triangles-class-10-similar-triangles-examples-solving-similar-triangles

Triangles class 10 | Similar triangles examples | Solving similar triangles

http://www.learncbse.in/ncert-solutions-class-10th-maths-chapter-6-triangles-exercise-6-1-question-1/

http://www.learncbse.in/rd-sharma-class-10-solutions/

http://www.learncbse.in/cbse-sample-papers/

AAA similarity criterion

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

SSS (Side–Side–Side) similarity criterion for two triangles

If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.

SAS (Side–Angle–Side) similarity criterion for two triangles

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar

00:02 Q5. S and T are points on sides PR and QR of triangle PQR such that angle P = angle RTS. Show that triangle RPQ ~ triangle RTS.

01:58 AAA similarity criterion

02:24 Q6. In Fig. 6.37, if triangle ABE ~ triangle ACD, show that triangle ADE ~ triangle ABC.

04:40 SAS (Side–Angle–Side) similarity criterion

05:13 Q7. In Fig. 6.38, altitudes AD and CE of triangle ABC intersect each other at the point P. Show that:

06:15 (i) triangle AEP ~ triangle CDP

06:54 Vertically opposite angles are equal

07:44 (ii) triangle ABD ~ triangle CBE

09:02 use AAA similarity criterion

09:32 (iii) triangle AEP ~ triangle ADB

10:38 AAA similarity criterion

11:00 (iv) triangle PDC ~ triangle BEC

12:03 AA similarity criterion

12:23 Q8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that triangle ABE ~ triangle CFB.

15:01 Apply AA similarity criterion.

15:25 Q9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

16:20 (i) triangle ABC ~ triangle AMP 31:22 (ii) CA/PA = BC/MP.

17:36 Q10. CD and GH are respectively the bisectors of triangle ACB and triangle EGF such that D and H lie on sides AB and FE of angle ABC and angle EFG respectively. If triangle ABC ~ triangle FEG, show that:

19:43 (i) CD/GH = AC/FG | 22:21 (ii) triangle DCB ~ triangle HGE | 24:04 (iii) triangle DCA ~ triangle HGF

18:50 Angular bisector

learncbse.in

CBSE solutions for class 10 maths Chapter 6 Triangles Exercise 6.3

CBSE class 10 maths NCERT Solutions chapter 6 Triangles Exercise 6.2 | SIMILAR TRIANGLES

NCERT solutions for CBSE class 10 maths Triangles

CBSE class 10 maths solutions Triangles

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