Objective Practice Questions (MCQ)
- In figure l1 || l2, the value of x is
(a) 80°
(b) 100°
(c) 110°
(d) 70°
2. An exterior angle of a triangle is 80° and the interior opposite angles are in the ratio 1 : 3, measure of interior opposite angles are
(a) 30°, 90°
(b) 40°, 120°
(c) 20°, 60°
(d) 30°, 60°
3. The angle of a triangle are in the ratio 5 : 3 : 7, the triangle is
(а) an acute-angled triangle
(b) an obtuse angled triangle
(c) an right angled triangle
(d) an isosceles triangle.
4. In ΔABC, ∠A = 50° and the external bisectors of ∠B and ∠C meet at O as shown in figure. The measure of ∠BOC is
(a) 40°
(b) 65°
(c) 115°
(d) 140°
5. In figure the value of x is
(a) 120°
(b) 130°
(c) 110°
(d) 100°
Subjective Practice Questions
- Show that the angles of an equilateral triangle are 60° each.
Solution:
In ∆ABC, we have
AB = BC = CA
[ABC is an equilateral triangle]
AB = BC
⇒ ∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal]
Similarly, AC = BC
⇒ ∠A = ∠B …(2)
From (1) and (2), we have
∠A = ∠B = ∠C = x (say)
Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A]
∴ x + x + x = 180o
⇒ 3x = 180°
⇒ x = 60°
∴ ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each
2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A
Solution:
(i) In right ∆ABD and ∆ACD, we have
AB =AC [Given]
∠ADB = ∠ADC [Each 90°]
AD = DA [Common]
∴ ∆ABD ≅ ∆ACD [By RHS congruency]
So, BD = CD [By C.P.C.T.]
⇒ D is the mid-point of BC or AD bisects BC.
(ii) Since, ∆ABD ≅ ∆ACD,
⇒ ∠BAD = ∠CAD [By C.P.C.T.]
So, AD bisects ∠A
Ex 7.3 Class 9 Maths Question 1 Solution
3. ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
- (i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Solution:
(i) In ∆ABD and ∆ACD, we have
AB = AC [Given]
AD = DA [Common]
BD = CD [Given]
∴ ∆ABD ≅ ∆ACD [By SSS congruency]
∠BAD = ∠CAD [By C.P.C.T.] …(1)
(ii) In ∆ABP and ∆ACP, we have
AB = AC [Given]
∠BAP = ∠CAP [From (1)]
∴ AP = PA [Common]
∴ ∆ABP ≅ ∆ACP [By SAS congruency]
(iii) Since, ∆ABP ≅ ∆ACP
⇒ ∠BAP = ∠CAP [By C.P.C.T.]
∴ AP is the bisector of ∠A.
Again, in ∆BDP and ∆CDP,
we have BD = CD [Given]
DP = PD [Common]
BP = CP [ ∵ ∆ABP ≅ ∆ACP]
⇒ A BDP = ACDP [By SSS congruency]
∴ ∠BDP = ∠CDP [By C.P.C.T.]
⇒ DP (or AP) is the bisector of ∠BDC
∴ AP is the bisector of ∠A as well as ∠D.
(iv) As, ∆ABP ≅ ∆ACP
⇒ ∠APS = ∠APC, BP = CP [By C.P.C.T.]
But ∠APB + ∠APC = 180° [Linear Pair]
∴ ∠APB = ∠APC = 90°
⇒ AP ⊥ BC, also BP = CP
Hence, AP is the perpendicular bisector of BC.
Ex 7.2 Class 9 Solution
6. ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C.
Solution:
In ∆ABC, we have AB = AC [Given]
∴ Their opposite angles are equal.
⇒ ∠ACB = ∠ABC
Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆]
⇒ 90° + ∠B + ∠C = 180° [∠A = 90°(Given)]
⇒ ∠B + ∠C= 180°- 90° = 90°
But ∠B = ∠C
∠B = ∠C = 90∘2 = 45°
Thus, ∠B = 45° and ∠C = 45°