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

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

**Solution:**

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

∴∠C+∠D = 180^{0}

**Q.2. In a parallelogram ABCD, if ∠B = 135 ^{0}, determine the measures of its other angles.**

**Solution:**

Given ∠B = 135^{0}

ABCD is a parallelogram

∴∠A = ∠C,∠B =∠Dand∠A+∠B = 180^{0}

⇒∠A+135^{0 }= 180^{0}

⇒∠A = 45^{0}

⇒∠A = ∠C = 45^{0} and ∠B = ∠C = 135^{0}

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

**Solution:**

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

∴ ∠AOB = 90^{0}

**Q.4. ABCD is a rectangle with ∠ABD = 40 ^{0}. Determine ∠DBC**

**Solution:**

We have,

∠ABC = 90^{0}

⇒∠ABD+∠DBC = 90^{0} [∵∠ABD = 40^{0}]

⇒40^{0}+∠DBC = 90^{0}

∴∠DBC = 50^{0}

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

**Solution:**

Since ABCD is a parallelogram

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:**

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:**

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 = 90^{0} and ∠2+∠A = 90^{0}

⇒ ∠1+∠3+∠2+∠A = 90^{0}+90^{0}

⇒ ∠1+∠4+∠1+∠4 = 180^{0}

⇒ 2(∠1+∠4) = 180^{0}

⇒ ∠1+∠4 = 90^{0}

HEF = 90^{0}

Similarly we have ∠F = ∠G = ∠H = 90^{0}

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:**

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

∴OA = OC,OB = OD,and∠AOD = ∠COD = 90^{0}

And ∠AOB = ∠COB = 90^{0}

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 = 90^{0}

**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.

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.

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