Maths RS Aggarwal Solutions Class 9

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C

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 13C.

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 base (r) = 35cm
and height (h) = 84cm.

Question 2.
Solution:
Height of cone (h) = 6cm
Slant height (l) = 10cm.

Question 3.
Solution:
Volume of right circular cone = (100 π) cm³
Height (h) = 12cm.
Let r be the radius of the cone

Question 4.
Solution:
Circumference of the base = 44cm

Question 5.
Solution:
Slant height of the cone (l) = 25cm
Curved surface area = 550 cm²
Let r be the radius
πrl = curved surface area

Question 6.
Solution:
Radius.of base (r) = 35cm.
Slant height (l) = 37cm.
We know that

Question 7.
Solution:
Curved surface area = 4070 cm²
Diameter of the base = 70cm

Question 8.
Solution:
Radius of the conical tent = 7m
and height = 24 m.

Question 9.
Solution:
Radius of the first cone (r) = 1.6 cm.
and height (h) = 3.6 cm.

Question 10.
Solution:
Ratio in their heights =1:3
and ratio in their radii = 3:1
Let h1,h2 he their height and r1,r2 be their radii, then

The ratio between their volumes is 3:1
hence proved

Question 11.
Solution:
Diameter of the tent = 105m

Question 12.
Solution:
No. of persons to be s accommodated =11
Area to be required for each person = 4m²

Question 13.
Solution:
Height of the cylindrical bucket (h) = 32cm
Radius (r) = 18cm
Volume of sand filled in it = πr²h
= π x 18 x 18 x 32 cm³
= 10368π cm³
Volume of conical sand = 10368 π cm³
Height of cone = 24 cm

Question 14.
Solution:
Let h be the height and r be the radius of the cylinder and cone.
Curved surface area of cylinder = 2πrh
and curved surface area of cone = πrl

Question 15.
Solution:
Diameter of the pillar = 20cm
Radius (r) = \(\frac { 20 }{ 2 } \) = 10cm

Question 16.
Solution:
Height of the bigger cone (H) = 30cm
By cutting a small cone from it, then volume of smaller cone = \(\frac { 1 }{ 27 } \) of volume of big cone

Let radius and height of the smaller cone be r and h
and radius and height of the bigger cone be R and H.


Hence at the height of 20cm from the base it was cut off. Ans.

Question 17.
Solution:
Height of the cylinder (h) = 10cm.
Radius (r) = 6cm.
Height of the cone = 10cm

Question 18.
Solution:
Diameter of conical vessel = 40cm
Radius (r) = \(\frac { 40 }{ 2 } \) = 20cm
and depth (h) = 24cm.
.’. Volume = \(\frac { 1 }{ 3 } \) πr²h

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C are helpful to complete your math homework.

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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.

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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.

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RS Aggarwal Class 9 Solutions Chapter 12 Geometrical Constructions Ex 12A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 12 Geometrical Constructions Ex 12A.

Question 1.
Solution:
Steps of Constructions :
(i) Draw a line segment AB = 5cm.
(ii) With A as centre and a radius equal to more than half of AB, drawn two arcs one above and other below of AB.

(iii) With centre B, and with same radius, draw two arcs intersecting the previously arcs at C and D respectively.
(iv) Join CD, intersecting AB at P.
Then CD is the perpendicular bisector of AB at the point P.

Question 2.
Solution:
Steps of constructions.
(i) Draw a line segment AB.
(ii) With A as centre and with small radius drawn arc cutting AB at P.
(iii) With P as centre and same radius draw another arc cutting the previous arc at Q and then R.

(iv) Bisect arc QR at S.
(v) Join AS and produce it to X such that ∠ BAX = 90°.
(vi) Now with centres P and S and with a suitable radius, draw two arcs intersecting each other at T.
(vii) Join AT and produced it to C Then ∠BAC = 45°.
(viii) Again with centres P and T and suitable radius draw two arcs intersecting each at D.
(ix) Join AD.
AD is the bisector of ∠ BAC

Question 3.
Solution:
Steps of construction.
(i) Draw a line segment AB.
(ii) With centre A and same radius draw an arc which meets AB at P.
(iii) With centre P and same radius, draw arcs first at Q and then at R.
(iv) With centres Q and R, draw arcs intersecting each other at C intersecting the first arc at T.

(v) Join AC
Then ∠BAC = 90°
(vi) Now with centres P and T and with some suitable radius, draw two arcs intersecting each other at L.
(vii) Join AL and produce it to D.
Then AD is the bisector of ∠ BAC.

Question 4.
Solution:
Steps of construction.
(i) Draw a line segment BC = 5cm.
(ii) With centres B and C and radius
5cm, draw two arcs intersecting each other at A.

(iii) Join AB and AC.
Then ∆ ABC is the required equilateral triangle.

Question 5.
Solution:
We know that altitudes of equilateral triangle are equal and each angle is 60°.
Steps of construction.
(i) Draw a line XY and take a point D on it.
(ii) At D, draw a perpendicular and cut off DA = 5.4cm.

(iii) At A draw angles of 30° on each side of AD which meet XY at B and C respectively.
Then ∆ ABC is the required triangle.

Question 6.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5cm
(ii) With centre B and radius 3.8 cm draw an arc.

(iii) With centre C and radius 2.6 cm draw another arc intersecting the first arc at A.
(iv) Join AB and AC.
Then ∆ ABC is the required triangle.

Question 7.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.7cm.
(ii) At B, draw a ray BX making an angle of 60° with BC.

(iii) At C, draw another ray, CY making an angle of 30° which intersects the ray BX at A ,
Then ∆ ABC is the required triangle On measuring ∠ A, it is 90°.

Question 8.
Solution:
Steps of Construction :
(i) Draw a line segment QR * 5cm.
(ii) With centres Q and R and radius equal to 4.5cm, draw arcs intersecting eachother at P.

(iii) Join PQ and PR.
Then ∆ PQR is the required triangle.

Question 9.
Solution:
We know that in an isosceles triangle, two sides are equal and so their opposite angles are also equal.

Question 10.
Solution:
Steps of constructions :
(i) Draw a line segment BC = 4.5cm.
(ii) At B, draw a ray BX making an angle of 90° with BC.

(iii) With centre C and radius 5.3 cm, draw an arc intersecting BX at A.
(iv) Join AC.
Then ∆ ABC is the required right angled triangle.

Question 11.
Solution:
Steps of constructions :
(i) Draw a line XY.
(ii) Take a point D on XY.

(iii) Draw a perpendicular at D and cut off DA = 4.8 cm
(iv) At A, draw a line LM parallel to XY.
(v) At A, draw an angle of 30° with LM on one side and an angle of 60° with LM on other side meeting XY at B and C respectively
Then ∆ ABC is the required triangle.

Question 12.
Solution:
Steps of constructions :
(i) Draw a line segment EF = 12cm.
(ii) At E, draw a ray EX making an acute angle with EF.

(iii)From EX,cut off 3+2+4=9 equal parts.
(iv) Join E9 F.
(v) From E5 and E3, draw lines parallel to E9 F meeting EF at C and B respectively.
(vi) With centre B and radius EB and with centre C and radius CF, draw arcs intersecting eachother at A.
(vii) Join AB and AC.
Then ∆ ABC is the required triangle.

Question 13.
Solution:
Steps of constructions :

(i) Draw a line segment BC = 4.5cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BD = 8cm.
(iii) Join DC.
(iv) Draw the perpendicular bisector of BD which intersects BX at A.
(v) Join AC.
Then ∆ ABC is the required triangle.

Question 14.
Solution:
Steps of Constructions :
(i) Draw a line segment BC = 5.2 cm.
(ii) At B draw a ray BX making an angle of 30°.

(iii) From BX, cut off BD = 3.5cm.
(iv) Join DC.
(v) Draw perpendicular bisector of DC which intersects BX at A.
(vi) Join AC.
Then ∆ ABC is the required triangle.

Hope given RS Aggarwal Class 9 Solutions Chapter 12 Geometrical Constructions Ex 12A 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 11 Circle Ex 11C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11C.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11A
• RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11B
• RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11C

Question 1.
Solution:
In cyclic quad. ABCD, ∠ DBC = 60° and ∠BAC = 40°
∴∠ CAD and ∠ CBD are in the same segment of the circle.
∴∠ CAD = ∠ CBD or ∠ DBC
=> ∠ CAD = 60°
∴∠BAD = ∠BAC + ∠CAD
= 40° + 60° = 100°
But in cyclic quad. ABCD,
∠BAD + ∠BCD = 180°
(Sum of opposite angles)
=> 100° + ∠BCD = 180°
=> ∠BCD = 180° – 100°
∴ ∠ BCD = 80°
Hence (i) ∠BCD = 80° and
(ii) ∠CAD = 60° Ans.

Question 2.
Solution:
In the figure, POQ is diameter, PQRS is a cyclic quad, and ∠ PSR =150° In cyclic quad. PQRS.
∠ PSR + ∠PQR = 180°
(Sum of opposite angles)
=> 150° + ∠PQR = 180°
=> ∠PQR = 180°- 150° = 30°
=> ∠PQR =180° – 150° = 30°
Now in ∆ PQR,
∴∠ PRQ = 90° (Angle in a semicircle)
∴∠ RPQ + ∠PQR = 90°
=> ∠RPQ + 30° = 90°
=> ∠RPQ = 90° – 30° = 60° Ans.

Question 3.
Solution:
In cyclic quad. ABCD,
AB || DC and ∠BAD = 100°
∠ ADC = ∠BAD =180°
(co-interior angles)

=> ∠ ADC + 100° = 180°
=> ∠ADC = 180° – 100° = 80°
∴ ABCD is a cyclic quadrilateral.
∴ ∠BAD + ∠BCD = 180°
=> 100° + ∠ BCD = 180°
=> ∠BCD = 180° – 100°
=> ∠BCD = 80°
Similarly ∠ABC + ∠ADC = 180°
=> ∠ABC + 80° = 180°
=> ∠ABC = 180° – 80° = 100°
Hence (i) ∠BCD = 80° (ii) ∠ADC = 80° and (iii) ∠ABC = 100° Ans.

Question 4.
Solution:
O is the centre of the circle and arc ABC subtends an angle of 130° at the centre i.e. ∠AOC = 130°. AB is produced to P
Reflex ∠AOC = 360° – 130° = 230°
Now, arc AC subtends reflex ∠ AOC at the centre and ∠ ABC at the remaining out of the circle.

Question 5.
Solution:
In the figure, ABCD is a cyclic quadrilateral in which BA is produced to F and AE is drawn parallel to CD.
∠ABC = 92° and ∠FAE = 20°
ABCD is a cyclic quadrilateral.

Question 6.
Solution:
In the figure, BD = DC and ∠CBD = 30°
In ∆ BCD,
BD = DC (given)
∠ BCD = ∠ CBD
(Angles opposite to equal sides)
= 30°
But ∠BCD + ∠CBD + ∠BDC = 180° (Angles of a triangle)
=> 30°+ 30°+ ∠BDC = 180°
=> 60°+ ∠BDC = 180°
=> ∠ BDC =180° – 60° = 120°
But ABDC is a cyclic quadrilateral
∠BAC + ∠BDC = 180°
=> ∠BAC + 120°= 180°
=> ∠ BAC = 180° – 120° = 60°
Hence ∠ BAC = 60° Ans.

Question 7.
Solution:
(i) Arc ABC subtends ∠ AOC at the centre , and ∠ ADC at the remaining part of the circle.
∠ AOC = 2 ∠ ADC

Question 8.
Solution:
In the figure, ABC is an equilateral triangle inscribed is a circle
Each angle is of 60°.
∠ BAC = ∠ BDC
(Angles in the same segment)
∠BDC = 60°
BECD is a cyclic quadrilateral.
∠BDC + ∠BEC = 180°
(opposite angles of cyclic quad.)
=> 60°+ ∠BEC = 180°
=> ∠BEC = 180° – 60°= 120°
Hence ∠BDC = 60° and ∠BEC = 120° Ans.

Question 9.
Solution:
ABCD is a cyclic quadrilateral.
∠BCD + ∠BAD = 180°
(opposite angles of a cyclic quad.)
=> 100°+ ∠BAD = 180°
so ∠BAD = 180° – 100° = 80°
Now in ∆ ABD,
∠BAD + ∠ABD + ∠ADB = 180° (Angles of a triangle)

=> 80° + 50° + ∠ADB = 180°
=> 130°+ ∠ADB = 180°
=> ∠ADB = 180° – 130° = 50°
Hence, ∠ADB = 50° Ans.

Question 10.
Solution:
Arc BAD subtends ∠ BOD at the centre and ∠BCD at the remaining part of the circle.

Question 11.
Solution:
In ∆ OAB,
OA = OB (radii of the same circle)
∠OAB = ∠OBA = 50°
and Ext ∠BOD = ∠OAB + ∠OBA
=>x° = 50° + 50° – 100°
ABCD is a cyclic quadrilateral
∠BAD + ∠BCD = 180°
(opposite angles of a cyclic quad.)
=> 50°+ y° = 180°
=> y° = 180° – 50° = 130°
Hence x = 100° and y = 130° Ans.

Question 12.
Solution:
Sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively.
∠CBF = 130°, ∠CDE = x.
∠CBF + ∠CBA = 180° (Linear pair)
=> 130°+ ∠CBA = 180°
=> ∠CBA = 180° – 130° = 50°
But Ext. ∠ CDE = Interior opp. ∠ CBA (In cyclic quad. ABCD)
=> x = 50° Ans.

Question 13.
Solution:
In a circle with centre O AB is its diameter and DO || CB is drawn. ∠BCD = 120°
To Find : (i) ∠BAD (ii) ABD
(iii) ∠CBD (iv) ∠ADC
(v) Show that ∆ AOD is an equilateral triangle.
(i) ABCD is a cyclic quadrilateral.
∠BCD + ∠BAD = 180°
120° + ∠BAD = 180°

Question 14.
Solution:
AB = 6cm, BP = 2cm, DP = 2.5cm
Let CD = xcm
Two chords AB and CD

Question 15.
Solution:
O is the centre of the circle
∠ AOD = 140° and ∠CAB = 50°
BD is joined.
(i) ABDC is a cyclic quadrilateral.

Question 16.
Solution:
Given : ABCD is a cyclic quadrilateral whose sides AB and DC are produced to meet each other at E.
To Prove : ∆ EBC ~ ∆ EDA
Proof : In ∆ EBC and ∆ EDA
∠ E = ∠ E (common)
∠ECB = ∠EAD
{Exterior angle of a cyclic quad, is equal to its interior opposite angle}
and ∠ EBC = ∠EDA
∆ EBC ~ ∆ EDA (AAS axiom)
Hence proved

Question 17.
Solution:
Solution Given : In an isosceles ∆ ABC, AB = AC
A circle is drawn x in such a way that it passes through B and C and intersects AB and AC at D and E respectively.
DE is joined.
To Prove : DE || BC
Proof : In ∆ ABC,
AB = AC (given)
∠ B = ∠ C (angles opposite to equal sides)
But ∠ ADE = ∠ C (Ext. angle of a cyclic quad, is equal E to its interior opposite angle)
∠ADE = ∠B
But, these are corresponding angles
DE || BC.
Hence proved.

Question 18.
Solution:
Given : ∆ ABC is an isosceles triangle in which AB = AC.
D and E are midpoints of AB and AC respectively.
DE is joined.
To Prove : D, B, C, E are concyclic.
Proof: D and E are midpoints of sides AB and AC respectively.
DE || BC
In ∆ ABC, AB = AC
∠B = ∠C
But ∠ ADE = ∠ B (alternate angles)
∠ADE =∠C
But ∠ADE is exterior angle of quad. DBCE which is equal to its interior opposite angle C.
DBCE is a cyclic quadrilateral.
Hence D, B, C, E are con cyclic.
Hence proved.

Question 19.
Solution:
Given : ABCD is a cyclic quadrilateral whose perpendicular bisectors l, m, n, p of the side are drawn

To prove : l, m, n and p are concurrent.
Proof : The sides AB, BC, CD and DA are the chords of the circle passing through the vertices’s of quad. A, B, C and D. and perpendicular bisectors of a chord always passes through the centre of the circle.
l,m, n and p which are the perpendicular bisectors of the sides of cyclic quadrilateral will pass through O, the same point Hence, l, m, n and p are concurrent.
Hence proved.

Question 20.
Solution:
Given : ABCD is a rhombus and four circles are drawn on the sides AB, BC, CD and DA as diameters. Diagonal AC and BD intersect each other at O.

Question 21.
Solution:
Given: ABCD is a rectangle whose diagonals AC and BD intersect each other at O.
To prove : O is the centre of the circle passing through A, B, C and D

Question 22.
Solution:
Construction.
(i) Let A, B and C are three points
(ii) With A as centre and BC as radius draw an arc
(iii) With centre C, and radius AB, draw another arc which intersects the first arc at D.
D is the required point.
Join BD and CD, AC and BA and CB

BC = BC (common)
AC = BD (const.)
AB = DC
∴ ∆ ABC ≅ ∆ DBC (SSS axiom)
∴ ∠BAC ≅ ∠BDC (c.p.c.t.)
But these are angles on the same sides of BC
Hence these are angles in the same segment of a circle
A, B, C, D are concyclic Hence D lies on the circle passing througtvA, B and C.
Hence proved.

Question 23.
Solution:
Given : ABCD is a cylic quadrilateral (∠B – ∠D) = 60°
To prove : The small angle of the quad, is 60°

Question 24.

Solution:
Given : ABCD is a quadrilateral in which AD = BC and ∠ ADC = ∠BCD
To prove : A, B, C and D lie on a circle

Question 25.
Solution:
Given : In the figure, two circles intersect each other at D and C
∠BAD = 75°, ∠DCF = x° and ∠DEF = y°

Question 26.
Solution:
Given : ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angle.

Question 27.
Solution:
In a circle, two chords AB and CD intersect each other at E when produced.
AD and BC are joined.

Question 28.
Solution:
Given : Two parallel chords AB and CD of a circle BD and AC are joined and produced to meet at E.

Question 29.
Solution:
Given : In a circle with centre O, AB is its diameter. ADE and CBE are lines meeting at E such that ∠BAD = 35° and ∠BED = 25°.
To Find : (i) ∠DBC (ii) ∠DCB (iii) ∠BDC
Solution. Join BD and AC,

Hope given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11C are helpful to complete your math homework.

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