 \text{ cm}}{3 \text{ cm}} = 2 \) Since both ratios are equal to 2, the rectangles are similar, and the constant of proportionality \( k = 2 \). **Example 2: Finding a Missing Side Length** Triangles PQR and XYZ are similar. Side PQ = 4 cm, QR = 6 cm, and XY = 10 cm. Find the length of YZ. Since the triangles are similar, the ratio of corresponding sides is constant. \[ \frac{XY}{PQ} = \frac{YZ}{QR} \] Substitute the known values: \[ \frac{10}{4} = \frac{YZ}{6} \] First, find the constant of proportionality: \( k = \frac{10}{4} = 2.5 \) Now, solve for YZ: \( YZ = 6 \times 2.5 \) \( YZ = 15 \text{ cm} \) **Example 3: Ratio of Areas** Two similar triangles have corresponding sides in the ratio 1:3. If the area of the smaller triangle is \( 5 \text{ cm}^2 \), what is the area of the larger triangle? The ratio of corresponding sides (constant of proportionality) is \( k = 3 \) (from smaller to larger). The ratio of their areas is \( k^2 \). \[ \frac{\text{Area of Larger}}{\text{Area of Smaller}} = k^2 \] \[ \frac{\text{Area of Larger}}{5 \text{ cm}^2} = 3^2 \] \[ \frac{\text{Area of Larger}}{5 \text{ cm}^2} = 9 \] Area of Larger = \( 9 \times 5 \text{ cm}^2 = 45 \text{ cm}^2 \). ### Exercises 1. Are two squares always similar? Explain your answer. 2. Triangle ABC is similar to triangle DEF. If AB = 8 cm, BC = 10 cm, CA = 6 cm, and DE = 12 cm, find the lengths of EF and FD. 3. Two similar polygons have perimeters of 20 cm and 30 cm. What is the constant of proportionality between the smaller and larger polygon? 4. The ratio of the areas of two similar circles is 4:25. If the radius of the larger circle is 10 cm, find the radius of the smaller circle. 5. A photograph is 15 cm by 10 cm. It is enlarged to make a poster that is similar to the photograph. If the length of the poster is 45 cm, what is its width? What is the ratio of the poster's area to the photograph's area? ### Assignment 1. A map has a scale of 1:50000. If a park on the map has an area of \( 4 \text{ cm}^2 \), what is the actual area of the park in square meters? (Hint: The scale 1:50000 means the constant of proportionality for lengths is \( \frac{1}{50000} \) or \( 50000 \) if going from map to actual). 2. Two similar trapezoids have corresponding heights of 7 inches and 14 inches. If the area of the smaller trapezoid is \( 49 \text{ sq inches} \), what is the area of the larger trapezoid? 3. A model car is built to a scale of 1:24. If the actual car is 4.8 meters long, how long is the model car in centimeters? 4. Consider two similar pentagons. The ratio of their corresponding sides is 2:5. If the perimeter of the larger pentagon is 75 cm, what is the perimeter of the smaller pentagon? 5. Explain in your own words why corresponding angles must be equal for two figures to be similar, even if their sides are proportional. ### Summary In this lesson, we've learned that similar plane figures have the same shape but can differ in size. Key conditions for similarity are equal corresponding angles and proportional corresponding sides. The constant ratio of corresponding side lengths is called the **constant of proportionality** or **scale factor** (\( k \)). We also discovered that for similar figures, the ratio of their areas is equal to the square of the constant of proportionality, i.e., \( k^2 \). Understanding these concepts allows us to solve for unknown dimensions and areas in real-world problems involving scaled objects and designs. ## Areas and volume of similar plane figures ### Introduction Welcome to this lesson on the areas and volumes of similar plane figures and 3D objects! Building upon our understanding of similar figures, we will now explore how their areas and volumes are related through scale factors. This concept is vital in many real-world applications, from scaling maps and designs to understanding the capacities of containers of different sizes. ### Key Concepts #### Understanding Similar Figures Recall that two figures are **similar** if they have the same shape but not necessarily the same size. For similar figures: * Corresponding angles are equal. * Corresponding sides are in proportion. #### Linear Scale Factor (k) When comparing two similar figures, the ratio of their corresponding lengths is called the **linear scale factor**, denoted by \( k \). * If Figure 1 has a length \( L_1 \) and Figure 2 has a corresponding length \( L_2 \), then the linear scale factor from Figure 1 to Figure 2 is: \[ k = \frac{L_2}{L_1} \] #### Area of Similar Figures If two figures are similar with a linear scale factor \( k \), then the ratio of their areas is equal to the square of the linear scale factor, \( k^2 \). * **Area Scale Factor:** \( k^2 \) * **Formula:** If \( A_1 \) and \( A_2 \) are the areas of two similar figures with corresponding linear dimensions \( L_1 \) and \( L_2 \) respectively, then: \[ \frac{A_2}{A_1} = \left( \frac{L_2}{L_1} \right)^2 = k^2 \] This means \( A_2 = k^2 A_1 \). #### Volume of Similar Figures (3D Objects) If two 3D objects are similar with a linear scale factor \( k \), then the ratio of their volumes is equal to the cube of the linear scale factor, \( k^3 \). * **Volume Scale Factor:** \( k^3 \) * **Formula:** If \( V_1 \) and \( V_2 \) are the volumes of two similar 3D objects with corresponding linear dimensions \( L_1 \) and \( L_2 \) respectively, then: \[ \frac{V_2}{V_1} = \left( \frac{L_2}{L_1} \right)^3 = k^3 \] This means \( V_2 = k^3 V_1 \). #### Summary of Scale Factors | Type of Ratio | Relationship to Linear Scale Factor (\(k\)) | | :------------ | :--------------------------------------------- | | Linear | \( \frac{L_2}{L_1} = k \) | | Area | \( \frac{A_2}{A_1} = k^2 \) | | Volume | \( \frac{V_2}{V_1} = k^3 \) | ### Vocabulary * **Similar Figures:** Figures that have the exact same shape but can be different in size. * **Linear Scale Factor (k):** The ratio of any two corresponding lengths in similar figures. * **Area Scale Factor (k²):** The ratio of the areas of two similar figures, which is the square of the linear scale factor. * **Volume Scale Factor (k³):** The ratio of the volumes of two similar 3D objects, which is the cube of the linear scale factor. * **Corresponding Sides/Lengths:** Sides or lengths that are in the same relative position in two or more similar figures. ### Examples #### Example 1: Area of Similar Rectangles Two rectangles are similar. The smaller rectangle has a length of 6 cm and an area of 24 cm². If the larger rectangle has a corresponding length of 9 cm, find the area of the larger rectangle. **Solution:** 1. **Find the linear scale factor (k):** \( k = \frac{\text{Length of larger}}{\text{Length of smaller}} = \frac{9\text{ cm}}{6\text{ cm}} = \frac{3}{2} \) 2. **Find the area scale factor:** Area scale factor = \( k^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \) 3. **Calculate the area of the larger rectangle:** Let \( A_1 \) be the area of the smaller rectangle and \( A_2 \) be the area of the larger rectangle. \( \frac{A_2}{A_1} = k^2 \) \( \frac{A_2}{24} = \frac{9}{4} \) \( A_2 = \frac{9}{4} \times 24 \) \( A_2 = 9 \times 6 \) \( A_2 = 54 \text{ cm}^2 \) The area of the larger rectangle is 54 cm². #### Example 2: Volume of Similar Spheres Two spheres are similar. The radius of the smaller sphere is 3 cm and its volume is 36\( \pi \) cm³. If the radius of the larger sphere is 6 cm, find the volume of the larger sphere. **Solution:** 1. **Find the linear scale factor (k):** \( k = \frac{\text{Radius of larger}}{\text{Radius of smaller}} = \frac{6\text{ cm}}{3\text{ cm}} = 2 \) 2. **Find the volume scale factor:** Volume scale factor = \( k^3 = 2^3 = 8 \) 3. **Calculate the volume of the larger sphere:** Let \( V_1 \) be the volume of the smaller sphere and \( V_2 \) be the volume of the larger sphere. \( \frac{V_2}{V_1} = k^3 \) \( \frac{V_2}{36\pi} = 8 \) \( V_2 = 8 \times 36\pi \) \( V_2 = 288\pi \text{ cm}^3 \) The volume of the larger sphere is 288\( \pi \) cm³. ### Exercises 1. Two similar triangles have corresponding heights in the ratio 4:5. If the area of the larger triangle is 75 cm², what is the area of the smaller triangle? 2. The ratio of the volumes of two similar cylindrical tanks is 8:27. If the height of the smaller tank is 10 m, find the height of the larger tank. 3. A photograph has an area of 300 cm². It is reduced such that its linear dimensions are halved. What is the area of the reduced photograph? 4. Two similar cones have base radii of 7 cm and 14 cm. If the volume of the smaller cone is 100 cm³, calculate the volume of the larger cone. 5. If the surface areas of two similar cubes are in the ratio 49:100, what is the ratio of their volumes? ### Assignment 1. The areas of two similar pentagons are 81 cm² and 144 cm². If the perimeter of the larger pentagon is 60 cm, find the perimeter of the smaller pentagon. 2. Two similar bottles have volumes of 64 cm³ and 216 cm³. If the height of the larger bottle is 18 cm, find the height of the smaller bottle. 3. A model car is built to a scale of 1:10 (meaning 1 cm on the model represents 10 cm on the real car). If the real car has a fuel tank with a capacity (volume) of 40,000 cm³, what is the capacity of the model car's fuel tank? 4. The ratio of the volumes of two similar pyramids is 1:8. If the base area of the larger pyramid is 32 cm², what is the base area of the smaller pyramid? 5. If the linear dimensions of a sculpture are tripled to create a larger version, and the original sculpture weighs 5 kg, how much will the new sculpture weigh if the material density remains the same? (Hint: Weight is proportional to volume). ### Summary In this lesson, we explored the crucial relationships between linear dimensions, areas, and volumes of similar figures and 3D objects. We learned that: * The **linear scale factor (k)** is the ratio of corresponding lengths. * The **area scale factor** is \( k^2 \), meaning the ratio of areas is the square of the linear scale factor. * The **volume scale factor** is \( k^3 \), meaning the ratio of volumes is the cube of the linear scale factor. These relationships allow us to calculate unknown areas or volumes if we know the scale factor and one of the corresponding measurements. Understanding these concepts is essential for solving various practical problems involving scaling and proportions. ## Thales property and its converse ### Introduction Welcome to this lesson on Thales' Property and its converse! Named after the ancient Greek mathematician Thales of Miletus, this property is a fundamental theorem in geometry that beautifully connects circles and right-angled triangles. Understanding Thales' Theorem will help you identify right angles formed within circles and understand the relationship between a right-angled triangle and its circumcircle. ### Key Concepts #### Thales' Property (Theorem) Thales' Property, often called Thales' Theorem, states a special relationship between a circle's diameter and any point on its circumference. * **Statement:** If A, B, and C are distinct points on a circle where the line segment AC is a diameter of the circle, then the angle \( \angle ABC \) is a right angle (i.e., \( 90^\circ \)). * **In simpler terms:** Any angle inscribed in a semicircle is a right angle. * **Visual Understanding:** Imagine a circle. Draw a straight line through the center of the circle from one edge to the other – this is your diameter (AC). Now, pick any other point (B) on the edge of the circle. If you connect A to B and C to B, the angle formed at point B \( (\angle ABC) \) will always be \( 90^\circ \). * **Implications:** This theorem is powerful because it tells us that if we draw a triangle with its vertices on a circle, and one of its sides is a diameter, then the angle opposite the diameter must be a right angle. #### Converse of Thales' Property The converse of a theorem essentially reverses the statement. For Thales' Property, the converse is also true. * **Statement:** If a right-angled triangle is inscribed in a circle, then its hypotenuse is the diameter of the circle. * **Alternatively:** If the hypotenuse of a right-angled triangle is the diameter of a circle, then the vertex opposite the hypotenuse lies on the circumference of that circle. * **Visual Understanding:** If you have a right-angled triangle, and you draw a circle that passes through all three of its vertices (this is called a circumcircle), then the longest side of the right-angled triangle (the hypotenuse) will always be the diameter of that circle. * **Implications:** This allows us to determine if a given chord in a circle is a diameter, or if a specific vertex of a triangle lies on a circle, based on the angle it forms. #### Connection to Right-Angled Triangles and Circles Thales' property and its converse create a strong link between the geometry of circles and the properties of right-angled triangles. They provide a simple method to: * Identify right angles within circles. * Determine if a triangle's hypotenuse is a circle's diameter. * Construct right angles accurately using a compass and straightedge. ### Vocabulary * **Thales' Theorem/Property:** A geometric theorem stating that if a triangle's vertices lie on a circle and one side is a diameter, the angle opposite the diameter is a right angle. * **Converse:** A statement formed by interchanging the hypothesis and conclusion of another statement. * **Semicircle:** Half of a circle, bounded by a diameter and an arc. * **Diameter:** A straight line segment that passes through the center of a circle and whose endpoints lie on the circle. It is the longest chord of a circle. * **Hypotenuse:** The longest side of a right-angled triangle, always opposite the right angle. * **Inscribed Angle:** An angle formed by two chords in a circle that have a common endpoint on the circle's circumference. ### Examples #### Example 1: Applying Thales' Property In the diagram, AB is the diameter of a circle with center O. Point C is on the circumference such that \( \angle BAC = 35^\circ \). Find the measure of \( \angle ABC \) and \( \angle ACB \). **Solution:** 1. Since AB is the diameter and C is a point on the circumference, by Thales' Property, the angle subtended by the diameter at C is a right angle. Therefore, \( \angle ACB = 90^\circ \). 2. In \( \triangle ABC \), the sum of angles is \( 180^\circ \). So, \( \angle ABC + \angle BAC + \angle ACB = 180^\circ \). 3. Substitute the known values: \( \angle ABC + 35^\circ + 90^\circ = 180^\circ \). 4. \( \angle ABC + 125^\circ = 180^\circ \). 5. \( \angle ABC = 180^\circ - 125^\circ = 55^\circ \). So, \( \angle ACB = 90^\circ \) and \( \angle ABC = 55^\circ \). #### Example 2: Applying the Converse of Thales' Property A triangle PQR has vertices P, Q, and R. The coordinates of P are \( (0, 0) \), Q are \( (5, 0) \), and R are \( (2.5, 4) \). Is this triangle inscribed in a semicircle with PQ as its diameter? **Solution:** To determine if R lies on the semicircle with diameter PQ, we need to check if \( \triangle PQR \) is a right-angled triangle at R. If it is, then by the converse of Thales' theorem, PQ would be the diameter of the circumcircle. We can do this by checking if the slopes of PR and QR are perpendicular (i.e., their product is -1). 1. Slope of PR \( (m_{PR}) = \frac{4 - 0}{2.5 - 0} = \frac{4}{2.5} = \frac{4}{\frac{5}{2}} = \frac{8}{5} \). 2. Slope of QR \( (m_{QR}) = \frac{4 - 0}{2.5 - 5} = \frac{4}{-2.5} = \frac{4}{-\frac{5}{2}} = -\frac{8}{5} \). Now, multiply the slopes: \( m_{PR} \times m_{QR} = \frac{8}{5} \times \left(-\frac{8}{5}\right) = -\frac{64}{25} \). Since the product of the slopes is not -1, \( \angle PRQ \) is not \( 90^\circ \). Therefore, \( \triangle PQR \) is not a right-angled triangle at R, and thus R does not lie on the semicircle with PQ as its diameter. *Self-check using Pythagorean theorem:* * Length \( PQ = \sqrt{(5-0)^2 + (0-0)^2} = \sqrt{5^2} = 5 \) * Length \( PR = \sqrt{(2.5-0)^2 + (4-0)^2} = \sqrt{(2.5)^2 + 4^2} = \sqrt{6.25 + 16} = \sqrt{22.25} \) * Length \( QR = \sqrt{(2.5-5)^2 + (4-0)^2} = \sqrt{(-2.5)^2 + 4^2} = \sqrt{6.25 + 16} = \sqrt{22.25} \) \( PR^2 + QR^2 = 22.25 + 22.25 = 44.5 \). \( PQ^2 = 5^2 = 25 \). Since \( PR^2 + QR^2 \neq PQ^2 \), it is not a right-angled triangle, confirming the previous conclusion. ### Exercises 1. In a circle with center O, AB is a diameter. Point D is on the circumference such that \( \angle ABD = 50^\circ \). What is the measure of \( \angle ADB \) and \( \angle DAB \)? 2. You are given a line segment MN. Describe, using Thales' Theorem, how you would construct a right-angled triangle where MN is the hypotenuse, using only a compass and a straightedge. 3. Triangle XYZ is inscribed in a circle. If XY is a diameter of the circle, and \( \angle YXZ = 20^\circ \), find the measures of \( \angle XYZ \) and \( \angle XZY \). ### Assignment 1. Draw a circle with a radius of 4 cm. Draw a diameter PQ. Choose three different points R, S, and T on the circumference of the circle. Measure the angles \( \angle PRQ \), \( \angle PSQ \), and \( \angle PTQ \). What do you observe? Explain your observations using Thales' Theorem. 2. A triangle DEF has side lengths DE = 13 cm, EF = 5 cm, and DF = 12 cm. a) Is \( \triangle DEF \) a right-angled triangle? Justify your answer using the Pythagorean theorem. b) If a circle passes through points D, E, and F, what can you conclude about the side DE based on your answer in (a)? 3. Points A, B, and C are on a circle with center O. The line segment AB is a diameter of the circle. If the coordinates of A are \( (-4, 0) \) and B are \( (4, 0) \), and C is \( (0, y) \), find the possible value(s) of y for C to be on the circle. (Hint: Use the converse of Thales' Theorem, which implies \( \angle ACB \) must be \( 90^\circ \)). ### Summary In this lesson, we explored Thales' Property and its converse, two essential concepts in geometry: * **Thales' Property** states that an angle inscribed in a semicircle is always a right angle \( (90^\circ) \). * The **Converse of Thales' Property** states that if a right-angled triangle is inscribed in a circle, its hypotenuse is the diameter of that circle. These properties are incredibly useful for understanding relationships between angles, diameters, and points on a circle, as well as for solving geometric problems involving right-angled triangles within circles. ## Application of similarities to areas and volume ### 1. Introduction Welcome to this lesson on the application of similarity! In previous lessons, you learned about similar figures and solids, where they have the same shape but different sizes. We explored how their corresponding linear dimensions (like sides or heights) are proportional. In this lesson, we will extend these concepts to understand how the areas of similar two-dimensional figures and the volumes of similar three-dimensional solids are related to their linear scale factor. This will allow us to solve practical problems involving scaling. ### 2. Key Concepts #### 2.1 Similarity Ratio (Linear Scale Factor) When two figures or solids are similar, the ratio of any two corresponding linear dimensions (e.g., sides, radii, heights, perimeters) is constant. This constant ratio is called the **similarity ratio** or **linear scale factor**, often denoted by \( k \). * If two figures have corresponding linear dimensions \( L_1 \) and \( L_2 \), then the linear ratio is \( k = \frac{L_1}{L_2} \). #### 2.2 Relationship between Similarity Ratio and Area Ratio If two figures are similar with a linear ratio of \( k \), then the ratio of their areas is the square of the linear ratio, \( k^2 \). * Consider two similar squares with sides \( L_1 \) and \( L_2 \). Their areas are \( A_1 = L_1^2 \) and \( A_2 = L_2^2 \). * The ratio of their areas is \( \frac{A_1}{A_2} = \frac{L_1^2}{L_2^2} = (\frac{L_1}{L_2})^2 = k^2 \). Therefore, if \( \frac{\text{Linear Dimension}_1}{\text{Linear Dimension}_2} = k \), then \[ \frac{\text{Area}_1}{\text{Area}_2} = k^2 \] #### 2.3 Relationship between Similarity Ratio and Volume Ratio If two solids are similar with a linear ratio of \( k \), then the ratio of their volumes is the cube of the linear ratio, \( k^3 \). * Consider two similar cubes with sides \( L_1 \) and \( L_2 \). Their volumes are \( V_1 = L_1^3 \) and \( V_2 = L_2^3 \). * The ratio of their volumes is \( \frac{V_1}{V_2} = \frac{L_1^3}{L_2^3} = (\frac{L_1}{L_2})^3 = k^3 \). Therefore, if \( \frac{\text{Linear Dimension}_1}{\text{Linear Dimension}_2} = k \), then \[ \frac{\text{Volume}_1}{\text{Volume}_2} = k^3 \] #### 2.4 Summary of Ratios for Similar Figures/Solids Let \( k \) be the linear ratio of two similar figures/solids. * **Linear Ratio:** \( L_1 : L_2 = k \) * **Area Ratio:** \( A_1 : A_2 = k^2 \) * **Volume Ratio:** \( V_1 : V_2 = k^3 \) ### 3. Vocabulary * **Similar Figures/Solids:** Geometric objects that have the same shape but may differ in size. Corresponding angles are equal, and corresponding linear dimensions are proportional. * **Similarity Ratio (Linear Scale Factor):** The constant ratio between any two corresponding linear dimensions (e.g., sides, heights, radii, perimeters) of two similar figures or solids. * **Area Ratio:** The ratio of the areas of two similar two-dimensional figures. It is equal to the square of the linear scale factor. * **Volume Ratio:** The ratio of the volumes of two similar three-dimensional solids. It is equal to the cube of the linear scale factor. * **Corresponding Dimensions:** Sides, angles, or other features that are in the same relative position in two similar figures or solids. ### 4. Examples #### Example 1: Calculating Area of a Similar Figure Two similar triangles have corresponding sides in the ratio 2:3. If the area of the smaller triangle is 20 cm², find the area of the larger triangle. **Solution:** 1. **Identify the linear ratio:** \( k = \frac{\text{Side}_S}{\text{Side}_L} = \frac{2}{3} \) 2. **Determine the area ratio:** \( k^2 = (\frac{2}{3})^2 = \frac{4}{9} \) 3. **Set up the proportion for areas:** \( \frac{\text{Area}_S}{\text{Area}_L} = k^2 \) \( \frac{20}{\text{Area}_L} = \frac{4}{9} \) 4. **Solve for the unknown area:** \( 4 \times \text{Area}_L = 20 \times 9 \) \( \text{Area}_L = \frac{180}{4} = 45 \text{ cm}^2 \) The area of the larger triangle is 45 cm². #### Example 2: Calculating Volume of a Similar Solid Two similar cylindrical tanks have heights of 6 m and 9 m respectively. If the smaller tank holds 200 litres of water, what is the capacity (volume) of the larger tank? **Solution:** 1. **Identify the linear ratio:** \( k = \frac{\text{Height}_S}{\text{Height}_L} = \frac{6}{9} = \frac{2}{3} \) 2. **Determine the volume ratio:** \( k^3 = (\frac{2}{3})^3 = \frac{8}{27} \) 3. **Set up the proportion for volumes:** \( \frac{\text{Volume}_S}{\text{Volume}_L} = k^3 \) \( \frac{200}{\text{Volume}_L} = \frac{8}{27} \) 4. **Solve for the unknown volume:** \( 8 \times \text{Volume}_L = 200 \times 27 \) \( \text{Volume}_L = \frac{5400}{8} = 675 \text{ litres} \) The capacity of the larger tank is 675 litres. #### Example 3: Finding Linear Ratio from Area The areas of two similar rectangles are 36 cm² and 81 cm². If the width of the smaller rectangle is 4 cm, find the width of the larger rectangle. **Solution:** 1. **Identify the area ratio:** \( \frac{\text{Area}_S}{\text{Area}_L} = \frac{36}{81} = \frac{4}{9} \) 2. **Determine the linear ratio:** \( k = \sqrt{\frac{\text{Area}_S}{\text{Area}_L}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \) 3. **Set up the proportion for widths (linear dimensions):** \( \frac{\text{Width}_S}{\text{Width}_L} = k \) \( \frac{4}{\text{Width}_L} = \frac{2}{3} \) 4. **Solve for the unknown width:** \( 2 \times \text{Width}_L = 4 \times 3 \) \( \text{Width}_L = \frac{12}{2} = 6 \text{ cm} \) The width of the larger rectangle is 6 cm. ### 5. Exercises Test your understanding with these questions: 1. Two similar pentagons have corresponding sides in the ratio 3:5. If the area of the larger pentagon is 75 cm², find the area of the smaller pentagon. 2. A model car is built to a scale of 1:100 (meaning the linear ratio is \( \frac{1}{100} \)). If the volume of the actual car's engine is 1500 cm³, what is the volume of the model car's engine? 3. Two similar cones have curved surface areas of 50\( \pi \) cm² and 200\( \pi \) cm². If the height of the smaller cone is 5 cm, find the height of the larger cone and the ratio of their volumes. ### 6. Assignment Continue practicing at home to reinforce your understanding: 1. The areas of two similar triangles are 49 cm² and 121 cm². If the base of the larger triangle is 22 cm, find the base of the smaller triangle. 2. Two similar spheres have radii 3 cm and 6 cm. If the volume of the smaller sphere is 36\( \pi \) cm³, what is the volume of the larger sphere? What is the ratio of their surface areas? 3. A painting is enlarged. The original painting has an area of 0.25 m². The enlarged painting has an area of 1 m². By what linear scale factor was the painting enlarged? ### 7. Summary In this lesson, we explored the powerful applications of similarity to areas and volumes. The key takeaway is understanding the relationships between the linear scale factor (similarity ratio), area ratio, and volume ratio of similar figures and solids. Remember these crucial points: * If the **linear ratio** of two similar figures/solids is \( k \), then: * The **area ratio** is \( k^2 \). * The **volume ratio** is \( k^3 \). By correctly identifying the linear ratio, you can easily determine the relationship between the areas and volumes, allowing you to solve a wide range of problems involving scaling in two and three dimensions. This concept is vital in fields like engineering, architecture, and map-making.