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.
(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 |
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 |
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.
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.
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
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
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.
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.
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.
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}
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.
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}
The perimeter of two similar triangles A BC and PQR are 25 cm and 15 cm respectively.If one side of ( riangle)ABC is 9 cm, find the corresponding side of ( riangle)PQR.
6.7 cm
5.5 cm
5.4 cm
2.5 cm
The perimeter of two similar triangles ABC and PQR are 36 cm and 24 cm respectively.If PQ = 10 cm, find AB.
10 cm
20 cm
5 cm
15 cm
The measures of sides of ( riangle)ABC are 2 cm,4 cm and 5 cm respectively.If ( riangle)ABC ∼( riangle)PQR and the perimeter of ( riangle)PQR = 22 cm,find the measure of sides of ( riangle)PQR.
4 cm,8 cm,10 cm
1 cm,9 cm,13 cm
2 cm,14 cm,12 cm
5 cm,7 cm,9 cm
( riangle)ABC∼( riangle)PQR and AB=6 cm,BC = 7 cm,CA = 8 cm are given. If ( riangle)PQR has perimeter 42 cm,find its sides.
17 cm,12 cm,20 cm
9 cm, 5 cm,16 cm
12 cm,13.5 cm,18 cm
11 cm,13.6 cm,11 cm
How many conditions are there for similarity of triangles?
You must login to reply
Sanjok Thapa
Are a similar triangles congruent to each other?
Mar 15, 2017
0 Replies
Successfully Posted ...
Please Wait...