AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.3 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals Exercise 8.3

Question 1.

The opposite angles of a parallelogram are (3x – 2)° and (x + 48)°. Find the measure of each angle of the parallelogram.

Solution:

Given that the opposite angles of a parallelogram are (3x – 2)° and (x + 48)°

Thus 3x – 2 = x + 48

(∵ opp. angles of a //gm are equal)

3x – x = 48 + 2

2x = 50

x = \(\frac{50}{2} \) = 25°

∴ The given angles are (3 x 25 – 2)° and (25 + 48) °

= (75 – 2)° and 73° = 73° and 73°

We know the consecutive angles are supplementary.

∴ The other two angles are (180°-73°) and (180°-73°)

= 107° and 107°

∴ The four angles are 73°, 107°, 73° and 107°.

Question 2.

Find the measure of all the angles of a parallelogram, if one angle is 24° less than the twice of the smallest angle.

Solution:

Let the smallest angle = x

Then its consecutive angle = 180 – x°

By problem (180 – x)° = (2x- 24)°

(∵ opp. angles are equal)

180 + 24 = 2x + x

3x = 204

x = \(\frac{204}{3} \) = 68°

∴ The angles are

68°; (2 x 68 – 24)°; 68°; (2 x 68 – 24)°

= 68°, 112°, 68°, 112°

Question 3.

In the given figure ABCD is a paral-lelogram and E is the mid point of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB.

Solution:

Given that □ABCD is a parallelogram.

E is the midpoint of BC.

Let G be the midpoint of AD.

Join G, E.

Now in ΔAFD, GE is the line joining the midpoints G, E of two sides AD and FD.

∴GE // AF and GE = \(\frac{1}{2}\)AF

But GE = AB [ ∵ ABEG is a parallelo¬gram and AB, GE forms a pair of opp. sides]

\(\frac{1}{2}\) = AB ⇒ AF = 2AB

Hence Proved.

Question 4.

In the given figure ABCD is a paral¬lelogram. P, Q are the midpoints of sides AB and DC respectively. Show that

Solution:

□ABCD is a parallelogram.

P, Q are the mid points of AB and CD.

Join Q, P.

Now AB = CD (Opp. sides of a //gm)

\(\frac{1}{2}\)AB = \(\frac{1}{2}\)CD

PB = QC

Also PB // QC.

Now in □PBCQ;

PB = QC; PB//QC

Hence □PBCQ is a parallelogram.

Question 5.

ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and CD//BA as shown in the figure. Show that i) ∠DAC = ∠BCA

ii) ABCD is a parallelogram.

Solution:

Given that AABC is isosceles; AB = AC

AD is bisector of ∠QAC

i) In ΔABC, AB = AC ⇒ ∠B = ∠ACB

(angles opp. to equal sides)

Also ∠QAC = ∠B + ∠ACB

∠QAC = ∠BCA + ∠BCA

(∵∠BCA = ∠B)

⇒ \(\frac{1}{2}\)∠QAC = \(\frac{1}{2}\) [2 ∠BCA]

⇒ ∠DAC = ∠BCA [ ∵ AD is bisector of ∠QAC]

ii) From (i) ∠DAC = ∠BCA

But these forms a pair of alt. int. angles for the pair of lines AD and BC; AC as a transversal.

∴ AD//BC

In □ABCD ; AB // DC; BC // AD

□ABCD is a parallelogram.

Question 6.

ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure). Show that 1) ΔAPB ≅ ΔCQD ii) AP = CQ.

Solution:

Given that □ABCD is a parallelogram.

BD is a diagonal.

AP ⊥ BD and CQ ⊥ BD

i) In ΔAPB and ΔCQD

AB = CD ( ∵ Opp. sides of //gm ABCD)

∠APB = ∠CQD (each 90°)

∠PBA = ∠QDC (alt. int. angles for the lines AB and DC)

∴ ΔAPB ≅ ΔCQD (AAS congruence)

ii) From (1) ΔAPB ≅ ΔCQD

⇒ AP = CQ (CPCT)

Question 7.

In Δs ABC and Δs DEF, AB = DC and AB//DE; BC = EF and BC//EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that

i) ABED is a parallelogram

ii) BCFE is a parallelogram

iii) AC = DF

iv) ΔABC = ΔDEF

Solution:

Given that in ΔABC and ΔDEF

AB = DE and AB // DE

BC = EF and BC//EF.

i) In □ABED AB//ED and AB = ED

Hence □ABED is a parallelogram.

ii) In □BCFE; BC = EF and BC//EF

Hence □BCFE is a parallelogram.

iii) ACFD is a parallelogram (In a paral-lelogram opposite sides are equal).

So, AC = DF.

iv) Consider ΔABC = ΔDEF

AB = DE (given);

AC = DF (proved)

BC = EF (given)

∴ ΔABC ≅ ΔDEF (SSS congruency rule).

Question 8.

ABCD is a parallelogram. AC and BD are the diagonals intersect at ‘O’. P and Q are the points of trisection of the diagonal BD. Prove that CQ//AP and also AC bisects PQ.

Solution:

Given □ABCD is a parallelogram;

BD is a diagonal.

P, Q are the points of trisection of BD.

In ΔAPB and ΔCQD

AB = CD (•.• Opp. sides of //gm ABCD)

BP = DQ (given)

∠ABP = ∠CDQ (alt. int. angles for the lines AB//DC, BD as a transversal)

ΔAPB = ΔCQD (SAS congruence)

Similarly in ΔAQD and ΔCPB

AD = BC (opp. sides of //gm ABCD)

DQ = BP (given)

∠ADQ = ∠CBP (all int. angles for the lines AD//BC, BD as a transversal)

ΔAQD ≅ ΔCPB

Now in □APCQ

AP = CQ (CPCT of AAPB, ACQD)

AQ = CP (CPCT of AAQD and ACPB)

∴ □APCQ is a parallelogram.

∴ CQ//AP (opp. sides of//gm APCQ)

Also AC bisects PQ. [ ∵ diagonals of //gm APCQ]

Question 9.

ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA respectively. Such that AE = BF = CG = DH. Prove that EFGH is a square.

Solution:

Given that ABCD is a square.

E, F, G, H are the mid points of AB, BC, CD and DA.

Also AE = BF = CG = DH

In ΔABC; E, F are the mid points of sides AB and BC.

∴ EF//AC and EF = \(\frac{1}{2}\) AC

Similarly GH//AC and GH = AC

GF//BD and GF = \(\frac{1}{2}\) BD

HE//BD and HE = \(\frac{1}{2}\) BD

But AC = BD (∵ diagonals of a square)

∴ EF = FG = GH = HE

Hence EFGH is a rhombus.

Also AC ⊥ BD

(∵ diagonals of a rhombus)

∴ In //gm OIEJ [ ∵ 0I // EJ; IE // OJ]

We have ∠IOJ = ∠E

[ ∵ Opp. angles of a //gm]

∴ ∠E – 90°

Hence in quad. EFGH; all sides are equal and one angle is 90°.

∴ EFGH is a square.