Category: CBSE Class 7

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

Question 1.
Solution:
Radius of the circle (r) = 28 cm

Question 2.
Solution:
(i) Diameter of circle (d) = 35 cm

Question 3.
Solution:
Radius of a circle = 15 cm

Circumference = 2πr = 2 x 3.14 x 15 = 94.20 cm = 94.2 cm

Question 4.
Solution:
Circumference of a circle (c) = 57.2 cm

Question 5.
Solution:
Circumference (c) = 63.8 m

Question 6.
Solution:
Let c be the circumference and d be the diameter of the circle.
c = d + 30
⇒ dπ = d + 30
⇒ dπ – d = 30
⇒ d (π – 1) = 30

Question 7.
Solution:
The ratio of the radii of the circles = 5 : 3
Let radius of first circle = 5x
and radius of second circle = 3x
Circumference of first circle = 2πr = 2π x 5x = 10πx
and circumference of second circle = 2π x 3x = 6πx
Ratio = 10πx : 6πx = 10 : 6 = 5 : 3

Question 8.
Solution:
Radius of circular field (r) = 21 m
Circumference = 2πr

Question 9.
Solution:
Outer circumference = 616 m

Width of track = R – r = 98 – 84 = 14m

Question 10.
Solution:
Inner circumference of the circular track = 330 m

Rate of fencing = Rs. 20 per metre
Total cost = Rs. 20 x 396 = Rs. 7920

Question 11.
Solution:
Radius of inner circle (r) = 98 cm
Inner circumference = 2πr = 2 x \(\frac { 22 }{ 7 }\) x 98 cm = 616 cm

Radius of the outer circle (R) = 1 m 26cm = 126 cm
Outer circumference = 2πR = 2 x \(\frac { 22 }{ 7 }\) x 126 cm = 792 cm

Question 12.
Solution:
Side of equilateral triangle = 8.8 cm
Its perimeter = 8.8 x 3 = 26.4 cm
By bending their wire into a circular shape,
the circumference = 26.4 cm
Let d be the diameter,
Then C = dπ

Question 13.
Solution:
Each side of rhombus = 33 cm
Perimeter = 4 x 33 = 132 cm
Perimeter of circle = 132 cm
Let r be the radius

Question 14.
Solution:
Length of rectangle (l) = 18.7 cm
Breadth (b) = 14.3 cm
Perimeter = 2 (l + b) = 2 (18.7 + 14.3) cm = 2 x 33 = 66 cm
Circumference of the so formed circle = 66 cm

Question 15.
Solution:
Radius of the circle (r) = 35 cm
Its circumference (c) = 2πr

Question 16.
Solution:
Diameter of well (d) = 140 cm
Outer circumference of parapet = 616 cm
Let D be the diameter, then

Question 17.
Solution:
Diameter of wheel (d) = 98 cm

Question 18.
Solution:
Diameter of cycle wheel (d) = 70 cm

Question 19.
Solution:
Diameter of car wheel (d) = 77 cm
Circumference = πd = \(\frac { 22 }{ 7 }\) x 77 cm

Question 20.
Solution:
No. of revolutions = 5000
Distance covered = 11 km = 11 x 1000 = 11000 m

Question 21.
Solution:
Length of hour hand (r) = 4.2 cm
and length of minutes hand (R) = 7cm

Distance covered by hour hand in 24 hours = 2πR
= 2 x \(\frac { 22 }{ 7 }\) x 4.2
= 26.4 cm
But distance covered by minute hand in one hour = 2πR = 2 x \(\frac { 22 }{ 7 }\) x 7 = 44 cm
and distance covered by minute hand in 24 hours = 44 x 24 cm = 1056 cm
Sum of distance covered by these hands = 26.4 + 1056 = 1082.4 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

Question 1.
Solution:
(i) Base of the triangle = 42 cm

Height = 25 cm
Area = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 42 x 25 = 525 cm²
(ii) Base of the triangle = 16.8 m
and height = 75 cm = 0.75 m
Area = \(\frac { 1 }{ 2 }\) x Base x height
= \(\frac { 1 }{ 2 }\) x 16.8 x 0.75 m2 = 6.3 m²
(iii) Base of a triangle (b) = 8 m = 80 cm
and height (h) = 35 cm
Area = \(\frac { 1 }{ 2 }\) bh = \(\frac { 1 }{ 2 }\) x 80 x 35 = 1400 cm²

Question 2.
Solution:
Base of triangle = 16 cm
area of the triangle = 72 cm²

Question 3.
Solution:
Area of triangular region = 224 m²
Base = 28 m

Question 4.
Solution:
Area of triangle = 90 cm²
and height (h) = 12 cm

Question 5.
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area = \(\frac { 1 }{ 2 }\) bh = \(\frac { 1 }{ 2 }\) x 3x x x

Question 6.
Solution:
Area of the right angled triangle = 129.5 cm²

Question 7.
Solution:
In right angled ∆ABC,
Base BC = 1.2 m

and hypotenuse AC = 3.7 m
But AC² = AB² + BC² (Pythagoras Theorem)
⇒ (3.7)² = AB² + (1.2)²
⇒ 13.69 = AB² + 1.44
⇒ AB² = 13.69 – 1.44
⇒ AB² = 12.25 = (3.5)²
⇒ AB = 3.5 m
Now, area of ∆ABC = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x 1.2 x 3.5 m² = 2.1 m²

Question 8.
Solution:
Legs of a right angled triangle = 3 : 4
Let one leg (base) = 3x

Then second leg (altitude) = 4x
Area = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x 3x x 4x = 6x²
6x² = 1014
⇒ x² = \(\frac { 1014 }{ 6 }\) = 169 = (13)²
x = 13
one leg'(Base) = 3x = 3 x 13 = 39 cm
and second leg (altitude) = 4x = 4 x 13 = 52 cm

Question 9.
Solution:
One side BC of a right triangular scarf = 80 cm

and longest side AC = 1 m = 100 cm
By Pythagoras Theorem,
AC² = AB² + BC²
⇒ (100)² = AB² + (80)²
⇒ 10000 = AB² + 6400
⇒ AB² = 10000 – 6400
⇒ AB² = 3600 = (60)²
⇒ AB = 60
Second side = 60 cm
Area of the scarf = \(\frac { 1 }{ 2 }\) x b x h
= \(\frac { 1 }{ 2 }\) x 80 x 60 cm2 = 2400 cm²
Rate of cost = Rs. 250 per m²
Total cost =\(\frac { 2400 }{ 100 x 100 }\) x 250 = Rs. 60

Question 10.
Solution:
(i) Side of the equilateral triangle (a) = 18 cm

Question 11.
Solution:
Area of equilateral triangle = 16√3 cm²
Let each side = a
then \(\frac { \surd 3 }{ 4 }\) a² = 16√3
⇒ a² = \(\frac { 16\surd 3\times 4 }{ \surd 3 }\)
⇒ a² = 64 = (8)²
a = 8 cm
Each side = 8 cm

Question 12.
Solution:
Each side of an equilateral triangle = 24cm
Length of altitude = \(\frac { \surd 3 }{ 2 }\) a = \(\frac { \surd 3 }{ 2 }\) x 24
= 12√3 cm = 12 (1.73) = 20.76 cm

Question 13.
Solution:
(i) a = 13 m, b = 14 m, c = 15 m

= 2 x 2 x 3 x 7 x 7 x 7 = 4116 m²

Question 14.
Solution:
Let a = 33 cm, b = 44 cm, c = 55 cm

Question 15.
Solution:
Perimeter of the triangle = 84 cm
Ratio in side = 13 : 14 : 15
Sum of ratios =13 + 14 + 15 = 42
Let then first side = \(\frac { 84 x 13 }{ 42 }\) = 26 cm

Question 16.
Solution:
Let a = 42 cm, b = 34 cm, c = 20 cm

Question 17.
Solution:
In isosceles ∆ABC
Base BC = 48 cm.
and AB = AC = 30cm.
Let AD ⊥ BC

Question 18.
Solution:
Perimeter of an isosceles triangle = 32 cm
Base = 12 cm
Sum of two equal sides = 32 – 12 = 20 cm
Length of each equal side = \(\frac { 20 }{ 2 }\) = 10cm
Let AD ⊥ BC
BD = DC = \(\frac { 12 }{ 2 }\) = 6 cm

Question 19.
Solution:
In quadrilateral ABCD,
diagonal AC = 26 cm.

and perpendiculars DL = 12.8cm, BM = 11.2 cm
Area of quadrilateral ABCD
= \(\frac { 1 }{ 2 }\) (Sum of perpendicular) x diagonal
= \(\frac { 1 }{ 2 }\) (12.8 + 11.2) x 26 cm²
= \(\frac { 1 }{ 2 }\) x 24 x 26 = 312 cm²

Question 20.
Solution:
In quad. ABCD,
AB = 28 cm, BC = 26 cm, CD = 50 cm, DA = 40 cm
and diagonal AC = 30 cm
In ∆ABC,

Question 21.
Solution:
ABCD is a rectangle in which AB = 36 m
and BC = 24m
In ∆AED,
EF = 15 m
AD = BC = 24 m.
Now area of rectangle ABCD = l x b = 36 x 24 cm² = 864 cm²
Area of ∆AED = \(\frac { 1 }{ 2 }\) x AD x EF
= \(\frac { 1 }{ 2 }\) x 24 x 15 cm² = 180 cm²
Area of shaded portion = 864 – 180 = 684 m²

Question 22.
Solution:
In the fig. ABCD is a rectangle in which AB = 40 cm, BC = 25 cm.
P, Q, R and S and the mid points of sides, PQ, QR, RS and SP respectively
Then PQRS is a rhombus.
Now, join PR and QS.
PR = BC = 25cm and QS = AB = 40cm
Area of PQRS = \(\frac { 1 }{ 2 }\) x PR x QS
= \(\frac { 1 }{ 2 }\) x 25 x 40 = 500 cm²

Question 23.
Solution:
(i) Length of rectangle (l) = 18 cm
and breadth (b) = 10 cm
Area = l x b = 18 x 10 = 180 cm²

Area of right ∆EBC = \(\frac { 1 }{ 2 }\) x 10 x 8 = 40 cm²
and area of right ∆EDF = \(\frac { 1 }{ 2 }\) x 10 x 6 = 30 cm²
Area of shaded region = 180 – (40 + 30) = 180 – 70 = 110 cm²
(ii) Side of square = 20 cm

Area of square = a² = (20)² = 400 cm²
Area of right ∆LPM = \(\frac { 1 }{ 2 }\) x 10 x 10 cm² = 50 cm²
Area of right ∆RMQ = \(\frac { 1 }{ 2 }\) x 10 x 20 = 100 cm²
and area of right ∆RSL = \(\frac { 1 }{ 2 }\) x 20 x 10 = 100 cm²
Area of shaded region = 400 – (50 + 100 + 100) cm2 = 400 – 250 = 150 cm²

Question 24.
Solution:
In the quadrilateral ABCD
BD = 24 cm
AL ⊥ BD and CM ⊥ BD
AL = 5 cm and CM = 8 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D are helpful to complete your math homework.

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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.

 

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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.

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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.

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

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 15B.

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 ∆ABC,
∠A = 75°, ∠B = 45°
side BC is produced to D

forming exterior ∠ ACD
Exterior ∠ACD = ∠A + ∠B (Exterior angle is equal to sum of its interior opposite angles)
= 75° + 45° = 120°

Question 2.
Solution:
In ∆ABC, BC is produced to D forming an exterior angle ACD
∠ B = 68°, ∠ A = x°, ∠ ACB = y° and ∠ACD = 130°

In triangle,
Exterior angles is equal to sum of its interior opposite angles
∠ACD = ∠A + ∠B
⇒ 130° = x + 68°
⇒ x = 130° – 68° = 62°
But ∠ACB + ∠ACD = 180° (Linear pair)
⇒ y + 130° = 180°
⇒ y = 180° – 130° = 50°
Hence x = 62° and y = 50°

Question 3.
Solution:
In ∆ABC, side BC is produced to D forming exterior angle ACD.
∠ACD = 65°, ∠A = 32°
∠B = x, ∠ACB = y

In a triangle, the exterior angles is equal to the sum of its interior opposite angles
∠ACD = ∠A + ∠B
⇒ 65° = 32° + x
⇒ x = 65° – 32° = 33°
But ∠ ACD + ∠ ACB = 180° (Linear pair)
⇒ 65° + y = 180°
⇒ y = 180°- 65° = 115°
x = 33° and y = 115°

Question 4.
Solution:
In ∆ABC, side BC is produced to D forming exterior ∠ ACD

∠ACD = 110°, and ∠A : ∠B = 2 : 3
In a triangle, exterior angles is equal to the sum of its interior opposite angles
⇒ ∠ACD = ∠A + ∠B
⇒ ∠A + ∠B = 110°
But ∠A : ∠B = 2 : 3

But ∠A + ∠B + ∠C = 180° (sum of angles of a triangle)
⇒ 44° + 66° + ∠C = 180°
⇒ 110° + ∠C = 180°
⇒ ∠C = 180° – 110° = 70°
Hence ∠ A = 44°, ∠ B = 66° and ∠ C = 70°

Question 5.
Solution:
In ∆ABC, side BC is produced to forming exterior angle ACD.
∠ACD = 100° and ∠A = ∠B
Exterior angle of a triangle is equal to the sum of its interior opposite angles.

∠ACD = ∠A + ∠B But ∠A = ∠B
∠A + ∠A = ∠ACD = 100°
⇒ 2 ∠A = 100°
⇒ ∠A = 50°
∠B = ∠A = 50°
But ∠A + ∠B + ∠ ACB = 180° (sum of angles of a triangle)
⇒ 50° + 50° + ∠ ACB = 180°
⇒ 100° + ∠ ACB = 180°
⇒ ∠ ACB = 180° – 100° = 80°
Hence ∠ A = 50°, ∠ B = 50° and ∠ C = 80°

Question 6.
Solution:
In ∆ABC, side BC is produced to D From D, draw a line meeting AC at E so that ∠D = 40°
∠A = 25°, ∠B = 45°
In ∆ABC,

Exterior ∠ACD = ∠A + ∠B = 25° + 45° = 70°
Again, in ∆CDE,
Exterior ∠ AED = ∠ ECD + ∠ D = ∠ACD + ∠D = 70° + 40° = 110°
Hence ∠ACD = 70° and ∠AED = 110°

Question 7.
Solution:
In ∆ABC, sides BC is produced to D and BA to E
∠CAD = 50°, ∠B = 40° and ∠ACB = 100°
∠ ACB + ∠ ACD = 180° (Linear pair)
⇒ 100° + ∠ ACD = 180°
⇒ ∠ ACD = 180° – 100° = 80°

In ∆ACD,
∠ CAD + ∠ ACD + ∠ ADC = 180° (sum of angles of a triangle)
⇒ 50° + 80° + ∠ ADC = 180°
⇒ 130° + ∠ ADC = 180°
⇒ ∠ ADC = 180° – 130° = 50°
Now, in ∆ABD, BA is produced to E
Exterior ∠DAE = ∠ACD + ∠ADC = 80° + 50° = 130°
Hence ∠ ACD = 80°, ∠ ADC = 50° and ∠DAE = 130°

Question 8.
Solution:
In ∆ABC, BC is produced to D forming exterior ∠ ACD

∠ACD = 130°, ∠A = y°, ∠B = x° and ∠ACB = z°.
x : y = 2 : 3
Now, in ∆ABC,
Exterior ∠ACD = ∠A + ∠B

But ∠A + ∠B + ∠ACB = 180° (sum of angles of a triangle)
⇒ 78° + 52° + ∠ACB = 180°
⇒ 130° + ∠ACB = 180°
⇒ ∠ACB = 180° – 130°
⇒ ∠ACB = 50°
⇒ ∠ = 50°
Hence x = 52°, y = 78° and z = 50°

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

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

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 15A.

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 ∆ABC,
∠A = 72°, ∠B = 63°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ 72° + 63° + ∠C = 180°
⇒ 135° + ∠C = 180°
⇒ ∠C= 180°- 135° = 45°

Question 2.
Solution:
In. ∆PQR,
∠E = 105°, and ∠F = 40°
But ∠D + ∠E+ ∠F= 180° (sum of angles of a triangle)
⇒ ∠D + 105°+ 40°= 180°
⇒ ∠ D + 145° = 180°
⇒ ∠D = 180°- 145°
⇒ ∠D = 35°

Question 3.
Solution:
In ∆XYZ,
∠ X = 90°, ∠ Z = 48°
But ∠X + ∠Y + ∠Z = 180° (Sum of angles of a triangle)
⇒ 90° + ∠ Y + 48° = 180°
⇒ 138°+ ∠ Y = 180°
⇒ ∠Y = 180° – 138° = 42°
⇒ ∠Y = 42°

Question 4.
Solution:
Sum of angles of a triangle = 180°
and ratio in the three angles = 4 : 3 : 2

Question 5.
Solution:
In a right triangle
Sum of the two acute angles = 90°
One angle = 30°
Second angle = 90° – 36° = 54°

Question 6.
Solution:
In a right triangle
Sum of two acute angles = 90°
and ratio of these two angles = 2 : 1
Let first angle = 2x
Then second angle = x
2x + x = 90°
⇒ 3x = 90°
⇒ x = \(\frac { 90 }{ 3 }\) = 30°
First angle = 2x = 2 x 30° = 60°
and second angle = x = 1 x 30° = 30°

Question 7.
Solution:
In a triangle,
Measure of one angle = 100°
Sum of other two angles = 180° – 100° = 80°
(Sum of angles of a triangles)
But, these two angles are equal.
Measure of each angle = \(\frac { 80 }{ 2 }\) = 40°

Question 8.
Solution:
Sum of angles of a triangle = 180°
Let third angle = x
then, each equal angles = 2x
x + 2x + 2x = 180°
⇒ 5x = 180°
⇒ x = \(\frac { 180 }{ 5 }\) = 36°
Each equal angle = 2x = 2 x 36° = 72°
and third angle = 36°

Question 9.
Solution:
In a triangle ABC,
Let ∠A = ∠B + ∠C
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + ∠A = 180° (∠B + ∠C = ∠A)
⇒ 2A = 180°
⇒ ∠ A = \(\frac { 180 }{ 2 }\) = 90°
∠ A = 90°
Hence, ∆ABC is a right triangle.

Question 10.
Solution:
In a ∆ABC,
2 ∠A = 3 ∠B = 6 ∠C = 1 (suppose)

Question 11.
Solution:
In an equilateral triangle,
All sides are equal.
All angles are also equal.
Each angle = \(\frac { 180 }{ 3 }\) = 60°
(Sum of angles of a triangle = 180°)

Question 12.
Solution:
In the given figure,
ABC is a triangle in which DE || BC,
∠A = 65° and ∠B = 55°
DE || BC and ADB is the transversal
⇒ ∠ ADE = ∠ B (corresponding angles) = 55° (∠B = 55°)
In ∆ADE,
∠A + ∠ADE + ∠AED = 180° (sum of angles of a triangle)

⇒ 65° + 55° + ∠AED = 180°
⇒ ∠ 120° + ∠AED = 180°
⇒ ∠AED = 180°- 120° = 60°
∠AED = 60°
D || BC and AEC is the transversal
∠ C = ∠ AED (A corresponding angles)
∠C = 60°
Hence ∠ADE = 55°, ∠AED = 60° and ∠ C = 60°

Question 13.
Solution:
(i) No. In a triangle, only one right angle is possible as if there are two right angles, then The third angle will be ∠ero which is not possible.
(ii) No. In a triangle only one obtuse angle is possible as if there are two obtuse angles, then the sum of these two angles will be greater than 180° which is not possible.
(iii) Yes. two acute can arc possible.
(iv) No. The sum of these three angles will be greater than 180° which is not possible in a triangle.
(v) No. The sum of these angles will be less than 180° which is not possible.
(vi) Yes. The sum of there three angle will be in 180° which is possible.

Question 14.
Solution:
(i) Yes, it can be a right triangle also
(ii) Yes, if right triangle has its sides different then it is possible.
(iii) No, a right triangle cannot be an equilateral triangle as an equilateral triangle has each side 60°.
(iv) Yes, it is possible, if its sides opposite to acute angles are equal.

Question 15.
Solution:
(i) A right triangle cannot have an obtuse angle.
(ii) The acute angles of a right triangle are complementary.
(iii) Each acute angle of an isosceles right triangle measures 45°.
(iv) Each angle of an equilateral triangle measures 60°.
(v) The side opposite the right angle of the right triangle is called the hypotenuse.
(vi) The sum of the lengths of the sides of a triangle is called its perimeter.

 

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RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C

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

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

Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(c) Supplement of 45° is 135°
135°+ 45° = 180°

Question 2.
Solution:
(b) Complement of 80° is 10°
10° + 80° = 90°

Question 3.
Solution:
(b) The angle is its own complement.
The measure of the angles will be 45° (45° + 45° = 90°)

Question 4.
Solution:
(a) The angle is one-fifth of its supplement
Let angle be x, then
x + 5x = 180°
⇒ 6x = 180°
⇒ x = 30°
Angle is 30°

Question 5.
Solution:
(b) Let angle is x
Then its complement angle=x-24° But x + x- 24° = 90°
⇒ 2x = 90° + 24° = 114°
⇒ x = 57°
The required angle is 57°

Question 6.
Solution:
(b) Let required angle = x
Then its supplement angle = x + 32
But x + x + 32° = 180°
⇒ 2x = 180° – 32 = 148°
⇒ x = 74°
Required angle = 74°

Question 7.
Solution:
(c) Two supplementary angle are in the ratio = 3 : 2
Let first angle = 3x
Second angle = 2x
But 3x + 2x = 180°
⇒ 5x = 180°
⇒ x = 36°
Smaller angle = 2x = 2 x 36° = 72°

Question 8.
Solution:
(b) In the figure ∠BOC = 132°
But ∠AOC + ∠BOC =180° (Linear pair)
⇒ ∠AOC + 132° = 180°
⇒ ∠AOC = 180° – 132° = 48°

Question 9.
Solution:
(c) In the figure, ∠AOC = 68°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 68° + x = 180°
⇒ x = 180° – 68° = 112°

Question 10.
Solution:
(b) In the figure,
AOB is a straight line
∠AOC + ∠BOC = 180° (Linear pair)
⇒ 2x – 10° + 3x + 15° = 180°
⇒ 5x = 180° + 10° – 15° = 175°
⇒ x = 35°
x = 35

Question 11.
Solution:
(d) In the figure,
AOB is a straight line
∠AOC + ∠COD + ∠DOB = 180°
⇒ 55° + x + 45° = 180°
⇒ x + 100° = 180°
⇒ x = 180° – 100° = 80°

Question 12.
Solution:
(a) AOB is a straight line
x + y = 180°
But 4x = 5y

Question 13.
Solution:
(b) AB and CD intersect each other at O and ∠AOC = 50°
∠BOD = ∠AOC = 50° (Vertically opposite angles)

Question 14.
Solution:
(a) AOB is a straight line
∠AOC + ∠COD + ∠DOB = 180°
⇒ 3x – 8° + 50° + x + 10° = 180°
⇒ 4x = 180° + 8° – 50° – 10°
⇒ 4x = 128°
⇒ x = 32°

Question 15.
Solution:
(b) In ∆ABC, side BC is produced to D
∠ACD = 132° and ∠A = 54°
Ext. ∠ACD = ∠A + ∠B
⇒ 132° = 54° + ∠B
⇒ ∠B = 132° – 54° = 78°

Question 16.
Solution:
(c) In ∆ABC,
Side BC is produced to D
∠A = 45°, ∠B = 55°
Ext. ∠ACD = ∠A + ∠B = 45° + 55° = 100°

Question 17.
Solution:
(b) In ∆ABC, side BC is produced to D
∠ABC = 70° and ∠ACD = 120°
Ext. ∠ACD = ∠BAC + ∠ABC
⇒ 120° = ∠BAC + 70°
⇒ ∠BAC = 120° – 70° = 50°

Question 18.
Solution:
(c) In the figure,
∠AOB = 50°, ∠BOC = 90°
∠COD = 70°, ∠AOD = x.
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (Angles at a point)
⇒ 50° + 90° + 70° + x = 360°
⇒ 210 + x = 360°
⇒ x = 360° – 210°
⇒ x = 150°

Question 19.
Solution:
(c) In the figure,
Side BC of ∆ABC is produced to D
CE || BA is drawn
∠A = 50° and ∠ECD = 60°
AB || CE
∠ABC = ∠ECD (corresponding angle) = 60°
But in ∆ABC,
∠A + ∠B + ∠ACB = 180° (Angles of a triangles)
⇒ 50° + 60° + ∠ACB = 180°
⇒ ∠ACB = 180° – 50° – 60° = 70°

Question 20.
Solution:
(b) In ∆ABC,
∠A = 65°, ∠C = 85°
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
⇒ 65° +∠B+ 85° = 180°
⇒ 150° + ∠B = 180°
⇒ ∠B = 180° – 150° = 30°

Question 21.
Solution:
(d) Sum of angles of a triangle = 180°

Question 22.
Solution:
(c) Sum of angles of a quadrilateral = 360°

Question 23.
Solution:
(b) In the figure, AB || CD
∠OAB = 150°, ∠OCD = 120°

From O, draw OE || AB or CD
AB || DE
∠OAB + ∠AOE =180°
⇒ 150° + ∠AOE = 180°
⇒ ∠AOE = 180° – 150° = 30°
Similarly DE || CD
∠EOC + ∠OCD = 180°
⇒ ∠EOC + 120° = 180°
⇒ ∠EOC = 180° – 120° = 60°
Now ∠AOC = ∠AOE + ∠EOC = 30° + 60° = 90°

Question 24.
Solution:
(a) In the given figure,
PQ || RS,
∠PAB = 60° and ∠ACS = 100°
PQ || RS
∠ABC = ∠PAB (alternate angles) = 60°
But Ext. ∠ACS = ∠BAC + ∠ABC
⇒ 100° = ∠BAC + 60°
⇒ ∠BAC = 100° – 60° = 40°

Question 25.
Solution:
(c) In the figure, AB || CD || EF
∠ABG =110° and ∠GCD = 100°
∠BGC = x°
AB || EF
∠ABG + ∠BGE = 180°
⇒ 110° + ∠BGE = 180°
⇒ ∠BGE = 180° – 110° = 70°
Similarly CD || EF
∠GCD + ∠CGF = 180°
⇒ 100° + ∠CGF = 180°
⇒ ∠CGF = 180° – 100° = 80°
But ∠BGE + ∠BGC + ∠CGF = 180°
⇒ 70° + x + 80° = 180°
⇒ 150° + x = 180°
⇒ x = 180° – 150° = 30°

Question 26.
Solution:
(d) Sum of any two sides of a triangle is always greater than the third side

Question 27.
Solution:
(d) The diagonals of a rhombus always bisect each other at right angles.

Question 28.
Solution:
(c) In ∆ABC, ∠B = 90°
AB = 5 cm and AC = 13 cm
But AC² = AB² + BC² (By Pythagoras Theorem)
⇒ (13)² = (5)² + BC²
⇒ 169 = 25 + BC2
⇒ BC² = 169 – 25 = 144 = (12)²
BC = 12 cm

Question 29.
Solution:
(c) In ∆ABC, ∠B = 37°, ∠G = 29°
But ∠A + ∠B + ∠C = 180° (angles of a triangle)
⇒ ∠A + 37° + 29° = 180°
⇒ ∠A + 66° = 180°
⇒ ∠A = 180° – 66° = 114°

Question 30.
Solution:
(c) The ratio of angles of a triangle is 2 : 3 : 7
But sum of angles of a triangle = 180°

Question 31.
Solution:
In ∆ABC,
Let 2∠A = 3∠B = 6∠C = x

Question 32.
Solution:
(a) In ∆ABC,
∠A + ∠B = 65°, ∠B + ∠C = 140°
∠A = 65°
∠C = 140° – ∠B
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
⇒ 65° – ∠B + 140° – ∠B + ∠B = 180°
⇒ 205° – ∠B = 180°
⇒ ∠B = 205° – 180° = 25°

Question 33.
Solution:
(b) In ∆ABC, ∠A – ∠B = 33°
and ∠B – ∠C = 18°
∠A = 33° + ∠B and ∠C = ∠B – 18°
But ∠A + ∠B + ∠C = 180°
⇒ 33° + ∠B + ∠B + ∠B – 18° = 180°
⇒ 3∠B = 180° – 33° + 18° = 165°
⇒ ∠B = 55°

Question 34.
Solution:
(c) In ∆ABC
∠A + ∠B + ∠C= 180° (Sum of angles of a triangle)
But angles are (3x)°, (2x – 7)° and (4x – 11)°
3x + (2x – 7) + (4x – 11)° = 180°
⇒ 3x + 2x – 7 + 4x – 11° = 180°
⇒ 9x – 18° = 180°
⇒ 9x = 180° + 18° = 198°
⇒ x = 22°

Question 35.
Solution:
(c) ∆ABC is a right angled, ∠A = 90°
AB = 24 cm, AC = 7 cm

but BC² = AB² + AC²
⇒ BC² = (24)² + (7)² = 576 + 49 = 625 = (25)²
BC = 25 cm

Question 36.
Solution:
(b) Let AB is a ladder and A is the window
BC = 15 m, AC = 20 m

Now in right ∆ABC
AB² = BC² + AC² = (15)² + (20)² = 225 + 400 = 625 = (25)²
AB = 25 m
Length of ladder = 25 m

Question 37.
Solution:
(a) Let AB and CD are two poles such that
AB = 6 m, CD = 11 m
and distance between two poles BD = 12m
From A, draw AE || BD
AE = BD = 12m
CE = CD – ED = 11 – 6 = 5 m

Now in right ∆AEC
AC² = AE² + CE² = (12)² + (5)² = 144 + 25 = 169 = (13)²
AC = 13 m
Distance between tops of poles = 13 m

Question 38.
Solution:
(d) ∆ABC is an isosceles triangle
∠C = 90°,
AC = 5 cm
BC = AC = 5 cm

In right ∆ABC
AB² = AC² + BC² = (5)² + (5)² = 25 + 25 = 50 = 2 x 25
AB = √(2 x 25) = 5√2 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C 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 12 Simple Interest CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12A
• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12B
• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:
Let the sum be ₹ x

Question 3.
Solution:
P = ₹ 3625, A = ₹ 4495, T = 2 years
S.I. = A – P = ₹ 4495 – ₹ 3625 = ₹ 870

Question 4.
Solution:
P = ₹ 3600, A = ₹ 4410, R = 9%
S.I. = A – P = ₹ 4410 – ₹ 3600 = ₹ 810

Question 5.
Solution:
Let the sum be ₹ x
Amount = ₹ 2x
S.I. = (2x – x) = ₹ x
Time = 12 years
P = x, S.I. = x, T = 12 years

Question 6.
Solution:
Let the sum be ₹ 4

Mark (✓) against the correct answer in each of the following :
Question 7.
Solution:
(c) 9%
Let the sum be ₹ x
A = \(\frac { 49x }{ 40 }\)
We know:
A = P + S.I.
S.I. = A – P

Question 8.
Solution:
(c) ₹ 3500
A = ₹ 3626, R = 6%,

Question 9.
Solution:
(a) 9 months
P = ₹ 6000, A = ₹ 6360
S.I. = A – P = 6300 – 6000 = ₹ 360

Question 10.
Solution:
(c) 12%
Let the sum be ₹ x

Question 11.
Solution:

P = ₹ \(\frac { 100 }{ x }\)

Question 12.
Solution:
(b) 10%
Let the sum be ₹ x
Amount ₹ 2x
Time =10 years
S.I. = A – P = 2x – x = ₹ x

Question 13.
Solution:

Question 14.
Solution:
(i) False

Hope given RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest CCE Test Paper 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 12 Simple Interest Ex 12B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12A
• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12B
• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest CCE Test Paper

Question 1.
Solution:
Principal (P) = Rs. 6250
Rate of (R) = 4% p.a.
Period (T) = 6 months = \(\frac { 1 }{ 2 }\) year

Question 2.
Solution:
Amount (A) = Rs. 3605
Rate (R) = 5% p.a.

Question 3.
Solution:
Let sum (P) = Rs. 100

Question 4.
Solution:
Principal (P) = Rs. 8000
Amount (A) – Rs. 8360
S.I. = A – P = Rs. 8360 – Rs. 8000 = Rs. 360
Rate (R) = 6% p.a.

Question 5.
Solution:
Let sum (P) = Rs. 100
Then amount (A) = Rs. 100 x 2 = Rs. 200
S.I. = A – P = Rs. 200 – Rs. 100 = Rs. 100
Period (T) = 10 years

Question 6.
Solution:
S.I. = Rs. x
Rate (R) = x% p.a.
Time (T) = x year

Rs. \(\frac { 100 }{ x }\) (c)

Question 7.
Solution:
Let sum (P) = Rs. 100

Question 8.
Solution:
A’s principal (P) = Rs. 8000
Rate (R) = 12% p.a.
B’s principal = Rs. 9100
Rate = 10%
Let after x years, then amount will be equal

Question 9.
Solution:
Amount (A) = Rs. 720
Principal (P) = Rs. 600
S.I. = A – P = Rs. 720 – 600 = Rs. 120
Period (T) = 4 years

Question 10.
Solution:

Question 11.
Solution:
Let sum (P) = Rs. 100
Their S.I. = 0.125 of Rs. 100

Question 12.
Solution:
S.I. = Rs. 210

Hope given RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12B 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.