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# Class 9 RD Sharma Solutions – Chapter 14 Quadrilaterals- Exercise 14.3

• Last Updated : 28 Mar, 2021

### Question 1. In a parallelogram ABCD, determine the sum of angles C and D.

Solution:

In the given ||gm ABCD,

âˆ C + âˆ D = 180Â°

Becauseâˆ C and âˆ D are consecutive interior angles on the same side of CD traversal, so their sum is equal to 180Â°.

### Question 2. In a parallelogram ABCD, if âˆ B = 135Â°, determine the measures of its other angles.

Solution:

In the given ||gm ABCD,

âˆ B = 135Â°

So âˆ D = âˆ B = 135Â° (Because opposite angles of a ABCD ||gm)

But âˆ A + âˆ B = 180Â° (Sum of consecutive angles) ….(i)

Now put the value of âˆ B in eq(i)

âˆ B + 135Â° = 180Â°

âˆ A = 180Â° â€“ 135Â° = 45Â°

But âˆ C = âˆ B = 45Â° (Opposite angles of a ||gm)

Here, the other angles are âˆ A = 45Â°, âˆ C = 45Â°, and âˆ D = 135Â°.

### Question 3. ABCD is a square, AC and BD intersect at O. State the measure of âˆ AOB.

Solution:

In the given square ABCD,

Given that diagonal AC and BD intersect each other at O

Hence, âˆ AOB = 90Â°, because in square diagonals are bisect each other at right angle.

### Question 4. ABCD is a rectangle with âˆ ABD = 40Â°. Determine âˆ DBC.

Solution:

In rectangle ABCD,

Given that âˆ ABD = 40Â°, âˆ B = 90Â°, BD is its diagonal

âˆ ABD + âˆ DBC = 90Â° ……….(i)

Put the value ofâˆ ABD = 40Â° in eq(i)

â‡’ 40Â° + âˆ DBC = 90Â°

â‡’ âˆ DBC = 90Â° â€“ 40Â° = 50Â°

Hence, the value of angle âˆ DBC = 50Â°

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

Solution:

In the given ||gm ABCD,

Given that E and F are the mid points of the line AB and CD

and DE and BF are joined

To prove: EBFD is a ||gm

Construction: Join EF

Proof: ABCD is a ||gm

So, AB = CD and AB || CD (Opposite sides of a ||gm are equal and parallel to each other)

EB || DF and EB = DF (Given that E and F are mid points of AB and CD)

Hence, proved EBFD is a ||gm.

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

Solution:

In the given ||gm ABCD

It is given that P and Q are the points of trisection of the diagonal BD

To prove: (i) CQ || AP

(ii)AC bisects PQ

Proof: As we know that the diagonals of a parallelogram bisect each other

So, AO = OC and BO = OD

According to figure P and Q are point of trisection of BD

So BP = PQ = QD â€¦(i)

BO = OD and BP = QD â€¦(ii)

Now subtract, eq(ii) from eq(i) we get

OB â€“ BP = OD â€“ QD

â‡’ OP = OQ

OA = OC and OP = OQ (proved above)

Diagonals AC and PQ bisect each other at O

So, APCQ is a parallelogram

Hence, AP || CQ

### Question 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:

In the given square ABCD

It is given that E, F, G and H are the points on AB, BC, CD and DA respectively

Such that AE = BF = CG = DH

To prove : EFGH is a square

Proof : It is given that E, F, G and H are points on the sides AB, BC, CA and DA respectively such that

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

Then BE = CF = DG = AH = y (suppose)

Now in âˆ†AEH and âˆ†BFE

AE = BF (given)

âˆ A = âˆ B (each 90Â°)

AH = BE (proved)

Hence, by SAS criterion

âˆ†AEH â‰… âˆ†BFE

âˆ´ âˆ 1 = âˆ 2 and âˆ 3 = âˆ 4 (C.P.C.T.)

But âˆ 1 + âˆ 3 = 90Â° and âˆ 2 + âˆ 4 = 90Â° (âˆ A = âˆ B = 90Â°)

â‡’ âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 = 90Â° + 90Â° = 180Â°

â‡’ âˆ 1 + âˆ 4 + âˆ 1 + âˆ 4 = 180Â°

â‡’ 2(âˆ 1 + âˆ 4) = 180Â°

â‡’ âˆ 1 + âˆ 4 = 180Â°/2 = 90Â°

âˆ´ âˆ HEF = 180Â° â€“ 90Â° = 90Â°

Similarly, we can prove that

âˆ F = âˆ G = âˆ H = 90Â°

Since sides of the EFGH is are equal and each angle is of 90Â°

Hence, EFGH is a square.

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

Solution:

Given that ABCD is a rhombus

EABF is a straight line such that

EA = AB = BF

Now join ED and FC

Which meet at point G

To prove: âˆ EGF = 90Â°

Proof: As we know that the diagonals of a rhombus bisect

each other at right angles

AO = OC, BO = OD

So, âˆ AOD = âˆ COD = 90Â°

and âˆ AOB = âˆ BOC = 90Â°

In âˆ†BDE,

A and O are the mid-points of BE and BD respectively.

AO || ED

Similarly, OC || DG

In âˆ† CFA, B and O are the mid-points of AF and AC respectively

OB || CF and OD || GC

Now in DOCG

OC || DG and OD || CG

Hence, DOCG is a parallelogram.

âˆ DGC = âˆ DOC (Opposite angles of ||gm are equal)

âˆ DGC = 90Â° (becauseâˆ DOC = 90Â°)

Hence, proved

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

Solution:

In the given ||gm ABCD,

AB is produced to E so

Also EC produced meets AB produced in F.

To prove : BF = BC

Proof: In âˆ†ACE,

O and D are the mid points of sides AC and AE

DO || EC and DB || FC

â‡’ BD || EF

AB = BF

But AB = DC (Opposite sides of ||gm are equal)

DC = BF

Now in âˆ†EDC and âˆ†CBF,

DC = BF (proved)

âˆ EDC = âˆ CBF

(because âˆ EDC = âˆ DAB corresponding angles)

âˆ DAB = âˆ CBF (corresponding angles)

âˆ ECD = âˆ CFB (corresponding angles)

So, by ASA criterion,

âˆ†EDC â‰… âˆ†CBF

DE = BC (By C.P.C.T.)

â‡’ DC = BC

â‡’ AB = BC

â‡’ BF = BC (âˆµ AB = BF proved)

Hence, proved

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