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.
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 ,
h2 = p2 + b2
or , h2 = 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 ,
p2 + b2 = h2
or , 82 + b2 = 132
or , b2 = 132 - 82 = 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 , GH2 + H I2 = GI2
or , GH2 + 242 = 252
or , GH2 = 252 - 242
or , GH2 = 625 - 57
or , GH2 = 625 - 576
or , GH2 = 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
AC2 = AB2 + BC2
or , AC2 = 122 + 82 = 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
AB2 + BC2 = AC2
or , 82 + BC2 = 122
or , BC2 = 122 - 82 = 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 ,
AB2 + BC2= AC2
or , 72 + BC2 = 7.62
or , BC2 = 7.62 - 72 = 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 , BC2 = (6.1)2 - (5.1)2
or , BC2 = 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 ,
DF2 = EF2 + DE2
or , DF2 = 122 + 92
or , DF = \(\sqrt{144 + 81}\) = \(\sqrt{225}\) = 15.
Now , in rt. angled triangle ADF ,
AF = 8cm , DF = 15 cm , DA = ?
Here , in ADF ,
DA2 = DF2 + AF2
or , x2 = 152 + 82
or , x2 = 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 ,
JH2 = HI2 + JI2
or , JH2 = 52 + 22
or , JH2 = 29
\(\therefore\) JH = \(\sqrt{29}\) cm.
Again ,
HG2 = JH2 + GJ2
or , HG2 = (\(\sqrt{29}\) )2 + 32
or , HG2 = 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 , AC2 + AB2 = (15)2 + (8)2 = 289
and BC2 = (17)2 = 289
AC2 + BC2 = AB2
h2 = p2 + b2 , so given triangle is right angled triangle in which BAC = 90o
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