Practice Set 5.1 Geometry 9th Std Maths Part 2 Answers Chapter 5 Quadrilaterals
ii. ∠YZW + ∠XYZ = 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠YZW + 135°= 180° [From (i)]
∴ ∠YZW = 180°- 135°
∴ ∠YZW = 45°
iii. l(OY) = 5 cm [Given]
l(OY) = l(WY) [Diagonals of a parallelogram bisect each other]
∴ l(WY) = 2 x l(OY)
= 2 x 5
∴ l(WY) = 10 cm
∴∠XWZ = 135°, ∠YZW = 45°, l(WY) = 10 cm
ii. ∠A = (3x + 12)°
= [3(40) + 12]°
=(120 +12)°= 132°
∠B = (2x – 32)°
= [2(40) – 32]°
= (80 – 32)° = 48°
∴ ∠C = ∠A = 132°
∠D = ∠B = 48° [Opposite angles of a parallelogram]
∴ The value of x is 40, and the measures of ∠C and ∠D are 132° and 48° respectively.
ii. Perimeter of □ABCD = 150 cm [Given]
∴ AB + BC + DC + AD = 150
∴ (x + 25) +x + (x + 25) + x – 150
∴ 4x + 50 = 150
∴ 4x = 150 – 50
∴ 4x = 100
∴ x =
∴ x = 25
iii. AD = BC = x = 25 cm
AB = DC = x + 25 = 25 + 25 = 50 cm
∴ The lengths of the sides of the parallelogram are 25 cm, 50 cm, 25 cm and 50 cm.
ii. ∠A = x° = 60°
∠B = 2x° = 2 x 60° = 120°
∠A = ∠C = 60°
∠B = ∠D= 120° [Opposite angles of a parallelogram]
∴ The measures of the angles of the parallelogram are 60°, 120°, 60° and 120°.
Question 6
Maharashtra Board Solution
In the adjoining figure, □PQRS and □ABCR are two parallelograms. If ∠P = 110°, then find the measures of all the angles of □ABCR.
Solution & Step-by-Step Answer:
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Solution & Step-by-Step Answer:
□PQRS is a parallelogram. [Given] ∴ ∠R = ∠P [Opposite angles of a parallelogram] ∴ ∠R = 110° …..(iii) □ABCR is a parallelogram. [Given] ∴ ∠A + ∠R= 180° [Adjacent angles of a parallelogram are supplementary] ∴ ∠A+ 110°= 180° [From (i)] ∴ ∠A= 180°- 110° ∴ ∠A = 70° ∴ ∠C = ∠A = 70° ∴ ∠B = ∠R= 110° [Opposite angles of a parallelogram] ∴ ∠A = 70°, ∠B = 110°, ∴ ∠C = 70°, ∠R = 110°
Question 7
Maharashtra Board Solution
In the adjoining figure, □ABCD is a parallelogram. Point E is on the ray AB such that BE = AB, then prove that line ED bisects seg BC at point F. Given: □ABCD is a parallelogram. BE = AB To prove: Line ED bisects seg BC at point F i.e. FC = FB
Solution & Step-by-Step Answer:
Proof: □ABCD is a parallelogram. [Given] ∴ seg AB ≅ seg DC …….(i) [Opposite angles of a parallelogram] seg AB ≅ seg BE ……..(ii) [Given] seg DC ≅ seg BE ……..(iii) [From (i) and (ii)] side DC || side AB [Opposite sides of a parallelogram] i.e. side DC || seg AE and seg DE is their transversal. [A-B-E] ∴ ∠CDE ≅ ∠AED ∴ ∠CDF ≅ ∠BEF …..(iv) [D-F-E, A-B-E] In ∆DFC and ∆EFB, seg DC = seg EB [From (iii)] ∠CDF ≅ ∠BEF [From (iv)] ∠DFC ≅ ∠EFB [Vertically opposite angles] ∴ ∆DFC ≅ ∆EFB [SAA test] ∴ FC ≅ FB [c.s.c.t] ∴ Line ED bisects seg BC at point F.
Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Practice Set 5.1 Intext Questions and Activities
Question 1
Maharashtra Board Solution
Write the following pairs considering □ABCD. (Textbook pg. no 57) Pairs of adjacent sides: i. AB, AD ii. AD, DC iii. DC, BC iv. BC, AB |
Pairs of adjacent angles:
i. ∠A, ∠B
ii. ∠C, ∠D
iii. ∠B, ∠C
iv. ∠D, ∠A
Pairs of opposite sides:
i. AB, DC
ii. AD, BC
Pairs of opposite angles:
i. ∠A, ∠C
ii. ∠B, ∠D
Question 2. Complete the following tree diagram. (Textbook pg. no 57) |
Question 3. In the above theorem, to prove ∠DAB ≅ ∠BCD, is any change in the construction needed? If so, how will you write the proof making the change? (Textbook pg. no. 60)
Solution & Step-by-Step Answer:
Yes Construction: Draw diagonal BD. Proof: side AB || side CD and diagonal BD is their transversal. [Given] ∴ ∠ABD ≅ ∠CDB ……..(i) [Alternate angles] side BC || side AD and diagonal BD is their transversal. [Given] ∴ ∠ADB ≅ ∠CBD ……..(ii) [Alternate angles] In ∆DAB and ∆BCD, ∠ABD ≅ ∠CDB [From (i)] seg BD ≅ seg DB [Common side] ∴ ∠ADB ≅ ∠CBD [From (ii)] ∴ ∆DAB ≅ ∆BCD [ASA test] ∴ ∠DAB ≅ ∠BCD [c.a.c.t.] Note: ∠DAB s ∠BCD can be proved using the same construction as in the above theorem. ∠BAC ≅ ∠DCA …..(i) ∠DAC ≅ ∠BCA ……(ii) ∴ ∠BAC + ∠DAC ≅ ∠DCA + ∠BCA [Adding (i) and (ii)] ∴ ∠DAB ≅ ∠BCD [Angle addition property]
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