EXERCISE 7.1
Introduction - Triangles Class 9th Maths
Q 1, Ex. 7.1, Page No 118 - Triangles - NCERT Class 9th Maths Solutions
Page No 118:
Question 1:
In quadrilateral ACBD, AC = AD and AB bisects ∠A (See the given figure). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
Answer:
In ΔABC and ΔABD,AC = AD (Given)
∠CAB = ∠DAB (AB bisects ∠A)
AB = AB (Common)
∴ ΔABC ≅ ΔABD (By SAS congruence rule)
∴ BC = BD (By CPCT)
Therefore, BC and BD are of equal lengths.
Q 2, Ex. 7.1, Page No 119 - Triangles - NCERT Class 9th Maths Solutions
Page No 119:
Question 2:
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (See the given figure). Prove that(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Answer:
In ΔABD and ΔBAC,AD = BC (Given)
∠DAB = ∠CBA (Given)
AB = BA (Common)
∴ ΔABD ≅ ΔBAC (By SAS congruence rule)
∴ BD = AC (By CPCT)
And, ∠ABD = ∠BAC (By CPCT)
Q 3, Ex. 7.1, Page No 119 - Triangles - NCERT Class 9th Maths Solutions
Question 3:
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.
Answer:
In ΔBOC and ΔAOD,∠BOC = ∠AOD (Vertically opposite angles)
∠CBO = ∠DAO (Each 90º)
BC = AD (Given)
∴ ΔBOC ≅ ΔAOD (AAS congruence rule)
∴ BO = AO (By CPCT)
⇒ CD bisects AB.
Q 4, Ex. 7.1, Page No 119 - Triangles - NCERT Class 9th Maths Solutions
Question 4:
l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that ΔABC ≅ ΔCDA.
Answer:
In ΔABC and ΔCDA,∠BAC = ∠DCA (Alternate interior angles, as p || q)
AC = CA (Common)
∠BCA = ∠DAC (Alternate interior angles, as l || m)
∴ ΔABC ≅ ΔCDA (By ASA congruence rule)
Q 5, Ex. 7.1, Page No 119 - Triangles - NCERT Class 9th Maths Solutions
Question 5:
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Answer:
In ΔAPB and ΔAQB,∠APB = ∠AQB (Each 90º)
∠PAB = ∠QAB (l is the angle bisector of ∠A)
AB = AB (Common)
∴ ΔAPB ≅ ΔAQB (By AAS congruence rule)
∴ BP = BQ (By CPCT)
Or, it can be said that B is equidistant from the arms of ∠A.
Page No 120:
Q 6, Ex. 7.1, Page No 120 - Triangles - NCERT Class 9th Maths Solutions
Question 6:
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Answer:
It is given that ∠BAD = ∠EAC∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE
In ΔBAC and ΔDAE,
AB = AD (Given)
∠BAC = ∠DAE (Proved above)
AC = AE (Given)
∴ ΔBAC ≅ ΔDAE (By SAS congruence rule)
∴ BC = DE (By CPCT)
Q 7, Ex. 7.1, Page No 120 - Triangles - NCERT Class 9th Maths Solutions
Question 7:
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that(i) ΔDAP ≅ ΔEBP
(ii) AD = BE
Answer:
It is given that ∠EPA = ∠DPB⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE
⇒ ∠DPA = ∠EPB
In
DAP and EBP,∠DAP = ∠EBP (Given)
AP = BP (P is mid-point of AB)
∠DPA = ∠EPB (From above)
∴ ΔDAP ≅ ΔEBP (ASA congruence rule)
∴ AD = BE (By CPCT)
Q 8, Ex. 7.1, Page No 120 - Triangles - NCERT Class 9th Maths Solutions
Question 8:
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = $\frac{1}{2}$AB
Answer:
(i) In ΔAMC and ΔBMD,AM = BM (M is the mid-point of AB)
∠AMC = ∠BMD (Vertically opposite angles)
CM = DM (Given)
∴ ΔAMC ≅ ΔBMD (By SAS congruence rule)
∴ AC = BD (By CPCT)
And, ∠ACM = ∠BDM (By CPCT)
(ii) ∠ACM = ∠BDM
However, ∠ACM and ∠BDM are alternate interior angles.
Since alternate angles are equal,
It can be said that DB || AC
⇒ ∠DBC + ∠ACB = 180º (Co-interior angles)
⇒ ∠DBC + 90º = 180º
⇒ ∠DBC = 90º
(iii) In ΔDBC and ΔACB,
DB = AC (Already proved)
∠DBC = ∠ACB (Each 90
)BC = CB (Common)
∴ ΔDBC ≅ ΔACB (SAS congruence rule)
(iv) ΔDBC ≅ ΔACB
∴ AB = DC (By CPCT)
⇒ AB = 2 CM
∴ CM =$\frac{1}{2}$AB
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