Read RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations

**Exercise 14.1**

**Question 1: Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.**

**Solution:**

It is given that, the three angles of a quadrilateral are 110°, 50° and 40°

Let the fourth angle be ‘a’

We know, sum of all angles of a quadrilateral = 360°

110° + 50° + 40° + a° = 360°

200° + a° = 360°

a = 360° – 200°

a = 160°

Hence, the fourth angle is 160°.

**Question 2: In a quadrilateral ABCD, the angles A, B, C and D are in the ratio of 1:2:4:5. Find the measure of each angles of the quadrilateral.**

**Solution:**

Let us assumed that, the angles of the quadrilaterals are

A = x

B = 2x

C = 4x

D = 5x

A + B + C + D = 360° (Sum of all angles of a quadrilateral is 360°)

x + 2x + 4x + 5x = 360°

12x = 360°

x = (360°)/12 = 30°

Hence, the value of ∠A, ∠B, ∠C and ∠D are as below:-

A = x

A = 30°

B = 2x

B = 2×30°

B = 60°

C = 4x

C = 4 × 30°

C = 120°

D = 5x

D = 5 × 30°

D = 150°

Hence, the angle A, B, C and D are 30°,60°,120° and 150° respectively.

**Question 3: In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = 1/2 (∠A + ∠B).**

**Solution:**

In ΔDOC,

∠D + ∠COD + ∠C = 180° (sum of all angles in a triangle is 180°)

1/2∠D + ∠COD + 1/2∠C = 180°

∠COD = 180°- 1/2∠D - 1/2∠C (OC and OD are bisectors of ∠C and ∠D)

∠COD = 180° – 1/2 (∠D + ∠C) _______(i)

In Quadrilateral ABCD

∠A + ∠B + ∠C + ∠D = 360° (Sum of all angles of a quadrilateral is 360°)

∠C + ∠D = 360° – (∠A + ∠B) ________(ii)

Put the value of ∠C + ∠D in equation (i)

From equation (i) and (ii) we get,

∠COD = 180° – 1/2{360° – (∠A + ∠B)}

∠COD = 180° – 1/2 × 360° + 1/2 × (∠A + ∠B)}

∠COD = 180° − 180° + 1/2 (∠A + ∠B)

∠COD = 1/2 (∠A + ∠B)

Hence Proved

**Question 4: The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.**

**Solution:**

Let us assumed that, the angles of the quadrilaterals are

A = 3x

B = 5x

C = 9x

D = 13x

A + B + C + D = 360° (sum of all interior angles of a quadrilateral is 360°)

3x + 5x + 9x + 13x = 360°

30x = 360°

x = (360°)/(30°)

x = 12°

Hence, the value of ∠A, ∠B, ∠C and ∠D are as below:-

A = 3x

A = 3 × (12°)

A = 36°

B = 5x

B = 5 × (12°)

B = 60°

C = 9x

C = 9 × (12°)

C = 108°

D = 13x

D = 13 × (12°)

D = 156°

**Exercise 14.2**

**Question 1: Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each angle of the parallelogram.**

**Solution:**

It is given that, the two opposite angles of a parallelogram are (3x – 2) ° and (50 – x) °.

(3x – 2)° = (50 – x)° (opposite sides of a parallelogram are equal)

3x – 2 = 50 – x

3x + x = 50 + 2

4x = 52

x = 13

The value of ∠x is 13°

(3x - 2)°

(3 × 13 – 2)

37°

(50 - x)°

(50 – 13)

37°

x + 37 = 180° (Adjacent angles of a parallelogram are supplementary)

x = 180° − 37° = 143°

Hence, the required angles are 37°, 143°, 37° and 143°.

**Question 2: If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.**

**Solution:**

Let the angle be x.

It is given that the adjacent angle is 2/3 x

x + 2x/3 = 180° (Adjacent angle of a parallelogram is supplementary)

Take the LCM of denominator

x/1 + 2x/3 = 180°

(3x + 2x)/3 = 180°

3x + 2x = 3 × 180°

3x + 2x = 540°

5x = 540°

x = (540°)/5

x = 108°

Hence, the first angle is 108°

The second angle = 2/3 x

The second angle = 2/3(108°)

The second angle = 72°

Similarly, the 3rd and 4th angles are 108° and 72°

**Question 3: Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle.**

**Solution:**

It is given that, one angle of a parallelogram is 24° less than twice the smallest angle.

Let x be the smallest angle, then

x + 2x – 24° = 180°

3x – 24° = 180°

3x = 180° + 24°

3x = 204°

x = (204°)/3

x = 68°

Another angle = 2x – 240

Another angle = 2(68°) – 24°

Another angle = 112°

Hence, four angles are 68°, 112°, 68°, 112°.

**Question 4: The perimeter of a parallelogram is 22cm. If the longer side measures 6.5cm what is the measure of the shorter side?**

**Solution:**

Let us assumed that x is the shorter side of a parallelogram.

It is given that the Perimeter = 22 cm and Longer side = 6.5 cm

We know that,

Perimeter = Sum of all sides

x + 6.5 + 6.5 + x = 22

2x + 13 = 22

2x = 22 - 13

2x = 9

x = 9/2

x = 4.5

So, the shorter side of a parallelogram is 4.5 cm.

**Question 5: In a parallelogram ABCD, ∠D = 135°. Determine the measures of ∠A and ∠B.**

**Solution:**

It is given that ABCD is a parallelogram.

We know that, adjacent angles are supplementary

∠D + ∠C = 180°

135° + ∠C = 180°

∠C = 180° − 135°

∠C = 45°

Also, opposite sides of a parallelogram are equal.

∠A = ∠C = 45°

∠B = ∠D = 135°

**Question 6: ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.**

**Solution:**

It is given that ABCD is a parallelogram

∠A = 70°

∠A + ∠B = 180° (adjacent angles are supplementary)

70° + ∠B = 180°

∠B = 180° − 70°

∠B = 110°

Also, opposite sides of a parallelogram are equal.

∠A = ∠C = 70°

∠B = ∠D = 110°

**Question 7: In Figure, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A, and ∠B meet at P, prove that AD = DP, PC = BC and DC = 2AD.**

**Solution:**

It is given that, AP bisects ∠A

∠DAP = ∠PAB = 30°

∠A + ∠B = 180° (Adjacent angles are supplementary)

∠B + 60° = 180°

∠B = 180° − 60°

∠B = 120°

Also, BP bisects ∠B

∠PBA = ∠PBC = 60°

∠PAB = ∠DPA = 30° (Alternate interior angles)

AD = DP (opposite sides and angles of a parallelogram are equal)

∠PBA = ∠BPC = 60° (Alternate interior angles)

PC = BC

DC = DP + PC

DC = AD + BC

DC = AD + AD (opposite sides of a parallelogram are equal AD = BC)

DC = 2AD

Hence Proved

**Question 8: In figure, ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB, and ∠ADB.**

**Solution:**

∠CBD = ∠ABD = 60° (Alternate interior angle)

In ∠BDC

∠A = ∠C = 75° (opposite sides and angles of a parallelogram are equal)

∠CBD + ∠C + ∠CDB = 180° (sum of all angles in a triangle is 180°)

60° + 75° + ∠CDB = 180°

∠CDB = 180° − 60° - 75°

∠CDB = 180° − 135°

∠CDB = 45°

Hence, the measure of ∠CDB and ∠ADB are 45°and 60° respectively.

**Question 9: In figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.**

**Solution:**

In ΔBEF and ΔCED

∠BEF = ∠CED (Vertically opposite angle)

BE = CE (E is the mid-point of BC)

∠EBF = ∠ECD (Alternate interior angles)

ΔBEF ≅ ΔCED (By ASA congruence)

BF = CD (CPCT)

BF = CD

AF = AB + BF

AF = AB + AB

AF = 2AB

Hence proved

**Question 10: Which of the following statements are true (T) and which are false (F)?**

**(i) In a parallelogram, the diagonals are equal.**

**(ii) In a parallelogram, the diagonals bisect each other.**

**(iii) In a parallelogram, the diagonals intersect each other at right angles.**

**(iv) In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.**

**(v) If all the angles of a quadrilateral are equal, it is a parallelogram.**

**(vi) If three sides of a quadrilateral are equal, it is a parallelogram.**

**(vii) If three angles of a quadrilateral are equal, it is a parallelogram.**

**(viii) If all the sides of a quadrilateral are equal, it is a parallelogram.**

**Solution:**

(i) In a parallelogram, the diagonals are equal. : False

(ii) In a parallelogram, the diagonals bisect each other. : True

(iii) In a parallelogram, the diagonals intersect each other at right angles. : False

(iv) In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram. : False

(v) If all the angles of a quadrilateral are equal, it is a parallelogram. : True

(vi) If three sides of a quadrilateral are equal, it is a parallelogram. : False

(vii) If three angles of a quadrilateral are equal, it is a parallelogram. : False

(viii) If all the sides of a quadrilateral are equal, it is a parallelogram. : True

**Exercise 14.3**

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

**Solution:**

It is given that ABCD is a parallelogram,

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

∠C + ∠D = 180°

**Question 2: In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.**

**Solution:**

It is given that ABCD is a parallelogram,

∠B = 135°

∠A = ∠C

∠B = ∠D

∠A + ∠B = 180° (consecutive interior angles)

∠A + 135° = 180°

∠A = 45°

∠A = ∠C = 45°

∠B = ∠D = 135°

Hence, the measure of ∠A, ∠B, ∠C and ∠D is 45°, 135°, 45° and 135°.

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

**Solution:**

∠AOB = 90°

We know that, diagonals of a square bisect each other at right angle.

**Question 4: ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC.**

**Solution:**

∠ABC = 90° (each angle of a rectangle is 90°)

It is given that ∠ABD = 40°

∠ABD + ∠DBC = 90°

40° + ∠DBC = 90°

∠DBC = 90° - 40°

∠DBC = 50°

Hence, the measure of ∠DBC is 50°.

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

**Solution:**

It is given that, ABCD is a parallelogram.

AB ∥ DC and AB = DC (sides of a parallelogram)

EB ∥ DF and 1/2 × AB = 1/2 × DC

EB ∥ DF and EB = DF (sides of a parallelogram)

EBFD is a parallelogram.

**Question 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:**

BD is a diagonal a parallelogram ABCD

OA = OC and OB = OD

P and Q are point of intersection of BD.

BP = PQ = QD

OB = OD are BP = QD

OB - BP = OD - QD

OP = OQ

In quadrilateral APCQ,

OA = OC and OP = OQ

Diagonals of Quadrilateral APCQ bisect each other.

APCQ is a parallelogram

AP ∥ CQ

Hence Proved

**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:**

It is given that,

ABCD is a square

AE = BF = CG = DH

BE = CF = DG = AH

In ΔAEH and ΔBEF

AE = BF

∠A = ∠B

AH = BE

ΔAEH ≅ ΔBFE (by SAS congruency criterion)

∠1 = ∠2 (CPCT)

∠3 = ∠4 (CPCT)

∠1 + ∠3 = 90° _______(1)

∠2 + ∠4 = 90° _______(2)

Adding both the equation

∠1 + ∠3 + ∠2 + ∠4 = 90° + 90° (∠1 = ∠2 proved above)

∠1 + ∠4 + ∠1 + ∠4 = 180°

2∠1 + 2∠4 = 180°

2(∠1 + ∠4) = 180°

∠1 + ∠4 = (180°)/2

∠1 + ∠4 = 90°

∠HEF = 90°

∠GFE = ∠FGH = ∠GHE = 90°

EFGH is a Square.

Hence Proved

**Question 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:**

It is given that ABCD is a rhombus.

OA = OC, OB = OD

∠AOD = ∠COD = 90°

∠AOB = ∠COB = 90°

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

In quadrilateral DOGC

OC ∥ DG

OD ∥ GC

DOCG is a parallelogram

∠DGC = ∠DOC

∠DGC = 90°

Hence Proved

**Question 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:**

ABCD is a parallelogram.

DC = AB

AD = BC

DE = DC (Given)

In ∆EAF

By converse if mid-point theorem

DC||AB

DC||AF

D and C are the mid points of AE and FE.

DE = AD = BC = DC = AB

In ∆ACE

O is mid-point of AC

D is mid-point of AE

DO||EC and DO||EF

DB||EF and OB||CF

In ∆ACF

O is mid-point of AC

OB||CF (mid-point theorem)

B is mid-point of AF

AB = BF

BC = AB = BF

BC = BF

Hence proved

**Exercise 14.4**

**Question 1: In a ΔABC, D, E and F are, respectively, the mid points of BC, CA and AB. If the lengths of sides AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ΔDEF.**

**Solution:**

It is given that,

AB = 7 cm, BC = 8 cm, AC = 9 cm

In ∆ABC,

D, E and F are, respectively, the mid points of BC, CA and AB.

EF = 1/2 BC (Midpoint Theorem)

DF = 1/2 AC (Midpoint Theorem)

DE = 1/2 AB (Midpoint Theorem)

Perimeter of Triangle = Side + Side + Side

Perimeter of ∆DEF = DE + EF + DF

Perimeter of ∆DEF = 1/2AB + 1/2BC + 1/2AC

Perimeter of ∆DEF = 1/2 (AB + BC + AC)

Perimeter of ∆DEF = 1/2 (7 + 8 + 9)

Perimeter of ∆DEF = 1/2 (24)

Perimeter of ∆DEF = 12cm

**Question 2: In a ΔABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.**

**Solution:**

In ΔABC,

D, E and F are mid points of AB, BC and AC

In a Quadrilateral DECF:

DE ∥ AC

DE = AC/2 (By Mid-point theorem)

CF = AC/2

DE = CF

DECF is a parallelogram.

∠C = ∠D = 70° (Opposite sides of a parallelogram)

Similarly,

ADEF is a parallelogram, ∠A = ∠E = 50°

BEFD is a parallelogram, ∠B = ∠F = 60°

Hence, the ∠D, ∠E and ∠F are 70°, 50° and 60°.

**Question 3: In a triangle, P, Q and R are the mid points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.**

**Solution:**

In ΔABC,

R and P are mid points of AB and BC

RP ∥ AC ⇒ RP = AC/2 (By Mid-point Theorem)

In a quadrilateral, ARPQ

RP ∥ AQ

RP = AQ (A pair of side is parallel and equal)

So, ARPQ is a parallelogram.

AR = AB/2

AR = 30/2

AR = 15 cm

AR = QP = 15 cm (Opposite sides are equal)

RP = AC/2

RP = 21/2

RP = 10.5 cm

RP = AQ = 10.5cm (Opposite sides are equal)

Perimeter of ARPQ = AR + QP + RP +AQ

Perimeter of ARPQ = 15 +15 +10.5 +10.5

Perimeter of ARPQ = 51

Hence, the perimeter of quadrilateral ARPQ is 51 cm.

**Question 4: In a ΔABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.**

**Solution:**

It is given that, ABXC is a quadrilateral

AD = DX (Given)

BD = DC (Given)

Diagonals AX and BC bisect each other.

ABXC is a parallelogram.

Hence Proved

**Question 5: In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.**

**Solution:**

It is given that,

In ΔABC

E and F are mid points of AC and AB

EF ∥ BC

EF = BC/2 (By mid-point theorem)

In ΔABP

F is the mid-point of AB, again by mid-point theorem

FQ ∥ BP

Q is the mid-point of AP

AQ = QP

Hence Proved

**Question 6: In a ΔABC, BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML = NL.**

**Solution:**

Given that, BM and CN are perpendiculars from B and C.

In ΔBLM and ΔCLN

∠BML = ∠CNL (each 90°)

BL = CL (L is the mid-point)

∠MLB = ∠NLC (Vertically opposite angle)

ΔBLM ≅ ΔCLN (By ASA criterion)

LM = LN (CPCT)

**Question 7: In figure, triangle ABC is a right-angled triangle at B. Given that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, calculate**

**(i) The length of BC**

**(ii) The area of ΔADE.**

**Solution:**

In ΔABC,

It is given that, ∠B = 90°, AB = 9 cm, AC = 15 cm

AC

^{2}= AB^{2}+ BC^{2}(By using Pythagoras theorem)15

^{2}= 9^{2}+ AC^{2}AC

^{2}= 225 – 81 = 144BC = 12

AD = DB = AB/2 (D is the mid−point of AB)

AD = DB = 9/2

AD = DB = 4.5 cm

D and E are mid-points of AB and AC

DE ∥ BC

DE = BC/2 (By mid−point theorem)

Area of ΔADE = 1/2 × AD × DE

Area of ΔADE = 1/2 × 4.5 × 6

Area of ΔADE = 13.5

Hence, the area of ΔADE is 13.5 cm

^{2}**Question 8: In figure, M, N and P are mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.**

**Solution:**

It is given that, MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm.

M and N are mid-points of AB and AC

By mid−point theorem, we have

MN||BC

MN = BC/2

MN = 3 cm (given)

BC = 2MN

BC = 6 cm

AC = 2MP

AC = 2(2.5)

AC = 5 cm

AB = 2 NP

AB = 2(3.5)

AB = 7 cm

**Question 9: ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of ΔABC.**

**Solution:**

ABCQ and ARBC are parallelograms.

BC = AQ and BC = AR

AQ = AR (A is the mid-point of QR)

B and C are the mid points of PR and PQ respectively.

AB = PQ/2 (By mid−point theorem)

BC = QR/2 (By mid−point theorem)

CA = PR/2 (By mid−point theorem)

PQ = 2AB

QR = 2BC

PR = 2CA

PQ + QR + RP = 2 (AB + BC + CA)

Perimeter of ΔPQR = 2 (Perimeter of ΔABC)

**Question 10: In figure, BE ⊥ AC, AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.**

**Solution:**

It is given that, BE ⊥ AC and P, Q and R are respectively mid-point of AH, AB and BC.

In ΔABC, Q and R are mid-points of AB and BC respectively.

By Mid-point theorem, we know

QR ∥ AC_________(i)

In ΔABH,

Q and P are the mid-points of AB and AH respectively

QP ∥ BH ________(ii)

But, BE ⊥ AC

From (i) and (ii) we have,

QP ⊥ QR

∠PQR = 90°

Hence Proved

**Question 11: In figure, AB = AC and CP ∥ BA and AP is the bisector of exterior ∠CAD of ∆ABC. Prove that**

**(i) ∠PAC = ∠BCA.**

**(ii) ABCP is a parallelogram.**

**Solution:**

It is given that, AB = AC and CD ∥ BA and AP is the bisector of exterior ∠CAD of ΔABC

(i) It is given that,

AB = AC

∠ACB = ∠ABC (Opposite angles of equal sides)

∠CAD = ∠ABC + ∠ACB

∠PAC + ∠PAD = 2∠ACB (∠PAC = ∠PAD)

2∠PAC = 2∠ACB

∠PAC = ∠ACB

(ii) It is given that,

∠PAC = ∠BCA

AP ∥ BC (Given)

CP ∥ BA (Given)

Thus, ABCP is a parallelogram.

**Question 12: ABCD is a kite having AB = AD and BC = CD. Prove that the figure found by joining the mid points of the sides, in order, is a rectangle.**

**Solution:**

It is given that, A kite ABCD having AB = AD and BC = CD. P, Q, R, S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.

In ∆ABC, P and Q are the mid-points of AB and BC respectively.

PQ ∥ AC

PQ = 1/2 AC_________(i)

In ∆ADC, R and S are the mid-points of CD and AD respectively.

RS ∥ AC

RS = 1/2 AC_________(ii)

From (i) and (ii) we get,

PQ ∥ RS

PQ = RS

PQRS is a quadrilateral.

A pair of opposite sides is equal and parallel.

Hence, PQRS is a parallelogram.

We shall prove that one angle of parallelogram PQRS is a right angle.

AB = AD

1/2AB = 1/2AD

AP = AS________(iii) (P and S are midpoints of AB and AD)

∠1 = ∠2________(iv)

In ΔPBQ and ΔSDR, we have

PB = SD (1/2AD = 1/2AB)

BQ = DR (PB = SD)

PQ = SR (PQRS is a parallelogram)

ΔPBQ ≅ ΔSDR (SSS criterion of congruence)

∠3 = ∠4 (CPCT)

∠3 + ∠SPQ + ∠2 = 180°

∠1 + ∠PSR + ∠4 = 180°

∠3 + ∠SPQ + ∠2 = ∠1 + ∠PSR + ∠4

∠SPQ = ∠PSR (∠1 = ∠2 and ∠3 = ∠4)

Here, transversal PS cuts parallel lines SR and PQ at S and P respectively.

∠SPQ + ∠PSR = 180°

2∠SPQ = 180°

∠SPQ = 90° (∠PSR = ∠SPQ)

Thus, PQRS is a parallelogram such that ∠SPQ = 90°.

Hence, PQRS is a parallelogram.

**Question 13: Let ABC be an isosceles triangle in which AB = AC. If D, E, F be the mid points of the, sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles.**

**Solution:**

It is given that, D, E and F is mid-points of sides BC, CA and AB respectively.

ABC is a Isosceles triangle

AB ∥ DE and AC ∥ DF

AF ∥ DE and AE ∥ DF

So, ABDE is a parallelogram.

AF = DE

(1 )/2AB = DE

AE = DF

(1 )/2 AC = DF

DE = DF (AB = AC)

AE = AF = DE = DF

ABDF is a rhombus.

Hence, AD and FE bisect each other at right angle.

**Question 14: ABC is a triangle. D is a point on AB such that AD = 1/4 AB and E is a point on AC such that AE = 1/4 AC. Prove that DE = 1/4BC.**

**Solution:**

By construction we know that, P and Q be the mid-points of AB and AC respectively.

In ∆ABC,

PQ || BC

PQ = 1/2BC________(i) (by mid-point theorem)

In ΔAPQ,

D and E are the mid-points of AP and AQ respectively.

DE || PQ

DE = 1/2PQ ________(ii) (by mid-point theorem)

From equation (i) and (ii) we get,

DE = 1/2PQ

DE = 1/2 (1/2 BC)

DE = 1/2BC

Hence proved

**Question 15: In Figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = (1 )/2 AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.**

**Solution:**

Join B and D

AC and BD intersect each other at O.

ABCD is a parallelogram

OC = 1/2AC

CQ = 1/2 (1/2 AC)

CQ = 1/2OC

ΔDCO, P and Q are mid points of DC and OC respectively.

PQ ∥ DO

In ΔCOB, Q is the mid-point of OC

QR ∥ OB

Hence, R is the mid-point of BC.

**Question 16: In figure, ABCD and PQRC are rectangles and Q is the mid-point of AC. Prove that**

**(i) DP = PC**

**(ii) PR = 1/2 AC**

**Solution:**

(i) In ΔADC,

Q is the mid-point of AC

PQ || AD

Thus, P is the mid-point of DC.

DP = DC (By mid-point theorem)

(ii) Similarly, R is the mid-point of BC

PR = 1/2 BD

Diagonal of rectangle are equal, BD = AC

PR = 1/2 AC

**Question 17: ABCD is a parallelogram; E and f are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH.**

**Solution:**

It is given that, E and F are mid-points of AB and CD respectively.

ABCD is a parallelogram

AE = BE = 1/2AB

CF = DF = 1/2CD

AB = CD

1/2AB = 1/2CD

BE = CF

Thus, BEFC is a parallelogram

BC ∥ EF and BE = PH_______(i)

BC ∥ EF

AD ∥ EF (ABCD is a parallelogram)

Thus, AEFD is a parallelogram.

AE = GP

But E is the mid-point of AB.

So, AE = BF

Thus, GP = PH.

**Question 18: BM and CN are perpendiculars to a line passing through the vertex A of triangle ABC. If L is the mid-point of BC, prove that LM = LN.**

**Solution:**

BM and CN are perpendiculars to a line passing through the vertex A of triangle ABC.

We have,

BL = LC [As L is the given mid-point of BC]

Using the intercept theorem, we get

MS = SN

In ΔMLS and ΔLSN

MS = SN (Proved above)

∠LSM = ∠LSN (each 90°)

SL = LS (common)

ΔMLS ≅ ΔLSN (SAS Congruency Theorem)

LM = LN (CPCT)

**Question 19: Show that, the line segments joining the mid-points of opposite sides of a quadrilateral bisects each other.**

**Solution:**

Let us assumed that,

ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.

SP ∥ BD

SP = 1/2 BD________(i) (mid-point theorem)

In ΔBCD

QR ∥ BD

QR = 1/2 BD________ (ii)

From equations (i) and (ii), we get

SP ∥ QR

SP = QR

Hence PR and QS bisect each other.

**Question 20: Fill in the blanks to make the following statements correct:**

**(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is ____________.**

**(ii) The triangle formed by joining the mid-points of the sides of a right triangle is ____________.**

**(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ____________.**

**Solution:**

(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is Isosceles.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is Right triangle.

(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is Parallelogram.

**Exercise VSAQs**

**Question 1: In a parallelogram ABCD, write the sum of angles A and B.**

**Solution:**

It is given that ABCD parallelogram

∠A + ∠B = 180° (Adjacent angles of a parallelogram are supplementary)

**Question 2: In a parallelogram ABCD, if ∠D = 115°, then write the measure of ∠A.**

**Solution:**

It is given that ABCD parallelogram,

∠D = 115°

Since, ∠A and ∠D are adjacent angles of parallelogram.

∠A + ∠D = 180° (Adjacent angles of a parallelogram are supplementary)

∠A = 180° – 115°

∠A = 65°

Hence, the measure of ∠A is 65°.

**Question 3: PQRS is a square such that PR and SQ intersect at O. State the measure of ∠POQ.**

**Solution:**

It is given that, PQRS is a square.

Such that PR and SQ intersect at O

∠POQ = 90° (Diagonals of a square bisects each other at 90°)

**Question 4: In a quadrilateral ABCD, bisectors of angles A and B intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.**

**Solution:**

It is given that, ∠AOB = 75°

In a quadrilateral ABCD, bisectors of angles A and B intersect at O, then

∠AOB = 1/2 (∠ADC + ∠ABC)

∠AOB = 1/2 (∠D + ∠C)

75° = 1/2 (∠D + ∠C)

∠C + ∠D = 150°

**Question 5: The diagonals of a rectangle ABCD meet at O. If ∠BOC = 44°, find ∠OAD.**

**Solution:**

It is given that, ABCD is a rectangle

∠BOC = 44°

∠AOD = ∠BOC (vertically opposite angles)

∠AOD = ∠BOC (each 44°)

∠OAD = ∠ODA (Angles on same sides are equal)

OD = OA

∠OAD = 1/2 (180° – 44°) (Sum of all the angles of a triangle is 180°)

∠OAD = 1/2 (136°)

∠OAD = 68°

Hence, the measure of is ∠OAD = 68°

**Question 6: If PQRS is a square, then write the measure of ∠SRP.**

**Solution:**

PQRS is a square.

∠SRP = 1/2 (90°)

∠SRP = 45°

**Question 7: If ABCD is a rectangle with ∠BAC = 32°, find the measure of ∠DBC.**

**Solution:**

It is given that, ABCD is a rectangle and ∠BAC = 32°

AO = BO

∠DBA = ∠BAC = 32° (Angles on same side)

∠DBC + ∠DBA = 90° (each angle of a rectangle is 90°)

∠DBC + 32° = 90°

∠DBC = 90° - 32°

∠DBC = 58°

**Question 8: If ABCD is a rhombus with ∠ABC = 56°, find the measure of ∠ACD.**

**Solution:**

In a rhombus ABCD,

∠ABC = 56°

∠BCD = 2 (∠ACD) (Diagonals of a rhombus bisect the interior angles)

∠ACD = 1/2 (∠BCD) ______(1)

∠BCD + ∠ABC = 180° (angles of a rhombus are supplementary)

∠BCD = 180° – 56°

∠BCD = 124°

Put the value of ∠BCD in equation in (1)

∠ACD = 1/2 × 124°

∠ACD = 62°

**Question 9: The perimeter of a parallelogram is 22 cm. If the longer side measure 6.5 cm, what is the measure of shorter side?**

**Solution:**

It is given that, Perimeter of a parallelogram = 22 cm.

Longer side of a parallelogram = 6.5 cm

Let x be the shorter side

Perimeter = 2(x + 6.5)

22 = 2x + 13

2x = 22 – 13

2x = 9

x = 4.5

Hence, the measure of shorter side is 4.5 cm.

**Question 10: If the angles of a quadrilateral are in the ratio 3:5:9:13, then find the measure of the smallest angle.**

**Solution:**

It is given that, Angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13.

Let the sides are

A = 3x

B = 5x

C = 9x

D = 13x

A + B + C + D = 360° (sum of all the angles of a quadrilateral is 360°)

3x + 5x + 9x + 13x = 360°

30 x = 360°

x = 12°

Shortest angle is 3 × 12° = 36°

Hence, the measure of smallest angle is 36°.