Problem Set 5 Geometry 9th Std Maths Part 2 Answers Chapter 5 Quadrilaterals
i. If all pairs of adjacent sides of a quadrilateral are congruent, then it is called ____.
(A) rectangle
(B) parallelogram
(C) trapezium
(D) rhombus
(D) rhombus
ii. If the diagonal of a square is 22√2 cm, then the perimeter of square is ____.
(A) 24 cm
(B) 24√2 cm
(C) 48 cm
(D) 48√2 cm
Answer:
In ∆ABC,
AC2= AB2+ BC2
∴ (122√2 )2= AB2+ AB2
∴
∴ AB = 12 cm
∴ Perimeter of □ABCD = 4 x 12 = 48 cm
(C) 48 cm
iii. If opposite angles of a rhombus are (2x)° and (3x – 40)°, then the value of x is ____.
(A) 100°
(B) 80°
(C) 160°
(D) 40°
Answer:
2x = 3x – 40 … [Pythagoras theorem]
∴ x = 40°
(D) 40°
Question 2
Maharashtra Board Solution
Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.
Solution & Step-by-Step Answer:
Let □ABCD be the rectangle. AB = 7 cm, BC = 24 cm In ∆ABC, ∠B = 90° [Angle of a rectangle] AC2 = AB2 + BC2 [Pythagoras theorem] = 72 + 242 = 49 + 576 = 625 AC = √625 [Taking square root of both sides] = 25 cm ∴ The length of the diagonal of the rectangle is 25 cm.
Question 3
Maharashtra Board Solution
If diagonal of a square is 13 cm, then find its side.
Solution & Step-by-Step Answer:
Let □PQRS be the square of side x cm. ∴ PQ = QR = x cm …..(i) [Sides of a square] ∴ In ∆PQR, ∠Q = 90° [Angle of a square] ∴ PR2 = PQ2 + QR2 [Pythagoras theorem] ∴ 13 = x + x [From (i)] ∴ 169 = 2x2 The length of the side of the square is 6.5√2 cm.
Question 4
Maharashtra Board Solution
Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm. Find the length of its each side.
Solution & Step-by-Step Answer:
Let □STUV be the parallelogram. Ratio of two adjacent sides of a parallelogram is 3 : 4. Let the common multiple be x. ST = 3x cm and TU = 4x cm ∴ ST = UV = 3x cm TU = SV = 4x cm …..(i) [Opposite sides of a parallelogram] Perimeter of □STUV = 112 [Given] ∴ ST + TU + UV + SV = 112 ∴ 3x + 4x + 3x + 4x = 112 [From (i)] ∴ 14x = 112 ∴ x = ∴ x = 8 ∴ ST = UV = 3x = 3 x 8 = 24 cm ∴ TU = SV = 4x = 4 x 8 = 32 cm [From (i)] ∴ The lengths of the sides of the parallelogram are 24 cm, 32 cm, 24 cm and 32 cm.
Question 5
Maharashtra Board Solution
Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.
Solution & Step-by-Step Answer:
□PQRS is a rhombus. [Given] PR = 20 cm and QS = 48 cm [Given] ∴ PT = PR [Diagonals of a rhombus bisect each other] = x 20 = 10 cm Also, QT = QS [Diagonals of a rhombus bisect each other] = x 48 = 24 cm
ii. In ∆PQT, ∠PTQ = 90° [Diagonals of a rhombus are perpendicular to each other]
Question 6
Maharashtra Board Solution
Diagonals of a rectangle PQRS are intersecting in point M. If ∠QMR = 50°, then find the measure of ∠MPS.
Solution & Step-by-Step Answer:
□PQRS is a rectangle. ∴ PM = PR …(i) MS = QS …(ii) [Diagonals of a rectangle bisect each other] Also, PR = QS …..(iii) [Diagonals of a rectangle are congruent] ∴ PM = MS ….(iv) [From (i), (ii) and (iii)] In ∆PMS, PM = MS [From (iv)] ∴ ∠MSP = ∠MPS = x° …..(v) [Isosceles triangle theorem] ∠PMS = ∠QMR = 50° ……(vi) [Vertically opposite angles] In ∆MPS, ∠PMS + ∠MPS + ∠MSP = 180° [Sum of the measures of the angles of a triangle is 180°] ∴ 50° +x + x = 180° [From (v) and (vi)] ∴ 50° + 2x= 180 ∴ 2x= 180-50 ∴ 2x= 130 ∴ x = = 65° ∴ ∠MPS = 65° [From (v)]
Question 7
Maharashtra Board Solution
In the adjoining figure, if seg AB || seg PQ, seg AB ≅ seg PQ, seg AC || seg PR, seg AC ≅ seg PR, then prove that seg BC || seg QR and seg BC ≅ seg QR.
Solution & Step-by-Step Answer:
Given: seg AB || seg PQ, seg AB ≅ seg PQ, seg AC || seg PR, seg AC ≅ seg PR To prove: seg BC || seg QR, seg BC ≅ seg QR Proof: Consider □ABQP, seg AB || seg PQ [Given] seg AB ≅ seg PQ [Given] ∴ □ABQP is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent] ∴ segAP || segBQ …..(i) ∴ seg AP ≅ seg BQ …..(ii) [Opposite sides of a parallelogram] Consider □ACRP, seg AC || seg PR [Given] seg AC ≅ seg PR [Given] ∴ □ACRP is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent] ∴ seg AP || seg CR …(iii) ∴ seg AP ≅ seg CR …….(iv) [Opposite sides of a parallelogram] Consider □BCRQ, seg BQ || seg CR seg BQ ≅ seg CR ∴ □BCRQ is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent] ∴ seg BC || seg QR ∴ seg BC ≅ seg QR [Opposite sides of a parallelogram]
Question 8
Maharashtra Board Solution
In the adjoining figure, □ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that PQ || AB and PQ = ( AB + DC). Given : □ ABCD is a trapezium. To prove: Construction: Join points A and Q. Extend seg AQ and let it meet produced DC at R. Proof: seg AB || seg DC [Given] and seg BC is their transversal. ∴ ∠ABC ≅ ∠RCB [Alternate angles] ∴ ∠ABQ ≅ ∠RCQ ….(i) [B-Q-C] In ∆ABQ and ∆RCQ, ∠ABQ ≅∠RCQ [From (i)] seg BQ ≅ seg CQ [Q is the midpoint of seg BC] ∠BQA ≅ ∠CQR [Vertically opposite angles] ∴ ∆ABQ ≅ ∆RCQ [ASA test] seg AB ≅ seg CR …(ii) [c. s. c. t.] seg AQ ≅ seg RQ [c. s. c. t.] ∴ Q is the midpoint of seg AR. ….(iii) |
In ∆ADR,
Points P and Q are the midpoints of seg AD and seg AR respectively. [Given and from (iii)]
∴ seg PQ || seg DR [Midpoint theorem]
i.e. seg PQ || seg DC ……..(iv) [D-C-R]
But, seg AB || seg DC …….(v) [Given]
∴ seg PQ || seg AB [From (iv) and (v)]
In ∆ADR,
Question 9
Maharashtra Board Solution
In the adjoining figure, □ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonals AC and DB respectively, then prove that MN || AB.
Solution & Step-by-Step Answer:
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Solution & Step-by-Step Answer:
Given: □ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonals AC and DB respectively. To prove: MN || AB Construction: Join D and M. Extend seg DM to meet seg AB at point E such that A-E-B. Proof: seg AB || seg DC and seg AC is their transversal. [Given] ∴ ∠CAB ≅ ∠ACD [Alternate angles] ∴ ∠MAE ≅ ∠MCD ….(i) [C-M-A, A-E-B] In ∆AME and ∆CMD, ∠AME ≅ ∠CMD [Vertically opposite angles] seg AM ≅ seg CM [M is the midpoint of seg AC] ∠MAE ≅∠MCD [From (i)] ∴ ∆AME ≅ ∆CMD [ASA test] ∴ seg ME ≅ seg MD [c.s.c.t] ∴ Point M is the midpoint of seg DE. …(ii) In ∆DEB, Points M and N are the midpoints of seg DE and seg DB respectively. [Given and from (ii)] ∴ seg MN || seg EB [Midpoint theorem] ∴ seg MN || seg AB [A-E-B]
Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Problem Set 5 Intext Questions and Activities
Question 1
Maharashtra Board Solution
Draw five parallelograms by taking various measures of lengths and angles. (Textbook page no. 59) |
Question 2
Maharashtra Board Solution
Draw a parallelogram PQRS. Draw diagonals PR and QS. Denote the intersection of diagonals by letter O. Compare the two parts of each diagonal with a divider. What do you find? (Textbook page no. 60)
Solution & Step-by-Step Answer:
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Solution & Step-by-Step Answer:
seg OP = seg OR, and seg OQ = seg OS Thus we can conclude that, point O divides the diagonals PR and QS in two equal parts.
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