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 âˆ C and âˆ D.**

**Solution:**

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

âˆ´âˆ C+âˆ D = 180^{0}

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

**Solution:**

Given âˆ B = 135^{0}

ABCD is a parallelogram

âˆ´âˆ A = âˆ C,âˆ B =âˆ Dandâˆ A+âˆ B = 180^{0}

â‡’âˆ A+135^{0 }= 180^{0}

â‡’âˆ A = 45^{0}

â‡’âˆ A = âˆ C = 45^{0} and âˆ B = âˆ C = 135^{0}

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

**Solution:**

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

âˆ´ âˆ AOB = 90^{0}

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

**Solution:**

We have,

âˆ ABC = 90^{0}

â‡’âˆ ABD+âˆ DBC = 90^{0} [âˆµâˆ ABD = 40^{0}]

â‡’40^{0}+âˆ DBC = 90^{0}

âˆ´âˆ 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

â‡’ OB â€“ BP = OD â€“ QD

â‡’ 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 APâˆ¥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 Î”AEH and Î”BEF,we have

AE = BF

âˆ A=âˆ B

And AH = BE

So, by SAS congruency criterion, we have

Î”AEHâ‰ƒÎ”BFE

â‡’âˆ 1 = âˆ 2 and âˆ 3 = âˆ 4

But âˆ 1+âˆ 3 = 90^{0} and âˆ 2+âˆ A = 90^{0}

â‡’ âˆ 1+âˆ 3+âˆ 2+âˆ A = 90^{0}+90^{0}

â‡’ âˆ 1+âˆ 4+âˆ 1+âˆ 4 = 180^{0}

â‡’ 2(âˆ 1+âˆ 4) = 180^{0}

â‡’ âˆ 1+âˆ 4 = 90^{0}

HEF = 90^{0}

Similarly we have âˆ F = âˆ G = âˆ 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.

âˆ´OA = OC,OB = OD,andâˆ AOD = âˆ COD = 90^{0}

And âˆ AOB = âˆ COB = 90^{0}

In Î”BDE, A and O are mid-points of BE and BD respectively.

OAâˆ¥DE

OCâˆ¥DG

In Î”CFA, B and O are mid-points of AF and AC respectively.

OBâˆ¥CF

ODâˆ¥GC

Thus, in quadrilateral DOGC, we have

OCâˆ¥DGandODâˆ¥GC

â‡’ DOCG is a parallelogram

âˆ DGC = âˆ DOC

âˆ 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 Î”DCE,

âˆ DCE =âˆ DECâ€¦..(i) [In a triangle, equal sides have equal angles]

ABâˆ¥CD [Opposite sides of the parallelogram are parallel]

âˆ´AEâˆ¥CD [AB lies on AF]

AFâˆ¥CD and EF is the Transversal.

âˆ DCE = âˆ BFCâ€¦..(ii) [Pair of corresponding angles]

From (i) and (ii) we get

âˆ DEC = âˆ BFC

In Î”AFE,

âˆ AFE = âˆ AEF [âˆ DEC = âˆ BFC]

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

â‡’ AD + DE = AB + BF

â‡’BC + AB = AB + BF [Since, AD = BC, DE = CD and CD = AB, AB = DE]

â‡’ BC = BF

Hence proved.