Category: CBSE Class 9

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

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

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

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%,

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

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

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

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

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]

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

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

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

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

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

• RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1
• RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.2
• RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.3
• RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4
• 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 VSAQS
• RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles MCQS

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.

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.

Solution:
Two congruent circles with centres O and O’ intersect at A and B

∠AO’B = 50°
∵ OA = OB = O’A = 04B (Radii of the congruent circles)

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.

Solution:
∵ ABCD is a cyclic quadrilateral,

∴ ∠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.

Solution:
In the figure, ∠AOB = 80°, ∠ABC = 30°
∵ Arc AB subtends ∠AOB at the centre and
∠ACB at the remaining part of the circle

∴ ∠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.

Solution:
In the figure, ABCD is a parallelogram and
CDE is a straight line

∵ 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

Question 6.
In the figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.

Solution:
In the figure, AB is the diameter of the circle such that ∠A = 35° and ∠Q = 25°, join OP.

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 =

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.

Solution:
In the figure, join OC

∵ 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.

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.

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

∵ ∠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°

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

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

• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1
• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2
• RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS
• RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

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

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,

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:

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

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

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

Question 7.
The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is

Solution:
Sides of a triangle are 11 cm, 15 cm, 16 cm

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

Question 9.
The length of each side of an equilateral triangle of area 4 \(\sqrt { 3 } \) cm², is

Solution:
Area of an equilateral triangle = 4\(\sqrt { 3 } \) cm²
Let each side be = a

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

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

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

Question 13.
If the length of a median of an equilateral triangle is x cm, then its area is

Solution:
∵ The median of an equilateral triangle is the perpendicular to the base also,
∴ Let side of the triangle = a

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

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

Solution:
A square and an equilateral triangle have equal perimeter

Hope given RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS 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.

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

• RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1
• RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.2
• RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.3
• RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4
• 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 VSAQS
• RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles MCQS

Question 1.
In the figure, ∆ABC is an equilateral triangle. Find m ∠BEC.

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.

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.

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.

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)

⇒ ∠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.

Solution:
Given : ABCD is a cyclic quadrilateral in which AD || BC

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.

Solution:
Arc AC subtends ∠AOC at the centre and ∠APC at the remaining part of the circle

∴ ∠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.

Solution:
Two diameters AB and CD intersect each other at O. AC, CB and BD are joined

∠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

∠ 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°

∵ 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)

∵ 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.

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.

Solution:
In ∆ABC,
∠BAC + ∠ABC + ∠ACB = 180° (Sum of angles of a triangle)

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.

Solution:
In a circle, ∆ABC is an equilateral triangle

∴ ∠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.

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.

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)

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.

Solution:
∠CDE + ∠CDA = 180° (Linear pair)
⇒ 80° + ∠CDA = 180°
⇒ ∠CDA = 180° – 80° = 100°

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,

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°

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°

∠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

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

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

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°

∵ 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.

Solution:
In the figure, ABCD is a cyclic quadrilateral whose diagonals AC and BD are drawn ∠DBC = 55° and ∠BAC = 45°

∵ ∠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

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

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

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

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

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

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

Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 are helpful to complete your math homework.

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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

• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1
• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2
• RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS
• RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

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

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

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

Question 4.
Find the area of an equilateral triangle having each side 4 cm.
Solution:
Each side of equilateral triangle (a) = 4cm

Question 5.
Find the area of an equilateral triangle having each side x cm.
Solution:

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

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

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

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

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

Hope given RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS are helpful to complete your math homework.

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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

• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1
• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2
• RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS
• RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

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

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,

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

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

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

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

∴ Length of each side = \(\frac { 32 }{ 4 }\) = 8m
and length of one diagonal AC = 10 m
In ∆ABC, sides are 8m, 8m, 10m

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°

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,

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

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

∵ The diagonal of a parallelogram divides into two triangles equal in area,
Now area of ∆ABC,

Question 10.
Find the area of the blades of the magnetic compass shown in figure. (Take \(\sqrt { 11 } \) = 3.32).

Solution:
ABCD is a rhombus with each side 5cm and one diagonal 1cm
Diagonal BD divides into two equal triangles Now area of ∆ABD,

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

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

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

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

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.

Hope given RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2 are helpful to complete your math homework.

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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

• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1
• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2
• RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS
• RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

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

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

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

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

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

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

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

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

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

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

Question 11.
Find the area of the shaded region in figure.

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

Hope given RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1 are helpful to complete your math homework.

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