Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths

RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q.1. In a parallelogram ABCD, determine the sum of angles ∠C and ∠D.

Solution:

RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

∠C and ∠D are consecutive interior angles on the same side of the transversal CD.

∴∠C+∠D = 1800

Q.2. In a parallelogram ABCD, if ∠B = 1350, determine the measures of its other angles.

Solution:

Given ∠B = 1350

ABCD is a parallelogram

∴∠A = ∠C,∠B =∠Dand∠A+∠B = 1800

⇒∠A+135= 1800

⇒∠A = 450

⇒∠A = ∠C = 450 and ∠B = ∠C = 1350

Q.3. ABCD is a square. AC and BD intersect at O. State the measure of ∠AOB.

Solution:

RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Since, diagonals of a square bisect each other at right angle.

∴ ∠AOB = 900

Q.4. ABCD is a rectangle with ∠ABD = 400. Determine ∠DBC

Solution:

RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

We have,

∠ABC = 900

⇒∠ABD+∠DBC = 900    [∵∠ABD = 400]

⇒400+∠DBC = 900

∴∠DBC = 500

Q.5. The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.

Solution:

RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Since ABCD is a parallelogram

RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

EBFD is a parallelogram.

Q.6. P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.

Solution:

RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

We know that,

Diagonals of a parallelogram bisect each other.

Therefore, OA = OC and OB = OD

Since P and Q are point of intersection of BD.

Therefore, BP = PQ = QD

Now, OB = OD are BP = QD

⇒ OB – BP = OD – QD

⇒ OP = OQ

Thus in quadrilateral APCQ, we have

OA = OC and OP = OQ

Diagonals of Quadrilateral APCQ bisect each other.

Therefore APCQ is a parallelogram.

Hence AP∥CQ.

Q.7. ABCD is a square. E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.

Solution:

RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

We have,

AE = BF = CG = DH = x (say)

BE = CF = DG = AH = y (say)

In ΔAEH and ΔBEF,we have

AE = BF

∠A=∠B

And AH = BE

So, by SAS congruency criterion, we have

ΔAEH≃ΔBFE

⇒∠1 = ∠2 and ∠3 = ∠4

But ∠1+∠3 = 900 and ∠2+∠A = 900

⇒ ∠1+∠3+∠2+∠A = 900+900

⇒ ∠1+∠4+∠1+∠4 = 1800

⇒ 2(∠1+∠4) = 1800

⇒ ∠1+∠4 = 900

HEF = 900

Similarly we have ∠F = ∠G = ∠H = 900

Hence, EFGH is a Square.

Q.8. ABCD is a rhombus, EAFB is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.

Solution:

RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

We know that the diagonals of a rhombus are perpendicular bisector of each other.

∴OA = OC,OB = OD,and∠AOD = ∠COD = 900

And ∠AOB = ∠COB = 900

In ΔBDE, A and O are mid-points of BE and BD respectively.

OA∥DE

OC∥DG

In ΔCFA, B and O are mid-points of AF and AC respectively.

OB∥CF

OD∥GC

Thus, in quadrilateral DOGC, we have

OC∥DGandOD∥GC

⇒ DOCG is a parallelogram

∠DGC = ∠DOC

∠DGC = 900

Q.9. ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC.

Solution:

Draw a parallelogram ABCD with AC and BD intersecting at O.

Produce AD to E such that DE = DC

Join EC and produce it to meet AB produced at F.

RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

In ΔDCE,

∠DCE =∠DEC…..(i)     [In a triangle, equal sides have equal angles]

AB∥CD              [Opposite sides of the parallelogram are parallel]

∴AE∥CD              [AB lies on AF]

AF∥CD and EF is the Transversal.

∠DCE = ∠BFC…..(ii)          [Pair of corresponding angles]

From (i) and (ii) we get

∠DEC = ∠BFC

In ΔAFE,

∠AFE = ∠AEF                    [∠ DEC = ∠ BFC]

Therefore, AE = AF          [In a triangle, equal angles have equal sides opposite to them]

⇒ AD + DE = AB + BF

⇒BC + AB = AB + BF          [Since, AD = BC, DE = CD and CD = AB, AB = DE]

⇒ BC = BF

Hence proved.

The document RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on RD Sharma Solutions Ex-14.3, Quadrilaterals, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What are the different types of quadrilaterals?
Ans. There are several types of quadrilaterals, namely square, rectangle, parallelogram, rhombus, trapezium, and kite. Each type has its own unique properties and characteristics.
2. How can I identify if a given quadrilateral is a square?
Ans. To identify if a given quadrilateral is a square, you need to check if all four sides are equal in length and all four angles are right angles (90 degrees). If these conditions are satisfied, then the quadrilateral is a square.
3. What are the properties of a parallelogram?
Ans. A parallelogram has the following properties: - Opposite sides are parallel - Opposite sides are equal in length - Opposite angles are equal - Consecutive angles are supplementary (add up to 180 degrees) - Diagonals bisect each other
4. How can I find the area of a trapezium?
Ans. To find the area of a trapezium, you need to use the formula: Area = (1/2) × (sum of parallel sides) × height. The height is the perpendicular distance between the two parallel sides.
5. What is the difference between a rhombus and a square?
Ans. The main difference between a rhombus and a square is in their angles. A rhombus has all four sides equal in length, but its angles are not necessarily right angles. On the other hand, a square has all four sides equal in length and all four angles are right angles (90 degrees).
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