Condition is given, here a square and a rectangular field have the same perimeter.
Given, side of a square field = 60 m
∴ Perimeter of square field = 4 Side = 4\(\times\) 60m = 240 m
and area of square field = (side)²= (60)² = 60m\(\times\) 60m
 = 3600 m²
Also, given lengrh of rectangular field = 80 m
Let breadth of rectangular field be x m.
According to the question,
Perimeter of square = Perimeter of rectangle
Perimeter of rectangular field = 240 m
Also, perimeter of a rectangular field = 2 (Length + Breadth)
∴  2(80 + x) = 240 ⇒ (80+x) = \(\frac { 240 }{ 2 } \)
⇒  80 + x = 120
⇒ x = 120- 80 ⇒ x = 40
Now, area of rectangular field = Length \(\times\) Breadth
= 80 m x 40m = 3200 m²
Here, it is clear that, 3600 > 3200
∴  Area of square field> Area of rectangular field
Hence, square field has a larger area.

(i) Given figure is a parallelogram in which base (b) is 14 cm and height (h) is 7 cm.
∴ Area of parallelogram = b\(\times\) h
= 14\(\times\) 7 = 98cm²

(ii) Given figure is a semi-circle, whose radius (r) is 7 cm.
∴  Area of semi CircIe = \(\frac { 1 }{ 2 } { \pi r }^{ 2 }\)
= \(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times (7)^{ 2 }\)        \(\left[ \because \pi =\frac { 22 }{ 7 } \right] \)
= 11 \(\times\)7 = 77 cm²

(iii) Given figure is a triangle in which base is 14 cm and height is 7 cm
∴  Area of rectangle = \(\frac { 1 }{ 2 } \) \(\times\)Base\(\times\) Height
= \(\frac { 1 }{ 2 } \) \(\times\)  14\(\times\) 7 = 49 cm²

(iv)Given figure is a rectangle whose length is 14 cm and breadth is 7 cm.
∴  Area of rectangle =Length \(\times\)Breadth
=  14\(\times\) 7 = 98 cm²

(v) Given figure is a square whose side is 7 cm
∴  Area of square = (Side)²
 = (7)²
 =49 cm²

Hence, the required matching is given below

Figures	Areas
	49 cm2
	77 cm2
	98 cm2

Let given figure be a quadrilateral a ABCD, which is a parallelogram and its diagonal divides it into two congruent triangles. So, base of these two congruent triangles is 8 ern and height is 2 crn (from the figure).

∴ Area of parallelogram = 2 \(\times\)(Area of ΔADC) = 2\(\times\) \(\left( \frac { 1 }{ 2 } \times 8\times 2 \right) \)
 [∵ area of triangle =\(\frac { 1 }{ 2 } \) \(\times\)base \(\times\)height]
= 2 \(\times\)8 = 16 cm²

We know that
Area of trapezium= \(\frac { 1 }{ 2 } \times \) (Sum of parallel sides) \(\times\) Height
Here, height of trapezium = 3 cm and parallel sides are 9 cm and 7 cm
∴ Required area of trapezium = \(\left[ \frac { 1 }{ 2 } \times (9+7)\times 3 \right] \)
= \(\frac { 1 }{ 2 } \times \)16 \(\times\)3 = 8 \(\times\)3 = 24 cm²

9 :4

240 m²

400 m²

Let ABCD be the rhombus as shown below:
DO= OB = 4 cm, since diagonals of a rhombus are perpendicular bisectors of each other. By using Pythagoras theorem in ΔAOB
AO²+OB² AB²
AO = \(\sqrt { { AB }^{ 2 }-OB^{ 2 } } =\sqrt { 5^{ 2 }-4^{ 2 } } \) =3 cm
∴  AC = 2 \(\times\) 3 = 6 cm      [∵ AC =2 A0]
Thus,area of the rhombus
=\(\frac { 1 }{ 2 } \times \) d₁\(\times\)d₂ = 8\(\times\)6 =2\(\times\)4 cm²

Volume of water = 160 m³
Area of rectangular field = 800 m²
Let h be the height of water level in the field.
Now, volume of water = Volume of cuboid formed on the field by water
⇒ 160= Area of base x Height
⇒ 160 = 800\(\times\)h
∴ h = \(\frac { 160 }{ 800 } \)=0.2 m
Hence, the required height is 0.2 m.

Given,diameter of base of cylindrical container = 14 cm
∴ Radius of base of cylindrical container, r =\(\frac { 14 }{ 7 } \) =7cm and height of cylindrical container = 20 cm
Now, label is placed 2 ern from top and bottom.
∴  Height of label, h = (20 - 2 - 2) = 16 cm
Now, area of label = Cured surface area of cylindrical portion of which label is placed
 =   \(2\pi rh=2\times \frac { 22 }{ 7 } \times 7\times 16=704\quad cm^{ 2 }\)  [∵ radius = 7 cm]
Hence, area of the label is 704 cm²

By joining two cubes end-to-end each with side b. We get, a cuboid whose length = b + b = 2b, breadth = b and height = b
∴  Total surface area of this cuboid =2(LB+BH+HL)
 = 2(2b\(\times\)b+b\(\times\)b+b\(\times\)2b) = 2(2b²+b²+2b²) = 2\(\times\)5b² =10b²
So, surface area of cuboid formed by joining two cubes is not equal to 12b² , it is 10b²
Now, by joining three cubes end-to-end each with side b.
We get, a cuboid whose length = b + b + b = 3b, breadth = b and height = b
∴ Total surface area of this cuboid = 2(LB + BH + HL)
= 2(3b\(\times\)b+b\(\times\)b+b\(\times\)3b)
=2\(\times\)(3b²+b²+3b²) = 2\(\times\)7b² =14b²
So, the surface area of cuboid formed by joining three such cuboid is, it is not equal to 18b², it is 14b²

Let ABCD be a rhombus.
Given, side (AB) = 6 cm,altitude (PD) = 4 cm and diagonal (AC) = 8 cm
Let other diagonal BD be x cm.Now, area of a rhombus = base \(\times\)altitude = 6cm\(\times\)4cm =24 cm²
Also,area of a rhombus =\(\frac { 1 }{ 2 } \times \) d₁ \(\times\)d₂
⇒ 24= \(\frac { 1 }{ 2 } \times \)x₁ \(\times\)x₂
⇒ 24 = 4\(\times\)d₂
⇒ x =\(\frac { 24 }{ 4 } \) =6 cm
Hence, length of the other diagonal of rhombus is 6 cm.

Side of both the cubes = c
On joining two cubes of side c each, we get length of the formed cuboid as c + c = 2c, breadth = c and height = c
∴ Surface area of the cuboid, so formed = 2(lb+bh+hl)
 = 2( 2c\(\times\)c + c\(\times\)c + c\(\times\)2c) = 2(2c² + c2 + 2c²) = 10c²

None: on cutting a cube into number of cubes, we have no change in volume of the cube. So, there will be no percentage increase in the volume.

Let the side of the first cube be a and other be b.
∴  \(\frac { { a }^{ 3 } }{ b^{ 3 } } =\frac { 1 }{ 64 } \Rightarrow \frac { { a }^{ 3 } }{ b^{ 3 } } =\left( \frac { 1 }{ 4 } \right) ^{ 3 }\)
⇒ \(\frac { a }{ b } =\frac { 1 }{ 4 } \Rightarrow a=\frac { b }{ 4 } \)
∴ Area of face of first cube = a² = \(\left( \frac { b }{ 4 } \right) ^{ 2 }=\frac { { b }^{ 2 } }{ 16 } \)
and area of face of other cube = b²
∴ Required ratio =\(​​\frac { { b }^{ 2 } }{ 16 } \)= b² =1:16