RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2

Other Exercises

Question 1.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? [NCERT]
Solution:
Length of water tank (l) = 6 m
Breadth (b) = 5 m
and depth (h) = 4.5 m
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q1.1
∴ Volume of water in it = lbh
= 6 x 5 x 4.5 m3 = 135 m3
Capacity of water in litres = 135 x 1000 litres (1 m3 = 1000 l)
= 135000 litres

Question 2.
A cubical vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid? [NCERT]
Solution:
Length of vessal (l) = 10 m
Breadth (b) = 8 m
Volume = 380 m3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q2.1

Question 3.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m3. [NCERT]
Solution:
Length of pit (l) = 8m
Width (b) = 6 m
and depth (h) = 3 m
∴ Volume of earth digout = lbh
= 8 x 6 x 3 = 144 m3
Cost of digging the pit at the rate of ₹30 per m3
= 144 x 30 = ₹4320

Question 4.
If the areas of three adjacent faces of a cuboid are 8 cm2, 18 cm2 and 25 cm2. Find the volume of the cuboid.
Solution:
Let x, y, z be the three adjacent faces of the cuboid, then
x = 8 cm2, y = 18 cm2, z = 25 cm2
and let l, b, h are the dimensions of the cuboid, then
x = lb = 8 cm2
y = bh = 18 cm2
z = hl = 25 cm2
∴ Volume = lbh
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q4.1

Question 5.
The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. dm. Find its dimensions.
Solution:
Let breadth of a room (b) = x
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q5.1

Question 6.
Three metal cubes with edges 6 cm, 8 cm and 10 cm respectively are melted together and formed into a single cube. Find the volume, surface area and diagonal of the new cube.
Solution:
Edge of first cube = 6 cm
Edge of second cube = 8 cm
and edge of third cube = 10 cm
∴ Volume of 3 cubes = (6)3 + (8)3 + (10)3 cm3
= 216 + 512 + 1000 cm3
= 1728 cm3
∴ Edge of so formed cube
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q6.1

Question 7.
Two cubes, each of volume 512 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of each volume = 512 cm3
∴ Side (edge) = \(\sqrt [ 3 ]{ 512 }\)
=\(\sqrt [ 3 ]{ { 8 }^{ 3 } }\)  = 8 cm
Now by joining two cubes, then Length of so formed cuboid (l)
= 8 + 8 = 16 cm
Breadth (b) = 8 cm
and height (h) = 8 cm
∴ Surface area = 2(lb + bh + hl)
= 2[16 x 8 + 8 x 8 + 8 x 16] cm2
= 2[128 + 64 + 128] cm2
= 2 x 320 = 640 cm2

Question 8.
A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.
Solution:
Edge of metal cube = 12 cm
∴ Its volume = (Edge)3 = (12)3 cm33
= 1728 cm3
It is melted and form 3 cubes
Edge of one smaller cube = 6 cm
and edge of second smaller cube = 8 cm
∴ Volume of two smaller cubes = (6)3 + (8)3 cm3
= 216 + 512 cm3 = 728 cm3
∴ Volume of third smaller cube = 1728 – 728 = 1000 cm3
∴ Edge of the third cube = \(\sqrt [ 3 ]{ 1000 }\)
= \(\sqrt [ 3 ]{ (10)3 }\)  cm = 10 cm

Question 9.
The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall, if each person requires 150 m3 of air?
Solution:
Length of cinema hall (l) = 100 m
Breadth (b) = 50 m
and height (h) = 18 m
∴ Volume of air in it = lbh
= 100 x 50 x 18 m= 90000 m3
Air required for one person = 150 m3
∴ Number of persons in the hall = \(\frac { 90000 }{ 150 }\) = 600 persons

Question 10.
Given that 1 cubic cm of marble weighs 0.25 kg, the weight of marble block 28 cm in width and 5 cm thick is 112 kg. Find the length of the block.
Solution:
Weight of 1 cm3 = 0.25 kg
Breadth of the block (b) = 28 cm
Thickness (h) = 5 cm
and total weight of the block = 112 kg
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q10.1

Question 11.
A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 25 cm, 18 cm and 15 cm respectively. How much cubic cm of a liquid can be placed in it? Also, find the volume of the wood used in it.
Solution:
Outer length of the closed wooden box (l) = 25 cm
Breadth (b) = 18 cm
and height (h) = 15 cm
Width of wood = 2 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q11.1
∴ Inner length = 25 – 2×2 = 25- 4 = 21cm
Breadth =18- 2×2 = 18-4 = 14 cm
and height =15- 2×2 = 15- 4=11 cm
Now outer volume = 25 x 18 x 15 cm3 = 6750 cm3
and inner volume = 21 x 14 x 11 cm3 = 3234 cm3
(i) Inner volume = 3234 cm3
(ii) Volume of wood = 6750 – 3234 = 3516 cm3

Question 12.
The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm x 3 cm x 0.75 cm can be put in this box?
Solution:
External length of a closed wooden box (L) = 48 cm
Width (B) = 36 cm
and height (H) = 30 cm
Thickness of wood = 1.5 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q12.1
∴ Internal length (l) = 48 – 2 x 1.5 cm = 48 – 3 = 45 cm
Width (b) = 36 – 2 x 1.5 cm
= 36 – 3 = 33 cm
and height (h) = 30 – 2 x 1.5 cm
= 30 – 3 = 27 cm
Now volume of internal box
= lbh = 45 x 33 x 27 cm3
Volume of one bricks = 6 x 3 x 0.75 cm3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q12.2

Question 13.
A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm, find the rise in water level in the vessel.
Solution:
Edge of a cube = 9 cm
Volume of cube = (9)3 cm3
= 729 cm3
Now length of vessel (l) = 15 cm
and breadth (b) = 12 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q13.1

Question 14.
A field is 200 m long and 150 m broad. There is a plot, 50 m long and 40 m broad, near the field. The plot is dug 7 m deep and the earth taken out is spread evenly on the field. By how many metres is the level of the field raised? Give the answer to the second place of decimal.
Solution:
Length of a field (l) = 200 m
Breadth (b) = 150 m
Length of plot = 50 m
and breadth = 40 m
Depth of plot = 7 m
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q14.1
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q14.2

Question 15.
A field is in the form of a rectangle of length 18 m and width 15 m. A pit, 7.5 m long, 6 m broad and 0.8 m deep, is dug in a comer of the field and the earth taken out is spread- over the remaining area of the field. Find out the extent to which the level of the field has been raised.
Solution:
Length of a field (L) = 18m
and width (B) = 15 m
Length of pit (l) = 7.5 m
Breadth (b) = 6 m
and depth (h) = 0.8 m
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q15.1
∴ Volume of earth dugout = lbh
= 7.5 x 6 x 0.8 m3
= 45 x 0.8 = \(\frac { 45 x 4 }{ 5 }\) = 36 m3
Total area of the field = L x B
= 18 x 15 = 270 m2
and area of pit = lb = 7.5 x 6 = 45 m2
∴ Remaining area of the field excluding pit
= 270 – 45 = 225 m2
Let by spreading the earth on the remaining part of the field, the height = h
= 225 x h = 36
⇒ h = \(\frac { 36 }{ 225 }\) = \(\frac { 4 }{ 25 }\)= 0.16 m = 16 cm
∴ Level of field raised = 16 cm

Question 16.
A village having a population of 4000 requires 150 litres of water per head per day. It has a tank measuring 20 m x 15m x 6 m. For how many days will the water of this tank last? [NCERT]
Solution:
Population of a village = 4000
Water required per head per day = 150 litres
∴ Total water required = 4000 x 150 litres = 600000 litres
Dimensions of a tank = 20mx 15mx6m
∴ Volume of tank = 20 x 15 x 6 m3 = 1800 m3
Capacity of water in litres = 1800 x 1000 litres (1 m3 = 1000 litres)
= 1800000 litres
The water will last for = \(\frac { 1800000 }{ 600000 }\) = 3 days

Question 17.
A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in the figure. If the edge of each cube is 3 cm, find the volume of the structure built by the child. [NCERT]
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q17.1
Solution:
No. of cubes at the given structure = 1+ 2 + 3+ 4 + 5 = 15
Edge of one cube = 3 cm
∴ Volume of one cube = (3)3 = 3 x 3 x 3 cm3 = 27 cm3
∴Volume of the structure = 27 x 15 cm3 = 405 cm3

Question 18.
A godown measures 40 m x 25 m x 10 m. Find the maximum number of wooden crates each measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown. [NCERT]
Solution:
Length of godown (L) = 40 m
Breadth (B) = 25 m
and height (H) = 10 m
∴ Volume of godown = LBH
= 40 x 25 x 10 = 10000 m3
Dimension of one wooden crates = 1.5 m x 1.25 m x 0.5 m
∴Volume of one crate = 1.5 x 1.25 x 0.5 m3 = 0.9375 m3
∴ Number of crates to be stored in the
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q18.1

Question 19.
A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm x 12 cm x 8 cm, how many bricks would be required? [NCERT]
Solution:
Length of wall (L) = 10 m = 1000 cm
Height (H) = 4 m = 400 cm
Thickness (B) = 24 cm = 24 cm
∴ Volume of wall = LBH = 1000 x 24 x 400 cm3 = 9600000 cm3
Dimensions of one brick = 24 cm x 12 cm x 8 cm = 2304 cm3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q19.1

Question 20.
If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q20.1
Solution:
a, b, c are the dimensions of a cuboid S is the surface area and V is the volume
∴ V = abc and S = 2(lb + bc + ca)
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q20.2

Question 21.
The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that V= xyz.
Solution:
Let a, b, c are the dimensions of a cuboid then,
x = ab, y = bc, z = ca
and V = abc
Now L.H.S. = V2
= (abc)= a2b2c2
= ab.bc.ca = xyz = R.H.S.
Hence V2 = xyz

Question 22.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute? [NCERT]
Solution:
Speed of water in a river = 2 km/hr
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q22.1

Question 23.
Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km per hour. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired?
Solution:
Width of canal (b) = 30 dm = 3 m
Depth (h) = 12 dm = 1.2 m
Speed of water = 100 km/hr
Length of water flow in 30 minutes = \(\frac { 1 }{ 2 }\) hr
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q23.1

Question 24.
Half cubic metre of gold-sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold- sheet.
Solution:
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q24.1

Question 25.
How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cubic cm of iron weighs 15 g, find the weight of the empty box in kg.
Solution:
External length of open box (L) = 36 cm
Breadth (B) = 25 cm
and Height (H) = 16.5 cm
Width of iron sheet used = 1.5 cm
∴ Inner length (l) = 36 – 1.5 x 2 = 36 – 3 = 33 cm
Breadth (b) = 25 – 2 x 1.5 = 25 – 3 = 22 cm
and Height (h) = 16.5 – 1.5 = 15 cm
∴ Volume of the iron used = Outer volume – Inner volume
= 36 x 25 x 16.5 – 33 x 22 x 15
= 14850 – 10890 = 3960 cm3
Weight of 1 cm3 = 15 g
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q25.1

Question 26.
A rectangular container, whose base is a square of side 5 cm, stands on a horizontal table, and holds water upto 1 cm from the top. When a cube is placed in the water it is completely submerged, the water rises to the top and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge.
Solution:
Base of the container = 5 cm x 5 cm
Level of water upto 1 cm from the top After placing a cube in it, the water rises to the top and 2 cubic cm of water overflows,
(i) ∴ Volume of water = 5 x 5 x 1 + 2 = 25 + 2 = 27 cm3
∴ Volume of cube = 27 cm3
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q26.1

Question 27.
A rectangular tank is 80 m long and 25 m broad. Water-flows into it through a pipe whose cross-section is 25 cm2, at the rate of 16 km per hour. How much the level of the water rises in the tank in 45 minutes.
Solution:
Length of tank (l) = 80 m
Breadth (b) = 25 m
Area of cross section of the month of pipe = 25 cm2
and speed of water-flow =16 km/h
∴ Volume of water is 45 minutes
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q27.1

Question 28.
Water in a rectangular reservoir having base 80 m by 60 m is 6.5 m deep. In what time can the water be emptied by a pipe of which the cross-section is a square of side 20 cm, if the water runs through the pipe at the rate of 15 km/hr.
Solution:
Length of reservoir (l) = 80 m
Breadth (b) = 60 m
and depth (h) = 6.5 m
∴ Volume of water in it = lbh = 80 x 60 x 6.5 m3 = 31200 m3
Area of cross-section of the month of pipe = 20 x 20 = 400 cm2
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 Q28.1

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RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1

Other Exercises

Question 1.
Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm.
Solution:
Length of cuboid (l) = 80 cm
Breadth (b) = 40 cm
Height (h) = 20 cm
(i) ∴ Lateral surface area = 2h(l + b)
= 2 x 20(80 + 40) cm²
= 40 x 120 = 4800 cm²
(ii) Total surface area = 2(lb + bh + hl)
= 2(80 x 40 + 40 x 20 + 20 x 80) cm²
= 2(3200 + 800 + 1600) cm²
= 5600 x 2 = 11200 cm²

Question 2.
Find the lateral surface area and total surface area of a cube of edge 10 cm.
Solution:
Edge of cube (a) = 10 cm
(i) ∴ Lateral surface area = 4a²
= 4 x (10)² = 4 x 100 cm²= 400 cm²
(ii) Total surface area = 6a² = 6 x(10)² cm²
= 6 x 100 = 600 cm²

Question 3.
Find the ratio of the total surface area and lateral surface area of a cube.
Solution:
Let a be the edge of the cube, then Total surface area = 6a2²
and lateral surface area = 4a²
Now ratio between total surface area and lateral surface area = 6a² : 4a² = 3 : 2

Question 4.
Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?   [NCERT]
Solution:
Length of box (l) = 80 cm
Breadth (b) = 40 cm
and height (h) = 20 cm
∴ Total surface area = 2(lb + bh + hl)
= 2[80 x 40 + 40 x 20 + 20 x 80] cm²
= 2[3200 + 800 + 1600] cm² = 2 x 5600 = 11200 cm²
Size of paper sheet = 40 cm
∴ Area of one sheet = (40 cm)² = 1600 cm²
∴ No. of sheets required for the box = 11200 = 1600 = 7 sheets

Question 5.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m².
Solution:
Length of a room (l) = 5m
Breadth (b) = 4 m
and height (h) = 3 m
∴ Area of 4 walls = 2(l + b) x h
= 2(5 + 4) x 3 = 6 x 9 = 54 m²
and area of ceiling = l x b = 5 x 4 = 20 m²
∴ Total area = 54 + 20 = 74 m2
Rate of white washing = 7.50 per m²
∴ Total cost = ₹74 x 7.50 = ₹555

Question 6.
Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Let each side of a cube = a cm
Then surface area = 6a² cm²
and surface area of 3 such cubes = 3 x 6a² = 18a² cm²
By placing three cubes side by side we get a cuboid whose,
Length (l) = a x 3 = 3a
Breadth (b) = a
Height (h) = a
∴ Total surface area = 2(lb + bh + hf)
= 2[3a x a+a x a+a x 3a] cm²
= 2[3a² + a² + 3a²] = 14 a²
∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9

Question 7.
A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes.
Solution:
Side of cube = 4 cm
But cutting into 1 cm cubes, we get = 4 x 4 x 4 = 64
Now surface area of one cube = 6 x (1)²
= 6 x 1=6 cm²
and surface area of 64 cubes = 6 x 64 cm² = 384 cm²

Question 8.
The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.
Solution:
Let h be the height of the room
Length (l) = 18 m
and width (b) = 12 m
Now surface area of floor and roof = 2 x lb = 2 x 18 x 12 m²
= 432 m²
and surface area of 4-walls = 2h (l + b)
= 2h(18 + 12) = 2 x 30h m² = 60h m²
∵ The surface are of 4-walls and area of floor and roof are equal
∴ 60h = 432
⇒ h = \(\frac { 432 }{ 60 }\) = \(\frac { 72 }{ 10 }\) m
∴ Height = 7.2m

Question 9.
Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the tiles, if the cost of tiles is ₹360 per dozen. [NCERT]
Solution:
Edge of cubical tank = 1.5 m
∴ Area of 4 walls = 4 (side)² = 4(1.5)² m² = 4 x 225 = 9 m²
Area of floor = (1.5)² = 2.25 m²
∴ Total surface area = 9 + 2.25 = 11.25 m²
Edge of square tile = 25 m = 0.25 m²
∴ Area of 1 tile = (0.25)2 = .0625 m²
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q9.1

Question 10.
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.
Solution:
Let edge of a cube = a
Total surface area = 6a2
By increasing edge at 50%,
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q10.1

Question 11.
A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of ₹5 per metre sheet, sheet being 2 m wide.
Solution:
Length of iron tank (l) = 12 m
Breadth (b) = 9 m
Depth (h) = 4 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q11.1

Question 12.
Ravish wanted to make a temporary shelter for his car by making a box-like structure with tarpaulin that covers all the four sides and the top of tire car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make (he shelter of height 2.5 m with base dimensions 4 m x 3 m? [NCERT]
Solution:
Length of base (l) = 4m
Breadth (b) = 3 m
Height (h) = 2.5 m
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q12.1

Question 13.
An open box is made of wood 3 cm thick. Its external length, breadth and height are 1.48 m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface of ₹50 per sq. metre.
Solution:
Length of open wood box (L) = 1.48 m = 148 cm
Breadth (B) = 1.16 m = 116 cm
and height (H) = 8.3 dm = 83 cm
Thickness of wood = 3 cm
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q13.1
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q13.2

Question 14.
The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at ₹3.50 per square metre.
Solution:
Length of room (l) = 12.5 m
Breadth (b) = 9 m
and height (h) = 7 m
∴ Total area of walls = 2h(l + b)
= 2 x 7[12.5 + 9] = 14 x 21.5 m² = 301 m²
Area of 2 doors of 2.5 m x 1.2 m
= 2 x 2.5 x 1.2 m² = 6 m²
and area of 4 window of 1.5 m x 1 m
= 4 x 1.5 x 1 = 6 m²
∴ Remaining area of walls = 301 – (6 + 6)
= 301 – 12 = 289 m²
Rate of painting the walls = ₹3.50 per m²
∴ Total cost = 289 x 3.50 = ₹1011.50

Question 15.
The paint in a certain container is sufficient to paint on area equal to 9.375 m2. How many bricks of dimension 22.5 cm x 10 cm x 7.5 cm can be painted out of this container? [NCERT]
Solution:
Area of place for painting = 9.375 m²
Dimension of one brick = 22.5 cm x 10 cm x 7.5 cm
∴ Surface area of one bricks = 2 (lb + bh + hl)
= 2[22.5 x 10 + 10 x 7.5 + 7.5 x 22.5] cm2
= 2[225 + 75 + 168.75]
= 2 x 468.75 cm² = 937.5 cm²
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q15.1

Question 16.
The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of ₹8 and ₹9.50 per m2 is ₹1248. Find the dimensions of the box.
Solution:
Ratio in the dimensions of a cuboidal box = 2 : 3 : 4
Let length (l) = 4x
Breadth (b) = 3.v
and height (h) = 2x
∴ Total surface area = 2 [lb + bh + hl]
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q16.1

Question 17.
The cost of preparing the walls of a room 12 m long at the rate of ₹1.35 per square metre is ₹340.20 and the cost of matting the floor at 85 paise per square metre is ₹91.80. Find the height of the room.
Solution:
Cost of preparing walls of a room = ₹340.20
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q17.1
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q17.2

Question 18.
The length and breadth of a hall are in the ratio 4 : 3 and its height is 5.5 metres. The cost of decorating its walls (including doors and windows) at ₹6.60 per square metre is ₹5082. Find the length and breadth of the room
Solution:
Ratio in length and breadth = 4:3
and height (h) = 5.5 m
Cost of decorating the walls of a room including doors and windows = ₹5082
Rate = ₹6.60 per m²
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q18.1

Question 19.
A wooden bookshelf has external dimensions as follows: Height =110 cm, Depth = 25 cm, Breadth = 85 cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2. Find the total expenses required for polishing and painting the surface of the bookshelf.  [NCERT]
Solution:
Length (l) = 85 cm
Breadth (b) = 25 cm
and height (h) = 110 cm
Thickness of plank = 5 cm
Surface area to be polished
= [(100 x 85) + 2 (110 x 25) + 2 (85 x 25) + 2 (110 x 5) + 4 (75 x 5)]
= (9350 + 5500 + 4250 + 1100 + 1500) cm² = 21700 cm²
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q19.1
RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 Q19.2

 

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RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
The sides of a triangle are 16 cm, 30 cm, 34 cm. Its area is
(a) 225 cm²
(b) 225\(\sqrt { 3 } \) cm²
(c) 225\(\sqrt { 2 } \) cm²
(d) 240 cm²
Solution:
Sides of triangle and 16 cm, 30 cm, 34 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q1.1

Question 2.
The base of an isosceles right triangle is 30 cm. Its area is
(a) 225 cm²
(b) 225\(\sqrt { 3 } \) cm²
(c) 225\(\sqrt { 2 } \) cm²
(d) 450 cm²
Solution:
Base of isosceles triangle ∆ABC = 30cm
Let each of equal sides = x
Then AB = AC = x
Now in right ∆ABC,
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q2.1

Question 3.
The sides of a triangle are 7cm, 9cm and 14cm. Its area is
(a) 12\(\sqrt { 5 } \) cm²
(b) 12\(\sqrt { 3 } \) cm²
(c) 24 \(\sqrt { 5 } \) cm²
(d) 63 cm²
Solution:
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q3.1

Question 4.
The sides of a triangular field are 325 m, 300 m and 125 m. Its area is
(a) 18750 m²
(b) 37500 m²
(c) 97500 m²
(d) 48750 m²
Solution:
Sides of a triangular field are 325m, 300m, 125m
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q4.1

Question 5.
The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is
(a) 20 cm
(b) 30 cm
(c) 40 cm
(d) 50 cm
Solution:
The sides of a triangle are 50 cm, 78 cm, 112cm
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q5.1

Question 6.
The sides of a triangle are 11m, 60m and 61m. Altitude to the smallest side is
(a) 11m
(b) 66 m
(c) 50 m
(d) 60 m
Solution:
Sides of a triangle are 11m, 60m and 61m
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q6.1

Question 7.
The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q7.1
Solution:
Sides of a triangle are 11 cm, 15 cm, 16 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q7.2

Question 8.
The base and hypotenuse of a right triangle are respectively 5cm and 13cm long. Its area is
(a) 25 cm²
(b) 28 cm²
(c) 30 cm²
(d) 40 cm²
Solution:
In a right triangle base = 5 cm
base hypotenuse = 13 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q8.1

Question 9.
The length of each side of an equilateral triangle of area 4 \(\sqrt { 3 } \) cm², is
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q9.1
Solution:
Area of an equilateral triangle = 4\(\sqrt { 3 } \) cm²
Let each side be = a
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q9.2

Question 10.
If the area of an isosceles right triangle is 8cm, what is the perimeter of the triangle.
(a) 8 + \(\sqrt { 2 } \) cm²
(b) 8 + 4\(\sqrt { 2 } \) cm²
(c) 4 + 8\(\sqrt { 2 } \) cm²
(b) 12\(\sqrt { 2 } \) cm²
Solution:
Let base = x
ABC an isosceles right triangle, which has 2 sides same
⇒ Height = x
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q10.1
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q10.2

Question 11.
The length of the sides of ∆ABC are consecutive integers. If ∆ABC has the same perimeter as an equilateral triangle with a side of length 9cm, what is the length of the shortest side of ∆ABC?
(a) 4
(b) 6
(c) 8
(d) 10
Solution:
Side of an equilateral triangle = 9 cm
Its perimeter = 3 x 9 = 27 cm
Now perimeter of ∆ABC = 27 cm
and let its sides be x, x + 1, x +2
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q11.1

Question 12.
In the figure, the ratio of AD to DC is 3 to 2. If the area of ∆ABC is 40cm2, what is the area of ∆BDC?
(a) 16 cm²
(b) 24 cm²
(c) 30 cm²
(d) 36 cm²
Solution:
Ratio in AD : DC = 3:2
and area ∆ABC = 40 cm²
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q12.1
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q12.2

Question 13.
If the length of a median of an equilateral triangle is x cm, then its area is
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q13.1
Solution:
∵ The median of an equilateral triangle is the perpendicular to the base also,
∴ Let side of the triangle = a
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q13.2

Question 14.
If every side of a triangle is doubled, then increase in the area of the triangle is
(a) 100\(\sqrt { 2 } \) %
(b) 200%
(c) 300%
(d) 400%
Solution:
Let the sides of the original triangle be a, b, c
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q14.1

Question 15.
A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 1272 cm, then area of the triangle is
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q15.1
Solution:
A square and an equilateral triangle have equal perimeter
RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS Q15.2

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RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS

RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS

Other Exercises

Question 1.
Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.
Solution:
In ∆ABC,
Base BC = 5cm
Altitude AD = 4cm
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q1.1

Question 2.
Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.
Solution:
Sides of triangle are 3 cm, 4cm and 5cm
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q2.1

Question 3.
Find the area of an isosceles triangle having the base x cm and one side y cm.
Solution:
In isosceles ∆ABC,
AB = AC = y cm
BC = x cm
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q3.1
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q3.2

Question 4.
Find the area of an equilateral triangle having each side 4 cm.
Solution:
Each side of equilateral triangle (a) = 4cm
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q4.1

Question 5.
Find the area of an equilateral triangle having each side x cm.
Solution:
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q5.1

Question 6.
The perimeter of a triangular field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.
Solution:
Perimeter of the field = 144 m
Ratio in the sides = 3:4:5
Sum of ratios = 3 + 4 + 5 = 12
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q6.1

Question 7.
Find the area of an equilateral triangle having altitude h cm.
Solution:
Altitude of an equilateral triangle = h
Let side of equilateral triangle = x
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q7.1

Question 8.
Let ∆ be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.
Solution:
Let a, b, c be the sides of the original triangle
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q8.1
Hence area of new triangle = 4 x area of original triangle.

Question 9.
If each side of a triangle is doubled, then find percentage increase in its area.
Solution:
Sides of original triangle be a, b, c
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q9.1

Question 10.
If each side of an equilateral triangle is tripled then what is the percentage increase in the area of the triangle?
Solution:
Let the sides of the original triangle be a, b, c and area ∆, then
RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS Q10.1

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RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2

RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2

Other Exercises

Question 1.
Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm (NCERT)
Solution:
In the quadrilateral, AC is the diagonal which divides the figure into two triangles
Now in ∆ABC, AB = 3 cm, BC = 4cm, AC = 5cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q1.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q1.2

Question 2.
The sides of a quadrangular field taken in order are 26 m, 27 m, 7 m and 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.
Solution:
In quad. ABCD, AB = 26 m, BC = 27 m CD = 7m, DA = 24 m, ∠CDA = 90°
Join AC,
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q2.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q2.2

Question 3.
The sides of a quadrilateral taken in order are 5, 12, 14 and 15 metres respectively, and the angle contained by the first two sides is a right angle. Find its area.
Solution:
In quad. ABCD,
AB = 5m, BC = 12 m, CD = 14m,
DA = 15 m and ∠ABC = 90°
Join AC,
Now in right ∆ABC,
AC² = AB² + BC² = (5)² + (12)²
= 25 + 144 = 169 = (13)²
∴ AC = 13 m
Now area of right ∆ABC
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q3.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q3.2

Question 4.
A park, in shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9m, BC = 12m, CD = 5m and AD = 8m. How much area does it occupy? (NCERT)
Solution:
In quadrilateral ABCD,
AB = 9m, BC = 12m, CD = 5m and
DA = 8m, ∠C = 90°
Join BD,
Now in right ∆BCD,
BD² = BC²+ CD² = (12)² + (5)²
= 144 + 25 = 169 = (13)²
∴ BD = 13m
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q4.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q4.2

Question 5.
Find the area of a rhombus whose perimeter is 80m and one of whose diagonal is 24m.
Solution:
Perimeter of rhombus ABCD = 80 m
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q5.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q5.2

Question 6.
A rhombus sheet whose perimeter = 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of ₹5 per m². Find the cost of painting.
Solution:
Perimeter of the rhombus shaped sheet = 32 m
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q6.1
∴ Length of each side = \(\frac { 32 }{ 4 }\) = 8m
and length of one diagonal AC = 10 m
In ∆ABC, sides are 8m, 8m, 10m
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q6.2

Question 7.
Find the area of a quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm. (Take \(\sqrt { 3 } \) = 1.73 )
Solution:
In quadrilateral ABCD, AD = 24cm, ∠BAD = 90°
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q7.1
BCD is an equilateral triangle with side 26cm
In right ∆ABD,
BD² = AB²+ AD²
(26)² = AB² + (24)²
⇒ 676 = AB² + 576
AB² = 676 – 576 = 100 = (10)²
∴ AB = 10cm
Now area of right ∆ABD,
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q7.2

Question 8.
Find the area of a quadrilateral ABCD in which AB = 42cm, BC = 21cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.
Solution:
In quadrilateral ABCD,
AB = 42 cm, BC = 21 cm, CD = 29cm DA = 34 cm, BD = 20 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q8.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q8.2

Question 9.
The adjacent sides of a parallelogram ABCD measures 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.
Solution:
In ||gm ABCD,
AB = 34cm, BC = 20 cm
and AC = 42 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q9.1
∵ The diagonal of a parallelogram divides into two triangles equal in area,
Now area of ∆ABC,
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q9.2

Question 10.
Find the area of the blades of the magnetic compass shown in figure. (Take \(\sqrt { 11 } \) = 3.32).
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q10.1
Solution:
ABCD is a rhombus with each side 5cm and one diagonal 1cm
Diagonal BD divides into two equal triangles Now area of ∆ABD,
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q10.2

Question 11.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.
Solution:
Area of a triangle with same base and area of a 11gm with equal sides of triangle are 13, 14, 15 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q11.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q11.2

Question 12.
Two parallel sides of a trapezium are 60cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.
Solution:
In trapezium ABCD, AB || DC
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q12.1
AB = 77cm, BC = 26 cm, CD 60cm DA = 25 cm
Through, C, draw CE || DA meeting AB at E
∴ AE = CD = 60 cm and EB = 77 – 60 = 17 cm,
CE = DA = 25 cm
Now area of ∆BCE, with sides 17 cm, 26 cm, 25 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q12.2
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q12.3

Question 13.
Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9cm, CD = 12cm, ∠ACB = 90° and AC = 15cm.
Solution:
In right ΔABC, ∠ACB = 90°
AB² = AC² + BC²
(17)² = (15)²+ BC² = 289 = 225 + BC²
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q13.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q13.2

Question 14.
A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown figure. The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each types of paper needed to make the hand fan.
Solution:
In the figure, a hand fan has 5 isosceles and triangle. With sides 25 cm, 25 cm and 14 cm each.
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 Q14.1

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RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1

RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1

Other Exercises

Question 1.
Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.
Solution:
Sides of triangle are 120 cm, 150 cm, 200 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q1.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q1.2

Question 2.
Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm.
Solution:
Sides of a triangle are 9 cpi, 12 cm, 15 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q2.1

Question 3.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
Perimeter of a triangle = 42 cm
Two sides are 18 cm and 10 cm
Third side = 42 – (18 + 10)
= 42 – 28 = 14 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q3.1

Question 4.
In a ∆ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ∆ABC and hence its altitude on AC.
Solution:
Sides of triangle ABC are AB = 15 cm, BC = 13 cm, AC = 14 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q4.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q4.2

Question 5.
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. [NCERT]
Solution:
Perimeter of a triangle = 540 m
Ratio in sides = 25 : 17 : 12
Sum of ratios = 25 + 17 + 12 = 54
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q5.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q5.2

Question 6.
The perimeter of a triangle is 300 m. If its sides are in the ratio 3:5:7. Find the area of the triangle. [NCERT]
Solution:
Perimeter of a triangle = 300 m
Ratio in the sides = 3 : 5 : 7
∴ Sum of ratios = 3 + 5 + 7= 15
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q6.1

Question 7.
The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.
Solution:
Perimeter of a triangular field = 240 dm
Two sides are 78 dm and 50 dm
∴ Third side = 240 – (78 + 50)
= 240 – 128 = 112 dm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q7.1

Question 8.
A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.
Solution:
Sides of a triangle are 35 cm, 54 cm, 61 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q8.1

Question 9.
The lengths of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.
Solution:
Ratio in the sides of a triangle = 3:4:5
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q9.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q9.2

Question 10.
The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.
Solution:
Perimeter of an isosceles triangle = 42 cm
Base = \(\frac { 3 }{ 2 }\) of its one of equal sides
Let each equal side = x, then 3
Base = \(\frac { 3 }{ 2 }\) x
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q10.1
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q10.2

Question 11.
Find the area of the shaded region in figure.
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q11.1
Solution:
In ∆ABC, AC = 52 cm, BC = 48 cm
and in right ∆ADC, ∠D = 90°
AD = 12 cm, BD = 16 cm
∴ AB²=AD² + BD² (Pythagoras Theorem)
(12)² + (16)² = 144 + 256 = 400 = (20)²
∴ AB = 20 cm
RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 Q11.2

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RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3

RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3

Other Exercises

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 Q1.1

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 Q2.1

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 Q3.1

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 Q4.1

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 Q5.1

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 Q6.1

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 Q7.1
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 Q8.1

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 Q9.1

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.3 Q10.1

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RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2

Other Exercises

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q1.1

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q2.1

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = \(\frac { 1 }{ 2 }\) x 108°
∴ We shall bisect it.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q3.1
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =\(\frac { 1 }{ 2 }\) x 90° = 45°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q4.1

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ \(\frac { 1 }{ 2 }\) ∠DCA + \(\frac { 1 }{ 2 }\) ∠DCB = 180° x \(\frac { 1 }{ 2 }\) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q5.1

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q6.1

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q7.1

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q8.1

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x \(\frac { 1 }{ 2 }\) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + \(\frac { 1 }{ 2 }\) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q9.1

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q10.1
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q10.2

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 \(\frac { 1 }{ 2 }\)°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q11.1
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q11.2
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q11.3
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q11.4
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q11.5
(vi) 22 \(\frac { 1 }{ 2 }\)°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 \(\frac { 1 }{ 2 }\)°
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.2 Q11.6

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RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.1

Other Exercises

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.1 Q1.1

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.1 Q2.1

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.1 Q3.1

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.1 Q4.1

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.1 Q5.1

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length \(\frac { 1 }{ 2 }\) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = \(\frac { 1 }{ 2 }\) (AB).
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.1 Q6.1

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length \(\frac { 3 }{ 4 }\) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = \(\frac { 3 }{ 4 }\) (AB).
RD Sharma Class 9 Solutions Chapter 16 Circles Ex 16.1 Q7.1

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RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS

RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS

Other Exercises

Question 1.
In the figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q1.1
Solution:
Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle
∴ ∠AOB = 2∠APB = 2 x 70° = 140°
Now in cyclic quadrilateral AOBC,
∠AOB + ∠ACB = 180° (Sum of the angles)
⇒ 140° +∠ACB = 180°
⇒ ∠ACB = 180° – 140° = 40°
∴ ∠ACB = 40°

Question 2.
In the figure, two congruent circles with centre O and O’ intersect at A and B. If ∠AO’B = 50°, then find ∠APB.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q2.1
Solution:
Two congruent circles with centres O and O’ intersect at A and B
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q2.2
∠AO’B = 50°
∵ OA = OB = O’A = 04B (Radii of the congruent circles)
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q2.3

Question 3.
In the figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = IT, AC and BD intersect at P. Then, find ∠DPC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q3.1
Solution:
∵ ABCD is a cyclic quadrilateral,
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q3.2
∴ ∠BAD + ∠BCD = 180°
⇒ 75° + ∠BCD – 180°
⇒ ∠BCD = 180°-75°= 105° and ∠ADC + ∠ABC = 180°
⇒ 77° + ∠ABC = 180°
⇒ ∠ABC = 180°-77°= 103°
∴ ∠DBC = ∠ABC – ∠ABD = 103° – 58° = 45°
∵ Arc AD subtends ∠ABD and ∠ACD in the same segment of the circle 3
∴ ∠ABD = ∠ACD = 58°
∴ ∠ACB = ∠BCD – ∠ACD = 105° – 58° = 47°
Now in ∆PBC,
Ext. ∠DPC = ∠PBC + ∠PCB
=∠DBC + ∠ACB = 45° + 47° = 92°
Hence ∠DPC = 92°

Question 4.
In the figure, if ∠AOB = 80° and ∠ABC = 30°, then find ∠CAO.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q4.1
Solution:
In the figure, ∠AOB = 80°, ∠ABC = 30°
∵ Arc AB subtends ∠AOB at the centre and
∠ACB at the remaining part of the circle
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q4.2
∴ ∠ACB = \(\frac { 1 }{ 2 }\)∠AOB = \(\frac { 1 }{ 2 }\) x 80° = 40°
In ∆OAB, OA = OB
∴ ∠OAB = ∠OBA
But ∠OAB + ∠OBA + ∠AOB = 180°
∴ ∠OAB + ∠OBA + 80° = 180°
⇒ ∠OAB + ∠OAB = 180° – 80° = 100°
∴ 2∠OAB = 100°
⇒ ∠OAB = \(\frac { { 100 }^{ \circ } }{ 2 }\)  = 50°
Similarly, in ∆ABC,
∠BAC + ∠ACB + ∠ABC = 180°
∠BAC + 40° + 30° = 180°
⇒ ∠BAC = 180°-30°-40°
= 180°-70°= 110°
∴ ∠CAO = ∠BAC – ∠OAB
= 110°-50° = 60°

Question 5.
In the figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD : ∠ABE.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q5.1
Solution:
In the figure, ABCD is a parallelogram and
CDE is a straight line
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q5.2
∵ ABCD is a ||gm
∴ ∠A = ∠C
and ∠C = ∠ADE (Corresponding angles)
⇒ ∠BCD = ∠ADE
Similarly, ∠ABE = ∠BED (Alternate angles)
∵ arc BD subtends ∠BAD at the centre and
∠BED at the remaining part of the circle
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q5.3

Question 6.
In the figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q6.1
Solution:
In the figure, AB is the diameter of the circle such that ∠A = 35° and ∠Q = 25°, join OP.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q6.2
Arc PB subtends ∠POB at the centre and
∠PAB at the remaining part of the circle
∴ ∠POB = 2∠PAB = 2 x 35° = 70°
Now in ∆OP,
OP = OB radii of the circle
∴ ∠OPB = ∠OBP = 70° (∵ ∠OPB + ∠OBP = 140°)
Now ∠APB = 90° (Angle in a semicircle)
∴ ∠BPQ = 90°
and in ∆PQB,
Ext. ∠PBR = ∠BPQ + ∠PQB
= 90° + 25°= 115°
∴ ∠PBR = 115°

Question 7.
In the figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q7.1
Solution:
In the figure, P and Q are the centres of two circles which intersect each other at C and B
ACD is a straight line ∠APB = 150°
Arc AB subtends ∠APB at the centre and
∠ACB at the remaining part of the circle
∴ ∠ACB = \(\frac { 1 }{ 2 }\) ∠APB = \(\frac { 1 }{ 2 }\) x 150° = 75°
But ∠ACB + ∠BCD = 180° (Linear pair)
⇒ 75° + ∠BCD = 180°
∠BCD = 180°-75°= 105°
Now arc BD subtends reflex ∠BQD at the centre and ∠BCD at the remaining part of the circle
Reflex ∠BQD = 2∠BCD = 2 x 105° = 210°
But ∠BQD + reflex ∠BQD = 360°
∴ ∠BQD+ 210° = 360°
∴ ∠BQD = 360° – 210° = 150°

Question 8.
In the figure, if O is circumcentre of ∆ABC then find the value of ∠OBC + ∠BAC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q8.1
Solution:
In the figure, join OC
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q8.2
∵ O is the circumcentre of ∆ABC
∴ OA = OB = OC
∵ ∠CAO = 60° (Proved)
∴ ∆OAC is an equilateral triangle
∴ ∠AOC = 60°
Now, ∠BOC = ∠BOA + ∠AOC
= 80° + 60° = 140°
and in ∆OBC, OB = OC
∠OCB = ∠OBC
But ∠OCB + ∠OBC = 180° – ∠BOC
= 180°- 140° = 40°
⇒ ∠OBC + ∠OBC = 40°
∴ ∠OBC = \(\frac { { 40 }^{ \circ } }{ 2 }\)  = 20°
∠BAC = OAB + ∠OAC = 50° + 60° = 110°
∴ ∠OBC + ∠BAC = 20° + 110° = 130°

Question 9.
In the AOC is a diameter of the circle and arc AXB = 1/2 arc BYC. Find ∠BOC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q9.1
Solution:
In the figure, AOC is diameter arc AxB = \(\frac { 1 }{ 2 }\) arc BYC 1
∠AOB = \(\frac { 1 }{ 2 }\) ∠BOC
⇒ ∠BOC = 2∠AOB
But ∠AOB + ∠BOC = 180°
⇒ ∠AOB + 2∠AOB = 180°
⇒ 3 ∠AOB = 180°
∴ ∠AOB = \(\frac { { 180 }^{ \circ } }{ 3 }\)  = 60°
∴ ∠BOC = 2 x 60° = 120°

Question 10.
In the figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q10.1
Solution:
In the figure, ABCD is a cyclic quadrilateral
CD is produced to E such that ∠ADE = 95°
O is the centre of the circle
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS Q10.2
∵ ∠ADC + ∠ADE = 180°
⇒ ∠ADC + 95° = 180°
⇒ ∠ADC = 180°-95° = 85°
Now arc ABC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle
∵ ∠AOC = 2∠ADC = 2 x 85° = 170°
Now in ∆OAC,
∠OAC + ∠OCA + ∠AOC = 180° (Sum of angles of a triangle)
⇒ ∠OAC = ∠OCA (∵ OA = OC radii of circle)
∴ ∠OAC + ∠OAC + 170° = 180°
2∠OAC = 180°- 170°= 10°
∴ ∠OAC = \(\frac { { 10 }^{ \circ } }{ 2 }\) = 5°

 

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RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5

RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5

Other Exercises

Question 1.
In the figure, ∆ABC is an equilateral triangle. Find m ∠BEC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q1.1
Solution:
∵ ∆ABC is an equilateral triangle
∴ A = 60°
∵ ABEC is a cyclic quadrilateral
∴ ∠A + ∠E = 180° (Sum of opposite angles)
⇒ 60° + ∠E = 180°
⇒ ∠E = 180° – 60° = 120°
∴ m ∠BEC = 120°

Question 2.
In the figure, ∆PQR is an isosceles triangle with PQ = PR and m ∠PQR = 35°. Find m ∠QSR and m ∠QTR.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q2.1
Solution:
In the figure, ∆PQR is an isosceles PQ = PR
∠PQR = 35°
∴ ∠PRQ = 35°
But ∠PQR + ∠PRQ + ∠QPR = 180° (Sum of angles of a triangle)
⇒ 35° + 35° + ∠QPR = 180°
⇒ 70° + ∠QPR = 180°
∴ ∠QPR = 180° – 70° = 110°
∵ ∠QSR = ∠QPR (Angle in the same segment of circles)
∴ ∠QSR = 110°
But PQTR is a cyclic quadrilateral
∴ ∠QTR + ∠QPR = 180°
⇒ ∠QTR + 110° = 180°
⇒ ∠QTR = 180° -110° = 70°
Hence ∠QTR = 70°

Question 3.
In the figure, O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q3.1
Solution:
In the figure, O is the centre of the circle ∠BOD =160°
ABCD is the cyclic quadrilateral
∵ Arc BAD subtends ∠BOD is the angle at the centre and ∠BCD is on the other part of the circle
∴ ∠BCD = \(\frac { 1 }{ 2 }\) ∠BOD
⇒ x = \(\frac { 1 }{ 2 }\) x 160° = 80°
∵ ABCD is a cyclic quadrilateral,
∴ ∠A + ∠C = 180°
⇒ y + x = 180°
⇒ y + 80° = 180°
⇒ y =180°- 80° = 100°
∴ x = 80°, y = 100°

Question 4.
In the figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q4.1
Solution:
In a circle, ABCD is a cyclic quadrilateral ∠BCD = 100° and ∠ABD = 70°
∵ ABCD is a cyclic quadrilateral,
∴ ∠A + ∠C = 180° (Sum of opposite angles)
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q4.2
⇒ ∠A + 100°= 180°
∠A = 180°- 100° = 80°
Now in ∆ABD,
∠A + ∠ABD + ∠ADB = 180°
⇒ 80° + 70° + ∠ADB = 180°
⇒ 150° +∠ADB = 180°
∴ ∠ADB = 180°- 150° = 30°
Hence ∠ADB = 30°

Question 5.
If ABCD is a cyclic quadrilateral in which AD || BC. Prove that ∠B = ∠C.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q5.1
Solution:
Given : ABCD is a cyclic quadrilateral in which AD || BC
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q5.2
To prove : ∠B = ∠C
Proof : ∵ AD || BC
∴ ∠A + ∠B = 180°
(Sum of cointerior angles)
But ∠A + ∠C = 180°
(Opposite angles of the cyclic quadrilateral)
∴ ∠A + ∠B = ∠A + ∠C
⇒ ∠B = ∠C
Hence ∠B = ∠C

Question 6.
In the figure, O is the centre of the circle. Find ∠CBD.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q6.1
Solution:
Arc AC subtends ∠AOC at the centre and ∠APC at the remaining part of the circle
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q6.2
∴ ∠APC = \(\frac { 1 }{ 2 }\) ∠AOC
= \(\frac { 1 }{ 2 }\) x 100° = 50°
∵ APCB is a.cyclic quadrilateral,
∴ ∠APC + ∠ABC = 180°
⇒ 50° + ∠ABC = 180° ⇒ ∠ABC =180°- 50°
∴ ∠ABC =130°
But ∠ABC + ∠CBD = 180° (Linear pair)
⇒ 130° + ∠CBD = 180°
⇒ ∠CBD = 180°- 130° = 50°
∴ ∠CBD = 50°

Question 7.
In the figure, AB and CD are diameiers of a circle with centre O. If ∠OBD = 50°, find ∠AOC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q7.1
Solution:
Two diameters AB and CD intersect each other at O. AC, CB and BD are joined
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q7.2
∠DBA = 50°
∠DBA and ∠DCA are in the same segment
∴ ∠DBA = ∠DCA = 50°
In ∆OAC, OA = OC (Radii of the circle)
∴ ∠OAC = ∠OCA = ∠DCA = 50°
and ∠OAC + ∠OCA + ∠AOC = 180° (Sum of angles of a triangle)
⇒ 50° + 50° + ∠AOC = 180°
⇒ 100° + ∠AOC = 180°
⇒ ∠AOC = 180° – 100° = 80°
Hence ∠AOC = 80°

Question 8.
On a semi circle with AB as diameter, a point C is taken so that m (∠CAB) = 30°. Find m (∠ACB) and m (∠ABC).
Solution:
A semicircle with AB as diameter
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q8.1
∠ CAB = 30°
∠ACB = 90° (Angle in a semi circle)
But ∠CAB + ∠ACB + ∠ABC = 180°
⇒ 30° + 90° + ∠ABC – 180°
⇒ 120° + ∠ABC = 180°
∴ ∠ABC = 180°- 120° = 60°
Hence m ∠ACB = 90°
and m ∠ABC = 60°

Question 9.
In a cyclic quadrilateral ABCD, if AB || CD and ∠B = 70°, find the remaining angles.
Solution:
In a cyclic quadrilateral ABCD, AB || CD and ∠B = 70°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q9.1
∵ ABCD is a cyclic quadrilateral
∴ ∠B + ∠D = 180°
⇒ 70° + ∠D = 180°
⇒ ∠D = 180°-70° = 110°
∵ AB || CD
∴ ∠A + ∠D = 180° (Sum of cointerior angles)
∠A+ 110°= 180°
⇒ ∠A= 180°- 110° = 70°
Similarly, ∠B + ∠C = 180°
⇒ 70° + ∠C- 180° ‘
⇒ ∠C = 180°-70°= 110°
∴ ∠A = 70°, ∠C = 110°, ∠D = 110°

Question 10.
In a cyclic quadrilateral ABCD, if m ∠A = 3(m ∠C). Find m ∠A.
Solution:
In cyclic quadrilateral ABCD, m ∠A = 3(m ∠C)
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q10.1
∵ ABCD is a cyclic quadrilateral,
∴ ∠A + ∠C = 180°
⇒ 3 ∠C + ∠C = 180° ⇒ 4∠C = 180°
⇒ ∠C = \(\frac { { 180 }^{ \circ } }{ 4 }\)  = 45°
∴ ∠A = 3 x 45°= 135°
Hence m ∠A =135°

Question 11.
In the figure, O is the centre of the circle and ∠DAB = 50°. Calculate the values of x and y.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q11.1
Solution:
In the figure, O is the centre of the circle ∠DAB = 50°
∵ ABCD is a cyclic quadrilateral
∴ ∠A + ∠C = 180°
⇒ 50° + y = 180°
⇒ y = 180° – 50° = 130°
In ∆OAB, OA = OB (Radii of the circle)
∴ ∠A = ∠OBA = 50°
∴ Ext. ∠DOB = ∠A + ∠OBA
x = 50° + 50° = 100°
∴ x= 100°, y= 130°

Question 12.
In the figure, if ∠BAC = 60° and ∠BCA = 20°, find ∠ADC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q12.1
Solution:
In ∆ABC,
∠BAC + ∠ABC + ∠ACB = 180° (Sum of angles of a triangle)
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q12.2
60° + ∠ABC + 20° = 180°
∠ABC + 80° = 180°
∴ ∠ABC = 180° -80°= 100°
∵ ABCD is a cyclic quadrilateral,
∴ ∠ABC + ∠ADC = 180°
100° + ∠ADC = 180°
∴ ∠ADC = 180°- 100° = 80°

Question 13.
In the figure, if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q13.1
Solution:
In a circle, ∆ABC is an equilateral triangle
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q13.2
∴ ∠A = 60°
∵ ∠BAC and ∠BDC are in the same segment
∴ ∠BAC = ∠BDC = 60°
∵ BECD is a cyclic quadrilateral
∴ ∠BDC + ∠BEC = 180°
⇒ 60° + ∠BEC = 180°
⇒ ∠BEC = 180°-60°= 120°
Hence ∠BDC = 60° and ∠BEC = 120°

Question 14.
In the figure, O is the centre of the circle. If ∠CEA = 30°, find the values of x, y and z.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q14.1
Solution:
∠AEC and ∠ADC are in the same segment
∴ ∠AEC = ∠ADC = 30°
∴ z = 30°
ABCD is a cyclic quadrilateral
∴ ∠B + ∠D = 180°
⇒ x + z = 180°
⇒ x + 30° = 180°
⇒ x = 180° – 30° = 150°
Arc AC subtends ∠AOB at the centre and ∠ADC at the remaining part of the circle
∴ ∠AOC = 2∠D = 2 x 30° = 60°
∴ y = 60°
Hence x = 150°, y – 60° and z = 30°

Question 15.
In the figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q15.1
Solution:
In the figure, two circles intersect each other at C and D
∠BAD = 78°, ∠DCF = x, ∠DEF = y
ABCD is a cyclic quadrilateral
∴ Ext. ∠DCF = its interior opposite ∠BAD
⇒ x = 78°
In cyclic quadrilateral CDEF,
∠DCF + ∠DEF = 180°
⇒ 78° + y = 180°
⇒ y = 180° – 78°
y = 102°
Hence x = 78°, and y- 102°

Question 16.
In a cyclic quadrilateral ABCD, if ∠A – ∠C = 60°, prove that the smaller of two is 60°.
Solution:
In cyclic quadrilateral ABCD,
∠A – ∠C = 60°
But ∠A + ∠C = 180° (Sum of opposite angles)
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q16.1
Adding, 2∠A = 240° ⇒ ∠A = \(\frac { { 62 }^{ \circ } }{ 2 }\)  = 120° and subtracting
2∠C = 120° ⇒ ∠C = \(\frac { { 120 }^{ \circ } }{ 2 }\)  = 60°
∴ Smaller angle of the two is 60°.

Question 17.
In the figure, ABCD is a cyclic quadrilateral. Find the value of x.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q17.1
Solution:
∠CDE + ∠CDA = 180° (Linear pair)
⇒ 80° + ∠CDA = 180°
⇒ ∠CDA = 180° – 80° = 100°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q17.2
In cyclic quadrilateral ABCD,
Ext. ∠ABF = Its interior opposite angle ∠CDA = 100°
∴ x = 100°

Question 18.
ABCD is a cyclic quadrilateral in which:
(i) BC || AD, ∠ADC =110° and ∠B AC = 50°. Find ∠DAC.
(ii) ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
(iii) ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.
Solution:
(i) In the figure,
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q18.1
ABCD is a cyclic quadrilateral and AD || BC, ∠ADC = 110°
∠BAC = 50°
∵ ∠B + ∠D = 180° (Sum of opposite angles)
⇒ ∠B + 110° = 180°
∴ ∠B = 180°- 110° = 70°
Now in ∆ABC,
∠CAB + ∠ABC + ∠BCA = 180° (Sum of angles of a triangle)
⇒ 50° + 70° + ∠BCA = 180°
⇒ 120° + ∠BCA = 180°
⇒ ∠BCA = 180° – 120° = 60°
But ∠DAC = ∠BCA (Alternate angles)
∴ ∠DAC = 60°
(ii) In cyclic quadrilateral ABCD,
Diagonals AC and BD are joined ∠DBC = 80°, ∠BAC = 40°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q18.2
Arc DC subtends ∠DBC and ∠DAC in the same segment
∴ ∠DBC = ∠DAC = 80°
∴ ∠DAB = ∠DAC + ∠CAB = 80° + 40° = 120°
But ∠DAC + ∠BCD = 180° (Sum of opposite angles of a cyclic quad.)
⇒ 120° +∠BCD = 180°
⇒ ∠BCD = 180°- 120° = 60°
(iii) In the figure, ABCD is a cyclic quadrilateral BD is joined
∠BCD = 100°
and ∠ABD = 70°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q18.3
∠A + ∠C = 180° (Sum of opposite angles of cyclic quad.)
∠A+ 100°= 180°
⇒ ∠A= 180°- 100°
∴ ∠A = 80°
Now in ∆ABD,
∠A + ∠ABD + ∠ADB = 180° (Sum of angles of a triangle)
⇒ 80° + 70° + ∠ADB = 180°
⇒ 150° +∠ADB = 180°
⇒ ∠ADB = 180°- 150° = 30°
∴ ∠ADB = 30°

Question 19.
Prove that the circles described on the four sides of a rhombus as diameter, pass through the point of intersection of its diagonals.
Solution:
Given : ABCD is a rhombus. Four circles are drawn on the sides AB, BC, CD and DA respectively
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q19.1
To prove : The circles pass through the point of intersection of the diagonals of the rhombus ABCD
Proof: ABCD is a rhombus whose diagonals AC and BD intersect each other at O
∵ The diagonals of a rhombus bisect each other at right angles
∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Now when ∠AOB = 90°
and a circle described on AB as diameter will pass through O
Similarly, the circles on BC, CD and DA as diameter, will also pass through O

Question 20.
If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that is diagonals are equal.
Solution:
Given : In cyclic quadrilateral ABCD, AB = CD
AC and BD are the diagonals
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q20.1
To prove : AC = BC
Proof: ∵ AB = CD
∴ arc AB = arc CD
Adding arc BC to both sides, then arc AB + arc BC = arc BC + arc CD
⇒ arc AC = arc BD
∴ AC = BD
Hence diagonal of the cyclic quadrilateral are equal.

Question 21.
Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).
Solution:
Given : In ∆ABC, circles are drawn on sides AB and AC
To prove : Circles drawn on AB and AC intersect at D which lies on BC, the third side
Construction : Draw AD ⊥ BC
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q21.1
Proof: ∵ AD ⊥ BC
∴ ∠ADB = ∠ADC = 90°
So, the circles drawn on sides AB and AC as diameter will pass through D
Hence circles drawn on two sides of a triangle pass through D, which lies on the third side.

Question 22.
ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
Solution:
In the figure, ABCD is a trapezium in which AD || BC and ∠B = 70°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q22.1
∵ AD || BC
∴ ∠A + ∠B = 180° (Sum of cointerior angles)
⇒ ∠A + 70° = 180°
⇒ ∠A= 180°- 70° = 110°
∴ ∠A = 110°
But ∠A + ∠C = 180° and ∠B + ∠D = 180° (Sum of opposite angles of a cyclic quadrilateral)
∴ 110° + ∠C = 180°
⇒ ∠C = 180°- 110° = 70°
and 70° + ∠D = 180°
⇒ ∠D = 180° – 70° = 110°
∴ ∠A = 110°, ∠C = 70° and ∠D = 110°

Question 23.
In the figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q23.1
Solution:
In the figure, ABCD is a cyclic quadrilateral whose diagonals AC and BD are drawn ∠DBC = 55° and ∠BAC = 45°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q23.2
∵ ∠BAC and ∠BDC are in the same segment
∴ ∠BAC = ∠BDC = 45°
Now in ABCD,
∠DBC + ∠BDC + ∠BCD = 180° (Sum of angles of a triangle)
⇒ 55° + 45° + ∠BCD = 180°
⇒ 100° + ∠BCD = 180°
⇒ ∠BCD = 180° – 100° = 80°
Hence ∠BCD = 80°

Question 24.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
Solution:
Given : ABCD is a cyclic quadrilateral
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q24.1
To prove : The perpendicular bisectors of the sides are concurrent
Proof : ∵ Each side of the cyclic quadrilateral is a chord of the circle and perpendicular of a chord passes through the centre of the circle
Hence the perpendicular bisectors of each side will pass through the centre O
Hence the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent

Question 25.
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
Solution:
Given : ABCD is a cyclic rectangle and diagonals AC and BD intersect each other at O
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q25.1
To prove : O is the point of intersection is the centre of the circle.
Proof : Let O be the centre of the circle- circumscribing the rectangle ABCD
Since each angle of a rectangle is a right angle and AC is the chord of the circle
∴ AC will be the diameter of the circle Similarly, we can prove that diagonal BD is also the diameter of the circle
∴ The diameters of the circle pass through the centre
Hence the point of intersection of the diagonals of the rectangle is the centre of the circle.

Question 26.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:
(i) AD || BC
(ii) EB = EC.
Solution:
Given : ABCD is a cyclic quadrilateral in which sides BA and CD are produced to meet at E and EA = ED
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q26.1
To prove :
(i) AD || BC
(ii) EB = EC
Proof: ∵ EA = ED
∴ In ∆EAD
∠EAD = ∠EDA (Angles opposite to equal sides)
In a cyclic quadrilateral ABCD,
Ext. ∠EAD = ∠C
Similarly Ext. ∠EDA = ∠B
∵ ∠EAD = ∠EDA
∴ ∠B = ∠C
Now in ∆EBC,
∵ ∠B = ∠C
∴ EC = EB (Sides opposite to equal sides)
and ∠EAD = ∠B
But these are corresponding angles
∴ AD || BC

Question 27.
Prove that the angle in a segment shorter than a semicircle is greater than a right angle.
Solution:
Given : A segment ACB shorter than a semicircle and an angle ∠ACB inscribed in it
To prove : ∠ACB < 90°
Construction : Join OA and OB
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q27.1
Proof : Arc ADB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle ∴ ∠ACB = \(\frac { 1 }{ 2 }\) ∠AOB But ∠AOB > 180° (Reflex angle)
∴ ∠ACB > \(\frac { 1 }{ 2 }\) x [80°
⇒ ∠ACB > 90°

Question 28.
Prove that the angle in a segment greater than a semi-circle is less than a right angle
Solution:
Given : A segment ACB, greater than a semicircle with centre O and ∠ACB is described in it
To prove : ∠ACB < 90°
Construction : Join OA and OB
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q28.1
Proof : Arc ADB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle
∴ ∠ACB =\(\frac { 1 }{ 2 }\) ∠AOB
But ∠AOB < 180° (A straight angle) 1
∴ ∠ACB < \(\frac { 1 }{ 2 }\) x 180°
⇒ ∠ACB <90°
Hence ∠ACB < 90°

Question 29.
Prove that the line segment joining the mid-point of the hypotenuse of a rijght triangle to its opposite vertex is half of the hypotenuse.
Solution:
Given : In a right angled ∆ABC
∠B = 90°, D is the mid point of hypotenuse AC. DB is joined.
To prove : BD = \(\frac { 1 }{ 2 }\) AC
Construction : Draw a circle with centre D and AC as diameter
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q29.1
Proof: ∵ ∠ABC = 90°
∴ The circle drawn on AC as diameter will pass through B
∴ BD is the radius of the circle
But AC is the diameter of the circle and D is mid point of AC
∴ AD = DC = BD
∴ BD= \(\frac { 1 }{ 2 }\) AC

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