R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 15 - Perimeter and Areas of Plane Figures
Chapter -
Chapter 17 - Perimeter and Areas of Plane Figures MCQ
Chapter 17 - Perimeter and Areas of Plane Figures FA
Chapter 17 - Perimeter and Areas of Plane Figures Ex. 15A
Let a = 42 cm, b = 34 cm and c = 20 cm
(i)Area of triangle =
(ii)Let base = 42 cm and corresponding height = h cm
Then area of triangle =
Hence, the height corresponding to the longest side = 16 cm
Let a = 18 cm, b = 24 cm, c = 30 cm
Then,2s = (18 + 24 + 30) cm = 72 cm
s = 36 cm
(s a) = 18cm, (s b) = 12 cm and (s c) = 6 cm
(i)Area of triangle =
(ii)Let base = 18 cm and altitude = x cm
Then, area of triangle =
Hence, altitude corresponding to the smallest side = 24 cm
On dividing 150 m in the ratio 5 : 12 : 13, we get
Length of one side =
Length of the second side =
Length of third side =
Let a = 25 m, b = 60 m, c = 65 m
(s a) = 50 cm, (s b) = 15 cm, and (s c) = 10 cm
Hence, area of the triangle = 750 m2
On dividing 540 m in ratio 25 : 17 : 12, we get
Length of one side =
Length of second side =
Length of third side = =120 m
Let a = 250m, b = 170 m and c = 120 m
Then, (s a) = 29 m, (s b) = 100 m, and (s c) = 150m
The cost of ploughing 100 area is = Rs. 18. 80
The cost of ploughing 1 is =
The cost of ploughing 9000 area =
= Rs. 1692
Hence, cost of ploughing = Rs 1692.
Let the length of one side be x cm
Then the length of other side = {40 (17 + x)} cm = (23 - x) cm
Hypotenuse = 17 cm
Applying Pythagoras theorem, we get
Hence, area of the triangle = 60 cm2
Let the sides containing the right - angle be x cm and (x - 7) cm
One side = 15 cm and other = (15 - 7) cm = 8 cm
perimeter of triangle (15 + 8 + 17) cm = 40 cm
Let the sides containing the right angle be x and (x 2) cm
One side = 8 cm, and other (8 2) cm = 6 cm
= 10 cm
Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm
Let each side of the equilateral triangle be a cm
Let each side of the equilateral triangle be a cm
Perimeter of equilateral triangle = 3a = (3 12) cm = 36 cm
Let each side of the equilateral triangle be a cm
area of equilateral triangle =
Height of equilateral triangle
Base of right angled triangle = 48 cm
Height of the right angled triangle =
Let the hypotenuse of right - angle triangle = 6.5 m
Base = 6 cm
Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm2
The circumcentre of a right - triangle is the midpoint of the hypotenuse
Hypotenuse = 2 × (radius of circumcircle)
= (2 × 8) cm = 16 cm
Base = 16 cm, height = 6 cm
Area of right angled triangle
Hence, area of the triangle= 48 cm2
Let each equal side be a cm in length.
Then,
Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm
Let each equal side be a cm and base = 80 cm
perimeter of triangle = (2a + b) cm
= (2 41 + 80) cm
= (82 + 80) cm = 162 cm
Hence, perimeter of the triangle = 162 cm
Let the height be h cm, then a= (h + 2) cm and b = 12 cm
Squaring both sides,
Therefore, a = h + 2 = (8 + 2)cm = 10 cm
Hence, area of the triangle = 48 cm2.
Area of shaded region = Area of ABC – Area of
DBC
First we find area of ABC
Second we find area of DBC which is right angled
Area of shaded region = Area of ABC – Area of
DBC
= (43.30 - 24) = 19. 30
Area of shaded region = 19.3
Let ABC is a isosceles triangle. Let AC, BC be the equal sides
Then AC = BC = 10cm. Let AB be the base of ABC right angle at C.
Area of right isosceles triangle ABC
Hence, area = 50 cm2 and perimeter = 34.14 cm
Chapter 17 - Perimeter and Areas of Plane Figures Ex. 15B
Area of floor = Length Breadth
Area of carpet = Length Breadth
=
Number of carpets =
= 216
Hence the number of carpet pieces required = 216
Area of verandah = (36 × 15) = 540
Area of stone = (0.6 × 0.5) [10 dm = 1 m]
Number of stones required =
Hence, 1800 stones are required to pave the verandah.
Perimeter of rectangle = 2(l + b)
2(l + b) = 56 Þ l + b = 28 cm
b = (28 l) cm
Area of rectangle = 192
l (28 l) = 192
28l - = 192
- 28l + 192 = 0
- 16l 12l + 192 = 0
l(l 16) 12(l 16) = 0
(l 16) (l 12) = 0
l = 16 or l = 12
Therefore, length = 16 cm and breadth = 12 cm
Length of the park = 35 m
Breadth of the park = 18 m
Area of the park = (35 18) = 630
Length of the park with grass =(35 5) = 30 m
Breadth of the park with grass = (18- 5) m = 13 m
Area of park with grass = (30 13) = 390
Area of path without grass = Area of the whole park area of park with grass
= 630 - 390
= 240
Hence, area of the park to be laid with grass = 240 m2
Length of the plot = 125 m
Breadth of the plot = 78 m
Area of plot ABCD = (125 78) = 9750
Length of the plot including the path= (125 + 3 + 3) m = 131 m
Breadth of the plot including the path = (78 + 3 + 3) m = 84 m
Area of plot PQRS including the path
= (131 84) = 11004
Area of path = Area of plot PQRS Area of plot ABCD
= (11004 9750)
= 1254
Cost of gravelling = Rs 75 per m2
Cost of gravelling the whole path = Rs. (1254 75) = Rs. 94050
Hence, cost of gravelling the path = Rs 94050
Area of rectangular field including the foot path = (54 35)
Let the width of the path be x m
Then, area of rectangle plot excluding the path = (54 2x) (35 2x)
Area of path = (54 35) (54 2x) (35 2x)
(54 35) (54 2x) (35 2x) = 420
1890 1890 + 108x + 70x - 4 = 420
178x - 4 = 420
4 - 178x + 420 = 0
2 - 89x + 210 = 0
2 - 84x 5x + 210 = 0
2x(x 42) 5(x 42) = 0
(x 42) (2x 5) = 0
Let the length and breadth of a rectangular garden be 9x and 5x.
Then, area of garden = (9x 5x)m = 45
Length of park excluding the path = (9x 7) m
Breadth of the park excluding the path = (5x 7) m
Area of the park excluding the path = (9x 7)(5x 7)
Area of the path =
(98x 49) = 1911
98x = 1911 + 49
Length = 9x = 9 20 = 180 m
Breadth = 5x = 5 20 = 100 m
Hence, length = 180 m and breadth = 100 m
Let the width of the carpet = x meter
Area of floor ABCD = (8 5)
Area of floor PQRS without border
= (8 2x)(5 2x)
= 40 16x 10x +
= 40 26x +
Area of border = Area of floor ABCD Area of floor PQRS
= [40 (40 26x + )]
=[40 40 + 26x - ]
= (26x - )
Area of the square =
Let diagonal of square be x
Length of diagonal = 16 cm
Side of square =
Perimeter of square = [4 side] sq. units
=[ 4 11.31] cm = 45.24 cm
Let d meter be the length of diagonal
Area of square field =
Time taken to cross the field along the diagonal
Hence, man will take 6 min to cross the field diagonally.
Rs. 14 is the cost of fencing a length = 1m
Rs. 28000 is the cost of fencing the length=
Perimeter = 4 side = 2000
side = 500 m
Area of a square =
= 250000
Cost of mowing the lawn =
Area of quad. ABCD = Area of ABD + Area of
DBC
For area of ABD
Let a = 42 cm, b = 34 cm, and c = 20 cm
For area of DBC
a = 29 cm, b = 21 cm, c = 20 cm
Area of quad. ABCD = Area of ABC + Area of
ACD
Now, we find area of a ACD
Area of quad. ABCD = Area of ABC + Area of
ACD
Perimeter of quad. ABCD = AB + BC + CD + AD
=(17 + 8 + 12 + 9) cm
= 46 cm
Perimeter of quad. ABCD = 46 cm
ABCD be the given quadrilateral in which AD = 24 cm, BD = 26 cm, DC = 26 cm and BC = 26 cm
By Pythagoras theorem
For area of equilateral DBC, we have
a = 26 cm
Area of quad. ABCD = Area of ABD + Area of
DBC
= (120 + 292.37) = 412.37
Perimeter ABCD = AD + AB + BC + CD
= 24 cm + 10 cm + 26 cm + 26 cm
= 86 cm
Area of the ||gm = (base height) sq. unit
= (25 16.8)
Longer side = 32 cm, shorter side = 24 cm
Distance between longer sides = 17.4 cm
Let the distance between the shorter sides be x cm
Area of ||gm = (longer side distance between longer sides)
= (shorter side distance between the short sides)
distance between the shorter side = 23.2 cm
Area ofparallelogram = 2 area of DABC
Opposite sides of parallelogram are equal
AD = BC = 20 cm
And AB = DC = 34 cm
In ABC we have
a = AC = 42 cm
b = AB = 34 cm
c = BC = 20 cm
Then, (s a) = 6 cm, (s b) = 14 cm and (s c) = 28 cm
We know that the diagonals of a rhombus, bisect each other at right angles
OA = OC = 15 cm,
And OB = OD = 8 cm
And AOB = 90
By Pythagoras theorem, we have
(i)Perimeter of rhombus = 4 side
4 side = 60 cm
By Pythagoras theorem
OB = 12 cm
OB = OD = 12 cm
BD = OB + OD = 12 cm + 12 cm = 24 cm
Length of second diagonal is 24 cm
(ii) Area of rhombus =
(i)Area of rhombus = 480
One of its diagonals = 48 cm
Let the second diagonal =x cm
Hence the length of second diagonal 20 cm
(ii)We know that the diagonals of a rhombus bisect each other at right angles
AC = 48, BD = 20 cm
OA = OC = 24 cm and OB = OD = 10 cm
By Pythagoras theorem , we have
(iii)Perimeter of the rhombus = (4 26) cm = 104 cm
Areaof cross section =
Let ABCD be a given trapezium in which
AB = 25, CD = 11
BC = 15, AD = 13
Draw CE || AD
In ||gm ADCE, AD || CE and AE || CD
AE = CD = 11 cm,
And BE = AB BE
= 25 11 = 14 cm
In BEC,
Area of BEC =
Let height of BEC is h
Area of BEC =
From (1) and (2), we get
7h = 84 h = 12 m
Area of trapezium ABCD
Other Chapters for CBSE Class 10 Maths
Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Linear equations in two variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- T-Ratios of Some Particular Angles Chapter 12- Trigonometric Ratios of Complementary Angles Chapter 13- Trigonometric Identities Chapter 14- Height and Distance Chapter 16- Areas of Circle, Sector and Segment Chapter 17- Volume and Surface Areas of Solids Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Chapter 19- ProbabilityR S AGGARWAL AND V AGGARWAL Solutions for CBSE Class 10 Subjects
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