Given:
Perimeter = 82 m and its area = 400 m2
Let the breadth of the rectangle be 'b' m.
As we know,
Perimeter of a rectangle = 2×(length + breadth)
82 = 2×(length + b)
41 = (length + b)
Length = (41 – b)m
As we know,
Area of the rectangle = length × breadth
400 = (41 – b)×(b)
400 = 41b – b2
b2 – 41b + 400 = 0
b2 – 25b – 16b + 400 = 0
b×(b – 25) – 16×(b – 25) = 0
(b – 16)(b – 25) = 0
Now,
either b – 16 = 0
⇒ b = 16
Or, b – 25 = 0
⇒ b = 25
Hence,
The breadth of the rectangle can be either 16 m or 25 m respectively.
Let the breadth of the rectangle be 'b' m
Then,
The length of the hall is 5 m more than its breadth i.e, = (b + 5) m
Given, area of the hall is = 84 m2
As the hall is rectangular,
Area of the rectangular hall = length × breadth
84 = b(b + 5)
b2 + 5b – 84 = 0
b2 + 12b – 7b – 84 = 0
b(b + 12) – 7(b + 12) = 0
(b + 12)(b – 7) = 0
Hence,
either b + 12 = 0
⇒ b = – 12 m (Side of a rectangle cannot be negative)
Or, b – 7 = 0
⇒ b = 7 m
So, only b = 7 m is considered.
⇒ b + 5 = 12
Therefore,
The length and breadth of the rectangle is 7 m and 12 m respectively.
Let A and B be the two squares.
And, let 's' cm be the side square A and (s + 4) cm be the side of the square B.
So,
Area of the square A = s2 cm2
Area of the square B =(s + 4)2 cm2
Given:
Area of the square A + Area of the square B = 656 cm2
⇒ s2 cm2 + (s + 4)2 cm2 = 656 cm2
s2+ s2 + 16 + 8s – 656 = 0
2s2 + 16 + 8s – 656 = 0
2(s2 + 4s – 320) = 0
s2 + 4s – 320 = 0
s2 + 20s – 16s – 320 = 0
s(s + 20) – 16(s + 20) = 0
(s + 20)(s – 16) = 0
Now,
either s + 20 = 0
⇒ s = -20 (Side of a square cannot be negative)
Or, s – 16 = 0 ⇒ s = 16
Hence, the value of s = 16 ⇒ s + 4 = 20
Therefore,
The side of the square A= 16 cm
The side of the square B = 20 cm
Let the altitude of the right angle triangle be 'a' m
Given:
The altitude exceeds the base by 7m ⇒ altitude = (a – 7)m
As we know,
Area of the triangle = 1/2 × base × altitude
⇒ 165 = 1/2 × (a − 7) × a
a(a – 7) = 330
a2 – 7a – 330 = 0
a2 – 22a + 15a – 330 = 0
a(a – 22) + 15(a – 22) = 0
(a – 22)(a + 15) = 0
Now,
either a – 22 = 0 ⇒ a = 22
Or, a + 15 = 0 ⇒ a = -15 (Altitude of a triangle cannot be negative)
So the value of a = 22 is only considered
⇒ a – 7 = 15
Hence,
The base and altitude of the right angled triangle are 15 cm and 22 cm respectively.
Let the breadth of the rectangular mango grove be 'b' m
Given:
The length of rectangle is twice of its breadth.
So, length = 2b
Area of the grove = 800 m2 (given)
As we know,
Area of the rectangle = length × breadth
800 = b × (2b)
2b2 – 800 = 0
b2 – 400 = 0
⇒ b = √400 = 20 (neglecting the negative sq. root as side can never be negative)
Therefore,
The breadth of the rectangular groove is 20 m.
And, the length of the rectangular groove is 40 m.
Yes, it is possible to design a rectangular groove whose length is twice of its breadth.
Perimeter of rectangular park = 80 m
Length + Breadth = 80/2 = 40 m
Let length = x m
Then breadth = 40 – x
According to the condition,
Area = Length x Breadth
x (40 – x) = 400
⇒ 40x – x² = 400
⇒ x² – 40x + 400 = 0
⇒ (x – 20)² = 0
⇒ x – 20 = 0
⇒ x = 20
Yes, it is possible
Length = 20 m
and breadth = 40 – x = 40 – 20 = 20 m
Let the side of the first square = x m and of second squares = y m
Given:
4x – 4y = 64
⇒ x – y = 16 ….(i)
and x² + y² = 640 ….(ii)
From (i), x = 16 + y
In (ii)
(16 + y)² + y² = 640
⇒ 256 + 32y + y² + y² = 640
⇒ 2y² + 32y + 256 – 640 = 0
⇒ y² + 16y – 192 = 0 (Dividing by 2)
⇒ y² + 24y – 8y – 192 = 0
⇒ y (y + 24) – 8 (y + 24) = 0
⇒ (y + 24)(y – 8) = 0
Either y + 24 = 0, then y = -24 (which is not possible as it is negative)
or y – 8 = 0, then y = 8
x = 16 + y = 16 + 8 = 24
Side of first square = 24 m
and side of second square = 8 m