Category: CBSE Class 9

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3

Other Exercises

• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Question 1.
In a parallelogram ABCD, determine the sum of angles ZC and ZD.
Solution:
In a ||gm ABCD,
∠C + ∠D = 180°
(Sum of consecutive angles)

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

∴ ∠D = ∠B = 135° (Opposite angles of a ||gm)
But ∠A + ∠B = 180° (Sum of consecutive angles)
⇒ ∠B + 135° = 180°
∴ ∠A = 180° – 135° = 45°
But∠C = ∠B = 45° (Opposite angles of a ||gm)
∴ Angles are 45°, 135°, 45°, 135°.

Question 3.
ABCD is a square, AC and BD intersect at O. State the measure of ∠AOB.
Solution:
In a square ABCD,
Diagonal AC and BD intersect each other at O
∵ Diagonals of a square bisect each other at right angle
∵∠AOB = 90°

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

∠B = 90°, BD is its diagonal
But ∠ABD = 40°
and ∠ABD + ∠DBC = 90°
⇒ 40° + ∠DBC = 90°
⇒ ∠DBC = 90° – 40° = 50°
Hence ∠DBC = 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:
Given : In ||gm ABCD, E and F are the mid points of the side AB and CD respectively
DE and BF are joined
To prove : EBFD is a ||gm
Construction : Join EF

Proof : ∵ ABCD is a ||gm
∴ AB = CD and AB || CD
(Opposite sides of a ||gm are equal and parallel)
∴ EB || DF and EB = DF (∵ E and F are mid points of AB and CD)
∴ 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:
Given : In ||gm, ABCD. P and Q are the points of trisection of the diagonal BD

To prove : (i) CQ || AP
AC bisects PQ
Proof: ∵ Diagonals of a parallelogram bisect each other
∴ AO = OC and BO = OD
∴ P and Q are point of trisection of BD
∴ BP = PQ = QD …(i)
∵ BO = OD and BP = QD …(ii)
Subtracting, (ii) from (i) we get
OB – BP = OD – QD
⇒ OP = OQ
In quadrilateral APCQ,
OA = OC and OP = OQ (proved)
Diagonals AC and PQ bisect each other at O
∴ 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:
Given : In square ABCD
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 : 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 ∆AEH and ∆BFE

AE = BF (given)
∠A = ∠B (each 90°)
AH = BE (proved)
∴ ∆AEH ≅ ∆BFE (SAS criterion)
∴ ∠1 = ∠2 and ∠3 = ∠4 (c.p.c.t.)
But ∠1 + ∠3 = 90° and ∠2 + ∠4 = 90° (∠A = ∠B = 90°)
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 90° + 90° = 180°
⇒ ∠1 + ∠4 + ∠1 + ∠4 = 180°
⇒ 2(∠1 + ∠4) = 180°
⇒ ∠1 + ∠4 = \(\frac { { 180 }^{ \circ } }{ 2 }\)  = 90°
∴ ∠HEF = 180° – 90° = 90°
Similarly, we can prove that
∠F = ∠G = ∠H = 90°
Since sides of quad. EFGH is are equal and each angle is of 90°
∴ 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 : ABCD is a rhombus, EABF is a straight line such that
EA = AB = BF
ED and FC are joined
Which meet at G on producing

To prove: ∠EGF = 90°
Proof : ∵ Diagonals of a rhombus bisect
each other at right angles
AO = OC, BO = OD
∠AOD = ∠COD = 90°
and ∠AOB = ∠BOC = 90°
In ∆BDE,
A and O are the mid-points of BE and BD respectively.
∴ AO || ED
Similarly, OC || DG
In ∆ CFA, B and O are the mid-points of AF and AC respectively
∴ OB || CF and OD || GC
Now in quad. DOCG
OC || DG and OD || CG
∴ DOCG is a parallelogram.
∴ ∠DGC = ∠DOC (opposite angles of ||gm)
∴ ∠DGC = 90° (∵ ∠DOC = 90°)
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:
Given : In ||gm ABCD,
AB is produced to E so that
DE = DA and EC produced meets AB produced in F.
To prove : BF = BC
Proof: In ∆ACE,

O and D are the mid points of sides AC and AE
∴ DO || EC and DB || FC
⇒ BD || EF
∴ AB = BF
But AB = DC (Opposite sides of ||gm)
∴ DC = BF
Now in ∆EDC and ∆CBF,
DC = BF (proved)
∠EDC = ∠CBF
(∵∠EDC = ∠DAB corresponding angles)
∠DAB = ∠CBF (corresponding angles)
∠ECD = ∠CFB (corresponding angles)
∴ AEDC ≅ ACBF (ASA criterion)
∴ DE = BC (c.p.c.t.)
⇒ DC = BC
⇒ AB = BC
⇒ BF = BC (∵AB = BF proved)
Hence proved.

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Mark the correct alternative in each of the following:
Question 1.
The opposite sides of a quadrilateral have
(a) no common point
(b) one common point
(c) two common points
(d) infinitely many common points
Solution:
The opposite sides of a quadrilateral have no common point. (a)

Question 2.
The consecutive sides of a quadrilateral have
(a) no common point
(b) one common point
(c) two common points
(d) infinitely many common points
Solution:
The consecutive sides of a quadrilateral have one common point. (b)

Question 3.
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following cases, PQRS is a parallelogram?
(a) ∠P = 100°, ∠Q = 80°, ∠R = 100°
(b) ∠P = 85°, ∠Q = 85°, ∠R = 95°
(c) PQ = 7 cm, QR = 7 cm, RS = 8 cm, SP = 8 cm
(d) OP = 6.5 cm, OQ = 6.5 cm, OR = 5.2 cm, OS = 5.2 cm
Solution:
PQRS is a quadrilateral, PR and QS intersect each other at O. PQRS is a parallelogram if ∠P = 100°, ∠Q = 80°, ∠R = 100° (a)

Question 4.
Which of the following quadrilateral is not a rhombus?
(a) All four sides are equal
(b) Diagonals bisect each other
(c) Diagonals bisect opposite angles
(d) One angle between the diagonals is 60°
Solution:
A quadrilateral is not a rhombus if one angle between the diagonals is 60°. (d)

Question 5.
Diagonals necessarily bisect opposite angles in a
(a) rectangle
(b) parallelogram
(c) isosceles trapezium
(d) square
Solution:
Diagonals necessarily bisect opposite angles in a square. (d)

Question 6.
The two diagonals are equal in a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) trapezium
Solution:
The two diagonals are equal in a rectangle. (c)

Question 7.
We get a rhombus by joining the mid-points of the sides of a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) triangle
Solution:
We get a rhombus by joining the mid points of the sides of a rectangle. (c)

Question 8.
The bisectors of any two adjacent angles of a parallelogram intersect at
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
The bisectors of any two adjacent angles of a parallelogram intersect at 90°. (d)

Question 9.
The bisectors of the angle of a parallelogram enclose a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) square
Solution:
The bisectors of the angles of a parallelogram enclose a rectangle. (c)

Question 10.
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a
(a) parallelogram
(b) rectangle
(c) square
(d) rhombus
Solution:
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. (a)

Question 11.
The figure formed by joining the mid-points of the adjacent sides of a rectangle is a
(a) square
(b) rhombus
(c) trapezium
(d) none of these
Solution:
The figure formed by joining the mid-points of the adjacent sides of a rectangle is a rhombus. (b)

Question 12.
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
(a) square
(b) rectangle
(c) trapezium
(d) none of these
Solution:
The figure formed by the joining the mid-points of the adjacent sides of a rhombus is a rectangle. (b)

Question 13.
The figure formed by joining the mid-points of the adjacent sides of a square is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram
Solution:
Tire figure formed by joining the mid-points of the adjacent sides of a square is a square. (b)

Question 14.
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
(a) rectangle
(b) parallelogram
(b) rhombus
(d) square
Solution:
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a parallelogram. (b)

Question 15.
If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is
(a) 176°
(b) 68°
(c) 112°
(d) 102°
Solution:
Let the smallest angle be x
The largest angle = 2x – 24°
But sum of two adjacent angles = 180°

Question 16.
In a parallelogram ABCD, If ∠DAB = 75° and ∠DBC = 60°, then ∠BDC =
(a) 75°
(b) 60°
(c) 45°
(d) 55°
Solution:
In ||gm ABC,

∠A = 75°, ∠DBC = 60°
But ∠A + ∠B = 180° (Sum of two consecutive angles)
⇒ 75° + ∠B = 180°
⇒ ∠B = 180°- 75“= 105°
But ∠DBC = 60°
∴ ∠DBA = 105°-60° = 45°
But ∠BDC = ∠DBA (Alternate angles)
∴ ∠BDC = 45° (c)

Question 17.
ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF =
(a) AE
(b) BE
(c) CE
(d) DE
Solution:
In ||gm ABCD, BD is joined forming two triangles ABD and BCD

E and F are the centroid of ∆ABD and ∆BCD
Now E and F trisect AC
i.e. AE = EF = FC
∴ EF = AE (a)

Question 18.
ABCD is a parallelogram, M is the mid¬point of BD and BM bisects ∠B. Then, ∠AMB =
(a) 45°
(b) 60°
(c) 90°
(d) 75°
Solution:
In ||gm ABCD, M is mid-point of BD and
BM bisects ∠B
AM is joined

∴AM bisects ∠A
But ∠A + ∠B = 180° (Sum of two consecutive angles)
∴ ∠AMB = 90° (c)

Question 19.
If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is
(a) 108°
(b) 54°
(c) 12°
(d) 81°
Solution:
Let adjacent angle of a ||gm = x
Then second angle = \(\frac { 2 }{ 3 }\) x
∴ x+ \(\frac { 2 }{ 3 }\) x= 180°
(Sum of two adjacent angles of a ||gm is 180°)

Question 20.
If the degree measures of the angles of quadrilateral are Ax, lx, 9x and 10JC, what is the sum of the measures of the smallest angle and largest angle?
(a) 140°
(b) 150°
(c) 168°
(d) 180°
Solution:
Sum of the angles of a quadrilateral = 360°
∴ 4x + 1x + 9x + 10x = 360°
⇒ 30x = 360°
⇒ x = \(\frac { { 360 }^{ \circ } }{ 30 }\)  = 12°
Now sum of smallest and largest angle = 4 x 12° + 10 x 12°
= 48° + 120° = 168° (c)

Question 21.
If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to
(a) 16 cm
(b) 15 cm
(c) 20 cm
(d) 17 cm
Solution:
Diagonals of a rhombus are 18 cm and 24 cm But diagonals of a rhombus bisect each other at right angles

Question 22.
ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC =
(a) 70°
(b) 110°
(c) 90°
(d) 120°
Solution:
In ||gm ABCD, AC is its diagonal which bisect ∠BAD

∠BAD = 35°
∴ ∠BAD = 2 x 35° = 70°
But ∠A + ∠B = 180° (Sum of consecutive angles)
⇒ 70° + ∠B = 180°⇒ ∠B = 180° – 70°
∴ ∠B = 110°
⇒ ABC = 110° (b)

Question 23.
In a rhombus ABCD, if ∠ACB = 40°, then ∠ADB =
(a) 70°
(b) 45°
(c) 50°
(d) 60°
Solution:
In rhombus ABCD, ∠ACB = 40°

∴ ∠BCD = 2 x ∠ACB
= 2 x 40° = 80°
But ∠BCD + ∠ADC = 180° (Sum of consecutive angles of ||gm)
⇒ 80° + ∠ADC = 180°
⇒ ∠ADC = 180° – 80° = 100°
∴ ∠ADB = \(\frac { 1 }{ 2 }\)∠ADC = \(\frac { 1 }{ 2 }\)x 100° = 50° (c)

Question 24.
In ∆ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are
(a) 70°, 70°, 40°
(b) 60°, 40°, 80°
(c) 30°, 40°, 110°
(d) 60°, 70°, 50°
Solution:
In ∆ABC,

∠A = 30°, ∠B = 40°, ∠C = 110°
D, E and F are mid-points of the sides of the triangle. By joining them in order,
DEF is a triangle formed
Now BDEF, CDFE and AFDE are ||gms
∴ ∠A = ∠D = 30°
∠B = ∠E = 40°
∠C = ∠F= 110°
∴ Angles are 30°, 40°, 110° (c)

Question 25.
The diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, then ∠OAB =
(a) 40°
(b) 50°
(c) 10°
(d) 90°
Solution:
In ||gm ABCD, diagonals AC and BD intersect each other at O

BOC = 90°, ∠BDC = 50°
∵ ∠BOC = 90°
∴ Diagonals of ||gm bisect each other at 90°
∴∠COD = 90°
In ∆COD,
∠OCD = 90° – 50° = 40°
But ∠OAB = ∠OCD (Alternate angles)
∴∠OAB = 40° (a)

Question 26.
ABCD is a trapezium in which AB || DC. M and N are the mid-points of AD and BC respectively. If AB = 12 cm, MN = 14 cm, then CD =
(a) 10 cm
(b) 12 cm
(c) 14 cm
(d) 16 cm
Solution:
In trapezium AB || DC
M and N are mid-points of sides AD and BC and MN are joined
AB = 12 cm, MN = 14 cm

∵ MN = \(\frac { 1 }{ 2 }\)(AB + CD)
⇒ 2MN = AB + CD
⇒ 2 x 14 = 12 + CD
CD = 2 x 14 – 12 = 28 – 12 = 16 cm (d)

Question 27.
Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 45°, then ∠B =
(a) 115°
(b) 120°
(c) 125°
(d) 135°
Solution:
Diagonals AC and BD of quadrilateral ABCD bisect each other at O

∴ AO = OC, BO = OD
∴ ABCD is a ||gm ∠A = 45°
But ∠A + ∠B = 180° (Sum of consecutive angles)
∴ ∠B = 180° – ∠A = 180° – 45°
= 135° (d)

Question 28.
P is the mid-point of side BC of a paralleogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =
(a) 5 cm
(b) 6 cm
(c) 8 cm
(d) 10 cm
Solution:
In ||gm ABCD, P is mid-point of BC
AD = 10cm

∠BAP = ∠DAP
Through P, draw PQ || AB
∴ ABPQ is rhombus
∴ AB = BP = AQ
= \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 10 = 5 cm
But CD = AB (Opposite sides of ||gm)
∴ CD = 5 cm (a)

Question 29.
In ∆ABC, E is the mid-point of median AD such that BE produced meets AC at E If AC = 10.5 cm, then AF =
(a) 3 cm
(b) 3.5 cm
(c) 2.5 cm
(d) 5 cm
Solution:
In ∆ABC, E is the mid-point of median AD
Such that BE produced meets AC at F
AC = 10.5 cm
Draw DG || AF

In ∆ADG
E is mid-point of AD and EF || DG
∴ F is mid-point of AG
⇒ AF = FG …(i)
In ∆BCF
D is mid-point of BC and DG || BF
∴ G is mid-point of FC
∴ FG = GC …(i)
From (i) and (ii)
AF = FG = GC = \(\frac { 1 }{ 3 }\) AC
But AC = 10.5 cm
∴ AF = \(\frac { 1 }{ 3 }\) AC = \(\frac { 1 }{ 3 }\) x 10.5 = 3.5 cm (b)

Question 30.
ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =

Solution:
In ||gm ABCD, E is mid-point of BC DE and AB are produced to meet at F

∵ E is mid point of BC
∴ BE = EC
In ∆BEF and ∆CDE
BE = EC
∠BEF = ∠CED (Vertically opposite angle)
and ∠EBF = ∠ECD (Alternate angles)
∴ ∆BEF ≅ ∆CDE (ASA criterian)
∴ DC = BF
But DC = AB
∴ AB = BF
AF = AB + BF = AB + AB
= 2AB (b)

Question 31.
In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B =
(a) 60°
(b) 80°
(c) 120°
(d) None of these
Solution:
In quadrilateral ABCD
⇒ ∠A + ∠C = 2(∠B + ∠D)
⇒ ∠A + ∠C = 2∠B + 2∠D
Adding 2∠A + 2∠C both sides
2∠A + 2∠C + ∠A + ∠C = 2∠A + 2∠B + 2∠C + 2∠D
⇒ 3∠A + 3∠C = 2(∠A + ∠B + ∠C + ∠D)
⇒ 3(∠A + ∠C) = 2 x 360° = 720°
∴ ∠A + ∠C = \(\frac { { 720 }^{ \circ } }{ 3 }\)  = 240°
⇒ 40° + ∠C = 240° (∵ ∠A = 40°)
∠C = 240° – 40° = 200°
Now 2(∠B + ∠D) = ∠A + ∠C = 240°
∠B + ∠D = \(\frac { { 240 }^{ \circ } }{ 2 }\)  = 120°
∴ ∠B = 60° = 120°
∴ ∠B = 60° (a)

Question 32.
The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC =
(a) 70°
(b) 90°
(c) 80°
(d) 100°
Solution:
In rectangle ABCD, diagonals AC and BD intersect each other at P

∠ABD = 50°
∴ ∠CAB = ∠ABD = 50° (∵ AP = BP)
Now in ∆APB
∠CAB + ∠ABD + ∠APB = 180° (Angles of a triangle)
⇒ ∠PAB + ∠PBA + ∠APB = 180°
⇒ 50° + 50° + ∠APB = 180°
⇒ ∠APB = 180° – 50° – 50° = 80°
But ∠DPC = ADB (Vertically opposite angles)
∴ ∠DPC = 80° (c)

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

Question 1.
If ABC and BDE are two equilateral triangles such that D is the mid-ponit of BC, then find ar(∆ABC) : ar(∆BDE).
Solution:
ABC and BDE are two equilateral triangles and D is the mid-point of BC

Let each side of AABC = a
Then BD = \(\frac { a }{ 2 }\)
∴ Each side of triangle BDE will be \(\frac { a }{ 2 }\)

Question 2.
In the figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.

Solution:
In rectangle ABCD,
CD = 6 cm, AD = 8 cm
∴ Area of rectangle ABCD = CD x AD
= 6 x 8 = 48 cm²
∵ DC || AB and AB is produced to F and DC is produced to G
∴ DG || AF
∵ Rectangle ABCD and ||gm CDEF are on the same base CD and between the same parallels
∴ ar(||gm CDEF) = ar(rect. ABCD)
= 48 cm²

Question 3.
In the figure of Q. No. 2, find the area of ∆GEF.

Solution:

Question 4.
In the figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ∆EFG.

Solution:
ABCD is a rectangle in which
AB = 10 cm, AD = 5 cm

∵ ABCD is a rectangle
∴DC || AB,
DC is produced to E and AB is produced to G
∴DE || AG
∵ Rectangle ABCD and ||gm ABEF are on the same base AB and between the same parallels
∴ ar(rect. ABCD) = ar(||gm ABEF)
= AB x AD = 10 x 5 = 50 cm²
Now ||gm ABEF and AEFG are on the same
base EF and between the same parallels
∴ area ∆EFG = \(\frac { 1 }{ 2 }\) ar(||gm ABEF)
= \(\frac { 1 }{ 2 }\) x 50 = 25 cm²

Question 5.
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find or(∆RAS).
Solution:
In quadrant PLRM, rectangle PQRS is in scribed

Radius of the circle = 13 cm
A is any point on PQ
AR and AS are joined, PS = 5 cm
In right ∆PRS,
PR² = PS² + SR²
⇒ (13² = (5)²+ SR²
⇒ 169 = 25 + SR²
⇒ SR² = 169 – 25 = 144 = (12)²
∴ SR = 12 cm
Area of rect. PQRS = PS x SR = 5x 12 = 60 cm²
∵ Rectangle PQRS and ARAS are on the same
base SR and between the same parallels
∴ Area ARAS = \(\frac { 1 }{ 2 }\) area rect. PQRS 1
= \(\frac { 1 }{ 2 }\) x 60 = 30 cm²

Question 6.
In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8 cm and PQ and BD intersect at O, then find area of ∆OPB.
Solution:
In sq. ABCD, P and Q are the mid points of sides AB and CD respectively PQ and BD are joined which intersect each other at O
Side of square AB = 8 cm

∴ Area of square ABCD = (side)²
∵ Diagonal BD bisects the square into two triangle equal in area
∴ Area ∆ABD = \(\frac { 1 }{ 2 }\) x area of square ABCD
= \(\frac { 1 }{ 2 }\) x 64 = 32 cm²
∵ P is mid point of AB of AABD, and PQ || AD
∴ O is the mid point of BD
∴ OP = \(\frac { 1 }{ 2 }\)AD = \(\frac { 1 }{ 2 }\) x 8 = 4 cm
and PB = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 8 = 4 cm
∴ Area ∆OPB = \(\frac { 1 }{ 2 }\)PB x OP
= \(\frac { 1 }{ 2 }\) x4x4 = 8 cm²

Question 7.
ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. If area of ∆ABC is 16 cm², find the area of ∆DEF.
Solution:
In ∆ABC, D is mid point of BC. E and F are the mid points of DC and AE respectively area of ∆ABC = 16 cm²
FD is joined

∵ D is mid point of BC
∴ AD is the median and median divides the triangle into two triangles equal in area
area ∆ADC = \(\frac { 1 }{ 2 }\) ar(∆ABC)
= \(\frac { 1 }{ 2 }\) x 16 = 8 cm²
Similarly, E is mid point of DC
∴ area (∆ADE) = \(\frac { 1 }{ 2 }\) ar(∆ADC)
= \(\frac { 1 }{ 2 }\) x 8 = 4 cm²
∵ F is mid point of AE of ∆ADE
∴ ar(∆DEF) = \(\frac { 1 }{ 2 }\)area (∆ADE)
= \(\frac { 1 }{ 2 }\) x 4 = 2 cm²

Question 8.
PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of ∆PBQ is 17 cm2, find the area of ∆ASR.
Solution:
In trapezium PQRS,
PS || QR
A and B are points on sides PQ and SR
Such that AB || QR
area of ∆PBQ = 17 cm²

∆ABQ and ∆ABR are on the same base AB and between the same parallels
∴ ar(∆ABQ) = ar(∆ABR) …(i)
Similarly, ∆ABP and ∆ABS are on the same base and between the same parallels
∴ ar(ABP) = ar(∆ABS) …(ii)
Adding (i) and (ii)
ar( ∆ABQ) + ar( ∆ABP)
= ar(∆ABR) + ar(∆ABS)
⇒ ar(∆PBQ) = ar(∆ASR)
Put ar(PBQ) = 17 cm²
∴ ar(∆ASR) = 17 cm²

Question 9.
ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CQ : QP = 3 : 1. If ar(∆PBQ) = 10 cm2, find the area of parallelogram ABCD.
Solution:
In ||gm ABCD, P is mid point on AB,
PC and BD intersect each other at Q

CQ : QP = 3 : 1
ar(∆PBQ) = 10 cm²
In ||gm ABCD,
BD is its diagonal
∴ ar(∆ABD) = ar(∆BCD) = \(\frac { 1 }{ 2 }\) ar ||gm ABCD
∴ ar(||gm ABCD) = 2ar(∆ABD) …(i)
In ∆PBC CQ : QP = 3 : 1
∵ ∆PBQ and ∆CQB have same vertice B
∴ 3 x area ∆PBQ = ar(∆CBQ)
⇒ area(∆CBQ) = 3 x 10 = 30 cm²
∴ ar(∆PBC) = 30 + 10 = 40 cm²
Now ∆ABD and ∆PBC are between the
same parallel but base PB = \(\frac { 1 }{ 2 }\) AB
∴ ar(∆ABD) = 2ar(∆PBC)
= 2 x 40 = 80 cm²
But ar(||gm ABCD) = 2ar(∆ABD)
= 2 x 80 = 160 cm²

Question 10.
P is any point on base BC of ∆ABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If ar(∆ABC) = 12 cm², then find area of ∆EPC.
Solution:
P is any point on base of ∆ABC
D is mid point of BC
DE || PA drawn which meet AC at E
ar(∆ABC) = 12 cm²
AD and PE are joined

∵ D is mid point of BC
∴ AD is median
∴ ar(∆ABD) = ar(∆ACD)
= \(\frac { 1 }{ 2 }\) (∆ABC) = \(\frac { 1 }{ 2 }\) x 12 = 6 cm² …(i)
∵ ∆PED and ∆ADE are on the same base DE and between the same parallels
∴ ar(∆PED) = ar(∆ADE)
Adding ar(∆DCE) to both sides,
ar(∆PED) + ar(∆DCE) = ar(∆ADE) + ar(∆DCE)
ar(∆EPC) = ar(∆ACD)
⇒ ar(∆EPC) = ar(∆ABD) = 6 cm2 [From (i)]
∴ ar(∆EPC) = 6 cm²

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3

Other Exercises

• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

Question 1.
In the figure, compute the area of the quadrilateral.

Solution:
In the quadrilateral ABCD,
∠A = 90°, ∠CBD = 90°, AD = 9 cm, BC = 8 cm and CD = 17 cm

In right ∆BCD,
CD = BC² + BD² (Pythagoras Theorem)
⇒ (17)² = (8)² + BD²
⇒ 289 = 64 + BD²
⇒ BD² = 289 – 64 = 225 = (15)²
∴ BD = 15 cm
Now in right ∆ABD,
BD² = AB² + AD²
⇒ (15)² = AB² + (9)²
⇒ 225 = AB² + 81
⇒ AB²= 225 – 81 = 144 = (12)²
∴ AB = 12 cm

Question 2.
In the figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of ∆OTS if PQ = 8 cm.

Solution:
In square PQRS, T and U are the mid-points of the sides PS and QR
TU, QS and US are joined
PQ = 8 cm
∴ T and U are mid-points of the opposites sides PS and QR
∴ TU || PQ TO || PQ
In RQS,
T is mid-point of PS and TO || PQ
∴ O is the mid point of SQ 1 1

Question 3.
Compute the area of trapezium PQRS in the figure.

Solution:
In ∆TQR, ∠RTQ = 90°
∴ QR² = TQ² + RT²
⇒ (17)² = (8)² + RT²
⇒ 289 = 64 + RT²
⇒ RT² = 289 – 64 = 225 = (15)²
∴ RT = 15 cm
and PQ = 8 + 8 = 16 cm

Question 4.
In the figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ∆AOB.

Solution:
In ∆AOB, ∠AOB = 90°
C is a point on AB such that AC = BC Join OC

Since C is the mid-point of hypotenuse of right ∆AOB
∴ AC = CB = OC = 6.5 cm
∴ AB = 6.5 + 6.5 = 13 cm
Now in right ∆AOB
⇒ AB² = AO² + OB² (Pythagoras Theorem)
⇒ (13)² = (12)² + OB²
⇒ 169 = 144 + OB²
⇒ OB² = 169 – 144 = 25 = (5)²
∴ OB = 5 cm

Question 5.
In the figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Solution:
In the trapezium ABCD,
AB = 7 cm
AL = BM = 4 cm
AD = BC = 5 cm

Question 6.
In the figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm.If OE = 2 \(\sqrt { 5 } \) , find the area of the rectangle.

Solution:
Radius of the quadrant of circle = 2\(\sqrt { 5 } \) units
∴ OD diagonal of rectangle = 10 units (∵ OD = OB = OA = 10 cm)
DE = 2 \(\sqrt { 5 } \) cm
∴ In right ∆OED,
OD² = OE² + DEv
(10)² = OE² + (2\(\sqrt { 5 } \))²
100 = OE2 + 20
OE² = 100 – 20 = 80
⇒ OE² = (4\(\sqrt { 5 } \))²
∴ OE = 4\(\sqrt { 5 } \) cm
∴ Area of rectangle = lxb
= DE x OE
= 2\(\sqrt { 5 } \) x 4\(\sqrt { 5 } \)
= 8 x 5 = 40 cm2

Question 7.
In the figure, ABCD is a trapezium in which AB || DC. Prove that ar( ∆AOD = ar(∆BOC).

Solution:
In trapezium ABCD, diagonals AC and BD intersect each other at O
∴ ∆ADB and ∆ACB are on the same base AB and between the same parallels
∴ ar(∆ADB = ar(∆ACD)
Subtracting, ar(AAOB) from both sides,
ar(∆ADB) – ar(∆AOB) = ar(∆ACD) – ar(∆AOB)
⇒ ar(∆AOD) = ar(∆BOC)

Question 8.
In the figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(∆ADE) = ar(∆BCF) [NCERT]

Solution:
Given : In the figure, ABCD, ABEF and CDEF are ||gms
To prove : ar(∆ADE) = ar(∆BCF)
Proof: ∴ ABCD is a ||gm
∴ AD = BC
Similarly, in ||gm ABEF
AE = BF
and in ||gm CDEF,
DE = CF
Now, in ∆ADE and ∆BCF
AD = BC (proved)
DE = CF (proved)
AE = BF (proved)
∴ ∆ADE ≅ ∆BCF
∴ ar(∆ADE) = ar(∆BCF) (∵ Congruent triangles are equal in area)

Question 9.
In the figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(∆ABC) = ar(∆ABC)

Solution:
Given : In the figure, ∆ABC and ∆ABD are on the same base AB and line CD is bisected by AB at O i.e. CO = OD
To prove : ar(∆ABC) = ar(∆ABD)
Construction : Draw CL ⊥ AB and DM ⊥ AB

Question 10.
If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid point of median AD, prove that ar(∆BGC) = 2ar(∆AGC).
Solution:
Given : In ∆ABC, AD is its median. G is mid point of AD. BG and CF are joined
To prove :
(i) ar(∆ADB) = ar(∆ADC)
(ii) ar(∆BGC) = 2ar(∆AGC)
Construction : Draw AL ⊥ BC

Question 11.
A point D is taken on the side BC of a AABC such that BD = 2DC. Prove that ar(∆ABD) = 2ar(∆ADC)
Solution:
Given : In ∆ABC, D is a point on BC such that
BD = 2DC
AD is joined
To prove : ar(∆ABD) = 2ar(∆ADC)
Construction : Draw AL ⊥ BC

Question 12.
ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that
(i) ar(∆ADO) = or (∆CDO)
(ii) ar(∆ABP) = ar(∆CBP).
Solution:
Given : In ||gm ABCD, Diagonals AC and BD intersect each other at O
P is any point on BO
AP and CP are joined

To prove :
(i) ar(∆ADO) = ar(∆CDO)
(ii) ar(∆ABP) = ar(∆CBP)
Proof:
(i) In ∆ADC,
O is the mid point of AC
∴ ar(∆ADO) = ar(∆CDO)
(ii) Since O is the mid point of AC
∴ PO is the median of ∆APC
∴ af(∆APO) = or(∆CPO) …(i)
Similarly, BO is the median of ∆ABC
∴ ar(∆ABO) = ar(∆BCO) …(ii)
Subtracting (i) from (ii),
ar(∆ABO) – ar(∆APO) = ar(∆BCO) – ar( ∆CPO)
⇒ ar(∆ABP) = ar(∆CBP)
Hence proved.

Question 13.
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
(i) Prove that ar(∆ADF) = ar(∆ECF)
(ii) If the area of ∆DFB = 3 cm2, find the area of ||^gm ABCD.
Solution:
Given : In ||gm ABCD, BC is produced to E such that CE = BC
AE intersects CD at F
To prove :
(i) ar(∆ADF) = ar(∆ECF)
(ii) If ar(∆DFB) = 3 cm2, find the area of (||gm ABCD)

Question 14.
ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar(∆POA) = ar(∆QOC).
Solution:
Given : In ||gm ABCD, diagonals AC and BD intersect at O
A line through O intersects AB at P and CD at Q

To prove : ar(∆POA) = ar(∆QOC)
Proof : In ∆POA and ∆QOC,
OA = OC (O is mid-point of AC)
∠AOD = ∠COQ (Vertically opposite angles)
∠APO = ∠CQO (Alternate angles)
∴ ar(∆POA) ≅ ar(∆QOC) (AAS criterian)
∴ ar(∆POA) = ar(∆QOC)

Question 15.
In the figure, D and E are two points on BC such that BD = DE = EC. Show that ar(∆ABD) = ar(∆ADE) = ar(∆AEC). [NCERT]

Solution:
Given : D and E are two points on BC such that BD = DE = EC
AD and AE are joined
To prove : ar(∆ABD) = ar(∆ADE) = ar(∆AEC)
Construction : From A, draw AL ⊥ BC and XAY || BC

Proof: ∵ BD = DE = EC
and ∆ABD, ∆ADE and ∆AEC have equal bases and from the common vertex A
∴ ar(∆ABD) = ar(∆ADE) = ar(∆AEC)

Question 16.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
Show that: ar(∆APB) x ar(∆CPD) = ar(∆APD) x ar(∆BPC)
Solution:
Given : In quadrilateral ABCD, diagonal AC and BD intersect each other as P
To prove : ar(∆APB) x ar(∆CPD) = ar(APD) x ar(∆BPC)
Construction : Draw AL and CN perpendiculars on BD

Question 17.
If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.
Solution:
Given : In ||gm ABCD, P is any point in the ||gm
AP and BP are joined
To prove : ar(∆APB) < \(\frac { 1 }{ 2 }\) ar(||gm ABCD)
Construction : Draw DN ⊥AB and PM ⊥ AM

Question 18.
ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is a point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Solution:
Given : In ||gm ABCD, E is a point on AB such that BE = 2EA and F is a point on CD such that DF = 2FC. AE and CE are joined

Question 19.
In a ∆ABC, P and Q are respectively, the mid-points of AB and BC and R is the mid-point of AP. Prove that

Solution:
Given : In ∆ABC,
P and Q are mid-pionts of AB and BC R is mid-point of AP, PQ, RC, RQ are joined

Question 20.
ABCD is a parallelogram, G is the point on AB such that AG = 2GB, E is a point of DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(v) Find what portion of the area of parallelogram is the area of AEFG.
Solution:
Given : ABCD is a parallelogram and AG = 2GB, CE = 2DE and BF = 2FC
To prove :

(v) Find what portion of the area of parallelogram is the area of AFEG.
Construction : Draw EP ⊥ AB and EQ ⊥ BC

Question 21.
In the figure, CD || AE and CY || BA.
(i) Name a triangle equal in area of ACBX.
(ii) Prove that or(∆ZDE) = ar(∆CZA).
(iii) Prove that ar(∆CZY) = ar(∆EDZ).

Solution:
Given : In the figure,
CP || AE and CY || BA

To prove :
(i) Name a triangle equal in area of ∆CBX
(ii) Prove that ar(∆ZDE) = ar(∆CZA)
(iii) ar(BCZY) = ar(∆EDZ)
Proof:
(i) ∆CBX and ∆CYX are on the same base BY and between same parallels.
∴ ar(∆CBX) = ar(∆CYX)
(ii) ∆ADE and ∆ACE are on the same base AE
and between the same parallels (AE || CD)
∴ ar(∆ADE) = ar(∆ACE)
Subtracting ar(∆AZE) from both sides
⇒ ar(∆ADE) – ar(∆AZE) = ar(∆ACE) – ar(∆AZE)
⇒ ar(∆ZDE) = ar(∆ACZ)
⇒ ar∆ZDE = ar∆CZA
(iii) ∵ As ACY and BCY are on the same base CY and between the same parallels
∴ ar(∆ACY) = ar(∆BCY)
Now ar(∆ACZ) = ar(∆ZDE) (Proved)
⇒ ar(∆ACY) + ar(∆CYZ) = ar(∆EDZ)
⇒ ar(∆BCY) + ar(∆CYZ) = ar(∆EDZ)
∴ ar quad. (BCZY) = ar(EDZ)
Hence proved.

Question 22.
In the figure, PSD A is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that ar(∆PQE) = ar(∆CFD).

Solution:
Given : In the figure, PSDA is a ||gm in
which PQ = QR = RS and
AP || BQ || CR || DS
To prove : ar(∆PQE) = ar(∆CFD)
Construction : Join PD
Proof : ∵ PA || BQ || CR || DS
and PQ – QR = RS (Given)
∴ AB = BC = CD
∴ PQ = CD
Now in ABED, F is mid point of ED
∴ EF = FD
Similarly, EF = PE
⇒ PE = FD
In ∆PQE and ∆CFD,
∴ ∠EPQ = ∠FDC (Alternate angles)
PQ = CD
PE = FD (Proved)
∴ APQE ≅ ACFD (SAS cirterion)
∴ ar(∆PQE) = ar(∆CFD)

Question 23.
In the figure, ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively the mid-points of AD and BC, prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar(trap. DCYX) = \(\frac { 9 }{ 11 }\) ar(trap. XYBA)

Solution:
Given : In the figure, ABCD is a trapezium in which AB || DC
DC = 40 cm, AB = 60 cm
X and Y are the mid-points of AD and BC respectively
To prove :
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar(trap. DCYX) = \(\frac { 9 }{ 11 }\) m(trap. XYBA)
Construction : Join DY and produce it to meet AB produced at P

Question 24.
D is the mid-point of side BC of ∆ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(∆BOE) = \(\frac { 1 }{ 8 }\) ar(∆ABC).
Solution:
Given : In ∆ABC, D is mid point of BC, E is mid point BE and O is the mid point of AE. BO, AE, AD are joined.

Question 25.
In the figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar(∆ABP) = ar(∆ACQ).

Solution:
Given : In ∆ABC, X and Y are the mid pionts of AC and AB respectively. Through A, a line parallel to BC is drawn. Join BX and CY and produce them to meet the parallel line through A, at P and Q respectively and intersect each other at O.
To prove : ar(∆ABP) = ar(∆ACQ)
Construction : Join XY and produce it to both sides
Proof : ∵ X and Y are mid points of sides AC and AB
∴ XY || BC
Similarly, XY || PQ
∆BXY and ∆CXY are on the same base XY and between the same parallels
∴ ar(∆BXY) = ar(∆CXY) …(i)
Now, trap. XYAP and trap. XYAQ are on the same base XY and between the same parallels
∴ ar(XYAP) = ar(XYAQ) …(ii)
Adding (i) and (ii),
∴ ar(∆BXY) + ar(∆YAP)
= ar(CXY) + ar(XYAQ)
⇒ ar(∆ABP) = ar(∆ACQ)

Question 26.
In the figure, ABCD and AEFD are two parallelograms. Prove that
(i) PE = FQ
(ii) ar(∆APE) : ar(∆PFA) = ar(∆QFD) : ar(∆PFD)
(iii) ar(∆PEA) = ar(∆QFD).

Solution:
Given : Two ||gm ABCD and ||gm AEFD are on the same base AD. EF is produced to meet CD at Q. Join AF and PD also
To prove :
(i) PE = FQ
(ii) ar(∆APE) : ar(∆PFA) = ar(∆QFD) : ar(∆PFD)
(iii) ar(∆PEA) = ar(∆QFD)

Proof:
(i) In ∆AEP and DFQ,
AE = DF (Opposite sides of a ||gm)
∠AEP = ∠DFQ (Corresponding angles)
∠APE = ∠DQF (Corresponding angles)
∴ ∆AEP ≅ ∆DFQ (AAS axiom)
∴ PE = QF (c.p.c.t.)
(ii) and ar(∆AEP) = ar(∆DFQ) …(i)
(iii) ∵ ∆PFA and ∆PFD are on the same base PF and between the same parallels
∴ ar(∆PFA) = ar(∆PFD) …(ii)
From (i) and (ii),

Question 27.
In the figure, ABCD is a ||gm. O is any point on AC. PQ || AB and LM || AD. Prove that ar(||gm DLOP) = ar(||gm BMOQ).

Solution:
Given : In ||gm ABCD, O is any point on diagonal AC. PQ || AB and LM || BC
To prove : ar(||gm DLOP) = ar(||gm BMOQ)
Proof : ∵ Since, a diagonal of a parallelogram divides it into two triangles of equal area.
∴ ar(∆ADC) = or(∆ABC)
⇒ ar(∆APO) + or(||gm DLOP) + ar(∆OLC)
= ar(∆AOM) + ar(||^gm BMOQ) + ar( ∆OQC) …(i)
Since, AO and OC are diagonals of parallelograms AMOP and OQCL respectively,
∴ ar(∆APO) = ar(∆AMO) …(ii)
And, ar(∆OLC) = ar(∆OQC) …(Hi)
Subtracting (ii) and (iii) from (i), we get ar(||^gm DLOP) = ar(||^gm BMOQ)

Question 28.
In a ∆ABC, if L and M are points on AB and AC respectively such that LM || BC.
Prove that:
(i) ar(∆LCM) = ar(∆LBM)
(ii) ar(∆LBC) = ar(∆MBC)
(iii) ar(∆ABM) = ar(∆ACL)
(iv) ar(∆LOB) = ar(∆MOC).
Solution:
Given : In ∆ABC,
L and M are mid points on AB and AC
LM, LC and MB are joined

To prove :
(i) ar(∆LCM) = or(∆LBM)
(ii) ar(∆LBC) = ar(∆MBC)
(iii) ar(∆ABM) = ar(∆ACL)
(iv) ar(∆LOB) = ar(∆MOC)
Proof: ∵ L and M are the mid points of AB and AC
∴ LM || BC
(i) Now ∆LBM and ∆LCM are on the same base LM and between the same parallels
∴ar(∆LBM) = ar(∆LCM) …(i)
⇒ ar(∆LCM) = ar(∆LBM)
(ii) ∵ ∆LBC and ∆MBC are on the same base
BC and between the same parallels
∴ ar(∆LBC) = ar(∆MBC) …(ii)
(iii) a(∆LMB) = ar(∆LMC) [From (i)]
⇒ ar(∆ALM) + ar(∆LMB)
= ar(∆ALM) + ar(∆LMC) [Adding or(∆ALM) to both sides]
⇒ ar(∆ABM) = ar(∆ACL)
(iv) ∵ ar(∆LBC) = ar(∆MBC) [From (ii)]
⇒ ar(∆LBC) – ar(∆BOC) = ar(∆MBC) – ar(∆BOC)
ar(∆LBO) = ar(∆MOC)

Question 29.
In the figure, ABC and BDC are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F.


Solution:
Given : ABC and BDE are two equilateral triangles and D is mid point of BC. AE intersects BC in F
To prove :

Question 30.
In the figure, ABC is a right triangle right angled at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that [NCERT]
(i) ∆MBC ≅ ∆ABD
(ii) ar(BYXD) = ar(∆MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2ar(∆FCB)
(vi) ar(CYXE) = ar(∆CFG)
(vii) ar(BCED) = ar(AMBN) + ar(ACFG)

Solution:
Given : In ∆ABC, ∠A = 90°
BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively
AX ⊥ DE meeting DE at X
To prove :
(i) ∆MBC ≅ ∆ABD
(ii) ar(BYXD) = 2ar(∆MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2ar(∆FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = or(AMBN) + ar(ACFG)
Construction : Join AD, AE, BF and CM
Proof:
(i) In ∆MBC and ∆ABD,
MB=AB (Sides of square)
BC = BD
∠MBC = ∠ABD (Each angle = 90° + ∠ABC)
∴ ∆MBC ≅ ∆ABD (SAS criterian)
∴ ar(∆MBC) = ar(∆ABD) …(i)
(ii) ∵ ∆ABD and rectangle BYXD are on the same base BD and between the same parallels
∴ ar(∆ABD) = \(\frac { 1 }{ 2 }\) ar(rect. BYXD)
⇒ ar(rect. BYXD) = 2ar(∆ABD)
⇒ ar(rect. BYXD) = 2ar(∆MBC) …(ii)
(iii) Similarly, ∆MBC and square MBAN are on the same base MB and between the same parallels
∴ ar(∆MBC) = ar(sq. ABMN) …(iii)
From (ii) and (iii)
ar(sq. ∆BMN) = ar(rect. BYXD)
(iv) In AFCB and ∆ACE,
FC = AC
CB = CE (Sides of squares)
∠FCB = ∠ACE (Each = 90° + ∠ACB)
∴ ∆FCB = ∆ACE (SAS criterian)
(v) ∵ ∆FCB ≅ ∆ACE (Proved)
∴ ar(∆FCB) = ar(∆ACE)
∵∆ACE and rectangle CYXE are on the same base and between the same parallels
∴ 2ar(∆ACE) = ar(CYXC)
⇒ 2ar(∆FCB) = ar(CYXE) …(iv)
(vi) ∵ AFCB and rectangle FCAG are on the base FC and between the same parallels
∴ 2ar(∆FCB) = ar(FCAG) …(v)
From (iv) and (v)
ar(CMXE) = ar(ACFG)
(vii) In ∆ACB.
BC2 = AB2 + AC2 (By Pythagoras Theorem)
⇒ BC x BD = AB x MB + AC x FC
⇒ ar(BCED) = ar(ABMN) + ar(ACFG)
Hence proved.

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3

Other Exercises

• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Question 1.
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Solution:
Given : In ∆ABC and ∆DEF,
∠B = ∠E = 90°
∠C = ∠F
AB = DE

To prove : ∆ABC = ∆DEF
Proof : In ∆ABC and ∆DEF,
∠B = ∠E (Each = 90°)
∠C = ∠F (Given)
AB = DE (Given)
∆ABC = ∆DEF (AAS axiom)

Question 2.
If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.
Solution:
Given : In ∆ABC, AE is the bisector of vertical exterior ∠A and AE \(\parallel\) BC

To prove : ∆ABC is an isosceles
Proof: ∵ AE \(\parallel\) BC
∴ ∠1 = ∠B (Corresponding angles)
∠2 = ∠C (Alternate angle)
But ∠1 = ∠2 (∵ AE is the bisector of ∠CAD)
∴ ∠B = ∠C
∴ AB = AC (Sides opposite to equal angles)
∴ ∆ABC is an isosceles triangle

Question 3.
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Solution:
Given : In ∆ABC, AB = AC
∠A = 2(∠B + ∠C)

To calculate: Base angles,
Let ∠B = ∠C = x
Then ∠A = 2(∠B + ∠C)
= 2(x + x) = 2 x 2x = 4x
∵ Sum of angles of a triangle = 180°
∴ 4x + x + x – 180° ⇒ 6x = 180°
⇒ x= \(\frac { { 180 }^{ \circ } }{ 6 }\)  = 30° o
∴ ∠B = ∠C = 30 and ∠A = 4 x 30° = 120

Question 4.
Prove that each angle of an equilateral triangle is 60°. [NCERT]
Solution:
Given : ∆ABC is an equilateral triangle

Proof: In ∆ABC,
AB = AC (Sides of an equilateral triangle)
∴ ∠C = ∠B …(i)
(Angles opposite to equal angles)
Similarly, AB = BC
∴ ∠C = ∠A …(ii)
From (i) and (ii),
∠A = ∠B = ∠C
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
∴ ∠A = ∠B = ∠C = \(\frac { { 180 }^{ \circ } }{ 3 }\)= 60°

Question 5.
Angles A, B, C of a triangle ABC are equal to each other. Prove that ∆ABC is equilateral.
Solution:
Given : In ∆ABC, ∠A = ∠B = ∠C

To prove : ∆ABC is an equilateral
Proof: In ∆ABC,
∴ ∠B = ∠C (Given)
∴ AC = AB …(i) (Sides opposite to equal angles)
Similarly, ∠C = ∠A
∴ BC =AB …(ii)
From (i) and (ii)
AB = BC = CA
Hence ∆ABC is an equilateral triangle

Question 6.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆ABC, ∠A = 90°

AB =AC (Given)
∴ ∠C = ∠B (Angles opposite to equal sides)
But ∠B + ∠C = 90° (∵ ∠B = 90°)
∴ ∠B = ∠C = \(\frac { { 90 }^{ \circ } }{ 2 }\) = 45°
Hence ∠B = ∠C = 45°

Question 7.
PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
Solution:
Given : In ∆PQR, PQ = PR
S is a point on PQ and PT || QR

To prove : PS = PT
Proof : ∵ST || QR
∴ ∠S = ∠Q and ∠T = ∠R (Corresponding angles)
But ∠Q = ∠R (∵ PQ = PR)
∴ PS = PT (Sides opposite to equal angles)

Question 8.
In a ∆ABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced, prove that ∠MOC = ∠ABC.
Solution:
Given : In ∆ABC, AB = AC the bisectors of ∠B and ∠C intersect at O. M is any point on BO produced.

To prove : ∠MOC = ∠ABC
Proof: In ∆ABC, AB = BC
∴ ∠C = ∠B
∵ OB and OC are the bisectors of ∠B and ∠C
∴ ∠1 =∠2 = \(\frac { 1 }{ 2 }\)∠B
Now in ∠OBC,
Ext. ∠MOC = Interior opposite angles ∠1 + ∠2
= ∠1 + ∠1 = 2∠1 = ∠B
Hence ∠MOC = ∠ABC

Question 9.
P is a point on the bisector of an angle ∠ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.
Solution:
Given : In ∆ABC, P is a point on the bisector of ∠B and from P, RPQ || AB is draw which meets BC in Q

To prove : ∆BPQ is an isosceles
Proof : ∵ BD is the bisectors of CB
∴ ∠1 = ∠2
∵ RPQ || AB
∴ ∠1 = ∠3 (Alternate angles)
But ∠1 == ∠2 (Proved)
∴ ∠2 = ∠3
∴ PQ = BQ (sides opposite to equal angles)
∴ ∆BPQ is an isosceles

Question 10.
ABC is a triangle in which ∠B = 2∠C, D is a point on BC such that AD bisects ∠BAC = 72°.
Solution:
Given: In ∆ABC,
∠B = 2∠C, AD is the bisector of ∠BAC AB = CD
To prove : ∠BAC = 72°
Construction : Draw bisector of ∠B which meets AD at O

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5

Other Exercises

• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Question 1.
ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.
Solution:
Given : In ∆ABC, D is mid-point of BC and DE ⊥ AB, DF ⊥ AC and DE = DF

To Prove : ∆ABC is an isosceles triangle
Proof : In right ∆BDE and ∆CDF,
Side DE = DF
Hyp. BD = CD
∴ ∆BDE ≅ ∆CDF (RHS axiom)
∴ ∠B = ∠C (c.p.c.t.)
Now in ∆ABC,
∠B = ∠C (Prove)
∴ AC = AB (Sides opposite to equal angles)
∴ AABC is an isosceles triangle

Question 2.
ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that ∆ABC is an isosceles.
Solution:
Given : In ∆ABC,
BE ⊥ AC and CF ⊥ AB
BE = CF

To prove : AABC is an isosceles triangle
Proof : In right ABCE and ABCF Side
BE = CF (Given)
Hyp. BC = BC (Common)
∴ ∆BCE ≅ ∆BCF (RHS axiom)
∴ ∠BCE = ∠CBF (c.p.c.t.)
∴ AB = AC (Sides opposite to equal angles)
∴ ∆ABC is an isosceles triangle

Question 3.
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
Solution:
Given : A point P lies in the angle ABC and PL ⊥ BA and PM ⊥ BC and PL = PM. PB is joined

To prove : PB is the bisector ∠ABC,
Proof : In right ∆PLB and ∆PMB
Side PL = PM (Given)
Hyp. PB = PB (Common)
∴ ∆PLB ≅ ∆PMB (RHS axiom)
∴ ∆PBL = ∆PBM (c.p.c.t.)
∴ PB is the bisector of ∠ABC

Question 4.
In the figure, AD ⊥ CD and CB ⊥ CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.

Solution:
Given : In the figure,
AD ⊥ CD and CB ⊥ CD, AQ = BP and DP = CQ

To prove : ∠DAQ = ∠CBP
Proof : ∵ DP = CQ
∴ DP + PQ = PQ + QC
⇒ DQ = PC
Now in right ∆ADQ and ∆BCP
Side DQ = PC (Proved)
Hyp. AQ = BP
∴ ∆ADQ ≅ ∆BCP (RHS axiom)
∴ ∠DAQ = ∠CBP (c.p.c.t.)

Question 5.
Which of the following statements are true (T) and which are false (F):
(i) Sides opposite to equal angles of a triangle may be unequal.
(ii) Angles opposite to equal sides of a triangle are equal.
(iii) The measure of each angle of an equilateral triangle is 60°.
(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.
(v) The bisectors of two equal angles of a triangle are equal.
(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.
(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.
(viii)If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.
(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.
Solution:
(i) False : Sides opposite to equal angles of a triangle are equal.
(ii) True.
(iii) True.
(iv) False : The triangle is an isosceles triangle.
(v) True.
(vi) False : The triangle is an isosceles.
(vii) False : The altitude an equal.
(viii) False : If one side and hypotenuse of one right triangle on one side and hypotenuse of the other right triangle are equal, then triangles are congruent.
(ix) True.

Question 6.
Fill in the blanks in the following so that each of the following statements is true.
(i) Sides opposite to equal angles of a triangle are …….
(ii) Angle opposite to equal sides of a triangle are …….
(iii) In an equilateral triangle all angles are …….
(iv) In a ∆ABC if ∠A = ∠C, then AB = …….
(v) If altitudes CE and BF of a triangle ABC are equal, then AB = ……..
(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is ……… CE.
(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then ∆ABC ≅ ∆……
Solution:
(i) Sides opposite to equal angles of a triangle are equal.
(ii) Angle opposite to equal sides of a triangle are equal.
(iii) In an equilateral triangle all angles are equal.
(iv) In a ∆ABC, if ∠A = ∠C, then AB = BC.
(v) If altitudes CE and BF of a triangle ABC are equal, then AB = AC.
(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is equal to CE.
(vii) In right triangles ABC and DEF, it hypotenuse AB = EF and side AC = DE, then ∆ABC ≅ ∆EFD.

Question 7.
ABCD is a square, X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
Solution:
Given : In square ABCD, X and Y are points on side AD and BC respectively and AY = BX

To prove : BY = AX
∠BAY = ∠ABX
Proof: In right ∆BAX and ∆ABY
AB =AB (Common)
Hyp. BX = AY (Given)
∴ ∆BAX ≅ ∆ABY (RHS axiom)
∴ AX = BY (c.p.c.t.)
∠ABX = ∠BAY (c.p.c.t.)
Hence, BY = AX and ∠BAY = ∠ABX.

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2

Other Exercises

• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Question 1.
Two opposite angles of a parallelogram are (3x- 2)° and (50 – x)°. Find the measure of each angle of the parallelogram.
Solution:
∵ Opposite angles of a parallelogram are equal
∴ 3x – 2 = 50 – x
⇒ 3x + x – 50 + 2
⇒ 4x = 52
⇒ x = \(\frac { 52 }{ 4 }\) = 13
∴ ∠A = 3x – 2 = 3 x 13 – 2 = 39° – 2 = 37°
∠C = 50° -x = 50° – 13 = 37°
But∠A + B = 180°
∴ 37° + ∠B = 180°
⇒ ∠B = 180° – 37° = 143°

and ∠D = ∠B (Opposite angles of a ||gm)
∴ ∠D = 143°
Hence angles and 37°, 143°, 37°, 143°

Question 2.
If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.
Solution:
Let in ||gm ABCD,
∠A =x
Then ∠B = \(\frac { 2 }{ 3 }\) x

But, ∠A + ∠B = 180° (Sum of two adjacent angles of a ||gm)
⇒ x + \(\frac { 2 }{ 3 }\)x = 180°
⇒ \(\frac { 5 }{ 3 }\)x = 180°
⇒ x = 180° x \(\frac { 3 }{ 5 }\) = 108°
∴ ∠A = 108°
and ∠B = 108° x \(\frac { 2 }{ 3 }\) = 72°
But, ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 108°, ∠D = 72°
Hence angles are 108°, 72°, 108°, 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:
Let smallest angle of a ||gm = x
Then second angle = 2x – 24°
But these are consecutive angles
∴ x + (2x- 24°) = 180°
⇒ x + 2x – 24° = 180°

⇒ 3x = 180° + 24° = 204°
⇒ x =\(\frac { { 204 }^{ \circ } }{ 3 }\)  = 68°
∴ Smallest angle = 68°
and second angle = 2x 68° – 24°
= 136°-24° = 112°
∵ The opposite angles of a ||gm are equal Other two angles will be 68° and 112°
∴ Hence angles are 68°, 112°, 68°, 112°

Question 4.
The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?
Solution:
In a ||gm ABC,

Perimeter = 22cm
and longest side = 6.5 cm
Let shorter side = x
∴ 2x (6.5 + x) = 22
⇒ 13 + 2x = 22
⇒ 2x = 22 – 13 = 9
⇒ x = \(\frac { 9 }{ 2 }\) = 4.5
∴ shorter side = 4.5cm

Question 5.
In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.
Solution:
In ||gm ABCD,
∠D = 135°
But, ∠A + ∠D = 180° (Sum of consecutive angles)
⇒∠A+ 135° = 180°

⇒ ∠A = 180° – 135° = 45°
∵ ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠B = 135°

Question 6.
ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.
Solution:
In ||gm ABCD,
∠A = 70°
But ∠A + ∠B = 180° (Sum of consecutive angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70° = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles of a ||gm)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

Question 7.
In the figure, ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.

Solution:
In ||gm ABCD,
∠A + ∠B = 180°
(Sum of consecutive angles) But, ∠A = 75°

∴ ∠B = 180° – ∠A = 180° – 75° = 105°
∴ DBA = 105° -60° = 45°
But ∠CDB = ∠DBA (alternate angles)
= 45°
and ∠ADB = ∠DBC = 60°

Question 8.
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) False, Diagonals of a parallelogram are not equal.
(ii) True.
(iii) False, Diagonals bisect each other at right angles is a rhombus or a square only.
(iv) False, In a quadrilateral, if opposite sides are equal and parallel, then it is a ||gm.
(v) False, If all angles are equal, then it is a square or a rectangle.
(vi) False, If opposite sides are equal and parallel then it is a ||gm
(vii) False, If opposite angles are equal, then it is a parallelogram.
(viii)False, If all the sides are equal then it is a square or a rhombus but not parallelogram.

Question 9.
In the 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:
Given : In ||gm ABCD,
∠A = 60°
Bisector of ∠A and ∠B meet at P.
To prove :
(i) AD = DP
(ii) PC = BC
(iii) DC = 2AD
Construction : Join PD and PC
Proof : In ||gm ABCD,
∠A = 60°
But ∠A + ∠B = 180° (Sum of excutive angles)
⇒ 60° + ∠B = 180°
∴ ∠B = 1809 – 60° = 120°
∵ DC || AB
∠PAB = ∠DPA (alternate angles)
⇒ ∠PAD = ∠DPA (∵ ∠PAB = ∠PAD)
∴ AB = DP
(PA is its angle bisector, sides opposite to equal angles)
(ii) Similarly, we can prove that ∠PBC = ∠PCB (∵ ∠PAB = ∠BCA alternate angles)
∴ PC = BC
(iii) DC = DP + PC
= AD + BC [From (i) and (ii)]
= AD + AB = 2AB (∵BC = AD opposite sides of the ||gm)
Hence DC = 2AD

Question 10.
In the figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove thatAF = 2AB.
Solution:
Given : In ||gm ABCD,
E a mid point of BC
DE is joined and produced to meet AB produced at F

To prove : AF = 2AB
Proof : In ∆CDE and ∆EBF
∠DEC = ∠BEF (vertically opposite angles)
CE = EB (E is mid point of BC)
∠DCE = ∠EBF (alternate angles)
∴ ∆CDE ≅ ∆EBF (SAS Axiom)
∴ DC = BF (c.p.c.t.)
But AB = DC (opposite sides of a ||gm)
∴ AB = BF
Now, AF = AB + BF = AB + AB = 2AB
Hence AF = 2AB

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Mark the correct alternative in each of the following:
Question 1.
In ∆ABC ≅ ∆LKM, then side of ∆LKM equal to side AC of ∆ABC is
(a) LX
(b) KM
(c) LM
(d) None of these
Solution:
Side AC of ∆ABC = LM of ∆LKM (c)

Question 2.
In ∆ABC ≅ ∆ACB, then ∆ABC is isosceles with
(a) AB=AC
(b) AB = BC
(c) AC = BC
(d) None of these
Solution:
∵ ∆ABC ≅ ∆ACB
∴ AB = AC (a)

Question 3.
In ∆ABC ≅ ∆PQR, then ∆ABC is congruent to ∆RPQ, then which of the following is not true:
(a) BC = PQ
(b) AC = PR
(c) AB = PQ
(d) QR = BC
Solution:
∵ ∆ABC = ∆PQR
∴ AB = PQ, BC = QR and AC = PR
∴ BC = PQ is not true (a)

Question 4.
In triangles ABC and PQR three equality relations between some parts are as follows: AB = QP, ∠B = ∠P and BC = PR State which of the congruence conditions applies:
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Solution:
In two triangles ∆ABC and ∆PQR,
AB = QP, ∠B = ∠P and BC = PR
The condition apply : SAS (a)

Question 5.
In triangles ABC and PQR, if ∠A = ∠R, ∠B = ∠P and AB = RP, then which one of the following congruence conditions applies:
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Solution:
In ∆ABC and ∆PQR,
∠A = ∠R
∠B = ∠P
AB = RP
∴ ∆ABC ≅ ∆PQR (ASA axiom) (b)

Question 6.
If ∆PQR ≅ ∆EFD, then ED =
(a) PQ
(b) QR
(c) PR
(d) None of these
Solution:
∵ ∆PQR = ∆EFD
∴ ED = PR (c)

Question 7.
If ∆PQR ≅ ∆EFD, then ∠E =
(a) ∠P
(b) ∠Q
(c) ∠R
(d) None of these
Solution:
∵ ∆PQR ≅ ∆EFD
∴ ∠E = ∠P (a)

Question 8.
In a ∆ABC, if AB = AC and BC is produced to D such that ∠ACD = 100°, then ∠A =
(a) 20°
(b) 40°
(c) 60°
(d) 80°
Solution:
In ∆ABC, AB = AC
∴ ∠B = ∠C
But Ext. ∠ACD = ∠A + ∠B

∠ACB + ∠ACD = 180° (Linear pair)
∴ ∠ACB + 100° = 180°
⇒ ∠ACB = 180°-100° = 80°
∴ ∠B = ∠ACD = 80°
But ∠A + ∠B 4- ∠C = 180°
∴ ∠A + 80° + 80° = 180°
⇒∠A+ 160°= 180°
∴ ∠A= 180°- 160° = 20° (a)

Question 9.
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is
(a) 100°
(b) 120°
(c) 110°
(d) 130°
Solution:
In ∆ABC,
∠A = 2(∠B + ∠C)
= 2∠B + 2∠C

Adding 2∠A to both sides,
∠A + 2∠A = 2∠A + 2∠B + 2∠C
⇒ 3∠A = 2(∠A + ∠B + ∠C)
⇒ 3∠A = 2 x 180° (∵∠A + ∠B + ∠C = 180° )
⇒ 3∠A = 360°
⇒∠A = \(\frac { { 360 }^{ \circ } }{ 3 }\)  = 120°
∴ ∠A = 120° (b)

Question 10.
Which of the following is not a criterion for congruence of triangles?
(a) SAS
(b) SSA
(c) ASA
(d) SSS
Solution:
SSA is not the criterion of congruence of triangles. (b)

Question 11.
In the figure, the measure of ∠B’A’C’ is
(a) 50°
(b) 60°
(c) 70°
(d) 80°

Solution:
In the figure,
∆ABC ≅ ∆A’B’C’
∴ ∠A = ∠A
⇒3x = 2x- + 20
⇒ 3x – 2x = 20
⇒ x = 20
∠B’A’C’ = 2x + 20 = 2 x 20 + 20
= 40 + 20 = 60° (b)

Question 12.
If ABC and DEF are two triangles such that ∆ABC ≅ ∆FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then, which of the following is true?
(a) DF = 5 cm, ∠F = 60°
(b) DE = 5 cm, ∠E = 60°
(c) DF = 5 cm, ∠E = 60°
(d) DE = 5 cm, ∠D = 40°
Solution:
∵ ∆ABC ≅ ∆FDE,
AB = 5 cm, ∠A = 80°, ∠B = 40°
∴ DF = 5 cm, ∠F = 80°, ∠D = 40°
∴ ∠C =180°- (80° + 40°) = 180° – 120° = 60°

∴ ∠E = ∠C = 60°
∴ DF = 5 cm, ∠E = 60° (c)

Question 13.
In the figure, AB ⊥ BE and FE ⊥ BE. If BC = DE and AB = EF, then ∆ABD is congruent to
(a) ∆EFC
(b) ∆ECF
(c) ∆CEF
(d) ∆FEC

Solution:
In the figure, AB ⊥ BE, FE ⊥ BE
BC = DE, AB = EF,
then CD + BC = CD + DE BD = CE
In ∆ABD and ∆CEF,
BD = CE (Prove)
AB = FE (Given)
∠B = ∠E (Each 90°)
∴ ∆ABD ≅ ∆FCE (b)

Question 14.
In the figure, if AE || DC and AB = AC, the value of ∠ABD is
(a) 70°
(b) 110°
(c) 120°
(d) 130°

Solution:
In the figure, AE || DC
∴ ∠1 = 70° (Vertically opposite angles)
∴ ∠1 = ∠2 (Alternate angles)
∠2 = ∠ABC (Base angles of isosceles triangle)

∴ ABC = 90°
But ∠ABC + ∠ABD = 180° (Linear pair)
⇒ 70° +∠ABD = 180°
⇒∠ABD = 180°-70°= 110°
∴ ∠ABD =110° (b)

Question 15.
In the figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is
(a) 52°
(b) 76°
(c) 156°
(d) 104°

Solution:
In ∆ABC, AB = AC
AC is produced to E

CD || BA is drawn
∠ABC = 52°
∴ ∠ACB = 52° (∵ AB = AC)
∴ ∠BAC = 180°-(52° +52°)
= 180°-104° = 76°
∵ AB || CD
∴ ∠ACD = ∠BAC (Alternate angles)
= 76°
and ∠BCE + ∠DCB = 180° (Linear pair)
∠BCE + 52° = 180°
⇒∠BCE = 180°-52°= 128°
∠x + ∠ACD = 380°
⇒ x + 76° = 180°
∴ x= 180°-76°= 104° (d)

Question 16.
In the figure, if AC is bisector of ∠BAD such that AB = 3 cm and AC = 5 cm, then CD =
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm

Solution:
In the figure, AC is the bisector of ∠BAD, AB = 3 cm, AC = 5 cm

In ∆ABC and ∆ADC,
AC = AC (Common)
∠B = ∠D (Each 90°)
∠BAC = ∠DAC (∵ AC is the bisector of ∠A)
∴ ∆ABC ≅ ∆ADC (AAS axiom)
∴ BC = CD and AB = AD (c.p.c.t.)
Now in right ∆ABC,
AC² = AB² + BC²
⇒ (5)² = (3)² + BC²
⇒25 = 9 + BC²
⇒ BC² = 25 – 9 = 16 = (4)²
∴ BC = 4 cm
But CD = BC
∴ CD = 4 cm (c)

Question 17.
D, E, F are the mid-point of the sides BC, CA and AB respectively of ∆ABC. Then ∆DEF is congruent to triangle
(a) ABC
(b) AEF
(c) BFD, CDE
(d) AFE, BFD, CDE
Solution:
In ∆ABC, D, E, F are the mid-points of the sides BC, CA, AB respectively
DE, EF and FD are joined

∵ E and F are the mid-points
AC and AB,
∴ EF = \(\frac { 1 }{ 2 }\) BC and EF || BC
Similarly,
DE = \(\frac { 1 }{ 2 }\) AB and DE || AB
DF = \(\frac { 1 }{ 2 }\) AC and DF || AC
∴ ∆DEF is congruent to each of the triangles so formed
∴ ∆DEF is congruent to triangle AFE, BFD, CDE (d)

Question 18.
ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, ∠BAD =
(a) 55°
(b) 70°
(c) 35°
(d) 110°

Solution:
In ∆ABC, AB = AC
AD is median to BC

∴ BD = DC
In ∆ADB, ∠D = 90°, ∠B = 35°
But ∠B + BAD + ∠D = 180° (Sum of angles of a triangle)
⇒ 35° + ∠BAD + 90° = 180°
⇒∠BAD + 125°= 180°
⇒ ∠BAD = 180°- 125°
⇒∠BAD = 55° (a)

Question 19.
In the figure, X is a point in the interior of square ABCD. AXYZ is also a square. If DY = 3 cm and AZ = 2 cm, then BY =
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm

Solution:
In the figure, ABCD and AXYZ are squares
DY = 3 cm, AZ = 2 cm

DZ = DY + YZ
= DY + Z = 3 + 2 = 5 cm
In ∆ADZ, ∠2 = 90°
AD² + AZ² + DZ² = 2² + 5² cm
= 4 + 25 = 29
In ∠ABX, ∠X = 90°
AB² = AX² + BX²
AD² = AZ² + BX²
(∵ AB = AD, AX = AZ sides of square)
29 = 2² + BX²
⇒ 29 = 4 + BX²
⇒ BX² = 29 – 4 = 25 = (5)²
∴ BX = 5 cm (a)

Question 20.
In the figure, ABC is a triangle in which ∠B = 2∠C. D is a point on side BC such that AD bisects ∠BAC and AB = CD. BE is the bisector of ∠B. The measure of ∠BAC is
(a) 72°
(b) 73°
(c) 74°
(d) 95°

Solution:
In the figure, ∠B = 2∠C, AD and BE are the bisectors of ∠A and ∠B respectively,
AB = CD

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS are helpful to complete your math homework.

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