CBSE Class 7

RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17A
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper

Question 1.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6cm

(ii) At B, draw an arc of the radius 5cm.
(iii) At C, draw another arc of the radius 5.4cm which intersects the first arc at A
(iv) Join AB and AC
(v) With centre B and C and radius more than half of BC, draw arcs intersecting each other at L and M.
(vi) Join LM which intersects BC at Q and produce it to P.
Then PQ is perpendicular bisector of side BC.

Question 2.
Solution:
Steps of construction :
(i) Draw a line segment QR = 6cm
(ii) With centre Q and radius 4.4 cm draw an arc.
(iii) With centre R and radius 5.3 cm, draw another arc intersecting the first arc at P.
(iv) Join PQ and PR.

(v) With centre P and a suitable radius, draw an arc meeting PR at E and PQ at F.
(vi) With centres E and F, with same radius, draw two arcs intersecting each other at G.
(vii) Join PG and produce it to meet QR at S. Then PS is the bisector of ∠P.

Question 3.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6.2 cm.
(ii) With centres B and C radius 6.2 cm, draw arcs intersecting each other at A.
(iii) Join AB and AC.

∆ABC is the required equilateral triangle.
On measuring, each angle is equal to 60°.

Question 4.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5.3 cm.
(ii) With centre B and C, and radius 4.8 cm, draw arcs intersecting each other at A
(iii) Join AB and AC, Then ∆ABC is the required triangle.
(iv) Now, with centre A and a suitable radius draw an arc intersecting BC at L and M.
(v) Then with centre L and M, draw two arcs intersecting eachother at E.
(vi) Join AE intersecting BC at D. Then AD is perpendicular to BC.

On measuring ∠B and ∠C, each is equal to 55°.

Question 5.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3.8cm.
(ii) At A draw a ray AX making an angle of 60°.

(iii) Cut off AC = 5 cm from AX
(iv) Join CB.
Then ∆ABC is the required triangle.

Question 6.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.3 cm.
(ii) At C, draw a ray CY making an angle equal to 45°.

(iii) With centre C, and radius 6cm, draw an arc intersecting CY at A.
(iv) Join AB.
Then ∆ABC is the required triangle.

Question 7.
Solution:
(i) Draw a line segment AB = 5.2cm.
(ii) At A, draw a ray AX making an angle equal to 120°.

(iii) From AX cut off AC = 5.2cm.
(iv) Join BC.
Then ∆ABC is the required triangle.
(v) With centre A and some suitable radius draw an arc intersecting BC at L and M.
(vi) With centres L and M, draw two arcs intersecting each other at E.
(vii) Join AE intersecting BC at D Then AD is the perpendicular to BC.

Question 8.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6.2 cm.
(ii) At B draw a ray BX making an angle of 60°.

(iii) At C draw another ray CY making an angle of 45° which intersect the ray BX at A. .
Then ∆ABC is the required triangle.

Question 9.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5.8 cm.
(ii) At B draw a ray BX making an angle of 30°

(iii) At C draw another ray CY making an angle of 30° intersecting the BX at A
Then ∆ABC is the required triangle.
On measuring AB and AC, AB = 3.5cm and AC = 3.5cm.
AB = AC
∆ABC is an isosceles triangle.

Question 10.
Solution:
Steps of construction :
In ∆ABC, ∠A = 45° and ∠C = 75°
But ∠A + ∠B + ∠C = 180°

⇒ 45° + ZB + 75° = 180°
⇒ ∠B = 180° – 45° – 75°
⇒ ∠B = 180° – 120° = 60°
(i) Draw a line segment AB = 7cm.
(ii) At A, draw a ray AX making an angle of 45°.
(iii) At B, draw another ray BY making an angle of 60° which intersects AX at C
Then ∆ABC is the required triangle.

Question 11.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.8 cm
(ii) At C, draw a ray CX making an angle of 90°.

(iii) With centre B, an radius 6.3cm draw an arc intersecting CX at A.
(iv) Join AB.
Then ∆ABC is the required triangle.

Question 12.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.5cm
(ii) At B, draw a ray BX making an angle of 90°.

(iii) With centre C and radius 6cm draw an arc intersecting BX at A
(iv) Join AC.
Then ∆ABC is the required triangle.

Question 13.
Solution:
One acute angle = 30°, then
second acute angle will be = 90° – 30° = 60° (Sum of acute angles = 90°)
Steps of construction :
(i) Draw a line segment BC = 6cm.
(ii) At B draw a ray BX making an angle of 30°.
(iii) At C draw another ray CY making an angle of 60° which intersects BX at A

Then ∆ABC is the required triangle.

Hope given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B 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.

RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17A
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper

Question 1.
Solution:
Steps of construction :
(i) Draw a line segment AB
(ii) From a point P outside AB, draw a line PQ meeting AB at Q.

(iii) At P, draw a line PQ making an angle
∠QPC equal to ∠PQB with the help of compass and ruler and produce it to D.
Then the line CD is parallel to AB. Which is the required line.

Question 2.
Solution:
Steps of construction :
(i) Draw a line AB and take a point P on it.
(ii) From P, draw a perpendicular PX and cut off PQ = 3.5 cm.
(iii) From Q, draw a perpendicular line CD and produce it to both sides.

Then, CD is the required line which is parallel to AB.

Question 3.
Solution:
Steps of construction :
(i) Draw a line / and take a point P on it.
(ii) At P, draw a perpendicular line PX and cut off PQ = 4.3 cm.
(iii) From Q, draw a line m which is perpendicular on PX, and produce it to both sides.
Then m is the required line which is parallel to l.

Hope given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17A are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 16 Congruence Ex 16

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 16 Congruence Ex 16.

Question 1.
Solution:
(i) ∆ABC ≅ ∆EFD, Then
A ↔ E, B ↔ F and C ↔ D
AB = EF, BC = FD and CA = DE
∠A = ∠E, ∠B = ∠F and ∠C = ∠D
(ii) ∆CAB ≅ ∆QRP
C ↔ Q, A ↔ R and B ↔ P
CA = QR, AB = RP and BC = PQ
∠C = ∠Q, ∠A = ∠R and ∠B = ∠P
(iii) ∆XZY ≅ ∆QPR
X ↔ Q, Z ↔ P, Y ↔ R
XZ = QP, ZY = PR and YX = RQ
∠X = ∠Q, ∠Z = ∠P and ∠Y = ∠R
(iv) ∆MPN ≅ ∆SQR
M ↔ S, P ↔ Q and N ↔ R
MP = SQ, PN = QR and NM = RS
∠M = ∠S, ∠P = ∠Q and ∠N = ∠R.

Question 2.
Solution:
(i) In fig (i)
In ∆ABC and ∆DEF
∠C = ∠E
CA = ED
CB = EF
∆ACB ≅ ∆DEF (SAS condition)
(ii) In fig (ii)
In ∆RPQ and ∆LNM
Side PQ = NM
Hyp. RQ = LM
∆RPQ ≅ ∆LNM (RHS condition)
(iii) In ∆YXZ and ∆TRS
XY = RT
∠X = SR and YZ = TS
∆YXZ ≅ ∆TRS (SSS condition)
(iv) In ∆DEF and ∆PNM
∠E = ∠N
∠F = ∠M
EF = NM
∆DEF ≅ ∆PNM (ASA condition)
(v) In ∆ABC and ∆ADC
AC = AC (common)
∠ CAB = ∠ CAD (each 50°)
∠ ACB = ∠ DCA (each 60°)
∆ABC ≅ ∆ADC (ASA condition)

Question 3.
Solution:
In fig,
PL ⊥ OA and PM ⊥ OB and PL = PM
Now in right ∆PLO and ∆PMO,
Side PL = PM (given)
Hypotenuse OP = OP (common)
∆PLO ≅ ∆PMO (RHS condition)
Yes ∆PLO ≅ ∆PMO
Hence proved.

Question 4.
Solution:
In the figure,
AD = BC and AD || BC.
In ∆ABC and ∆ADC,
AC = AC (common)

BC = AB (given)
∠ACB = ∠CAD (Alternate angles)
∆ABC ≅ ∆ADC (SAS condition)
AB = DC (c.p.c.t)
Hence proved.

Question 5.
Solution:
In ∆ABD and ∆ACD,
AD = AD (common)
AB = AC (given)
BD = CD (given)
∆ABD ≅ ∆ADC (SSS condition)
∠BAD = ∠CAD (c.p.c.t.)
and ∠ADB = ∠ADC (c.p.c.t.)
But ∠ADB + ∠ADC = 180° (Linear pair)
∠ADB = ∠ADC = 90°
Hence proved.

Question 6.
Solution:
given : In ∆ABC, AD is the bisector of ∠A i.e. ∠BAD = ∠CAD
AD ⊥ BC.
To prove : ∆ABC is an isosceles
Proof : In ∆ADB and ∆ADC.
AD = AD (common)
∠ BAD = ∠ CAD (AD is the bisector of ∠A)
∠ ADB = ∠ ADC (each = 90°, AD ⊥ BC)
∆ADM ≅ ∆ADC (ASA condition)
AB = AC (c.p.c.t)
Hence ∆ABC is an isosceles triangle.
Hence proved.

Question 7.
Solution:
In the figure,
AB = AD, CB = CD
To prove : ∆ABC ≅ ∆ADC
Proof : In ∆ABC and ∆ADC
AC = AC (common)
AB = AD (given)
CB = CD (given)
∆ABC ≅ ∆ADC (SSS condition)
Hence proved.

Question 8.
Solution:
Given : In the figure,
PA ⊥ AB, QB ⊥ AB and PA = QB.
To prove : ∆OAP ≅ ∆OBQ,
Is OA = QB ?
Proof : In ∆OAP and ∆OBQ,
∠ A = ∠ B (each 90°)
AP = BQ (given)
∠AOP = ∠BOQ (vertically opposite angles)

∆OAP ≅ ∆OBQ (AAS condition)
OA = OB (c.p.c.t.)
Hence proved.

Question 9.
Solution:
Given : In right triangles ABC and DCB right angled at A and D respectively and AC = DB

To prove : ∆ABC ≅ ∆DCB.
Proof: In right angled ∆ABC and ∆DCB,
Hypotenuse BC = BC (common)
side AC = DB (given)
∆ABC ≅ ∆DCB (RHS condition)
Hence proved.

Question 10.
Solution:
Given: ∆ABC is an isosceles triangle in which AB = AC.
E and F are the midpoints of AC and AB respectively.

To prove : BE = CF
Proof : In ∆BCF and ∆CBE,
BC = BC (common)
BF = CE (Half of equal sides AB and AC)
∠CBF = ∠BCF (Angles opposite to equal sides)
∆BCF ≅ ∆CBE (SAS condition)
CF = BE (c.p.c.t.)
or BE = CF
Hence proved.

Question 11.
Solution:
Given : In isosceles ∆ABC,
AB = AC.
P and Q are the points on AB and AC respectively such that AP = AQ.
To prove : BQ = CP

Proof : In ∆ABQ and ∆ACP,
AB = AC (given)
AQ = AP (given)
∠ A = ∠ A (common)
∆ABQ ≅ ∆ACP (SAS condition)
BQ = CP (c.p.c.t.)
Hence proved.

Question 12.
Solution:
Given : ∆ABC is an isosceles triangle in which AB = AC.
AB and AC are produced to D and E respectively such that BD = CE.
BE and CD are joined.
To prove : BE = CD.
Proof : AB = AC and BD = CE
Adding we get:
AB + BD = AC + CE
AD = AE
Now, in ∆ACD and ∆ABE
AC = AB (given)
AD = AE (proved)
∠ A = ∠ A (common)
∆ACD ≅ ∆ABE (SSA condition)
CD = BE (c.p.c.t.)
Hence, BE = CD.

Question 13.
Solution:
Given : In ∆ABC,
AB = AC.
D is a point such that BD = CD.
AD, BD and CD are joined.
To prove : AD bisects ∠A and ∠D.
Proof : In ∆ABD and ∆CAD,
AD = AD (common)
AB = AC (given)
BD = CD (given)
∆ABD ≅ ∆CAD (SSS condition)
∠BAD = ∠CAD (c.p.c.t.)
and ∠BDA = ∠CDA (c.p.c.t.)
Hence AD is the bisector of ∠A and Z D.
Hence proved.

Question 14.
Solution:
Two triangles whose corresponding angles are equal, it is not necessarily that they should be congruent. It is possible if atleast one side must be equal. Below given a pair of triangles whose angles are equal but these are not congruent.

Question 15.
Solution:
In two triangles, if two sides and and included angle of the one equal to the corresponding two sides and included angle, then the two triangles are congruent.

If another angle except included angles are equal to each other and two sides are also equal these are not congruent. In the above figures, in ∆ABC and ∆PQR, two corresponding sides and one angle are equal, but these are not congruent.

Question 16.
Solution:
In ∆ABC,

Area = \(\frac { 1 }{ 2 }\) x BC x AL = \(\frac { 1 }{ 2 }\) x 5 x 4 = 10 cm²
and in ∆PQR
Area = \(\frac { 1 }{ 2 }\) x QR x PR = \(\frac { 1 }{ 2 }\) x 5 x 4 = 10 cm²
In these triangles
Areas of both triangles are equal but are not congruent to each other

Question 17.
Solution:
(i) Two line segments are congruent if they have the same length.
(ii) Two angles are congruent if they have equal measure.
(iii) Two squares are congruent if they have same side length.
(iv) Two circles are congruent if they have equal radius.
(v) Two rectangles are congruent if they have the same length and same breadth.
(vi) Two triangles are congruent if they have all parts equal.

Question 18.
Solution:
(i) False : Only those squares are congruent which have the same side.
(ii) True :
(iii) False : It is not necessarily, that those figures which have equal areas, must be congruent.
(iv) False : It is not necessarily that those triangles whose areas are equal, must be congruent.
(v) False : It is not necessarily that such triangles must be congruent.
(vi) True : It two angles and one side of a triangle are equal to the corresponding two angles and one side of the other are equal they are congruent.
(vii) False : Only three angles of one are equal the three angles of is not necessarily that these must be congruent.
(viii) True.
(ix) False : Only hypotenuse and one right angle of the one are equal to the hypotenuse and one right angles of the other, the triangles are not necessarily congruent, one side except them, must be equal.
(x) True : It is the definition of congruency of two triangles.

Hope given RS Aggarwal Solutions Class 7 Chapter 16 Congruence Ex 16 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.

RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15A
• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B
• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15C
• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15D

Question 1.
Solution:
In right triangle ABC, ∠B = 90° AB = 9cm, BC = 12cm

By Pythagoras Theorem,
AC² = AB² + BC² = (9)² + (12)² = 81 + 144 = 225
AC = √225 = 15 cm

Question 2.
Solution:
In right ∆ABC, ∠B = 90°
AC = 26cm, AB = 10cm
By Pythagoras Theorem
AC² = AB² + BC²

⇒ (26)² = (10)² + BC²
⇒ 676 = 100 + BC²
⇒ BC² = 676 – 100 = 576 = (24)²
⇒ BC = 24 cm

Question 3.
Solution:
In right ∆ABC, ∠C = 90°,
AB = 7.5cm, BC = 4.5cm
By Pythagoras Theorem
AB² = BC² + AC²

⇒ (7.5)² = (4.5)² + AC²
⇒ 56.25 = 20.25 + AC²
⇒ AC² = 56.25 – 20.25 = 36.00 = (6)²
⇒ AC = 6cm

Question 4.
Solution:
In ∆ABC, ∠B = 90°
Let each leg = x cm

By Pythagoras Theorem,
x² + x² = AC²
⇒ 2x² = 50
⇒ x² = 25 = (5)²
⇒ x = 5
Length of each equal leg = 5cm

Question 5.
Solution:
A triangle is a right-angled,
If (Hypotenuse)² = sum of squares or other two sides
If (39)² = (15)² + (36)² (Hypotenuse is the longest side)
If 1521 = 225 + 1296
If 1521 = 1521 Which is true.
It is a right-angled triangle.

Question 6.
Solution:
In ∆ABC, ∠C = 90°

a = 6cm, b = 4.5cm.
By Pythagoras Theorem
c² = a² + b² = (6)² + (4.5)² = 36.00 + 20.25 = 56.25 = (7.5)²
c = 7.5 cm

Question 7.
Solution:
A triangle will be a right angled
if (longest side)² = Sum of squares of other two sides
(i) a = 15cm, b = 20cm, c = 25cm.
Here, longest side = c ,
The triangle will be right angled
if c² = a² + b²
if (25)² = (15)² + (20)²
if 625 = 225 + 400 = 625 Which is true.
It is a right angled triangle.
(ii) a = 9cm, b = 12cm, c = 16cm
∆ABC is a right angled triangle if
c² = a² + b²
if (16)² = (9)² + (12)²
if 256 = 81 + 144 = 225
⇒ 256 = 225
Which is not true
Triangle is not a right angled triangle.
(iii) a = 10cm, b = 24cm, c = 26cm
The triangle ABC is a right angled triangle
if c² = a² + b²
if (26)² = (10)² + (24)²
if 676 = 100 + 576
if 676 = 676 Which is true.
The triangle is a right angled triangle.

Question 8.
Solution:
In ∆ABC,
∠B = 35° and ∠C = 55°
∠A = 180°- (∠B + ∠C) = 180° – (35° + 55°) = 180° – 90° = 90°
∆ABC is a right angled triangle
By Pythagoras Theorem,
BC² = AB² + AC²
(iii) is hue

Question 9.
Solution:
AB is a ladder and it is 15 m long B is window and BC = 12 m
In right ∆ABC
AB² = AC² + BC² (By Pythagoras Theorem)
⇒ (15)² = x² + (12)²
⇒ (15)² = x² + (12)²
⇒ 225 = x² + 144
⇒ x² = 225 – 144
⇒ x² = 81 = (9)²
x = 9 m
Distance of the foot of ladder from the wall = 9 m

Question 10.
Solution:
Let AB be the ladder and AC be the height.

Length of ladder AB = 5m
and height CA = 4.8m
Let distance of the ladder from the wall BC = x
Now in right angled ∆ABC, ∠C = 90°
AB² = AC² + BC² (By Pythagoras Theorem)
⇒ (5)² = (4.8)² + x²
⇒ 25 = 23.04 + x²
⇒ x² = 25.00 – 23.04 = 1.96 = (1.4)²
⇒ x = 1.4
The foot of ladder are 1.4m away from the wall.

Question 11.
Solution:
Let AB be the tree which broke at D and its top A touches the ground at C
their BD = 5m, BC = 12m,
Let AD = x m, then CD = x m
Now, in right ∆ABC,
CD² = BD² + BC²
(By Pythagoras Theorem)

CD² = (9)² + (12)² = 81 + 144 = 225 = (15)²
CD = 15m,
AD = x = 15m
Height of the tree AB = AD + BD = 15 + 9 = 24m

Question 12.
Solution:
AB and CD are two poles and they are 12,m apart
AB = 18 m, CD = 13m and BD = 12 m

From C, draw CE || BD Then
CE = BD = 12 m
and AE = AB – EB = AB – CD = 18 – 13 = 5 m
Join AC
Now in right ∆ACE
AC² = CE² + AE²
(By Pythagoras Theorem)
AC² = (12)² + (5)² = 144 + 25 = 169 = (13)²
AC = 13 m
Distance between their tops = 13 m

Question 13.
Solution:
A man starts from O and goes 35m due west and then 12m due north, then
In rights ∆OAB,
OA = 35 m
AB = 12 m

OB² = OA² + AB² (By Pythagoras Theorem)
= (35)² + (12)² = 1225 + 144 = 1369 = (37)²
OB = 37
Hence he is 37m away from the starting point

Question 14.
Solution:
A man goes 3km due north and then 4km east.

In right angled ∆OAB,
OA = 3km.
AB = 4km.
OB² = OA² + AB² (By Pythagoras Theorem)
= (3)² + (4)² = 9 + 16 = 25 = (5)²
OB = 5km
Hence he is 5km from the initial position.

Question 15.
Solution:
ABCD is a rectangle whose sides
AB = 16cm and BC = 12cm.
AC is its diagonal
In right angled ∆ABC
AC² = AB² + BC²
(By Pythagoras Theorem)
= (16)² + (12)² = 256 + 144 = 400 = (20)²
AC = 20cm
Hence length of diagonal AC = 20 cm

Question 16.
Solution:
ABCD is a rectangle and AC is its diagonal

AB = 40 cm and AC = 41 cm
Now in right ∆ABC
AC² = AB² + BC² (By Pythagoras Theorem)
⇒ (41)² = (40)² + BC²
⇒ 1681 = 1600 + BC²
⇒ BC² = 1681 – 1600 = 81 = (9)²
⇒ BC = 9 cm
Now perimeter of rectangle ABCD = 2 (AB + BC)
= 2 (40 + 9) = 2 x 49 = 98 cm

Question 17.
Solution:
Perimeter of rhombus ABCD = 4 x Side
Diagonal AC = 30 cm and BD = 16 cm
The diagonals of rhombus bisect each other at right angles
AO = OC = \(\frac { 30 }{ 2 }\) = 15 cm
and BO = OD = \(\frac { 16 }{ 2 }\) = 8 m
Now in right ∆AOB,
AB² = AO² + BO² = (15)² + (8)² = 225 + 64 = 289 = (17)²
AB = 17 cm
Now perimeter = 4 x side = 4 x 17 = 68 cm

Question 18.
Solution:
(i) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
(ii) If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right angled.
(iii) Of all the line segments that can be drawn to a given line from a given point outside it, the perpendicular is the shortest.

Hope given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15D 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.

RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15A
• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B
• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15C
• RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15D

Question 1.
Solution:
We know that in a triangle, sum of any two sides is greater than the third side. Therefore :
(i) 1cm, 1cm, 1cm
It is possible to draw a triangle
(1 + 1) cm > 1cm (sum of two sides is greater than the third)
(ii) 2cm, 3cm, 4cm
It is also possible to draw the triangle
(2 + 3) cm > 4cm (sum of two sides is greater than third side)
(iii) 7cm, 8cm, 15cm
It is not possible to draw the triangle
(7 + 8)cm not > 15cm
But (7 + 8) cm = 15 cm
(iv) 3.4 cm, 2.1 cm, 5.3 cm
It is possible to draw the triangle
(3.4 + 2.1) cm > 5.3 cm
⇒ 5.5cm > 5.3 cm
(v) 6cm, 7cm, 14cm
It is not possible to draw
(6 + 7) cm not > 14cm
i.e. 13cm not > 14cm (13cm < 14cm)

Question 2.
Solution:
Two sides of a triangle are 5 cm and 9 cm long
Then the third side will be less then (5 + 9) or less than 14 cm

Question 3.
Solution:
(i) In ∆APB,
PA + PB > AB (sum of two sides is greater than its third side)
(ii) In ∆PBC,
PB + PC > BC (sum of two sides is greater than its third side)
(iii) In ∆PAC,
AC < PA + PC (PA + PC > AC)

Question 4.
Solution:
Proof: AM is the median of ∆ABC
M is mid-point of BC

In ∆ABM,
AB + BM > AM ….(i)
(Sum of any two sides of a triangle is greater than its third side)
Similarly in ∆ACM,
AC + MC > AM ….(ii)
Adding (i) and (ii)
AB + BM + AC + MC > 2 AM
⇒ AB + AC + BM + MC > 2AM
⇒ AB + AC + BC > 2AM
Hence proved.

Question 5.
Solution:
Given: In ∆ABC, P is a point on BC.
AP is joined.
To prove :
(AB + BC + AC) > 2AP
Proof : In ∆ABP,
AB + BP > AP …(i) (Sum of two sides is greater than third)
Similarly in ∆ACP,

AC + PC > AP …(ii)
Adding (i) and (ii)
AB + BP + AC + PC > AP + AP
⇒ AB + BP + PC + CA > 2AP
⇒ AB + BC + CA > 2AP
Hence proved.

Question 6.
Solution:
ABCD is a quadrilateral AC and BD are joined.
Proof: Now in ∆ABC
AB + BC > AC ….(i)

(Sum of any two sides of a triangle is greater than its third side)
Similarly in ∆ADC,
AD + CD > AC ….(ii)
In ∆ABD,
AB + AD > BD ….(iii)
and in ∆BCD,
BC + CD > BD ……..(iv)
Adding (i), (ii), (iii) and (iv)
AB + BC + CD + AD + AB + AD + BC + CD > AC + AC + BD + BD
⇒ 2 (AB + BC + CD + AD) > 2(AC + BD)
⇒ AB + BC + CD + AD > AC + BD
Hence proved.

Question 7.
Solution:
Given : O is any point outside of the ∆ABC
To prove : 2(OA + OB + OC) > (AB + BC + CA)
Construction : Join OA, OB and DC.
Proof: In ∆AOB,

OA + OB > AB ….(i) (Sum of two sides of a triangle is greater than its third side)
Similarly in ∆BOC,
OB + OC > BC …(ii)
and in ∆COA
OC + OA > CA …(iii)
Adding (i), (ii) and (iii), we get:
OA + OB + OB + OC + OC + OA > AB + BC + CA
2 (OA + OB + OC) > (AB + BC + CA)
Hence proved.

Hope given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15C are helpful to complete your math homework.

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