Class 9 Exam  >  Class 9 Notes  >  Extra Documents & Tests for Class 9  >  NCERT Solutions Chapter 7 - Triangles (I), Class 9, Maths

NCERT Solutions for Class 9 Maths Chapter 7 - Chapter 7 - Triangles (I),

Exercise 7.3

1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that
 (i) ΔABD ≅ ΔACD
 (ii) ΔABP ≅ ΔACP
 (iii) AP bisects ∠A as well as ∠D.
 (iv) AP is the perpendicular bisector of BC. 

NCERT Solutions for Class 9 Maths Chapter 7 - Chapter 7 - Triangles (I),

Answer

Given,
ΔABC and ΔDBC are two isosceles triangles.

(i) In ΔABD and ΔACD,
AD = AD (Common)
AB = AC (ΔABC is isosceles)
BD = CD (ΔDBC is isosceles)
Therefore, ΔABD ≅ ΔACD by SSS congruence condition.

(ii) In ΔABP and ΔACP,
AP = AP (Common)
∠PAB = ∠PAC (ΔABD ≅ ΔACD so by CPCT) 
AB = AC (ΔABC is isosceles)
Therefore, ΔABP ≅ ΔACP by SAS congruence condition.

(iii) ∠PAB = ∠PAC by CPCT as ΔABD ≅ ΔACD.
AP bisects ∠A. --- (i)
also,
In ΔBPD and ΔCPD,
PD = PD (Common)
BD = CD (ΔDBC is isosceles.)
BP = CP (ΔABP ≅ ΔACP so by CPCT.)
Therefore, ΔBPD ≅ ΔCPD by SSS congruence condition. 
Thus, ∠BDP = ∠CDP by CPCT. --- (ii)
By (i) and (ii) we can say that AP bisects ∠A as well as ∠D.

(iv) ∠BPD = ∠CPD (by CPCT as ΔBPD ≅ ΔCPD)
and BP = CP --- (i) 
also,
∠BPD + ∠CPD = 180° (BC is a straight line.)
⇒ 2∠BPD = 180°
⇒ ∠BPD = 90° ---(ii) 
From (i) and (ii),
AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
 (i) AD bisects BC                      

(ii) AD bisects ∠A.

Answer

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

Given,

AD is an altitude and AB = AC

(i) In ΔABD and ΔACD,

∠ADB = ∠ADC = 90°
 AB = AC (Given)
AD = AD (Common) 
Therefore, ΔABD ≅ ΔACD by RHS congruence condition.
Now,
BD = CD (by CPCT)
Thus, AD bisects BC

(ii) ∠BAD = ∠CAD (by CPCT)
Thus, AD bisects ∠A.

3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig. 7.40). Show that:
 (i) ΔABM ≅ ΔPQN
 (ii) ΔABC ≅ ΔPQR

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

Answer

Given,
AB = PQ, BC = QR and AM = PN

(i) 1/2 BC = BM and 1/2QR = QN (AM and PN are medians)
also,
BC = QR
⇒ 1/2 BC = 1/2QR
⇒ BM = QN 
In ΔABM and ΔPQN,
AM = PN (Given)
AB = PQ (Given)
BM = QN (Proved above) 
Therefore, ΔABM ≅ ΔPQN by SSS congruence condition.

(ii) In ΔABC and ΔPQR,
AB = PQ (Given)
∠ABC = ∠PQR (by CPCT)
BC = QR (Given)

Therefore, ΔABC ≅ ΔPQR by SAS congruence condition.

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

 Answer 

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

Given,
BE and CF are two equal altitudes. 
In ΔBEC and ΔCFB,
∠BEC = ∠CFB = 90° (Altitudes)
 BC = CB (Common)
BE = CF (Common) 
Therefore, ΔBEC ≅ ΔCFB by RHS congruence condition.
Now,
∠C = ∠B (by CPCT)
Thus, AB = AC as sides opposite to the equal angles are equal.

5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

 Answer

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

Given,
AB = AC
In ΔABP and ΔACP,
∠APB = ∠APC = 90° (AP is altitude)
AB = AC (Given)
AP = AP (Common) 
Therefore, ΔABP ≅ ΔACP by RHS congruence condition.
Thus, ∠B = ∠C (by CPCT)

Exercise 7.4

1. Show that in a right angled triangle, the hypotenuse is the longest side.

 Answer

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

ABC is a triangle right angled at B.
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90° and ∠B is 90°.
Since, B is the largest angle of the triangle, the side opposite to it must be the largest.
So, BC is the hypotenuse which is the largest side of the right angled triangle ABC.

2. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

NCERT Solutions for Class 9 Maths Chapter 7 - Chapter 7 - Triangles (I),

Answer

Given,
∠PBC < ∠QCB
Now, 
∠ABC + ∠PBC = 180°
⇒ ∠ABC = 180° - ∠PBC
also,
∠ACB + ∠QCB = 180°
⇒ ∠ACB = 180° - ∠QCB 
Since, 
∠PBC < ∠QCB therefore, ∠ABC > ∠ACB
Thus, AC > AB as sides opposite to the larger angle is larger.


3. In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

NCERT Solutions for Class 9 Maths Chapter 7 - Chapter 7 - Triangles (I),

Answer

Given,
∠B < ∠A and ∠C < ∠D
Now,
AO <  BO --- (i) (Side opposite to the smaller angle is smaller)
OD < OC ---(ii) (Side opposite to the smaller angle is smaller)
Adding (i) and (ii) 
AO + OD < BO + OC
⇒ AD < BC

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).
 Show that ∠A > ∠C and ∠B > ∠D.

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

Answer

In ΔABD,
AB < AD < BD
∴ ∠ADB < ∠ABD --- (i) (Angle opposite to longer side is larger.)
Now,
In ΔBCD,
BC < DC < BD
∴ ∠BDC < ∠CBD --- (ii)
Adding (i) and (ii) we get,
∠ADB + ∠BDC < ∠ABD + ∠CBD 
⇒ ∠ADC < ∠ABC
⇒ ∠B > ∠D
Similarly, 
In ΔABC,
∠ACB < ∠BAC --- (iii) (Angle opposite to longer side is larger.)
Now,
In ΔADC,
∠DCA < ∠DAC --- (iv)
Adding (iii) and (iv) we get,
∠ACB + ∠DCA < ∠BAC + ∠DAC 
⇒ ∠BCD < ∠BAD
⇒ ∠A > ∠C


5. In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

Answer

Given,
PR > PQ and PS bisects ∠QPR
To prove,
∠PSR > ∠PSQ
Proof,
∠PQR > ∠PRQ --- (i) (PR > PQ as angle opposite to larger side is larger.)
∠QPS = ∠RPS --- (ii) (PS bisects ∠QPR)
∠PSR = ∠PQR + ∠QPS --- (iii) (exterior angle of a triangle equals to the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS --- (iv) (exterior angle of a triangle equals to the sum of opposite interior angles)
Adding (i) and (ii)
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [from (i), (ii), (iii) and (iv)]


6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer

Class IX, Mathematics, NCERT, CBSE, Questions and Answer, Q and A, Important

Let l is a line segment and B is a point lying o it. We drew a line AB perpendicular to l. Let C be any other point on l.
To prove,
AB < AC
Proof,
In ΔABC, 
∠B = 90°
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90°
∴ ∠C mustbe acute angle. or ∠C < ∠B
⇒ AB < AC (Side opposite to the larger angle is larger.) 

The document NCERT Solutions for Class 9 Maths Chapter 7 - Chapter 7 - Triangles (I), is a part of the Class 9 Course Extra Documents & Tests for Class 9.
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FAQs on NCERT Solutions for Class 9 Maths Chapter 7 - Chapter 7 - Triangles (I),

1. What are the properties of a triangle?
Ans. A triangle is a closed figure formed by three line segments. The properties of a triangle are: 1. The sum of the angles of a triangle is always 180 degrees. 2. The sum of the lengths of any two sides of a triangle is always greater than the length of the third side. 3. The longest side of a triangle is opposite the largest angle, and the shortest side is opposite the smallest angle. 4. The sum of the lengths of any two sides of a triangle is always greater than the length of the third side.
2. What is the Pythagorean theorem and how is it used in triangles?
Ans. The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. It can be represented as a^2 + b^2 = c^2, where 'a' and 'b' are the lengths of the legs of the triangle and 'c' is the length of the hypotenuse. The Pythagorean theorem is used to find missing side lengths in right-angled triangles and to determine if a triangle is a right-angled triangle.
3. How can we classify triangles based on their sides and angles?
Ans. Triangles can be classified based on their sides and angles as follows: Based on sides: - Equilateral triangle: All three sides are equal in length. - Isosceles triangle: Two sides are equal in length. - Scalene triangle: All three sides are different in length. Based on angles: - Acute triangle: All three angles are less than 90 degrees. - Right triangle: One angle is exactly 90 degrees. - Obtuse triangle: One angle is greater than 90 degrees.
4. How can we determine if two triangles are similar?
Ans. Two triangles are similar if their corresponding angles are equal and their corresponding sides are in proportion. This can be determined using the following tests: 1. Angle-Angle (AA) similarity: If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. 2. Side-Side-Side (SSS) similarity: If the corresponding sides of two triangles are in the same ratio, then the triangles are similar. 3. Side-Angle-Side (SAS) similarity: If one pair of corresponding sides are in the same ratio and the included angles are equal, then the triangles are similar.
5. How can we find the area of a triangle?
Ans. The area of a triangle can be found using the formula: Area = (1/2) * base * height, where the base is the length of the triangle's base and the height is the perpendicular distance from the base to the opposite vertex. If the lengths of the three sides of the triangle are known, the area can also be found using Heron's formula: Area = √(s(s-a)(s-b)(s-c)), where 's' is the semi-perimeter of the triangle and 'a', 'b', and 'c' are the lengths of its sides.
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