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NCERT Solutions for Class 9 Maths Chapter 7 Triangles
NCERT Solutions for Class 9 Maths Chapter 7 Triangles are prepared by CBSE student eCARE expert to score good marks in class 9. This chapter contain many topics which are very important to score good in class 9. There are some important topic/Activity mentioned below:
ABC is a triangle. Locate a point in the interior of ? ABC which is equidistant from all the vertices of ? ABC.
Solution:
Let us consider a ?ABC.
Draw l, the perpendicular bisector of AB.
Draw m, the perpendicular bisector of BC.
Let the two perpendicular bisectors l and m meet at O.
O is the required point which is equidistant from A, B and C.
Note: If we draw a circle with centre O and radius OB or OC, then it will pass through A, B and C. The point O is called circumcentre of the triangle.
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Some questions are asked by class 9 students related to Maths
Question 1.
In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ?ABC ≅ ?ABD. What can you say about BC and BD?
Solution:
In quadrilateral ACBD, we have AC = AD and AB being the bisector of ∠A.
Now, In ?ABC and ?ABD,
AC = AD (Given)
∠ CAB = ∠ DAB ( AB bisects ∠ CAB)
and AB = AB (Common)
∴ ? ABC ≅ ?ABD (By SAS congruence axiom)
∴ BC = BD (By CPCT)
Ex 7.1 Class 9 Maths Question 2.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that
(i) ?ABD ≅ ?BAC
(ii) BD = AC
(iii) ∠ABD = ∠ BAC
Solution:
In quadrilateral ACBD, we have AD = BC and ∠ DAB = ∠ CBA
(i) In ? ABC and ? BAC,
AD = BC (Given)
∠DAB = ∠CBA (Given)
AB = AB (Common)
∴ ? ABD ≅ ?BAC (By SAS congruence)
(ii) Since ?ABD ≅ ?BAC
⇒ BD = AC [By C.P.C.T.]
(iii) Since ?ABD ≅ ?BAC
⇒ ∠ABD = ∠BAC [By C.P.C.T.]