## Maharashtra State Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Problem Set 5

Question 1.

Choose the correct alternative answer and fill in the blanks.

i. If all pairs of adjacent sides of a quadrilateral are congruent, then it is called ____.

(A) rectangle

(B) parallelogram

(C) trapezium

(D) rhombus

Answer:

(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,

AC^{2} = AB^{2} + BC^{2}

∴ (12^{2} √2 )^{2} = AB^{2} + AB^{2}

∴ \( A B^{2}=\frac{12^{2} \times 2}{2}=12^{2}\)

∴ 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.

Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.

Solution:

Let □ABCD be the rectangle.

AB = 7 cm, BC = 24 cm

In ∆ABC, ∠B = 90° [Angle of a rectangle]

AC^{2} = AB^{2} + BC^{2} [Pythagoras theorem]

= 7^{2} + 24^{2}

= 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.

If diagonal of a square is 13 cm, then find its side.

Solution:

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]

∴ PR^{2} = PQ^{2} + QR^{2} [Pythagoras theorem]

∴ 13 = x + x [From (i)]

∴ 169 = 2x^{2}

The length of the side of the square is 6.5√2 cm.

Question 4.

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:

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 = \(\frac { 112 }{ 14 }\)

∴ 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.

Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.

Solution:

□PQRS is a rhombus. [Given]

PR = 20 cm and QS = 48 cm [Given]

∴ PT = \(\frac { 1 }{ 2 }\) PR [Diagonals of a rhombus bisect each other]

= \(\frac { 1 }{ 2 }\) x 20 = 10 cm

Also, QT = \(\frac { 1 }{ 2 }\) QS [Diagonals of a rhombus bisect each other]

= \(\frac { 1 }{ 2 }\) x 48 = 24 cm

ii. In ∆PQT, ∠PTQ = 90° [Diagonals of a rhombus are perpendicular to each other]

∴ PQ^{2} = PT^{2} + QT^{2} [Pythagoras- theorem]

= 10^{2} + 24^{2}

= 100 + 576

∴ PQ^{2} = 676

∴ PQ = \(\sqrt {676 }\) [Taking square root of both sides]

= 26 cm

∴ The length of side PQ is 26 cm.

Question 6.

Diagonals of a rectangle PQRS are intersecting in point M. If ∠QMR = 50°, then find the measure of ∠MPS.

Solution:

□PQRS is a rectangle.

∴ PM = \(\frac { 1 }{ 2 }\) PR …(i)

MS = \(\frac { 1 }{ 2 }\) 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 = \(\frac { 130 }{ 2 }\) = 65°

∴ ∠MPS = 65° [From (v)]

Question 7.

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:

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.

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 = \(\frac { 1 }{ 2 }\) ( 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.

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:

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.

Draw five parallelograms by taking various measures of lengths and angles. (Textbook page no. 59)

Question 2.

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)

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.

Question 3.

To verify the different properties of quadrilaterals.

Material: A piece of plywood measuring about 15 cm x 10 cm, 15 thin screws, twine, scissor.

Note: On the plywood sheet, fix five screws in a horizontal row keeping a distance of 2 cm between any two adjacent screws. Similarly make two more rows of screws exactly below the first one. Take care that the vertical distance between any two adjacent screws is also 2 cm.

With the help of the screws, make different types of quadrilaterals of twine. Verify the properties of sides and angles of the quadrilaterals. (Textbook page no. 75)