RBSE Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 9 Quadrilaterals Ex 9.2.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Quadrilaterals |

Exercise |
Ex 9.2 |

Number of Questions Solved |
17 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 9 Quadrilaterals Ex 9.2

Question 1.

In figure, ABCD and AEFG are (RBSESolutions.com) both parallelograms. If ∠C = 55° then find the value of ∠F.

Solution.

∵ ABCD is a parallelogram.

⇒ ∠A = ∠C = 55°

(in a parallelogram, opposite angles are equal)

Also AEFG is a parallelogram.

∴ ∠A = ∠F = 55°.

Question 2.

Can all the angles of a quadrilateral be acute angles? Give reason for your answer.

Solution.

No, angle sum property of a quadrilateral is always equal to 360°. If all the angles would be acute then sum of all the angles could not reach to 360°.

Question 3.

Can all the angles of a quadrilateral be (RBSESolutions.com) right angles? Give reason for your answer.

Solution.

Yes, angle sum of a quadrilateral is 360°. In this situation quadrilateral will become either a square or a rectangle.

Question 4.

Diagonals of a quadrilateral bisect each other. If ∠A = 35° then find ∠B.

Solution.

∵Diagonals of a quadrilateral bisect each other. It means it is a parallelogram.

∴ ∠A + ∠B = 180°

(sum of interior angles on the same side of a transversal is 180°)

⇒ 35° + ∠B = 180°

⇒ ∠B = 145°.

Question 5.

Opposite angles of a quadrilateral (RBSESolutions.com) are equal.If AB = 4 cm then determine the length of CD.

Solution.

∵Opposite angles of a quadrilateral are equal. It means it is parallelogram and in parallelogram opposite sides are equal.

∴ CD = AB = 4 cm.

Question 6.

ABCD is a rhombus in which altitude drawn from D to AB bisects AB. Find the angles of a rhombus.

Solution.

Let side of rhombus be x

∴ ∆ABD is an equilateral triangle.

⇒ ∠A = 60°

⇒ ∠A + ∠D = 180°

⇒ ∠D = 120°

⇒ ∠A = 60°, ∠B = 120°,

∠C = 60°, ∠D = 120°.

Question 7.

Through A, B and C, lines RQ, PR and QP have been (RBSESolutions.com) drawn respectively parallel to sides BC, CA and AB of a ∆ABC as shown in figure. Show that BC = \(\frac { 1 }{ 2 }\)QR.

Solution.

Given: PQ || AB, PR || AC and RQ || BC.

To show that

BC = \(\frac { 1 }{ 2 }\)QR

Proof: In quadrilateral BCAR,

∵ BR || CA and BC || RA

⇒ BCAR is a parallelogram.

⇒ BC = AR …(i)

Now, in quadrilateral BCQA,

BC || AQ and AB || QC

⇒ BCQA is (RBSESolutions.com) a parallelogram.

⇒ BC = AQ …(ii)

On adding (i) and (ii), we get

2BC = AR + AQ

⇒ 2BC = RQ

⇒ BC = \(\frac { 1 }{ 2 }\)QR

Hence proved.

Question 8.

D, E and F are the mid-points of the sides BC, CA and AB respectively of a ∆ABC. Show that ∆DEF is (RBSESolutions.com) also an equilateral triangle.

Solution.

∵ D, E and F are mid-points of sides BC,CA and AB respectively

∴ Using mid point theorem

By using (i), (ii) and (iii), we get

DE = FE = FD

⇒ ∆DEF is an equilateral triangle.

Question 9.

The points P and Q have been taken on opposite sides AB and CD respectively of (RBSESolutions.com) a parallelogram ABCD in such a way that AP = CQ (see figure).

Show that AC and PQ bisect each other.

Solution.

Given: ABCD is a parallelogram such that AP = CQ.

To prove: AC and PQ bisect each other.

∠MAP = ∠MCP

(alternate angles)

AP = CQ (given)

∠APM = ∠CQM

(alternate angles)

∴ ∆AMP ≅ ∆CMQ

(by ASA congruency rule)

⇒ AM = CM

and PM = QM (by c.p.c.t)

⇒ AC and PQ bisect each other.

Question 10.

E is the mid-point of the side AD of (RBSESolutions.com) a trapezium ABCD in which AB || DC. Through E parallel to AB a line intersects BC at F. Show that F will be the mid-point of sides BC.

Solution.

Given: ABCD is a trapezium in which AB || DC and E is mid-point of AD.

Construction: Draw EF || AB and join A to C.

To prove: F is mid-point of BC.

Proof: In ∆ADC, E is mid-point of AD and EO is drawn parallel to AB i.e., parallel to DC also. Then (RBSESolutions.com) by using converse of mid- point theorem, O will be the mid-point of AC.

Similarly, in ∆ABC, O is the mid-point of AC and OF || AB.

⇒ F will be the mid-point of BC.

(by converse of mid-point theorem)

Question 11.

In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are the mid-points of sides AB and BC respectively, then find the length of side DE.

Solution.

In ∆ABC, D and E are mid-points (RBSESolutions.com) of sides AB and BC respectively.

∴ Using mid-point theorem,

DE = \(\frac { 1 }{ 2 }AC\) = \(\frac { 1 }{ 2 }X7cm\) = 3.5 cm

Question 12.

In figure, it is given that BDEF and FDCE are (RBSESolutions.com) parallelograms. Can you say that BD = CD? Why or why not?

Solution.

∵ It is given that BDEF and FDCE are parallelograms.

⇒ BD = FE ….(i)

(opposite side of parallelogram are equal)

Also DC = FE …(ii)

(reason as above)

⇒ BD = CD [from (i) and (ii)]

Question 13.

In figure, D, E and F are the mid-points (RBSESolutions.com) of the sides BC, CA and AB respectively. If AB = 4.3 cm, BC = 5.6 cm and AC = 3.9 cm, then find the perimeter of the following:

(i) ADEF and

(ii) quad. BDEF

Solution.

(i) Using mid-point theorem i.e., the line joining the mid-point of two sides of (RBSESolutions.com) a triangle is parallel to third side and half of it.

∴ Perimeter of ∆DEF

= DE + FE + DF

= 2.15 + 2.8 + 1.95 = 6.9 cm

(ii) Perimeter of quad. BDEF

= BD + DE + EF + FB

= 2.8 + 2.15 + 2.8 + 2.15

= 9.9 cm.

Question 14.

Show that the line segment joining the mid-points (RBSESolutions.com) of the consecutive sides of a square is also a square.

Solution.

Given: ABCD is a square in which E, F, G and H are respectively the mid-points of sides AB, BC, CD and DA.

To prove: EFGH is a square.

Construction: Join A to C and similarly B to D.

Proof: In ∆ABC

∵ E and F are mid points of sides AB and BC

∴ From mid-point theorem

EF = \(\frac { 1 }{ 2 }\)AC …(i)

Similarly, HG = \(\frac { 1 }{ 2 }\) AC …(ii)

From (i) and (ii) EF = HG

Similarly, by joining B to D.

We can say HE = GF

Again by using (RBSESolutions.com) mid-point theorem,

Since ABCD is a square

∴ AB = BC = CD = DA

⇒ \(\frac { 1 }{ 2 }\)AB = \(\frac { 1 }{ 2 }\)DA

⇒ AE = AH

⇒ ∠AHE = ∠AEH = 45°

Similarly, ∠DHG = ∠DGH = 45°

∴∠AHE + ∠GHE + ∠DHG = 180°

(straight angle)

⇒ 45° + ∠GHE + 45° = 180°

∠GHE = 90°

⇒ EFGH is a square.

Question 15.

The diagonals of a quadrilateral are perpendicular to each other. Show (RBSESolutions.com) that the quadrilateral, formed by joining the mid-points of its sides is a rectangle.

Solution.

Given: A quadrilateral whose diagonals AC = BD are perpendicular to each other P, Q, R, S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.

To prove: PQRS is a rectangle.

Proof: In ∆ABC, P and Q are the mid-points of AB and BC respectively.

(Using the mid-point theorem)

In ∆ADC, R and S are the (RBSESolutions.com) mid-points of CD and AD respectively.

RS || AC and RS = \(\frac { 1 }{ 2 }\)AC …(ii)

(same reason as above)

From (i) and (ii)

PQ || RS and PQ = RS

(pair of opposite sides are equal and parallel)

So, PQRS is a parallelogram.

Suppose diagonals AC and BD of quadrilateral ABCD intersect at O. Now in ∆ABD, P is the mid-point of AB and S is the mid-point of AD.

∴ PS || BD => PN || OM

Also from (i) PQ || AC => PM || ON

Hence, quadrilateral PMON is (RBSESolutions.com) a parallelogram.

⇒ ∠MPN = ∠MON

(opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (∵ ∠BOA = ∠MON)

⇒ ∠MPN = 90°

⇒ ∠QPS = 90°

Thus, PQRS is a parallelogram whose one angle ∠QPS = 90°.

Hence, PQRS is a rectangle.

Question 16.

Prove that in a right-angled triangle, the median bisecting (RBSESolutions.com) the hypotenuse is half of the hypotenuse.

Solution.

Given: ∆ABC is a right angled triangle, right angled at B and BD is its median.

To prove: BD = \(\frac { 1 }{ 2 }\)AC

Construction: Draw DE perpendicular on BC.

Proof: DE ⊥ BC (by construction)

and D is the mid-point of hypotenuse AC.

∴BE = EC

(by converse of mid-point theorem)

Now, ∆’s DBE and DEC

BE = EC

DE = DE (common side)

∠DEB = ∠DEC = 90°

(by construction)

∴∆DBE ≅ ∆DEC

(by SAS congruency)

⇒ DC = DB …(i)

Again, D is mid-point of AC

2DC = AC …(ii)

From (i) and (ii), we get

⇒ 2DB = AC

⇒ BD = \(\frac { 1 }{ 2 }\)AC Hence proved.

Question 17.

Prove that the quadrilateral formed by joining (RBSESolutions.com) the mid-points of the consecutive sides of a rectangle is a rhombus.

Solution.

Given: A rectangle ABCD in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively.

Join PQ, QR, RS and SP respectively.

To prove: PQRS is a rhombus.

Construction: Join AC

Proof: In ∆ABC, P and Q are the mid-point of sides AB and BC respectively.

Therefore, PQ || AC and PQ = \(\frac { 1 }{ 2 }\)AC …(i)

(by mid-point theorem)

Also in ∆ADC

SR || AC and SR = \(\frac { 1 }{ 2 }\)AC …(ii)

From (i) and (ii), we get

PQ || SR and PQ = SR …(iii)

⇒ PQRS is (RBSESolutions.com) a parallelogram.

Now ABCD is a rectangle. (given)

⇒ AD = BC

⇒ \(\frac { 1 }{ 2 }\)AD = \(\frac { 1 }{ 2 }\)BC => AS = BQ ,..(iv)

⇒ In ∆’s APS and BPQ, we have

AP = BP

(∵ P is the mid-point of AB)

∠PAS = ∠PBQ = 90°

and AS = BQ from (iv)

So, ∆APS ≅ ∆BPQ

(by SAS congruency property)

⇒ PS = PQ …(v)

From (iii) and (v), we obtain that PQRS is a parallelogram such that

PS = PQ

Hence, PQRS is a rhombus.

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