Q1: If AD = BC and ∠ BAD = ∠ ABC, then ∠ ACB is equal to
(a) ∠ABD
(b) ∠ BAD
(c) ∠BAC
(d) ∠BDA
Ans: (d)
In △ABC and △ABD
AD =BC (given)
∠ BAD = ∠ ABC (Given)
AB = AB (Common side)
∴ △ABC ≅ △ABD
By CPCT theorem, ∠ACB=∠BDA ( By SAS Congruency)
Q2: If O is a midpoint of AB and ∠BQO = ∠APO, then ∠OAP is equal to
(a) ∠QPA
(b) ∠OQB
(c) ∠QBO
(d) ∠BOQ
Ans: (c)
In △AOP and △BOQ
AO = BO….. (O is the midpoint of AB)
∠APO = ∠BQO (Given)
∠AOP = ∠BOQ (Vertically opposite Angles)
∴△AOP ≅ △BOQ (By AAS Congruency)
By CPCT ∠OAP = ∠QBO
Q3: If △ABC is an isosceles triangle, ∠ B = 650, find ∠ A.
(a) 60º
(b) 70º
(c) 50º
(d) none of these
Ans: (c)
Since △ABC is on an isosceles triangle
∴ ∠B = ∠C
∴ ∠B = 65º
∴ ∠C = 65º
∴ ∠A+∠B +∠C = 180º
∴ ∠A + 130º = 180º
∴ ∠A = 180º − 130º
∴∠A = 50º
Q4: An angle is 140 more than its complement. Find its measure.
(a) 42
(b) 32
(c) 52
(d) 62
Ans: (c)
Two angles whose sum equals 90 degrees are called complementary angles.
let first angle = x
it’s Complement = 90º − x
According to the question,
x = 14º + 90º − x
x = 104 − x
⇒ 2x = 104º
⇒ x = 1042
∴ x = 52º
Q5: If ABCD is a quadrilateral where AD = CB, AB = CD, and ∠ D = ∠ B, then ∠CAB is equal to
(a) ∠ACD
(b) ∠CAD
(c) ∠ACD
(d) ∠BAD
Ans: (c)
In △ABC and △CDA
CB = AD (Given)
AB = CD (Given)
∠B = ∠D (Given)
∴△ABC ≅ △CDA (By SAS Congruency)
By CPCT theorem
∠CAB = ∠ACD
Q6: If AB ⊥BC and ∠A =∠C, then the correct statement will
(a) AB ≠ AC
(b) AB = BC
(c) AB = AD
(d) AB = AC
Ans: (b)
In △ABC, ∠A = ∠C
Opposite sides to equal angles are also equal
AB = BC
Q7: If AB = AC and ∠ ACD = 1200, find ∠A.
(a) 500
(b) 600
(c) 700
(d) none of these
Ans: (b)
Since AB = AC
⇒∠ABC = ∠ACB = x (say)
Let ∠BAC = y
We know,
Exterior angles = sum of interior opposite angles
120º = ∠ABC + ∠BAC
120º = x + y−−−(1)
Again, ∠ACB + ∠ACD = 180º
x+120º = 180º
∴ x = 60º
From(1),
60º + y = 120º
⇒ y = 60º
⇒ ∠A = 60º
Q1: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Ans: In ΔAOD and ΔBOC,
BC = AD given
∠OAD = ∠OBC.(90 º)
BC  AD (alternate angles are equal)
∠ODA = ∠OCB.
So, ΔAOD ≅ ΔBOC (by the ASA congruence rule) OD = OC (CPCT)
and these form a pair of alternate angles for line segments AD and BC.
Therefore, CD bisects AB.
Q2: AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that
(i) ΔDAP ≌ ΔEBP
(ii) AD = BE
Ans. (i) ΔIn DAP and ΔEBP
P is its midpoint of AB
∴ AP = PB
So, ΔDAP ≌ ΔEBP(by the ASA congruence rule)
(ii) AD = BE(CPCT)
Q3: In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD. Show that AD = AE.
Ans: In ΔABD and ΔACE,
AB = AC (Given) ..(1)
∠B = ∠C (Angles opposite to equal sides) ..(2)
Also, BE = CD..(3)
So, BE – DE = CD – DE
That is, BD = CE (3)
So, ΔABD ≌ ΔACE
(Using (1), (2), (3) and SAS rule).
This gives AD = AE (CPCT)
Q4: In Figure OA = OB and OD = OC.
Show that
(i) ΔAOD ≅ ΔBOC
(ii) AD  BC
Ans: (i) In ΔAOD and ΔBOC,
∠AOD = ∠BOC.(pair of vertically opposite angles)
So, ΔAOD ≅ ΔBOC(by the SAS congruence rule)
(ii) ∠OAD = ∠OBC (CPCT)
and these form a pair of alternate angles for line segments AD and BC.
Therefore, AD  BC.
Q5: In Fig, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Ans. In ΔABC and ΔADE,
∠ADC+ ∠BAD = ∠ADC + ∠EAC
∠BAC = ∠DAE
So, ΔABC ≌ ΔADE(by the SAS congruence rule)
BC = DE (CPCT)
Q6: In ΔABC, the bisector AD of ∠A is perpendicular to side BC. Show that AB = AC and ΔABC is isosceles.
Ans: In ΔABD and ΔADC
AD bisects ∠A
⇒ ∠BAD = ∠CAD
∠ADC = ∠ADB 90º
AD = AD (Common Side)
So, ΔABD ≌ ΔADC (by the SAS congruence rule)
AB = AC(CPCT)
∴ ΔABC is isosceles
Q7: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that
(i) ΔABE ≌ ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Ans: In ΔABE and ΔACF,
BE = FC (Given) ..(1)
∠A = ∠A (Common Angle) ..(2)
Also, ∠ABE = ∠ACF ....(3)
So, ΔABE ≌ ΔACF(Using (1), (2), (3) and ASA rule).
This gives AB = AC (CPCT)
∴ ABC is an isosceles triangle
48 videos387 docs65 tests

NCERT Exemplar Solutions: Triangles Doc  25 pages 
ValueBased Questions: Triangles Doc  2 pages 
NCERT Solutions: Triangles (Exercise 7.17.3) Doc  11 pages 
1. What are the different types of triangles? 
2. How do you find the area of a triangle? 
3. What is the Pythagorean theorem? 
4. How can you determine if three given side lengths form a triangle? 
5. What are the properties of similar triangles? 
48 videos387 docs65 tests

NCERT Exemplar Solutions: Triangles Doc  25 pages 
ValueBased Questions: Triangles Doc  2 pages 
NCERT Solutions: Triangles (Exercise 7.17.3) Doc  11 pages 

Explore Courses for Class 9 exam
