Subject: Compulsory Maths

\(\triangle\)ABC and \(\triangle\)XYZ are similar i.e, \(\triangle\)ABC and \(\traingle\)XYZ have same shape but the size is different. If any two angles are equal to each other, then they are similar.

There are three conditions for similarity of triangles:

**i) Angle, Angle similarity test**

^{Fig:Angle Angle}

If two angles of one triangles are respectively equal to two angles of another triangle, then two triangles are similar.

For example:

Here,∠B =∠Y and∠C =∠Z. The remaining angles∠A and∠X are also equal.

∴ \(\triangle\)ABC∼ \(\triangle\)XYZ

**ii) Side, Side, Side similarity test**

^{Fig: SSS}

If the corresponding sides of two triangles are proportional, then the triangles are similar.

For example:

Here, PQ/XY = QR/YZ = PR/XZ

∴ \(\triangle\)PQR∼ \(\triangle\)XYZ

**iii) Side, Angle, Side similarity test**

^{Fig: SAS}

If two corresponding sides of two triangles are proportional and the angle contained by these sides are equal, then the triangles are similar.

For example:

Here, XY/AB = YZ/BC and∠Y =∠B

∴ \(\triangle\)XYZ∼ \(\triangle\)ABC

Two polygons are similar under following conditions:

**i) When two or more polygons are equiangular, they are similar.**

In the figure, ∠A =∠P,∠B =∠Q,∠C =∠R,∠D =∠S

∴ quad ABCD∼ quad PQRS

**ii)** **When the corresponding sides of two polygons are proportional, they are similar.**

In the figure, AB/PQ = BC/QR = CD/RS = DA/SP

∴ quad ABCD∼ quad PQRS

**iii) When the corresponding diagonals of the polygons are proportional to their corresponding sies, they are similar.**

In the figure, AC/PR = BD/QS = AB/PQ

∴ quad ABCD∼ quad PQRS

**iv) When the corresponding diagonals divide the polygons into the equal number of similar triangles, the polygons are equal.**

\(\triangle\)ABC∼ \(\triangle\)PQR, \(\triangle\)ACD∼ \(\triangle\)PRS, \(\triangle\)ADE∼ \(\triangle\)PSV

∴ polygon ABCDE∼ polygon PQRS

Note: Theorem with '*' in similarity chapter do not need proof or experimental verification but the problems related to them are included in the curriculum.

- Two geometrical objects are called similar if they both have the same shape, or one has the same shape as the mirror image of the other. More precisely, one can be obtained from the other by uniformly scaling (enlarging or shrinking), possibly with additional translation, rotation and reflection.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

In the figure , BE || CD. Find the values of x and y with reasons.

Statements | Reasons |

1. In ABE and ACD (i) ABE = ACD (ii) BEA = CDA (iii) ABE ~ ACD | 1. (i) Corresponding angles ; BE || CD. (ii) Corresponding angles ; BE || CD. (iii) By A.A similarity |

2. \(\frac{AC}{AB}\) = \(\frac{CD}{BE}\) or , \(\frac{x + 4}{4}\) = \(\frac{3}{2}\) or , 2x + 18 = 12 oe , 2x = 12 - 8 = 4 \(\therefore\) \(\frac{4}{2}\) = 2 cm. | 2. In similar triangles ACD and ABE corresponding sides are in proportion. |

3. \(\frac{AD}{AE}\) = \(\frac{CD}{BE}\) or , \(\frac{y + 3}{y}\) = \(\frac{8.1}{5.4}\) or , \(\frac{y + 3}{y}\) = \(\frac{3}{2}\) or , 2y + 6 = 3y or , 6 = 3y - 3y \(\therefore\) y = 6 cm. | 3. From statement (2). |

In the figure LPM = MNQ. Find the measure of NM.

Here , in OLN and LPM

1. LNO = LPM | 1. From given. |

2. OLN = PLM | 2. Common Angle. |

3. OLN ~ LPM | 3. Two equiangular triangles are similar. |

4. \(\frac{LN}{LP}\)= \(\frac{LO}{LM}\) or , \(\frac{LM + MN}{LP}\) = \(\frac{LP + PQ}{LM}\) or , \(\frac{5 + MN}{6}\) = \(\frac{6 + 1}{5}\) = \(\frac{7}{5}\) or , 25 + 5MN = 42 or , 5MN = 42 - 25 = 17 \(\therefore\) MN = \(\frac{17}{5}\) = 3.4 cm | Corresponding sides of similar triangles are proportional. |

In the figure , CBD = CEA. Find ED with geometric reasons.

(i) CEA = CBD | Given |

(ii) ACE = BCD | Common Angle |

(iii) CAE = BDC | Remaining Angle |

(iv) AEC~BDC | By A.A.A similarity |

(v) \(\frac{CE}{BC}\) =\(\frac{CA}{CD}\) or ,\(\frac{ED + 5}{6}\) =\(\frac{3 + 6}{5}\) or ,\(\frac{ED + 5}{6}\) =\(\frac{9}{5}\) or , ED + 5 =\(\frac{54}{5}\) or , ED = 10.8 -5 = 5.8 cm | Being AEC ~ BDC |

Find the measures of Ab in the adjoining figure.

Here , DCO and OAB

1. CDO = OBA | 1. Being DC || AB , DB is transversal and alternate angles. |

2/ DCO = OAB | 2. Being DC || AB , AC is transversal and alternate angles. |

3/ DCO ~ OAB | 3. By A.A.A similarity |

4.\(\frac{AB}{DC}\) =\(\frac{BO}{OD}\) or ,\(\frac{x}{3}\) =\(\frac{5}{2}\) or , x =\(\frac{15}{2}\) = 7.5 x = 7.5 cm | 4. Corresponding sides of similar triangle are proportional |

Using Pythagoras theorem , find the missing side in each right angled triangle given below :

(i).

Here , p = 7cm , b = 5cm , h= ?

We know that ,

h^{2} = p^{2} + b^{2}or , h^{2} = 72 + 52 = 49 + 25 = 74

\(\therefore\) h = \(\sqrt{74}\) = 8.6 cm Ans.

Using Pythagoras theroem , find the missing side in each in each right angled triangle given below :

(ii).

Here , h = 13cm , p = 8cm , b = ?

We know that ,

p^{2} + b^{2} = h^{2}

or , 8^{2} + b^{2} = 13^{2}

or , b^{2} = 13^{2} - 8^{2} = 169 - 64 = 105

b = \(\sqrt{105}\) = 10.25 cm Ans.

Using Pythagoras theorem , find the missing side in each right angled triangle given below :

(iii).

Here , GH^{2} + H I^{2} = GI^{2}

or , GH^{2} + 24^{2} = 252

or , GH^{2} = 25^{2} - 242

or , GH^{2} = 625 - 57

or , GH^{2} = 625 - 576

or , GH^{2} = 49

or , GH= \(\sqrt{49}\) = 7 cm

What is the length of the diagonal of a rectangle of dimensions 8 cm \(\times\) 12 cm ?

In given rectangle ABCD , AB = 12cm , BC= 8cm , diagonal (AC) = ?

In rt. angled triangle ABC

AC^{2} = AB^{2} + BC^{2}

or , AC^{2} = 12^{2} + 8^{2} = 144 + 64

= 208

Diagonal AC = \(\sqrt{208}\) = 14.42 cm

What is the length of the diagonal of a square of length 6 m ?

In square ABCD , BC = 6cm , AB = 6cm , diagonal(AC) =?

In rt. angled triangle ABC ,

AC2 = Ab2 + BC2

or , AC2 =62 + 62 = 36 + 36 = 72

diagonal AC = \(\sqrt{72}\) =\(\sqrt[6]{2}\) = 8.48 cm.

Find the length of the sides of a rectangle if its diagonal is 12 cm and one of the sides is 8 cm.

In rectangle ABCD , AB = 8cm , diagonal AC = 12cm ,BC = ?

In rt. angled triangle ABC

AB^{2} + BC^{2} = AC^{2}

or , 8^{2} + BC^{2} = 12^{2}

or , BC2 = 12^{2} - 8^{2} = 144 - 64 = 80

or , BC = \(\sqrt{80}\) = 8.94 cm Ans.

A 7m high telephone post is supported by a 7.6 m long wire tied to the top of the pole and fixed at the ground. How far is the rope fixed at the ground from the foot of the pole ?

Let us consider AB be the telephone post and CA be wire.

Here CB is the perpendicular distance from the rope fixed at ground C to the post AB.

Here , AB = 7m and AC = 7.6 m

Now , in rt.angled triangle ABC ,

AB^{2} + BC^{2}= AC^{2}

or , 7^{2} + BC^{2} = 7.6^{2}

or , BC2 = 7.6^{2} - 7^{2} = 57.76 - 49 = 8.76

\(\therefore\) BC = \(\sqrt{8.76}\) = 2.96 m Ans.

Find the area of the rectangle having length 5.1 cm and diagonal 6.1 cm.

In rectangle ABCD ,

Length (AB)= 5.1 cm , diagonal (AC) = 6.1 cm and breadth (BC) = ?

Here , in rt., angled triangle ABC ,

(5.1)^{2} + (BC)^{2} = (6.1)^{2}

or , BC^{2} = (6.1)^{2} - (5.1)^{2}

or , BC^{2} = 37.21 - 26.01 = 11.2

\(\therefore\) BC = \(\sqrt{11.2}\) = 3.35 cm

Area of rectangle ABCD = length \(\times\) breadth = 5.1 \(\times\) 3.35 cm ^{2}

= 17.085 cm ^{2}

Find the value of x in the following figures :

(ii)

In DEF , EF =12cm , DE = 9cm , AD = ?

In rt. angled triangle DEF , using pythagoras theorem ,

DF^{2} = EF^{2} + DE^{2}

or , DF^{2} = 12^{2} + 9^{2}

or , DF = \(\sqrt{144 + 81}\) = \(\sqrt{225}\) = 15.

Now , in rt. angled triangle ADF ,

AF = 8cm , DF = 15 cm , DA = ?

Here , in ADF ,

DA^{2} = DF^{2} + AF^{2}

or , x^{2} = 15^{2} + 8^{2}

or , x^{2} = 225 + 64

\(\therefore\) x = \(\sqrt{289}\) = 17cm.

Find the value of x.

(iii),

IN JHI ,HI = 5cm , JI = 2cm , HJ = ?

In rt. angled triangle JHI ,

JH^{2} = HI^{2} + JI^{2}

or , JH^{2} = 5^{2} + 2^{2}

or , JH^{2} = 29

\(\therefore\) JH = \(\sqrt{29}\) cm.

Again ,

HG2 = JH^{2} + GJ^{2}

or , HG^{2} = (\(\sqrt{29}\) )^{2} + 32

or , HG^{2} = 29 + 9 = 38

or ,HG = \(\sqrt{38}\) 6.16

\(\therefore\)x = 6.16 cm Ans.

By using Pythagoras theorem , determine which of the folllowing triangles are right angled triangles ?

(i).

Here , AC^{2} + AB^{2} = (15)^{2} + (8)^{2} = 289

and BC^{2} = (17)^{2} = 289

AC^{2} + BC^{2} = AB^{2}

h^{2} = p^{2} + b^{2} , so given triangle is right angled triangle in which BAC = 90^{o}

© 2019-20 Kullabs. All Rights Reserved.