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RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4A
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4D
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4E
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself

MCQ
Question 1.
Solution:
(b) ∆ABC ~ ∆DEF

Question 2.
Solution:
(a) In the given figure,

Question 3.
Solution:
(b) Length of pole AB = 6 m
and CD = 11 m
and distance between the BD = 12 m
Draw AE || BD, then
ED = AB = 6 m, AE = BD = 12 m
CE = CD – ED = 11 – 6 = 5 m

Now, in right ∆ACE,
AC² = AE² + CE² (Pythagoras Theorem)
= (12)² + (5)² = 144 + 25 = 169 = 13²
AC = 13
Distance between their tops = 13 m

Question 4.
Solution:
(c) Area of two similar triangles ABC and DEF are 25 cm² and 36 cm².
Altitude of first ∆ABC is AL = 3.5 m.
Let DM be the altitude of second triangle.

Short-Answer Questions
Question 5.
Solution:
∆ABC ~ ∆DEF
and 2AB = DE

Question 6.
Solution:
In the given figure,

DE || BC, AL ⊥ BC
AD = x cm,
DB = (3x + 4) cm AE = (x + 3) cm, EC = (3x + 19) cm
In ∆ABC, DE || BC

Question 7.
Solution:
Let AB be the ladder, and A is the window.
Length of ladder AB = 10 m and AC = 8 m
Distance between the foot of the ladder from the house = x m

In right ∆ABC,
AB² = AC² + BC²
⇒ (10)² = (8)² + (x)²
⇒x² = (10)² – (8)²
⇒ x² = 100 – 64 = 36
⇒ x² = (6)²
⇒ x = 6
Distance between the foot of ladder and foot of the house = 6m.

Question 8.
Solution:
In ∆ABC, AB = BC = CA = 2a cm
AD ⊥ BC which bisects the base BC at D

BD = DC = \(\frac { 2a }{ 2 }\) = a cm
Now, in right ∆ABD
AB² = AD² + BD²
(2a)² = AD² + a²
⇒ AD² = 4a² – a² = 3a²
AD = √3a² = √3 a
Height of altitude = √3 a cm

Question 9.
Solution:
Let EF = x cm
∆ABC ~ ∆DEF

Question 10.
Solution:
ABCD is a trape∠ium in which AB || DC
AB = 2CD
Diagonals AC and BD intersect each other at O.

Question 11.
Solution:
Let corresponding sides of two similar triangles ABC and DEF are in the ratio 2 : 3.

Question 12.
Solution:
In the given figure,

Question 13.
Solution:
Given : In ∆ABC, AD is the bisector of ∠A which meets BC at D.

Construction : Produce BA and draw CE || DA meeting BA produced at E.
DA || CE
∠3 = ∠1 (corresponding angles)
∠4 = ∠2 (alternate angles)
But ∠3 = ∠4 (AD is the bisector of ∠A)
∠1 = ∠2
AC = AE (Sides opposite to equal angles)
Now, in ∆AEC,
AD || EC

Question 14.
Solution:
Given : In ∆ABC,
AB = BC = CA = a cm

Question 15.
Solution:
In rhombus ABCD, diagonals AC = 24 cm and BD = 10 cm.

The diagonals of a rhombus bisect each other at right angles.
AO = OC = \(\frac { 24 }{ 2 }\) = 12 cm
and BO = OD = \(\frac { 10 }{ 2 }\) = 5 cm
Now, in right ∆AOB,
AB² = AO² + BO² = 12² + 5² = 144 + 25 = 169 = (13)²
AB = 13
Each side of a rhombus = 13 cm

Question 16.
Solution:
Given : ∆ABC ~ ∆DEF

Long-Answer Questions
Question 17.
Solution:
Given: In the given figure, ∆ABC and ∆DBC are on the same base BC but in opposite sides.
AD and BC intersect at O.

Question 18.
Solution:
In the given figure,
XY || AC and XY divides
∆ABC into two regions equal in area

Question 19.
Solution:
In the given figure, ∆ABC is an obtuse triangle, obtuse angle at B.
AD ⊥ CB produced.

To prove : AC² = AB² + BC² + 2BC x BD
Proof: In ∆ADB, ∠D = 90°
AB² = AD² + DB² (Pythagoras Theorem)
⇒ AD² = AB² – DB² ……(i)
Similarly, in right ∆ADC,
AC² = AD² + DC²
= AB² – DB² + (DB + BC)² [From (i)]
= AB² – DB² + DB² + BC² + 2BC x BD
= AB² + BC² + 2 BC x BD

Question 20.
Solution:
In the given figure,
PA, QB and RC is perpendicular to AC.
AP = x, QB = z, RC =y, AB = a and BC = b.

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself 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 9 Solutions Chapter 13 Volume and Surface Area Ex 13B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13A
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13B
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13C
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13D

Question 1.
Solution:
Radius of the base of a cylinder (r) = 5cm.
and height (h) = 21cm

Question 2.
Solution:
Diameter of the base of the cylinder = 28cm
Radius = \(\frac { 1 }{ 2 } \) x 28 = 14 cm
Height (h) = 40cm.

Question 3.
Solution:
Radius of cylinder (r) = 10.5cm
Height (h) = 60cm.

Question 4.
Solution:
Diameter of cylinder = 20cm
Radius (r) = \(\frac { 20 }{ 2 } \) = 10cm

Question 5.
Solution:
Curved surface area of cylinder = 4400 cm²
Circumference of its base = 110 cm

Question 6.
Solution:
The ratio of the radius and height of a cylinder = 2:3
Volume =1617 cm³
Let radius = 2x
and height = 3x.

Question 7.
Solution:
Total surface area of the cylinder = 462 cm²
Curved surface area = \(\frac { 1 }{ 3 } \) x 462 = 154

Question 8.
Solution:
Total surface area of solid
cylinder = 231 cm²

Question 9.
Solution:
Sum of radius and height = 37m.
and total surface area = 1628 m²
Let r be the radius

Question 10.
Solution:
Total surface area = 616 cm²
Curved surface area = \(\frac { 616X1 }{ 2 } \) = 308

Question 11.
Solution:
Volume of gold = 1 cm³
diameter of wire = 0.1 mn

Question 12.
Solution:
Ratio in the radii of two cylinders = 2:3
and ratio in the heights = 5:3
If r1 and r2 and the radii and h1 and h2 are the heights, then

Question 13.
Solution:
Side of square = 12cm
and height = 17.5cm

Question 14.
Solution:
Diameter of cylindrical bucket = 28cm
Radius (r) = \(\frac { 28 }{ 8 } \) = 14cm
Height (h) = 72cm.

Question 15.
Solution:
Length of pipe (l) = 1m = 100cm
diameter of pipe = 3cm.
Inner radius = \(\frac { 3 }{ 2 } \) cm

Question 16.
Solution:
Internal diameter of cylindrical tube = 10.4 cm
Radius (r) = \(\frac { 10.4 }{ 2 } \) = 5.2cm.

Question 17.
Solution:
Length of barrel (h) = 7cm
Diameter = 5mm.

Question 18.
Solution:
Diameter of pencil = 7mm
.’. Radius (R) = \(\frac { 7 }{ 2 } \) mm = \(\frac { 7 }{ 20 } \) cm.
and diameter of graphite in it = 1mm

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 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 9 Solutions Chapter 13 Volume and Surface Area Ex 13A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13A
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13B
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13C
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13D

Question 1.
Solution:
(i) Length of cuboid (l) = 12cm
Breadth (b) = 8cm
and height (h) = 4.5cm

Question 2.
Solution:
Length of closed rectangular cistern (l) = 8m
breadth (b) = 6m
and depth (b) = 2.5m.
(i) .’. Volume of cistern = l.b.h.
= 8 x 6 x 2.5 m³ = 120m³
(ii) Total surface area = 2(lb + bh + hl)
= 2(8 x 6 + 6 x 2.5 + 2.5 x 8) cm²
= 2(48 + 15 + 20)
= 2 x 83 m²
= 166 m² Ans.

Question 3.
Solution:
Length of room (l) = 9m
Breadth (b) = 8m
and height (h) = 6.5m

Question 4.
Solution:
Length of pit (l) = 20m
Breadth (b) = 6m

Question 5.
Solution:
Length of wall (l) = 8m.
Width (b) = 22.5 cm = \(\frac { 225 }{ 10X100 } =\frac { 9 }{ 40 } m\)
and height (h) = 6m.
Volume of wall = l.b.h.

Question 6.
Solution:
Length of wall (l) = 15m.
Width (b) = 30cm = \(\frac { 30 }{ 100 } =\frac { 3 }{ 10 } m\)
Height (h) = 4m

Question 7.
Solution:
Outer length of opened cistern = 1.35m = 135 cm
Breadth = 1.08 m = 108 cm
Depth = 90cm
Thickness of iron = 2.5cm.

Question 8.
Solution:
Depth of river = 2m
width = 45m.
Length of current in 60 minutes = 3km

Question 9.
Solution:
Total cost of box = Rs. 1620
Rate per sq. m = Rs. 30

Question 10.
Solution:
Length of room (l) = 10m
Breadth (b) = 10m
Height (h) = 5m

Question 11.
Solution:
Length of hall (l) = 20m
Breadth (b) = 16m
and height (h) = 4.5m.
Volume of the air inside the hall

Question 12.
Solution:
Length of class room (l) = 10m
Width (b) = 6.4 m
Height (h) = 5m.

Question 13.
Solution:
Volume of cuboid = 1536 m³
Length (l) = 16m
Ratio in breadth and height = 3:2
Let breadth (b) = 3x
their height (h) = 2x

Question 14.
Solution:
Length of cuboid (l) = 14 cm
Breadth (b) = 11 cm .
Let height (h) =x cm
Surface area = 2(lb + bh + hl)

Question 15.
Solution:
(a) Edge of cube (a) = 9m .
(i) volume = a³ = (9)³ m³ = 729 m³
(ii) Lateral surface area = 4a²

Question 16.
Solution:
Total surface area of a cube = 1176 cm²
Let each edge he ‘a’
then 6a² =1176

Question 17.
Solution:
Lateral surface area of a cube = 900 cm²
Let ‘a’ be the edge of the cube

Question 18.
Solution:
Volume of a cube = 512 cm³
Let ‘a’ be its edge, then

Question 19.
Solution:
Edge of first-cube = 3 cm.
Volume = (3)³ = 27 cm³

Question 20.
Solution:
Area of ground = 2 hectares
= 2 x 10000 = 20000 m²
Height of rain falls 5cm = \(\frac { 5 }{ 100 } \)m
∴ Volume of rain water = 20000 x \(\frac { 5 }{ 100 } \) m³
= 1000 m³ Ans.

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A 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.