(B) Multiple Choice Questions Exercise 7.1 Class 9...Class 9 NCERT Exemplar Solutions Mathematics...

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Class 9 NCERT Exemplar Solutions Mathematics

Chapter 7 Triangles (B) Multiple Choice Questions

Exercise 7.1 In each of the following, write the correct answer:

Q.1) Which of the following is not a criterion for congruence of triangles? (A) SAS (B) ASA (C) SSA (D) SSS

Sol.1) (c) We know that, two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle. Also, criterion for congruence of triangles are SAS (Side-Angle-Side), ASA (Angle-Side- Angle), SSS (Side-Side-Side) and RHS (right angle-hypotenuse-side). So, SSA is not a criterion for congruence of triangles.

Q.2) If 𝐴𝐵 = 𝑄𝑅, 𝐵𝐶 = 𝑃𝑅 and 𝐶𝐴 = 𝑃𝑄, then (a) 𝛥𝐴𝐵𝐶 ≅ 𝛥𝑄𝑅𝑃 (b) 𝛥𝐶𝐵𝐴 ≅ 𝛥𝑃𝑅𝑄 (c) 𝛥𝐵𝐴𝐶 ≅ 𝛥𝑅𝑄𝑃 (d) 𝛥𝑃𝑄𝑅 ≅ 𝛥𝐵𝐶𝐴

Sol.2) (b) We know that, if 𝛥𝑅𝑆𝑇 is congruent to 𝛥𝑈𝑉𝑊 i.e., 𝛥𝑅𝑆𝑇 = 𝛥𝑈𝑉𝑊, then sides of 𝛥𝑅𝑆𝑇 fall on corresponding equal sides of 𝛥𝑈𝑉𝑊 and angles of 𝛥𝑅𝑆𝑇 fall on corresponding equal angles of 𝛥𝑈𝑉𝑊. Here, given 𝐴𝐵 = 𝑄𝑅, 𝐵𝐶 = 𝑃𝑅 and 𝐶𝐴 = 𝑃𝑄, which shows that 𝐴𝐵 covers QR, BC covers 𝑃𝑅 and 𝐶𝐴 covers 𝑃𝑄 i.e., 𝐴 correspond to 𝑄, 𝐵 correspond to R and C correspond to P. or 𝐴 ↔ 𝑄, 𝐵 ↔ 𝑅, 𝐶 ↔ 𝑃 Under this correspondence, 𝛥𝐴𝐵𝐶 ≅ 𝛥𝑄𝑅𝑃, so option (a) is incorrect, or 𝛥𝐵𝐴𝐶 ≅ 𝛥𝑅𝑄𝑃, so option (c) is incorrect, or 𝛥𝐶𝐵𝐴 ≅ 𝛥𝑃𝑅𝑄, so option (b) is correct, or 𝛥𝐵𝐶𝐴 ≅ 𝛥𝑅𝑃𝑄, so option (d) is incorrect.

Q.3) In ∠A𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 and ∠𝐵 = 50°. Then ∠𝐶 is equal to (A) 40° (B) 50° (C) 80° (D) 130°

Sol.3) (b) Given, ∆𝐴𝐵𝐶 such that 𝐴𝐵 = 𝐴𝐶 and ∠B = 50°

In ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 ∠C = ∠B ∠C = 50°

Q.4) In ∆𝐴𝐵𝐶, 𝐵𝐶 = 𝐴𝐵 and ∠𝐵 = 80°. Then ∠A is equal to (A) 80° (B) 40° (C) 50° (D) 100°

Sol.4) (c) Given, ∆𝐴𝐵𝐶 such that 𝐵𝐶 = 𝐴𝐵 and ∠B = 80° In ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐵𝐶 ∠C = ∠A The sum of all angles of a triangle is 180°

∴ ∠A + ∠B + ∠C = 180° ∠A + 80° + ∠A = 180°

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2∠A = 180° − 80° = 100°

∠A =100°

2

∠A = 50° Q.5) In ∆𝑃𝑄𝑅, ∠𝑅 = ∠𝑃 and 𝑄𝑅 = 4 𝑐𝑚 and 𝑃𝑅 = 5 𝑐𝑚. Then the length of 𝑃𝑄 is

(A) 4 cm (B) 5 cm (C) 2 cm (D) 2.5 cm

Sol.5) (A) Given, ∆𝑃𝑄𝑅 such that ∠R = ∠P, QR = 4cm and 𝑃𝑅 = 5𝑐𝑚

In ∆𝑃𝑄𝑅, ∠R = ∠P 𝑃𝑄 = 𝑄𝑅 𝑃𝑄 = 4𝑐𝑚 Hence, the length of PWQ is 4cm.

Q.6) D is a point on the side BC of a ∆𝐴𝐵𝐶 such that AD bisects ∠𝐵𝐴𝐶. Then (A) 𝐵𝐷 = 𝐶𝐷 (B) 𝐵𝐴 > 𝐵𝐷 (C) BD > BA (D) 𝐶𝐷 > 𝐶𝐴

Sol.6) (b) Given, ∆𝐴𝐵𝐶 such that AD bisects ∠BAC ∠BAD = ∠CAD

In ∆𝐴𝐶𝐷, ∠BDA is an exterior angle ∠BDA > ∠CAD ∠BDA > ∠BAD 𝐵𝐴 > 𝐵𝐷

Q.7) It is given that ∆𝐴𝐵𝐶 ≅ ∆𝐹𝐷𝐸 and 𝐴𝐵 = 5 𝑐𝑚, ∠𝐵 = 40° and ∠𝐴 = 80°. Then which of the following is true? (A) 𝐷𝐹 = 5 𝑐𝑚, ∠𝐹 = 60° (B) 𝐷𝐹 = 5 𝑐𝑚, ∠𝐸 = 60° (C) 𝐷𝐸 = 5 𝑐𝑚, ∠𝐸 = 60° (D) 𝐷𝐸 = 5 𝑐𝑚, ∠𝐷 = 40°

Sol.7) (b) Given, ∆𝐴𝐵𝐶 ≅ ∆𝐹𝐷𝐸 and 𝐴𝐵 = 5𝑐𝑚, ∠B = 40°, ∠A = 80° Since, ∆𝐹𝐷𝐸 ≅ ∆𝐴𝐵𝐶 𝐷𝐹 = 𝐴𝐵 𝐷𝐹 = 5𝑐𝑚 & ∠E = ∠C

∠E = ∠C = 180° − (∠A + ∠B)

∠E = 180° − (80° + 40°)

∠E = 60° Q.8) Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the

triangle cannot be (A) 3.6 cm (B) 4.1 cm (C) 3.8 cm (D) 3.4 cm

Sol.8) (d) Given, the length of two sides of a triangle are 5 𝑐𝑚 and 1.5 𝑐𝑚, respectively. Let sides 𝐴𝐵 = 5 𝑐𝑚 and 𝐶𝐴 = 1.5 𝑐𝑚 We know that, a closed figure formed by three intersecting lines (or sides) is called a triangle, if difference of two sides < third side and sum of two sides > third side ∴ 5 − 1.5 < 𝐵𝐶 and 5 + 1.5 > 𝐵𝐶 ⇒ 3.5 < 𝐵𝐶 and 6.5 > 𝐵𝐶



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Here, we see that options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

Q.9) In ∆𝑃𝑄𝑅, if ∠𝑅 > ∠𝑄, then (A) 𝑄𝑅 > 𝑃𝑅 (B) 𝑃𝑄 > 𝑃𝑅 (C) 𝑃𝑄 < 𝑃𝑅 (D) 𝑄𝑅 < 𝑃𝑅

Sol.9) (b) Given, ∠R > ∠Q 𝑃𝑄 > 𝑃𝑅

Q.10) In triangles 𝐴𝐵𝐶 and 𝑃𝑄𝑅, 𝐴𝐵 = 𝐴𝐶, ∠𝐶 = ∠𝑃 𝑎𝑛𝑑 ∠𝐵 = ∠𝑄. The two triangles are

(A) isosceles but not congruent (B) isosceles and congruent (C) congruent but not isosceles (D) neither congruent nor isosceles

Sol.10) (a) In ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 ∠C = ∠B So, ∆𝐴𝐵𝐶 is an isosceles triangle. But given that, ∠B = ∠Q ∠C = ∠P ∠P = ∠Q 𝑄𝑅 = 𝑃𝑅 So, ∆𝑃𝑄𝑅 is also an isosceles triangle. Therefore, both triangles are isosceles but not congruent. As we know that, AAA is not a criterion for congruence of triangles.

Q.11) In triangles ∆𝐴𝐵𝐶 and ∆𝐷𝐸𝐹, 𝐴𝐵 = 𝐹𝐷 and ∠𝐴 = ∠𝐷. The two triangles will be congruent by SAS axiom if (A) 𝐵𝐶 = 𝐸𝐹 (B) 𝐴𝐶 = 𝐷𝐸 (C) 𝐴𝐶 = 𝐸𝐹 (D) 𝐵𝐶 = 𝐷𝐸

Sol.11) (b) Given, in 𝛥𝐴𝐵𝐶 and 𝛥𝐷𝐸𝐹, 𝐴𝐵 = 𝐷𝐹 and ∠𝐴 = ∠𝐷 We know that, two triangles will be congruent by 𝐴𝑆𝐴 rule, if two angles and the included side of one triangle are equal to the two angles and the included side of other triangle. ∴ 𝐴𝐶 = 𝐷𝐸

(C) Short Answer Questions with Reasoning Exercise 7.2

Q.1) In triangles ∆𝐴𝐵𝐶 and ∆𝑃𝑄𝑅, ∠𝐴 = ∠𝑄 and ∠𝐵 = ∠𝑅. Which side of ∆𝑃𝑄𝑅 should be equal to side 𝐴𝐵 of ∆𝐴𝐵𝐶 so that the two triangles are congruent? Give reason for your answer.

Sol.1) We have given, in 𝛥𝐴𝐵𝐶 and 𝛥𝑃𝑄𝑅, ∠𝐴 = ∠𝑄 and ∠𝐵 = ∠𝑅 Since, AB and QR are included between equal angles.

Hence, the side of 𝛥𝑃𝑄𝑅 is QR which should be equal to side 𝐴𝐵 of 𝛥𝐴𝐵𝐶, so that the triangles are congruent by the rule ASA.

Q.2) In triangles ∆𝐴𝐵𝐶 and ∆𝑃𝑄𝑅, ∠A = ∠𝑄 𝑎𝑛𝑑 ∠B = ∠𝑅. Which side of ∆𝑃𝑄𝑅 should be equal to side BC of ∆𝐴𝐵𝐶 so that the two triangles are congruent? Give reason for your answer.

Sol.2) Given, In ∆𝐴𝐵𝐶 and ∆𝑃𝑄𝑅 , ∠𝐴 = ∠𝑄 and ∠𝐵 = ∠𝑅



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Since, two pairs of angles are equal in two triangles. We know that, two triangles will be congruent by AAS rule, if two angles and the side of one triangle are equal to the two angles and the side of other triangle. ∴ 𝐵𝐶 = 𝑅𝑃

Q.3) “If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?

Sol.3) No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

Q.4) “If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent.” Is the statement true? Why?

Sol.4) No, because sides must be corresponding sides.

Q.5) Is it possible to construct a triangle with lengths of its sides as 4 𝑐𝑚, 3 𝑐𝑚 and 7 𝑐𝑚? Give reason for your answer.

Sol.5) No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 = 7. As we know that, the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

Q.6) It is given that ∆𝐴𝐵𝐶 ≅ ∆𝑅𝑃𝑄. Is it true to say that 𝐵𝐶 = 𝑄𝑅? Why?

Sol.6) No, we know that two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding side and angles of other triangle. Here 𝛥𝐴𝐵𝐶 ≅ 𝛥𝑅𝑃𝑄 𝐴𝐵 = 𝑅𝑃, 𝐵𝐶 = 𝑃𝑄 and 𝐴𝐶 = 𝑅𝑄 Hence, it is not true to say that 𝐵𝐶 = 𝑄𝑅.

Q.7) If ∆𝑃𝑄𝑅 ≅ ∆𝐸𝐷𝐹, then is it true to say that 𝑃𝑅 = 𝐸𝐹? Give reason for your answer.

Sol.7) Yes, if 𝛥𝑃𝑄𝑅 ≅ 𝛥𝐸𝐷𝐹, then it means that corresponding angles and their sides are equal because we know that, two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding sides and angles of other triangle. Here, 𝛥𝑃𝑄𝑅 ≅ 𝛥𝐸𝐷𝐹 ∴ 𝑃𝑄 = 𝐸𝐷, 𝑄𝑅 = 𝐷𝐹 and 𝑃𝑅 = 𝐸𝐹 Hence, it is true to say that 𝑃𝑅 = 𝐸𝐹.

Q.8) In ∆𝑃𝑄𝑅, ∠𝑃 = 70°and ∠𝑅 = 30° Which side of this triangle is the longest? Give reason for your answer.

Sol.8) Given, in 𝛥𝑃𝑄𝑅, ∠𝑃 = 70° and ∠𝑅 = 30°. We know that, sum of all the angles of a triangle is 180°. ∠𝑃 + ∠𝑄 + ∠𝑅 = 180°

∠𝑄 = 180° − (70° + 30°)

∠𝑄 = 80° We know that here ∠𝑄 is longest, so side 𝑃𝑅 is longest. [∴ since in a triangle, the side opposite to the largest angle is the longest]

Q.9) AD is a median of the triangle ∆𝐴𝐵𝐶. Is it true that 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐴 > 2 𝐴𝐷? Give reason for your answer.

Sol.9) Yes, In ∆𝐴𝐵𝐷, we have 𝐴𝐵 + 𝐵𝐷 > 𝐴𝐷 …(i)



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In ∆𝐴𝐶𝐷, we have 𝐴𝐶 + 𝐶𝐷 > 𝐴𝐷 …(ii) On adding eq. (i) & (ii), we get (𝐴𝐵 + 𝐵𝐷 + 𝐴𝐶 + 𝐶𝐷) > 2𝐴𝐷 (𝐴𝐵 + 𝐵𝐷 + 𝐶𝐷 + 𝐴𝐶) > 2𝐴𝐷 Hence, 𝐴𝐵 + 𝐵𝐶 + 𝐴𝐶 > 2𝐴𝐷 [𝐵𝐶 = 𝐵𝐷 + 𝐶𝐷]

Q.10) 𝑀 is a point on side 𝐵𝐶 of a triangle ∆𝐴𝐵𝐶 such that AM is the bisector of ∠𝐵𝐴𝐶. Is it true to say that perimeter of the triangle is greater than 2 𝐴𝑀? Give reason for your answer.

Sol.10) Yes, In ∆𝐴𝐵𝐶, 𝑀 is a point of 𝐵𝐶 such that 𝐴𝑀 is the bisector of ∠𝐵𝐴𝐶. In ∆𝐴𝐵𝑀, 𝐴𝐵 + 𝐵𝑀 > 𝐴𝑀 …(i) In ∆𝐴𝐶𝑀, 𝐴𝐶 + 𝐶𝑀 > 𝐴𝑀 …(ii)

On adding eq. (i) & (ii), we get (𝐴𝐵 + 𝐵𝑀 + 𝐴𝐶 + 𝐶𝑀) > 2𝐴𝑀 (𝐴𝐵 + 𝐵𝑀 + 𝑀𝐶 + 𝐴𝐶) > 2𝐴𝑀 𝐴𝐵 + 𝐵𝐶 + 𝐴𝐶 > 2𝐴𝑀 Perimeter of ∆𝐴𝐵𝐶 > 2𝐴𝑀


Sol.11) No, Here, we see that 9 + 7 = 16 < 17 i.e., the sum of two sides of a triangle is less than the third side. Hence, it contradicts the property that the sum of two sides of a triangle is greater than the third side. Therefore, it is not possible to construct a triangle with given sides.


Sol.12) Yes, because in each case the sum of two sides is greater than the third side. i.e., 7 + 4 > 8, 8 + 4 > 7, 7 + 8 > 4 Hence, it is possible to construct a triangle with given sides.

(D) Short Answer Questions Exercise 7.3

Q.1) 𝐴𝐵𝐶 is an isosceles triangle with 𝐴𝐵 = 𝐴𝐶 and 𝐵𝐷 and 𝐶𝐸 are its two medians. Show that 𝐵𝐷 = 𝐶𝐸.

Sol.1) Given 𝛥𝐴𝐵𝐶 is an isosceles triangle in which 𝐴𝐵 = 𝐴𝐶 and 𝐵𝐷, 𝐶𝐸 are its two medians. To show 𝐵𝐷 = 𝐶𝐸. Proof: In ∆𝐴𝐵𝐷 and ∆𝐴𝐶𝐸, 𝐴𝐵 = 𝐴𝐶 [given] ∠𝐴 = ∠𝐴 [common angle] & 𝐴𝐷 = 𝐴𝐸 𝐴𝐵 = 𝐴𝐶

1

2𝐴𝐵 =

1

2𝐴𝐶

𝐴𝐸 = 𝐴𝐷 As 𝐷 is the mid-point of 𝐴𝐶 and 𝐸 is the mid-point of 𝐴𝐵. ∆𝐴𝐵𝐷 ≅ ∆𝐴𝐶𝐸 [by SAS congruence rule]



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𝐵𝐷 = 𝐶𝐸 [by CPCT]

Q.2) In Fig.7.4, 𝐷 and 𝐸 are points on side BC of a ∆𝐴𝐵𝐶 such that 𝐵𝐷 = 𝐶𝐸 and 𝐴𝐷 = 𝐴𝐸. Show that ∆𝐴𝐵𝐷 ≅ ∆𝐴𝐶𝐸.

Sol.2) Given, 𝐷 and 𝐸 are the points on side 𝐵𝐶 of a ∆𝐴𝐵𝐶 such that 𝐵𝐷 = 𝐶𝐸 and 𝐴𝐷 = 𝐴𝐸

To show: ∆𝐴𝐵𝐷 ≅ ∆𝐴𝐶𝐸 Proof: We have, 𝐴𝐷 = 𝐴𝐸 [given] ∠𝐴𝐷𝐸 = ∠𝐴𝐸𝐷 … (i)

We have, ∠𝐴𝐷𝐵 + ∠𝐴𝐷𝐸 = 180° [linear pair axiom]

∠𝐴𝐷𝐵 = 180° − ∠𝐴𝐷𝐸 = 180° − ∠𝐴𝐸𝐷 [from eq. (i)] In ∆𝐴𝐵𝐷 and ∆𝐴𝐶𝐸, ∠𝐴𝐷𝐵 = ∠𝐴𝐸𝐶 𝐵𝐷 = 𝐶𝐸 [given] & 𝐴𝐷 = 𝐴𝐸 [given] ∆𝐴𝐵𝐷 ≅ ∆𝐴𝐶𝐸 [by SAS congruence rule]

Q.3) ∆𝐶𝐷𝐸 is an equilateral triangle formed on a side 𝐶𝐷 of a square 𝐴𝐵𝐶𝐷 (Fig.7.5). Show that ∆𝐴𝐷𝐸 ≅ ∆𝐵𝐶𝐸.

Sol.3) Given, In figure ∆𝐶𝐷𝐸 is an equilateral triangle formed on a side 𝐶𝐷 of a square 𝐴𝐵𝐶𝐷.

To show ∆𝐴𝐷𝐸 ≅ ∆𝐵𝐶𝐸 Proof In ∆𝐴𝐷𝐸 and ∆𝐵𝐶𝐸 𝐷𝐸 = 𝐶𝐸 [sides of an equilateral triangles] ∠𝐴𝐷𝐸 = ∠𝐵𝐶𝐸 & 𝐴𝐷 = 𝐵𝐶 [sides of a square] ∆𝐴𝐷𝐸 ≅ ∆𝐵𝐶𝐸 [by SAS congruence rule]

Q.4) In Fig.7.6, 𝐵𝐴 ⊥ 𝐴𝐶, 𝐷𝐸 ⊥ 𝐷𝐹 such that 𝐵𝐴 = 𝐷𝐸 and 𝐵𝐹 = 𝐸𝐶. Show that ∆𝐴𝐵𝐶 ≅∆𝐷𝐸𝐹.

Sol.4) Given, In figure, 𝐵𝐴 ⊥ 𝐴𝐶, 𝐷𝐸 ⊥ 𝐷𝐹 such that 𝐵𝐴 = 𝐷𝐸 and 𝐵𝐹 = 𝐸𝐶

To show ∆𝐴𝐵𝐶 ≅ ∆𝐷𝐸𝐹 Proof Since, 𝐵𝐹 = 𝐸𝐶 On adding 𝐶𝐹 both sides, we get 𝐵𝐹 + 𝐶𝐹 = 𝐸𝐶 + 𝐶𝐹 𝐵𝐶 = 𝐸𝐹 … (i)

In ∆𝐴𝐵𝐶 and ∆𝐷𝐸𝐹, ∠𝐴 = ∠𝐷 = 90°



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𝐵𝐶 = 𝐸𝐹 from eq. (i) & 𝐵𝐴 = 𝐷𝐸 [given] ∆𝐴𝐵𝐶 ≅ ∆𝐷𝐸𝐹 [by RHS congruence rule]

Q.5) Q is a point on the side SR of a ∆𝑃𝑆𝑅 such that 𝑃𝑄 = 𝑃𝑅. Prove that 𝑃𝑆 > 𝑃𝑄.

Sol.5) Given, In ∆𝑃𝑆𝑅, 𝑄 is a point on the side 𝑆𝑅 such that 𝑃𝑄 = 𝑃𝑅 To prove 𝑃𝑆 > 𝑃𝑄 Proof In ∆𝑃𝑅𝑄, 𝑃𝑄 = 𝑃𝑅 [given] ∠𝑅 = ∠𝑃𝑄𝑅 …(i) But ∠𝑃𝑄𝑅 > ∠𝑆 …(ii) From eqs. (i) & (ii), we get ∠𝑅 > ∠𝑆 𝑃𝑆 > 𝑃𝑅 [side opposite to greater angle is longer] 𝑃𝑆 > 𝑃𝑄 [𝑃𝑄 = 𝑃𝑅]

Q.6) S is any point on side QR of a ∆𝑃𝑄𝑅. Show that: 𝑃𝑄 + 𝑄𝑅 + 𝑅𝑃 > 2 𝑃𝑆.

Sol.6) Given, In ∆𝑃𝑄𝑅, 𝑆 is any point on side 𝑄𝑅. To show In ∆𝑃𝑄𝑆, 𝑃𝑄 + 𝑄𝑆 > 𝑃𝑆 … (i) Similarly, in ∆𝑃𝑅𝑆, 𝑆𝑅 + 𝑅𝑃 > 𝑃𝑆 … (ii) On adding eq. (i) & (ii), we get 𝑃𝑄 + 𝑄𝑆 + 𝑆𝑅 + 𝑅𝑃 > 2𝑃𝑆 𝑃𝑄 + (𝑄𝑆 + 𝑆𝑅) + 𝑅𝑃 > 2𝑃𝑆 𝑃𝑄 + 𝑄𝑅 + 𝑅𝑃 > 2𝑃𝑆

Q.7) 𝐷 is any point on side 𝐴𝐶 of a ∆𝐴𝐵𝐶 with 𝐴𝐵 = 𝐴𝐶. Show that 𝐶𝐷 < 𝐵𝐷.

Sol.7) Given, In ∆𝐴𝐵𝐶, 𝐷 is any point on side 𝐴𝐶 such that 𝐴𝐵 = 𝐴𝐶. To show 𝐶𝐷 < 𝐵𝐷 or 𝐵𝐷 > 𝐶𝐷 Proof In ∆𝐴𝐵𝐶, 𝐴𝐶 = 𝐴𝐵 [given] ∠𝐴𝐵𝐶 = ∠𝐴𝐶𝐵 … (i) In ∆𝐴𝐵𝐶 and ∆𝐷𝐵𝐶, ∠𝐴𝐵𝐶 > ∠𝐷𝐵𝐶 ∠𝐴𝐶𝐵 > ∠𝐷𝐵𝐶 from eq. (i) 𝐵𝐷 > 𝐶𝐷 [side opposite to greater angle is longer] 𝐶𝐷 < 𝐵𝐷

Q.8) In Fig. 7.7, 𝑙 || 𝑚 and 𝑀 is the mid-point of a line segment 𝐴𝐵. Show that 𝑀 is also the mid-point of any line segment 𝐶𝐷, having its end points on 𝑙 and 𝑚, respectively.

Sol.8) Given, In the figure, 𝑙 || 𝑚 and 𝑀 is the mid point of a line segment 𝐴𝐵 i.e., 𝐴𝑀 = 𝐵𝑀

To show 𝑀𝐶 = 𝑀𝐷 Proof: 𝑙 || 𝑚 [given] ∠𝐵𝐴𝐶 = ∠𝐴𝐵𝐷 [alternate interior angle] ∠𝐴𝑀𝐶 = ∠𝐵𝑀𝐷 [vertically opposite angle] In ∆𝐴𝑀𝐶 and ∆𝐵𝑀𝐷, ∠𝐵𝐴𝐶 = ∠𝐴𝐵𝐷 [proved above] 𝐴𝑀 = 𝐵𝑀 [given]



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& ∠𝐴𝑀𝐶 = ∠𝐵𝑀𝐷 [proved above] ∆𝐴𝑀𝐶 ≅ ∆𝐵𝑀𝐷 [by ASA congruence rule] 𝑀𝐶 = 𝑀𝐷 [by CPCT]

Q.9) Bisectors of the angles B and C of an isosceles triangle with 𝐴𝐵 = 𝐴𝐶 intersect each other at 𝑂. 𝐵𝑂 is produced to a point 𝑀. Prove that ∠𝑀𝑂𝐶 = ∠𝐴𝐵𝐶.

Sol.9) Given, Lines, 𝑂𝐵 and 𝑂𝐶 are the angles bisectors of ∠𝐵 and ∠𝐶 of an isosceles ∆𝐴𝐵𝐶 such that 𝐴𝐵 = 𝐴𝐶 which intersect each otherat 𝑂 and 𝐵𝑂 is produced to 𝑀. To prove ∠𝑀𝑂𝐶 = ∠𝐴𝐵𝐶 Proof In ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 [given] ∠𝐴𝐶𝐵 = ∠𝐴𝐵𝐶 [angles opposite to equal sides are equal]

1

2∠𝐴𝐶𝐵 =

1

2∠𝐴𝐵𝐶 [dividing both sides 2]

∠𝑂𝐶𝐵 = ∠𝑂𝐵𝐶 … (i) Now, ∠𝑀𝑂𝐶 = ∠𝑂𝐵𝐶 + ∠𝑂𝐶𝐵 ∠𝑀𝑂𝐶 = ∠𝑂𝐵𝐶 + ∠𝑂𝐵𝐶 from eq. (i) ∠𝑀𝑂𝐶 = 2∠𝑂𝐵𝐶 [since 𝑂𝐵 is the bisector of ∠𝐵] ∠𝑀𝑂𝐶 = ∠𝐴𝐵𝐶 Hence proved.

Q.10) Bisectors of the angles 𝐵 and 𝐶 of an isosceles ∆𝐴𝐵𝐶 with 𝐴𝐵 = 𝐴𝐶 intersect each other at 𝑂. Show that external angle adjacent to ∠𝐴𝐵𝐶 is equal to ∠𝐵𝑂𝐶.

Sol.10) Given, ∆𝐴𝐵𝐶 is an isosceles triangle in which 𝐴𝐵 = 𝐴𝐶, 𝐵𝑂 and 𝐶𝑂 are the bisectors of ∠𝐴𝐵𝐶 and ∠𝐴𝐶𝐵 respectively at 𝑂. To show ∠𝐷𝐵𝐴 = ∠𝐵𝑂𝐶 Construction Line 𝐶𝐵 produced to D Proof In ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 [given] ∠𝐴𝐶𝐵 = ∠𝐴𝐵𝐶

1

2∠𝐴𝐶𝐵 =

1

2∠𝐴𝐵𝐶 [on dividing both sides by 2]

∠𝑂𝐶𝐵 = ∠𝑂𝐵𝐶 … (i)

In ∆𝐵𝑂𝐶, ∠𝑂𝐵𝐶 + ∠𝑂𝐶𝐵 + ∠𝐵𝑂𝐶 = 180° by angle sum property of a triangle

∠𝑂𝐵𝐶 + ∠𝑂𝐵𝐶 + ∠𝐵𝑂𝐶 = 180° from eq. (i)

2∠𝑂𝐵𝐶 + ∠𝐵𝑂𝐶 = 180° ∠𝐴𝐵𝐶 + ∠𝐵𝑂𝐶 = 180° [BO is the bisector of ∠𝐴𝐵𝐶]

180° − ∠𝐷𝐵𝐴 + ∠𝐵𝑂𝐶 = 180° [DBC is a straight line] −∠𝐷𝐵𝐴 + ∠𝐵𝑂𝐶 = 0 ∠𝐷𝐵𝐴 = ∠𝐵𝑂𝐶

Q.11) In Fig. 7.8, 𝐴𝐷 is the bisector of ∠𝐵𝐴𝐶. Prove that 𝐴𝐵 > 𝐵𝐷.

Sol.11) Given, 𝐴𝐵𝐶 is a triangle such that 𝐴𝐷 is the bisector of ∠𝐵𝐴𝐶.

To prove 𝐴𝐵 > 𝐵𝐷. Proof Since, AD is the bisector of ∠𝐵𝐴𝐶. But ∠𝐵𝐴𝐷 = ∠𝐶𝐴𝐷 …(i) ∴ ∠𝐴𝐷𝐵 > ∠𝐶𝐴𝐷 [exterior angle of a triangle is greater than each of the opposite interior angle] ∴ ∠𝐴𝐷𝐵 > ∠𝐵𝐴𝐷 [from Eq. (i)]



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𝐴𝐵 > 𝐵𝐷 [side opposite to greater angle is longer] Hence proved.

(E) Long Answer Questions Exercise 7.4

Q.1) Find all the angles of an equilateral triangle.

Sol.1) Let 𝐴𝐵𝐶 ne an equilateral triangle such that 𝐴𝐵 = 𝐵𝐶 = 𝐶𝐴 We have, 𝐴𝐵 = 𝐴𝐶 ∠𝐶 = ∠𝐵 Let ∠𝐶 = ∠𝐵 = 𝑥° … (i) Now, 𝐵𝐶 = 𝐵𝐴 ∠𝐴 = ∠𝐶 … (ii) From eq. (i) & (ii), we get ∠𝐴 = ∠𝐵 = ∠𝐶 = 𝑥 Now, in ∆𝐴𝐵𝐶, ∠𝐴 + ∠𝐵 + ∠𝐶 = 180° [by angle sum property]

𝑥 + 𝑥 + 𝑥 = 180° 3𝑥 = 180°

𝑥 =180°

3= 60°

Hence, ∠𝐴 = ∠𝐵 = ∠𝐶 = 60°

Q.2) The image of an object placed at a point 𝐴 before a plane mirror 𝐿𝑀 is seen at the point B by an observer at D as shown in Fig. 7.12. Prove that the image is as far behind the mirror as the object is in front of the mirror. [Hint: CN is normal to the mirror. Also, angle of incidence = angle of reflection]

Sol.2) Given, An object 𝑂𝐴 placed at a point 𝐴, 𝐿𝑀 be a plane mirror, 𝐷 be an observer and

𝑂𝐵 is the image. To prove The image is as far behind the mirror as the object is in the front of the mirror i.e., 𝑂𝐵 = 𝑂𝐴. Proof 𝐶𝑁 ⊥ LM and 𝐴𝐵 ⊥ LM 𝐴𝐵|| 𝐶𝑁 ∠𝐴 = ∠𝑖 …(i) [alternate interior angles] ∠𝐵 = ∠𝑟 …(ii) [corresponding angles] Also, ∠𝑖 = ∠𝑟 …(iii) [incident angle = reflected angle] From eqs. (i), (ii) & (iii), we get ∠𝐴 = ∠𝐵 In ∆𝐶𝑂𝐵 and ∆𝐶𝑂𝐴, ∠𝐵 = ∠𝐴 [proved above] ∠1 = ∠2 [each 90 degree] & 𝐶𝑂 = 𝐶𝑂 [common side] ∆𝐶𝑂𝐵 ≅ ∆𝐶𝑂𝐴 [by AAS congruence rule] 𝑂𝐵 = 𝑂𝐴 [by CPCT] Hence proved.

Q.3) ABC is an isosceles triangle with 𝐴𝐵 = 𝐴𝐶 and D is a point on BC such that 𝐴𝐷 ⊥ 𝐵𝐶 (Fig. 7.13). To prove that ∠𝐵𝐴𝐷 = ∠𝐶𝐴𝐷, a student proceeded as follows: In ∠𝐴𝐵𝐷 and ∠𝐴𝐶𝐷, 𝐴𝐵 = 𝐴𝐶 (Given) ∠𝐵 = ∠𝐶 (because 𝐴𝐵 = 𝐴𝐶)



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and ∠𝐴𝐷𝐵 = ∠𝐴𝐷𝐶 Therefore, ∠𝐴𝐵𝐷 ≅ ∠𝐴𝐶𝐷 (AAS) So, ∠𝐵𝐴𝐷 = ∠𝐶𝐴𝐷 (CPCT) What is the defect in the above arguments? [Hint: Recall how ∠𝐵 = ∠𝐶 is proved when 𝐴𝐵 = 𝐴𝐶].

Sol.3) In ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 ∠𝐴𝐶𝐵 = ∠𝐴𝐵𝐶 In ∆𝐴𝐵𝐷 and ∆𝐴𝐶𝐷, 𝐴𝐵 = 𝐴𝐶 [given] ∠𝐴𝐵𝐷 = ∠𝐴𝐶𝐷 [proved above] ∠𝐴𝐷𝐵 = ∠𝐴𝐷𝐶 [each 90 degree] ∆𝐴𝐵𝐷 ≅ ∆𝐴𝐶𝐷 [by AAS] So, ∠𝐵𝐴𝐷 = ∠𝐶𝐴𝐷 [by CPCT] So, the defect in the given argument is that firstly proof ∠𝐴𝐵𝐷 = ∠𝐴𝐶𝐷 Hence, ∠𝐴𝐵𝐷 = ∠𝐴𝐶𝐷 is defect.

Q.4) 𝑃 is a point on the bisector of ∠𝐴𝐵𝐶. If the line through 𝑃, parallel to 𝐵𝐴 meet 𝐵𝐶 at Q, prove that 𝐵𝑃𝑄 is an isosceles triangle.

Sol.4) Given we have P is a point on the bisector of ∠𝐴𝐵𝐶 and draw the line through P parallel to 𝐵𝐴 and meet 𝐵𝐶 at Q. To prove: ∆𝐵𝑃𝑄 is an isosceles triangle. Proof ∠1 = ∠2 [𝐵𝑃 is bisector of ∠𝐵] [given] Now, ∠1 = ∠3 [alternate interior angles] ∠2 = ∠3

𝑃𝑄 = 𝐵𝑄 [sides opposite to equal angle are equal] Hence, ∆𝐵𝑃𝑄 is an isosceles triangle. Hence proved.

Q.5) 𝐴𝐵𝐶𝐷 is a quadrilateral in which 𝐴𝐵 = 𝐵𝐶 and 𝐴𝐷 = 𝐶𝐷. Show that BD bisects both the angles 𝐴𝐵𝐶 and 𝐴𝐷𝐶.

Sol.5) Given 𝐴𝐵𝐶𝐷 is a quadrilateral in which 𝐴𝐵 = 𝐵𝐶 and 𝐴𝐷 = 𝐶𝐷 To show 𝐵𝐷 bisects both the angles, 𝐴𝐵𝐶 and 𝐴𝐷𝐶 Proof Since, 𝐴𝐵 = 𝐵𝐶 [given] ∠2 = ∠1 … (i) 𝐴𝐷 = 𝐶𝐷 [given] ∠4 = ∠3 …(ii) On adding eqs. (i) & (ii), we get ∠2 + ∠4 = ∠1 + ∠3 ∠𝐵𝐶𝐷 = ∠𝐵𝐴𝐷 … (iii) In ∆𝐵𝐴𝐷 and ∆𝐵𝐶𝐷, 𝐴𝐵 = 𝐵𝐶 ∠𝐵𝐴𝐷 = ∠𝐵𝐶𝐷 & 𝐴𝐷 = 𝐶𝐷 ∆𝐵𝐴𝐷 ≅ ∆𝐵𝐶𝐷 Hence, ∠𝐴𝐵𝐷 = ∠𝐶𝐵𝐷 and ∠𝐴𝐷𝐵 = ∠𝐶𝐷𝐵 i.e., 𝐵𝐷 bisects the angles 𝐴𝐵𝐶 and 𝐴𝐷𝐶.

Q.6) 𝐴𝐵𝐶 is a right triangle with 𝐴𝐵 = 𝐴𝐶. Bisector of ∠𝐴 meets BC at D. Prove that 𝐵𝐶 =2𝐴𝐷.

Sol.6) Given ∆𝐴𝐵𝐶 is a right triangle with 𝐴𝐵 = 𝐴𝐶, 𝐴𝐷 is the bisector of ∠𝐴.



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To prove 𝐵𝐶 = 2𝐴𝐷 Proof In ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 [given] ∠𝐶 = ∠𝐵 … (i)

Now, in right angled ∆𝐴𝐵𝐶, ∠𝐴 + ∠𝐵 + ∠𝐶 = 180° [angle sum property]

90° + ∠𝐵 + ∠𝐵 = 180° [from eq. (i)]

2∠𝐵 = 90° ∠𝐵 = 45° ∠𝐵 = ∠𝐶 = 45° ∠3 = ∠4 = 45° ∠1 = ∠2 = 45° [𝐴𝐷 is bisector of ∠𝐴] ∠1 = ∠3, ∠2 = ∠4 𝐵𝐷 = 𝐴𝐷, 𝐷𝐶 = 𝐴𝐷 …(ii) Hence, 𝐵𝐶 = 𝐵𝐷 + 𝐶𝐷 = 𝐴𝐷 + 𝐴𝐷 from eq. (ii) 𝐵𝐶 = 2𝐴𝐷 Hence proved

Q.7) 𝑂 is a point in the interior of a square 𝐴𝐵𝐶𝐷 such that ∆𝑂𝐴𝐵 is an equilateral triangle. Show that ∠𝑂𝐶𝐷 is an isosceles triangle.

Sol.7) Given, 𝑂 is a point in the interior of a square 𝐴𝐵𝐶𝐷 such that ∆𝑂𝐴𝐵 is an equilateral triangle. Construction Join 𝑂𝐶 and 𝑂𝐷 To show ∆𝑂𝐶𝐷 is an isosceles triangle Proof Since, 𝐴𝑂𝐵 is an equilateral triangle

∠OAB = ∠OBA = 60° ….(i)

Also, ∠DAB = ∠CBA = 90° … (ii) [each angle of a square is 90°] On subtracting eq. (i) from (ii), we get

∠DAB − ∠OAB = ∠CBA − ∠OBA = 90° − 60° i.e., ∠𝐷𝐴𝑂 = ∠𝐶𝐵𝑂 = 30° in ∆𝐴𝑂𝐷 and ∆𝐵𝑂𝐶, 𝐴𝑂 = 𝐵𝑂 [given] ∠𝐷𝐴𝑂 = ∠𝐶𝐵𝑂 [proved above] and 𝐴𝐷 = 𝐵𝐶 [sides of a square are equal] ∆𝐴𝑂𝐷 ≅ ∆𝐵𝑂𝐶 [by SAS congruence rule] Hence, 𝑂𝐷 = 𝑂𝐶 [by CPCT] In ∆𝐶𝑂𝐷, 𝑂𝐶 = 𝑂𝐷 Hence, ∆𝐶𝑂𝐷 is an isosceles triangle.

Q.8) ∆𝐴𝐵𝐶 and ∆𝐷𝐵𝐶 are two triangles on the same base 𝐵𝐶 such that 𝐴 and 𝐷 lie on the opposite sides of 𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 and 𝐷𝐵 = 𝐷𝐶. 𝑆how that 𝐴𝐷 is the perpendicular bisector of 𝐵𝐶.

Sol.8) Given Two triangles ∆𝐴𝐵𝐶 and ∆𝐷𝐵𝐶 on the same base 𝐵𝐶 such that 𝐴 and 𝐷 lie on the opposite sides of 𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 and 𝐷𝐵 = 𝐷𝐶. Also, 𝐴𝐷 intersects BC at O. To show 𝑨𝑫 is the perpendicular bisector of 𝐵𝐶. Proof In ∆𝐴𝐵𝐷 and ∆𝐴𝐶𝐷, 𝐴𝐵 = 𝐴𝐶 [given] 𝐴𝐷 = 𝐴𝐷 [common side] & 𝐵𝑂 = 𝐷𝐶 [given] ∆𝐴𝐵𝐷 ≅ ∆𝐴𝐶𝐷 ∠𝐵𝐴𝐷 = ∠𝐶𝐴𝐷 ∠𝐵𝐴𝑂 = ∠𝐶𝐴𝑂 In ∆𝐴𝑂𝐵 and ∆𝐴𝑂𝐶,



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𝐴𝐵 = 𝐴𝐶 [given] 𝐴𝑂 = 𝑂𝐴 [common side] & ∠𝐵𝐴𝑂 = ∠𝐶𝐴𝑂 [proved above] ∆𝐴𝑂𝐵 ≅ ∆𝐴𝑂𝐶 [by SAS congruence rule] 𝐵𝑂 = 𝑂𝐶 & ∠𝐴𝑂𝐵 = ∠𝐴𝑂𝐶 [by CPCT] …(i)

But ∠𝐴𝑂𝐵 + ∠𝐴𝑂𝐶 = 180° [linear pair axiom]

∠𝐴𝑂𝐵 + ∠𝐴𝑂𝐵 = 180° [from eq. (i)]

2∠𝐴𝑂𝐵 = 180°

∠𝐴𝑂𝐵 =180°

2= 90°

Hence, 𝐴𝐷 ⊥ BC and 𝐴𝐷 bisects 𝐵𝐶, i.e., 𝐴𝐷 is the perpendicular bisector of 𝐵𝐶.

Q.9) 𝐴𝐵𝐶 is an isosceles triangle in which 𝐴𝐶 = 𝐵𝐶. 𝐴𝐷 and 𝐵𝐸 are respectively two altitudes to sides 𝐵𝐶 and 𝐴𝐶. Prove that 𝐴𝐸 = 𝐵𝐷.

Sol.9) Given ∆𝐴𝐵𝐶 is an isosceles triangle in which 𝐴𝐶 = 𝐵𝐶. Also, 𝐴𝐷 and 𝐵𝐸 are two altitudes to sides 𝐵𝐶 and 𝐴𝐶, respectively. To prove 𝐴𝐸 = 𝐵𝐷 Proof In ∆𝐴𝐵𝐶, 𝐴𝐶 = 𝐵𝐶 ∠𝐴𝐵𝐶 = ∠𝐶𝐴𝐵 i.e., ∠𝐴𝐵𝐷 = ∠𝐸𝐴𝐵 … (i) In ∆𝐴𝐸𝐵 and ∆𝐵𝐷𝐴,

∠𝐴𝐸𝐵 = ∠𝐴𝐷𝐵 = 90° ∠𝐸𝐴𝐵 = ∠𝐴𝐵𝐷 [from eq. (i)] & 𝐴𝐵 = 𝐴𝐵 [common side] ∆𝐴𝐸𝐵 ≅ ∆𝐵𝐷𝐴 [by AAS congruence rule] 𝐴𝐸 = 𝐵𝐷 [by CPCT] Hence proved

Q.10) Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Sol.10) Given In ∆𝐴𝐵𝐶, 𝐴𝐷 is a median Construction Produce 𝐴𝐷 to a point 𝐸 such that 𝐴𝐷 = 𝐷𝐸 and join 𝐶𝐸. To prove 𝐴𝐶 + 𝐴𝐵 > 2𝐴𝐷 Proof In ∆𝐴𝐵𝐷 and ∆𝐸𝐶𝐷, 𝐴𝐷 = 𝐷𝐸 [by construction] 𝐵𝐷 = 𝐶𝐷 [given AD is the median] & ∠𝐴𝐷𝐵 = ∠𝐶𝐷𝐸 [vertically opposite angle] ∆𝐴𝐵𝐷 ≅ ∆𝐸𝐶𝐷 [by SAS congruence rule] 𝐴𝐵 = 𝐶𝐸 … (i) [by CPCT] Now, in ∆𝐴𝐸𝐶, 𝐴𝐶 + 𝐸𝐶 > 𝐴𝐸 [sum of 2 sides of a triangle is greater than the third side] 𝐴𝐶 + 𝐴𝐵 > 2𝐴𝐷 [from eq. (i) & also taken that 𝐴𝐷 = 𝐷𝐸] Hence proved.

Q.11) Show that in a quadrilateral 𝐴𝐵𝐶𝐷, 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 < 2 (𝐵𝐷 + 𝐴𝐶).

Sol.11) Given 𝐴𝐵𝐶𝐷 is a quadrilateral To show 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 < 2(𝐵𝐷 + 𝐴𝐶) Construction Join diagonals 𝐴𝐶 and 𝐵𝐷. Proof In ∆𝑂𝐴𝐵, 𝑂𝐴 + 𝑂𝐵 > 𝐴𝐵 … (i) In ∆𝑂𝐵𝐶, 𝑂𝐵 + 𝑂𝐶 > 𝐵𝐶 … (ii) In ∆𝑂𝐶𝐷, 𝑂𝐶 + 𝑂𝐷 > 𝐶𝐷 … (iii)



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In ∆𝑂𝐷𝐴, 𝑂𝐷 + 𝑂𝐴 > 𝐷𝐴 … (iv) On adding eqs.(i), (ii), (iii) & (iv), we get 2[(𝑂𝐴 + 𝑂𝐵 + 𝑂𝐶 + 𝑂𝐷)] > 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 2[(𝑂𝐴 + 𝑂𝐶) + (𝑂𝐵 + 𝑂𝐷)] > 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 2(𝐴𝐶 + 𝐵𝐷) > 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 < (𝐵𝐷 + 𝐴𝐶)

Q.12) Show that in a quadrilateral 𝐴𝐵𝐶𝐷, 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 > 𝐴𝐶 + 𝐵𝐷.

Sol.12) Given, 𝐴𝐵𝐶𝐷 is a quadrilateral Construction Join diagonals 𝐴𝐶 and 𝐵𝐷 To show 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 > 𝐴𝐶 + 𝐵𝐷 In ∆𝐴𝐵𝐶, 𝐴𝐵 + 𝐵𝐶 > 𝐴𝐶 … (i) In ∆𝐵𝐶𝐷, 𝐵𝐶 + 𝐶𝐷 > 𝐵𝐷 … (ii) In ∆𝐶𝐷𝐴, 𝐶𝐷 + 𝐷𝐴 > 𝐴𝐶 … (iii) In ∆𝐷𝐴𝐵, 𝐷𝐴 + 𝐴𝐵 > 𝐵𝐷 … (iv) On adding eqs. (i), (ii), (iii) & (iv), we get 2(𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴) > 2(𝐴𝐶 + 𝐵𝐷) 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 > 𝐴𝐶 + 𝐵𝐷

Q.13) In a triangle 𝐴𝐵𝐶, 𝐷 is the mid-point of side 𝐴𝐶 such that 𝐵𝐷 = ½ 𝐴𝐶. Show that ∠𝐴𝐵𝐶 is a right angle.

Sol.13) Given, In ∆𝐴𝐵𝐶, 𝐷 is the mid point of 𝐴𝐶 i.e., 𝐴𝐷 = 𝐶𝐷 such that 𝐵𝐷 =1

2𝐴𝐶

To show ∠𝐴𝐵𝐶 = 90°

Proof We have, 𝐵𝐷 =1

2𝐴𝐶 … (i)

Since, 𝐷 is the mid point of 𝐴𝐶

𝐴𝐷 = 𝐶𝐷 =1

2𝐴𝐶 … (ii)

From eqs. (i) & (ii), we get 𝐴𝐷 = 𝐶𝐷 = 𝐵𝐷 In ∆𝐷𝐴𝐵, 𝐴𝐷 = 𝐵𝐷 [proved above] ∠𝐴𝐵𝐷 = ∠𝐵𝐴𝐷 … (iii) [angle opposite to equal side are equal] In ∆𝐷𝐵𝐶, 𝐵𝐷 = 𝐶𝐷 [proved above] ∠𝐵𝐶𝐷 = ∠𝐶𝐵𝐷 … (iv) [angle opposite to equal sides are equal]

In ∆𝐴𝐵𝐶, ∠𝐴𝐵𝐶 + ∠𝐵𝐴𝐶 + ∠𝐴𝐶𝐵 = 180° [by angle sum property]

∠𝐴𝐵𝐶 + ∠𝐵𝐴𝐷 + ∠𝐷𝐶𝐵 = 180° ∠𝐴𝐵𝐶 + ∠𝐴𝐵𝐷 + ∠𝐶𝐵𝐷 = 180° [from eqs. (iii) & (iv)]

∠𝐴𝐵𝐶 + ∠𝐴𝐵𝐶 = 180° 2∠𝐴𝐵𝐶 = 180° ∠𝐴𝐵𝐶 = 90°

Q.14) In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

Sol.14) Given In ∆𝐴𝐵𝐶, ∠𝐵 = 90° and 𝐷 is the mid point of 𝐴𝐶 Construction Produce 𝐵𝐷 to 𝐸 such that 𝐵𝐷 = 𝐷𝐸 and join 𝐸𝐶.

To prove 𝐵𝐷 =1

2𝐴𝐶

Proof In ∆𝐴𝐷𝐵 and ∆𝐶𝐷𝐸, 𝐴𝐷 = 𝐷𝐶 𝐵𝐷 = 𝐷𝐸 & ∠𝐴𝐷𝐵 = ∠𝐶𝐷𝐸 ∆𝐴𝐷𝐵 ≅ ∆𝐶𝐷𝐸 𝐴𝐵 = 𝐸𝐶



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& ∠𝐵𝐴𝐷 = ∠𝐷𝐶𝐸 But ∠𝐵𝐴𝐷 and ∠𝐷𝐶𝐸 are alternate angles. So, 𝐸𝐶 parallel to 𝐴𝐵 and 𝐵𝐶 is a transversal.

∠𝐴𝐵𝐶 + ∠𝐵𝐶𝐸 = 180° [cointerior angles]

90° + ∠𝐵𝐶𝐸 = 180° [∠𝐴𝐵𝐶 = 90°, given]

∠𝐵𝐶𝐸 = 180° − 90° = 90° In ∆𝐴𝐵𝐶 and ∠𝐸𝐶𝐵, 𝐴𝐵 = 𝐸𝐶 [proved above] 𝐵𝐶 = 𝐶𝐵 [common side]

& ∠𝐴𝐵𝐶 = ∠𝐸𝐶𝐵 [each 90°] ∆𝐴𝐵𝐶 ≅ ∆𝐸𝐶𝐵 [by SAS congruence rule] 𝐴𝐶 = 𝐸𝐵 [by CPCT]

1

2𝐸𝐵 =

1

2𝐴𝐶 [dividing both sides by 2]

𝐵𝐷 =1

2𝐴𝐶

Hence proved.

Q.15) Two lines 𝑙 and 𝑚 intersect at the point 𝑂 and 𝑃 is a point on a line 𝑛 passing through the point 𝑂 such that 𝑃 is equidistant from 𝑙 and 𝑚. Prove that 𝑛 is the bisector of the angle formed by 𝑙 and 𝑚.

Sol.15) Given, Two lines 𝑙 and 𝑚 intersect at the point 𝑂 and 𝑃 is a point on a line 𝑛 passing through the point 𝑂 such that 𝑃 is equidistant from 𝑙 and 𝑚 i.e., 𝑃𝑄 = 𝑃𝑅 To prove 𝑛 is the bisector of the angle formed by 𝑙 and 𝑚. Proof In ∆𝑂𝑄𝑃 and ∆𝑂𝑅𝑃,

∠𝑃𝑄𝑂 = ∠𝑃𝑅𝑂 = 90° 𝑂𝑃 = 𝑂𝑃 [common side] 𝑃𝑄 = 𝑃𝑅 [given] ∆𝑂𝑄𝑃 ≅ ∆𝑂𝑅𝑃 [by RHS congruence rule] ∠𝑃𝑂𝑄 = ∠𝑃𝑂𝑅 [by CPCT] Hence, 𝑛 is the bisector of ∠𝑄𝑂𝑅.

Q.16) Line segment joining the mid-points 𝑀 and 𝑁 of parallel sides 𝐴𝐵 and 𝐷𝐶, respectively of a trapezium 𝐴𝐵𝐶𝐷 is perpendicular to both the sides 𝐴𝐵 and 𝐷𝐶. Prove that 𝐴𝐷 =𝐵𝐶.

Sol.16) Given, In trapezium 𝐴𝐵𝐶𝐷, points 𝑀 and 𝑁 are the mid points of parallel sides 𝐴𝐵 and 𝐷𝐶 respectively and join 𝑀𝑁, which is perpendicular to 𝐴𝐵 and 𝐷𝐶. To prove 𝐴𝐷 = 𝐵𝐶Proof Since, 𝑀 is the mid point of 𝐴𝐵 𝐴𝑀 = 𝑀𝐵 Now, in ∆𝐴𝑀𝑁 and ∆𝐵𝑀𝑁, 𝐴𝑀 = 𝑀𝐵 [proved above]

∠3 = ∠4 [each 90°] 𝑀𝑁 = 𝑀𝑁 [common side] ∆𝐴𝑀𝑁 = ∆𝐵𝑀𝑁 [by SAS congruence rule] ∠1 = ∠2 [by CPCT]

On multiplying both sides of above equation by -1 & then adding 90° both sides, we get

90° − ∠1 = 90° − ∠2 ∠𝐴𝑁𝐷 = ∠𝐵𝑁𝐶 … (i) Now, ∆𝐴𝐷𝑁 and ∆𝐵𝐶𝑁, ∠𝐴𝑁𝐷 = ∠𝐵𝑁𝐶 from eq. (i) 𝐴𝑁 = 𝐵𝑁 [∆𝐴𝑀𝑁 ≅ ∆𝐵𝑀𝑁] & 𝐷𝑁 = 𝑁𝐶 [𝑁 is the mid point of 𝐶𝐷 (given)]



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∆𝐴𝐷𝑁 ≅ ∆𝐵𝐶𝑁 [by SAS congruence rule] Hence, 𝐴𝐷 = 𝐵𝐶 [by CPCT]

Q.17) 𝐴𝐵𝐶𝐷 is a quadrilateral such that diagonal 𝐴𝐶 bisects the angles 𝐴 and 𝐶. Prove that 𝐴𝐵 = 𝐴𝐷 and 𝐶𝐵 = 𝐶𝐷.

Sol.17) Given, In a quadrilateral 𝐴𝐵𝐶𝐷,diagonal 𝐴𝐶 bisects the angles A and C. To prove 𝐴𝐵 = 𝐴𝐷 and 𝐶𝐵 = 𝐶𝐷 Proof In ∆𝐴𝐷𝐶 and ∆𝐴𝐵𝐶, ∠𝐷𝐴𝐶 = ∠𝐵𝐴𝐶 [AC is the bisector of ∠𝐴 & ∠𝐶 ] ∠𝐷𝐶𝐴 = ∠𝐵𝐶𝐴 [𝐴𝐶 is the bisector of ∠𝐴 and ∠𝐶] & 𝐴𝐶 = 𝐴𝐶 [common side] ∴ ∆𝐴𝐷𝐶 ≅ ∆𝐴𝐵𝐶 [by ASA congruence rule] 𝐴𝐷 = 𝐴𝐵 [by CPCT] & 𝐶𝐷 = 𝐶𝐵 [by CPCT] Hence proved.

Q.18) 𝐴𝐵𝐶 is a right triangle such that 𝐴𝐵 = 𝐴𝐶 and bisector of angle C intersects the side 𝐴𝐵 at 𝐷. Prove that 𝐴𝐶 + 𝐴𝐷 = 𝐵𝐶.

Sol.18) Given, In right angled ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 and 𝐶𝐷 is the bisector of ∠𝐶 Construction Draw 𝐷𝐸 ⊥ 𝐵𝐶 To prove 𝐴𝐶 + 𝐴𝐷 = 𝐵𝐶 Proof In right angled ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 and 𝐵𝐶 is a hypotenuse [given]

∠𝐴 = 90° In ∆𝐷𝐴𝐶 and ∆𝐷𝐸𝐶,

∠𝐴 = ∠3 = 90° ∠1 = ∠2 [given] 𝐷𝐶 = 𝐷𝐶 [common side] ∆𝐷𝐴𝐶 ≅ ∆𝐷𝐸𝐶 [by AAS congruence rule] 𝐷𝐴 = 𝐷𝐸 … (i) [by CPCT] & 𝐴𝐶 = 𝐸𝐶 … (ii) In ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 ∠𝐶 = ∠𝐵 … (iii) Again, In ∆𝐴𝐵𝐶, ∠𝐴 + ∠𝐵 + ∠𝐶 = 180 [by angle sum property]

90° + ∠𝐵 + ∠𝐵 = 180° from eq. (iii)

2∠𝐵 = 180° − 90° 2∠𝐵 = 90°

∠𝐵 =90°

2= 45°

In ∆𝐵𝐸𝐷, ∠5 = 180° − (∠𝐵 + ∠4) [by angle sum property]

= 180° − (45° + 90°)

= 180° − 135° = 45° ∠𝐵 = ∠5 𝐷𝐸 = 𝐵𝐸 [sides opposite to equal angles are equal] … (iv) From eqs. (i) & (iv), we get 𝐷𝐴 = 𝐷𝐸 = 𝐵𝐸 𝐵𝐶 = 𝐶𝐸 + 𝐸𝐵 = 𝐶𝐴 + 𝐷𝐴 from eqs. (ii) & (v) 𝐴𝐷 + 𝐴𝐶 = 𝐵𝐶 Hence proved.

Q.19) 𝐴𝐵 and 𝐶𝐷 are the smallest and largest sides of a quadrilateral 𝐴𝐵𝐶𝐷. Out of ∠B and ∠𝐷 decide which is greater.

Sol.19) Given, In quadrilateral 𝐴𝐵𝐶𝐷, 𝐴𝐵 is the smallest and 𝐶𝐷 is the largest side To find ∠𝐵 > ∠𝐷 or ∠𝐷 > ∠𝐵



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Construction Join BD Now, in ∆𝐴𝐵𝐷, 𝐴𝐷 > 𝐴𝐵 ∠1 > ∠3 … (i) In ∆𝐵𝐶𝐷, 𝐶𝐷 > 𝐵𝐶 ∠2 > ∠4 … (ii) On adding eqs. (i) & (ii), we get ∠1 + ∠2 > ∠3 + ∠4 Hence, ∠𝐵 > ∠𝐷

Q.20) Prove that in a triangle, other than an equilateral triangle, angle opposite the longest

side is greater than 2

3 of a right angle.

Sol.20) Consider ∆𝐴𝐵𝐶 in which 𝐵𝐶 is the longest side.

To prove ∠𝐴 =2

3 right angled

Proof In ∆𝐴𝐵𝐶, 𝐵𝐶 > 𝐴𝐵 ∠𝐴 > ∠𝐶 … (i) & 𝐵𝐶 > 𝐴𝐶 ∠𝐴 > ∠𝐵 … (ii) On adding eqs. (i) & (ii), we get 2∠𝐴 > ∠𝐵 + ∠𝐶 2∠𝐴 + ∠𝐴 > ∠𝐴 + ∠𝐵 + ∠𝐶 3∠𝐴 > ∠𝐴 + ∠𝐵 + ∠𝐶 3∠𝐴 > 180°

∠𝐴 >2

3× 90°

i.e., ∠𝐴 >2

3 of a right angled. Hence proved.

Q.21) 𝐴𝐵𝐶𝐷 is quadrilateral such that 𝐴𝐵 = 𝐴𝐷 and 𝐶𝐵 = 𝐶𝐷. Prove that 𝐴𝐶 is the perpendicular bisector of BD.

Sol.21) Given, In quadrilateral 𝐴𝐵𝐶𝐷, 𝐴𝐵 = 𝐴𝐷 and 𝐶𝐵 = 𝐶𝐷. Construction Join AC and BD. To prove AC is the perpendicular bisector of BD. Proof In ∆𝐴𝐵𝐶 and ∆𝐴𝐷𝐶, 𝐴𝐵 = 𝐴𝐷 [given] 𝐵𝐶 = 𝐶𝐷 [given] & 𝐴𝐶 = 𝐴𝐶 [common side] ∆𝐴𝐵𝐶 ≅ ∆𝐴𝐷𝐶 [by SAS congruence rule] ∠1 = ∠2 [by CPCT] Now, In ∆𝐴𝑂𝐵 and ∆𝐴𝑂𝐷, 𝐴𝐵 = 𝐴𝐷 [given] ∠1 = ∠2 [proved above] & 𝐴𝑂 = 𝐴𝑂 [common side] ∆𝐴𝑂𝐵 ≅ ∆𝐴𝑂𝐷 [by SAS congruence rule] 𝐵𝑂 = 𝐷𝑂 [by CPCT] & ∠3 = ∠4 [by CPCT]… (i)

But ∠3 + ∠4 = 180° [linear pair axiom]

∠3 + ∠3 = 180° [from eq. (i)]

2∠3 = 180°

∠3 =180°

2= 90°

i.e., 𝐴𝐶 is the perpendicular bisector of 𝐵𝐷.



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