NCERT Solution Class 9th Maths Chapter – 7 Triangles Examples

NCERT Solution Class 9th Maths Chapter – 7 Triangles Examples

NCERT Solution Class 9th Maths Chapter – 7 Triangles

TextbookNCERT
Class9th
SubjectMathematics
Chapter7th
Chapter NameTriangles
CategoryClass 9th Math Solutions 
Medium English
SourceLast Doubt

NCERT Solution Class 9th Maths Chapter – 7 Triangles

Chapter – 7

Triangles

Examples

Example 1 : In Fig. 7.8, OA = OB and OD = OC. Show that
(i) ∆ AOD ≅ ∆ BOC and (ii) AD || BC.

Solution – (i) You may observe that in ∆ AOD and ∆ BOC,
OA = OB 〉 (Given)
OD = OC 〉
Also, since ∠ AOD and ∠ BOC form a pair of vertically opposite angles, we have
∠ AOD = ∠ BOC.
So, ∆ AOD ≅ ∆ BOC (by the SAS congruence rule)
(ii) In congruent triangles AOD and BOC, the other corresponding parts are also
equal.
So, ∠ OAD = ∠ OBC and these form a pair of alternate angles for line segments
AD and BC.
Therefore, AD || BC

Example 2 : AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, show that P is equidistant from A and B.

Solution – Line l ⊥ AB and passes through C which is the mid-point of AB (see Fig. 7.9). You have to show that PA = PB. Consider ∆ PCA and ∆ PCB.
We have AC = BC (C is the mid-point of AB)
∠ PCA = ∠ PCB = 90° (Given)
PC = PC (Common)
So, ∆ PCA ≅ ∆ PCB (SAS rule)
and so, PA = PB, as they are corresponding sides of
congruent triangles.

Example 3 : Line-segment AB is parallel to another line-segment CD. O is the mid-point of AD (see Fig. 7.15). Show that (i) ∆AOB ≅ ∆DOC (ii) O is also the mid-point of BC.

Solution – (i) Consider ∆ AOB and ∆ DOC.
∠ ABO = ∠ DCO
(Alternate angles as AB || CD
and BC is the transversal)
∠ AOB = ∠ DOC
(Vertically opposite angles)
OA = OD (Given)
Therefore, ∆AOB ≅ ∆DOC (AAS rule)
(ii) OB = OC (CPCT)
So, O is the mid-point of BC

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