SELINA Solution Class 9 Chapter 20 Area and perimeter of plane figures Exercise 20B

Question 1

Find the area of a quadrilateral one of whose diagonals is 30 cm long and the perpendiculars from the other two vertices are 19 cm and 11 cm respectively. 

Sol:

Area = ${\displaystyle \frac{1}{2}}$ x one diagonal x sum of the length of the perpendicular drawn from it on the remaining two vertices. 

       = ${\displaystyle \frac{1}{2}\times 30\times (11+19)}$

       =  450 sq .cm

Question 2

The diagonals of a quadrilateral are 16 cm and 13 cm. If they intersect each other at right angles; find the area of the quadrilateral.

Sol:

Area of the quadrilateral = ${\displaystyle \frac{1}{2}}$ x the product of the diagonals.
                                        = ${\displaystyle \frac{1}{2}\times 16\times 13}$
                                        = 104 cm² 

Question 3

Calculate the area of quadrilateral ABCD, in which ∠ABD = 90^o, triangle BCD is an equilateral triangle of side 24 cm and AD = 26 cm.

Sol:

Consider the figure : 

From the right triangle ABD we have 
AB  = ${\displaystyle \sqrt{{26}^2-{24}^2}}$

       =${\displaystyle 2\sqrt{{13}^2-{12}^2}}$

       =  2( 5 )

       = 10

The area of right triangle ABC will be :

ΔABD = ${\displaystyle \frac{1}{2}\left(\text{AB}\right)\left(\text{BD}\right)}$

          = ${\displaystyle \frac{1}{2}}$${\displaystyle \left(10\right)\left(24\right)}$

          =  120

Again from the equilateral triangle BCD, we have  CP ⊥ BD
PC = ${\displaystyle \sqrt{{24}^2-{12}^2}}$

     = ${\displaystyle 12\sqrt{2^2-1^1}}$

     = ${\displaystyle 12\sqrt{3}}$

Therefore the area of the triangle BCD will be :

ΔBCD =${\displaystyle \frac{1}{2}\left(\text{BD}\right)\left(\text{PC}\right)}$

          = ${\displaystyle \frac{1}{2}\left(24\right)\left(12\sqrt{13}\right)}$

          = ${\displaystyle 144\sqrt{3}}$

Hence the area of the quadrilateral will be :
ΔABD + ΔBCD = 120 + ${\displaystyle 144\sqrt{3}}$
                      = 369.41 cm²  

Question 4

Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm ∠A = 90° and BC = CD = 52 cm.

Sol:

The figure can be drawn as follows :

Here ABD is a right triangle. So the area will be :

ΔABD = ${\displaystyle \frac{1}{2}\left(24\right)\left(32\right)}$
          = 384
Again

BD = ${\displaystyle \sqrt{{24}^2+{32}^2}}$

     =  ${\displaystyle 8\sqrt{3^2+4^2}}$

     =  8 ( 5 )

     = 40

Now BCD is an isosceles triangle and BP is perpendicular to BD, therefore

DP =${\displaystyle \frac{1}{2}\text{BD}}$

     = ${\displaystyle \frac{1}{2}\left(40\right)}$

     = 20

From the right triangle DPC we have

PC = sqrt(52^2 - 20^2)` 

     = ${\displaystyle 4\sqrt{{13}^2-5^2}}$
     = 4(12)
     = 48

So
ΔDPC =${\displaystyle \frac{1}{2}\left(40\right)\left(48\right)}$
          = 960

Hence the area of the quadrilateral will be :
ΔABD + ΔDPC = 960 + 384
                        = 1344 cm²  

Question 5

The perimeter of a rectangular field is ${\displaystyle \frac{3}{5}}$km. If the length of the field is twice its width; find the area of the rectangle in sq. meters.

Sol:

Let the width be x and length 2x km.
Hence

${\displaystyle 2(x+2x)=\frac{3}{5}}$

             x = ${\displaystyle \frac{1}{10}}$km

                = 100 m

Hence the width is 100m and length is 200m
The required area is given by
A = length x width
    = 100 x 200
    = 20,000 sq .m

Question 6

A rectangular plot 85 m long and 60 m broad is to be covered with grass leaving 5 malls around. Find the area to be laid with grass. 

Sol:

Length of the laid with grass = 85 - 5 - 5 = 75 m

Width of the laid with grass = 60 - 5 - 5 = 50 m

Hence the area of the laid with grass is given by

A =  75 x 50

   = 3750 sq . m

Question 7

The length and the breadth of a rectangle are 6 cm and 4 cm respectively. Find the height of a triangle whose base is 6 cm and the area is 3 times that of the rectangle.

Sol:

Area of the rectangle is given by

A = l x b

   = 6 x 4

   = 24 sq . cm

Let h be the height of the triangle ,then 

${\displaystyle \frac{1}{2}\times \text{base}\times h=3\text{A}}$

${\displaystyle \frac{1}{2}\times 6\times h=3\times 24}$ 

                  h = 24 cm

 

Question 8

How many tiles, each of area 400 cm², will be needed to pave a footpath which is 2 m wide and surrounds a grass plot 25 m long and 13 m wide? 

Sol:

Consider the following figure. 

Thus the required area = area shaded in blue + area shaded in red

= Area ABPQ +  Area TUDC +  Area A'PUD' +  Area QB'C'T

= 2Area ABPQ + 2Area QB'C'T

=2( Area ABPQ + Area QB'C'T )

Area of the footpath is given by
A =  2 x ( 25 + 25 + 17 + 17 )
   = 168 sq . cm
   = 1680000 sq . cm

Hence number of tiles required = ${\displaystyle \frac{1680000}{400}}$ = 4200

Question 9

The cost of enclosing a rectangular garden with a fence all around, at the rate of 75 paise per metre, is Rs. 300. If the length of the garden is 120 metres, find the area of the field in square metres. 

Sol:

The perimeter of the garden

s =  ${\displaystyle \frac{300}{0.75}}$ = 400 sq .m

Again, the length of the garden is given to be 120 m. hence breadth of the garden b is given by

2( l + b ) = S
2 ( 120 + b ) = 400
               b = 80m

Hence area of the field
A = 120 x 80
   = 9600 sq . m

Question 10

The width of a rectangular room is ${\displaystyle \frac{4}{7}}$of its length, x, and its perimeter is y. Write an equation connecting x and y. Find the length of the room when the perimeter is 4400 cm.

Sol:

Length of the rectangle = x

Width of the rectangle  = ${\displaystyle \frac{4}{7}}$ x

Hence its perimeter is given by 

2${\displaystyle (x+\frac{4}{7}x)}$ = y 

2${\displaystyle \left(\frac{11x}{7}\right)}$ =y

${\displaystyle \frac{22x}{7}}$ = y

Again it is given that the perimeter is 4400cm. 

Hence

${\displaystyle \frac{22x}{7}}$ = 4400

x = 1400

Length of the rectangle = 1400 cm = 14 m

Question 11

The length of a rectangular verandah is 3 m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter.

(i) Taking x as the breadth of the verandah, write an equation in x that represents the above statement

(ii) Solve the equation obtained in (i) above and hence find the dimensions of the verandah. 

Sol:

(i) The breadth of the verandah = x
Length of the verandah = x + 3
According to the question

2( x + ( x + 3 )) = x ( x + 3)
4x + 6 = x² + 3x 
x² - x - 6 = 0

(ii) From the above equation
           x² - x - 6 = 0
( x - 3 ) ( x + 2 ) = 0
                      x = 3
Hence breadth = 3m
Length = 3 + 3 = 6m

Question 12

The diagram, given below, shows two paths drawn inside a rectangular field 80 m long and 45 m wide. The widths of the two paths are 8 m and 15 m as shown. Find the area of the shaded portion. 

Sol:

Consider the following figure.

Thus, the area of the shaded portion
= Area( ABCD ) + Area( EFGH ) - Area( IJKL )           …(1)

Dimensions of ABCD: 45m x 15 m
Thus, the area of ABCD = 45 x 15 = 675 m²

Dimensions of EFGH : 80 m x 8 m
Thus, the area of EFGH = 80 x 8 = 640m²

Dimensions of IJKL: 15 m x 8 m
Thus, the area of IJKL = 80 x 8 = 120 m²

Therefore, from equation (1),

the area of the shaded portion = 675 + 640 - 120 = 1195 m²

Question 13

The rate for a 1.20 m wide carpet is Rs. 40 per meter; find the cost of covering a hall 45 m long and 32 m wide with this carpet. Also, find the cost of carpeting the same hall if the carpet, 80 wide, is at Rs. 25. Per meter. 

Sol:

First, we have to calculate the area of the hall.

Area = 45 x 32
        = 1440 m²

Cost = ${\displaystyle \frac{40}{1.20}\times 1440}$

        = 48,000

We need to find the cost of carpeting of 80 cm = 0.8 m wide carpet if the rate of carpeting is Rs. 25. Per meter.

Then

Cost = ${\displaystyle \frac{25}{0.8}\times 1440}$

        = Rs. 45,000

Question 14

Find the area and perimeter of a square plot of land, the length of whose diagonal is 15 meters. Given your answer correct to 2 places of decimals. 

Sol:

Let a be the length of each side of the square.
Hence
2a² = ( diagonal )²

a² = ${\displaystyle \frac{{15}^2}{2}}$

a² = 112.5

a = 10.60

Hence

Area = a²
        = 112.5 sq . m

And
Perimeter = 4a = 42.43 m

Question 15

The shaded region of the given diagram represents the lawn in the form of a house. On the three sides of the lawn, there are flowerbeds having a uniform width of 2 m.

(i) Find the length and the breadth of the lawn.
(ii) Hence, or otherwise, find the area of the flower-beds.

Sol:

Consider the following figure

(i) The length of the lawn = 30 - 2 - 2 = 26 m
The breadth of the lawn = 12 - 2 = 10 m

(ii) The orange shaded area in the figure is the required area.
Area of the flower bed is calculated as follows:
A = 10 x 2 + 10 x 2 + 30 x 2
   = 20 + 20 + 60 
   = 100 sq.m.

Question 16

A floor that measures 15 m 8 m is to be laid with tiles measuring 50 m 25 cm. Find the number of tiles required.
Further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered?

Sol:

Area of the floor = 15 x 8 = 120 sq.m

Area of one tiles = 0.50 x 0.25 = 0.125 sq.m

Number of tiles required
n = ${\displaystyle \frac{\text{Area of floor}}{\text{Area of tiles}}}$

= ${\displaystyle \frac{120}{0.125}}$

= 960

Area of carpet uncovered = 2( 1 x 15 + 1 x 6 ) = 42 sq.m

Fraction of floor uncovered = ${\displaystyle \frac{42}{120}=\frac{7}{20}}$

Question 17

Two adjacent sides of a parallelogram are 24 cm and 18 cm. If the distance between the longer sides is 12 cm; find the distance between the shorter sides.

Sol:

Since
Area = Base x Height
∴ 24 x 12 = 18 x h

h = ${\displaystyle \frac{24\times 12}{18}}$

   = 16 m
Hence the distance between the shorter sides is 16m.

Question 18

Two adjacent sides of a parallelogram are 28 cm and 26 cm. If one diagonal of it is 30 cm long; find the area of the parallelogram. Also, find the distance between its shorter sides.

Sol:

At first, we have to calculate the area of the triangle having sides, 28 cm, 26 cm, and 30 cm.

Let the area be S.

S = ${\displaystyle \frac{28+26+30}{2}}$

  = ${\displaystyle \frac{84}{2}}$

  = 42 cm
By Heron's Formula,
Area of a triangle = ${\displaystyle \sqrt{s(s-a)(s-b)(s-c)}}$

= ${\displaystyle \sqrt{42(42-28)(42-26)(42-30)}}$

= ${\displaystyle \sqrt{42\times 14\times 16\times 12}}$

= ${\displaystyle \sqrt{112896}}$
= 336 cm²

Area of a Parallelogram = 2 × Area of a triangle
= 2 × 336 = 672 cm²

We know that,
Area of a parallelogram = Height × Base

⇒ 672 = Height × 26

⇒ Height = 25.84 cm

∴ the distance between its shorter sides is 25.84 cm.

Question 19

The area of a rhombus is 216 sq. cm. If it's one diagonal is 24 cm; find:
(i) Length of its other diagonal,
(ii) Length of its side,
(iii) The perimeter of the rhombus.

Sol:

(i) We know that,
Area of Rhombus = ${\displaystyle \frac{1}{2}}$ x AC x BD
Here
A = 216 sq.cm
AC = 24 cm
BD = ?

Now,
A = ${\displaystyle \frac{1}{2}}$ x AC x BD

216 = ${\displaystyle \frac{1}{2}}$ x 24 x BD

BD = 18 cm.

(ii) Let a be the length of each side of the rhombus.

a² = ${\displaystyle {\left(\frac{\text{AC}}{2}\right)}^2+{\left(\frac{\text{BD}}{2}\right)}^2}$

a² = 12² + 9²

a² = 225
a = 15 cm

(iii) Perimeter of the rhombus = 4a = 60 cm.

Question 20

The perimeter of a rhombus is 52 cm. If one diagonal is 24 cm; find:
(i) The length of its other diagonal,
(ii) Its area.

Sol:

Let a be the length of each side of the rhombus.
4a = perimeter
4a = 52
a = 13 cm

(i) It is given that,
AC= 24 cm
We have to find BD.
Now
${\displaystyle a^2={\left(\frac{\text{AC}}{2}\right)}^2+{\left(\frac{\text{BD}}{2}\right)}^2}$

${\displaystyle {13}^2={12}^2+{\left(\frac{\text{BD}}{2}\right)}^2}$

${\displaystyle {\left(\frac{\text{BD}}{2}\right)}^2=5^2}$

BD = 10 cm
Hence the other diagonal is 10cm.

(ii) Area of the rhombus = ${\displaystyle \frac{1}{2}}$ x AC x BD

                                       = ${\displaystyle \frac{1}{2}}$ x 24 x 10

                                       = 120 sq.cm.

Question 21

The perimeter of a rhombus is 46 cm. If the height of the rhombus is 8 cm; find its area.

Sol:

Let a be the length of each side of the rhombus.
4a = perimeter
4a = 46
a = 11.5 cm

We know that,
Area = Base x Height
         = 11.5 x 8 
         = 92 sq.cm

Question 22

The figure given below shows the cross-section of a concrete structure. Calculate the area of cross-section if AB = 1.8 cm, CD = 0.6 m, DE = 0.8 m, EF = 0.3 m and AF = 1.2 m.

Sol:

The diagram is redrawn as follows:

Here
AF = 1.2 m, EF = 0.3m, DC = 0.6m, BK = 1.8 - 0.6 - 0.3 = 0.9 m
Hence,
Area of ABCDEF = Area of AHEF + Area of HKCD + Δ KBC

= 1.2 x 0.3 + 2 x 0.6 + ${\displaystyle \frac{1}{2}}$ x 2 x 0.9

= 2.46 sq.m

Question 23

Calculate the area of the figure given below: which is not drawn scale.

Sol:

Here we found two geometrical figures, one is a triangle and the other is the trapezium.
Now,
Area of the Triangle = ${\displaystyle \frac{1}{2}}$ x 12 x 25
                                = 150 sq.cm

Area of Trapezium = ${\displaystyle \frac{1}{2}\times (25+15)\times \left(\sqrt{{26}^2-{(25-15)}^2}\right)}$
= 20 x 24
= 480 sq.cm
hence area of the whole figure = 150 + 480 = 630 sq.cm

Question 24

The following diagram shows a pentagonal field ABCDE in which the lengths of AF, FG, GH, and HD are 50 m, 40 m, 15 m and 25 m respectively; and the lengths of perpendiculars BF, CH and EG are 50 m, 25 m and 60 m respectively. Determine the area of the field.

Sol:

We can divide the field into three triangles and one trapezium.
Let A, B, C be the three triangular regions and D be the trapezoidal region.
Now,
Area of A = ${\displaystyle \frac{1}{2}}$ x AD x GE

                = ${\displaystyle \frac{1}{2}}$ x ( 50 + 40 + 15 + 25 ) x 60

                = 3900 sq.m 

Area of B = ${\displaystyle \frac{1}{2}}$ x AF x BF

                = ${\displaystyle \frac{1}{2}}$ x 50 x 50

                = 1250 sq.m

Area of B = ${\displaystyle \frac{1}{2}}$ x HD x CH

                = ${\displaystyle \frac{1}{2}}$ x 25 x 25

                = 312.5 sq.m.

Area of D = ${\displaystyle \frac{1}{2}}$ x ( BF + CH ) x ( FG + GH )

                = ${\displaystyle \frac{1}{2}}$ x ( 50 + 25 ) x ( 40 + 15 )

               = ${\displaystyle \frac{1}{2}}$ x 75 x 55 
               = 2062.5 sq.cm     

Area of the figure = Area of A + Area of B + Area of C + Area of D
= 3900 + 1250 + 312.5 + 2062.5
= 7525 sq.m     

Question 25

A footpath of uniform width runs all around the outside of a rectangular field 30 m long and 24 m wide. If the path occupies an area of 360 m², find its width.

Sol:

Let x be the width of the footpath.
Then
Area of footpath = ${\displaystyle 2\times (30+24)x+4x^2}$
                            = 4x² + 108x

Again it is given that the area of the footpath is 360sq.m.
Hence,
4x² + 108x = 360
x² + 27x - 90 = 0
( x - 3 )( x + 30 ) = 0
x = 3
Hence width of the footpath is 3m.

Question 26

A wire when bent in the form of a square encloses an area of 484 m². Find the largest area enclosed by the same wire when bent to from:
(i) An equilateral triangle.
(ii) A rectangle of length 16 m.

Sol:

The area of the square is 484.

Let a be the length of each side of the square.
Now
a² = 484 
a = 22 m
Hence the length of the wire is = 4 x 22 = 88 m.

(i) Now, this 88 m wire is bent in the form of an equilateral triangle.
Side of the triangle = ${\displaystyle \frac{88}{3}}$

                               = 29.3 m

Area of the triangle = ${\displaystyle \frac{\sqrt{3}}{4}}$ x ( Side )²

                                = ${\displaystyle \frac{\sqrt{3}}{4}}$ x ( 29.3 )²

                                = 372.58 m²

(ii) Let x be the breadth of the rectangle.
Now,
2( l + b ) = 88
16 + x = 44
x = 28 m
Hence area = 16 x 28 = 448 m².

Question 27.1

Trapezium given below; find its area.

Sol:

Area of ΔEBC = ${\displaystyle \frac{1}{4}\times 8\times \sqrt{4\times {10}^2-8^2}}$

= ${\displaystyle \frac{1}{4}\times 8\times 18.3}$
= 36.6 cm²

Again
Area of ΔEBC = ${\displaystyle \frac{1}{2}}$ x 8 x h

              36.6 = 4h
                  h = 9.15

Area of ABCD = ${\displaystyle \frac{1}{2}\times (12+20)\times 9.15}$

                       = ${\displaystyle \frac{1}{2}}$ x 32 x 9.15

                       = 146.64 sq.cm

Question 27.2

Trapezium given below; find its area.

Sol:

Area of ABCD = ${\displaystyle \frac{1}{2}\times (8+14)\times \left(\sqrt{{10}^2-6^2}\right)}$

= ${\displaystyle \frac{1}{2}}$ x 22 x 8
= 88 sq.cm

Question 27.3

 Trapezium given below; find its area.

Sol;

For the triangle EBC,
S = 19 cm

Area of ΔEBC = ${\displaystyle \sqrt{19\times (19-16)\times (19-12)\times (19-10)}}$

= ${\displaystyle \sqrt{19\times 3\times 7\times 9}}$

= 59.9 sq.cm

Let h be the height.
Area of ΔEBC= ${\displaystyle \frac{1}{2}}$ x 12 x h
⇒ 59.9 = 6h
⇒ h = ${\displaystyle \frac{59.9}{6}}$ = 9.98 cm

Area of ABCD = ${\displaystyle \frac{1}{2}}$ x ( 20 + 32 ) x 9.98

                       = ${\displaystyle \frac{1}{2}}$ x 52 x 9.98
                       = 259.48 cm²

Question 27.4

Trapezium given below; find its area.

Sol:

In the given figure, we can observe that the non-parallel sides are equal and hence it is an isosceles trapezium.

Therefore, let us draw DE and CF perpendiculars to AB.

Thus, the area of the parallelogram is given by
AB = AE + EF + FB and CD = EF = 18 cm, we have
30 = AE + 18 + FB
⇒ 30 = AE + 18 + AE
⇒  2AE + 18 = 30
⇒ 2AE = 30 - 18
⇒ 2AE = 12
⇒ AE = 6 cm

Now, consider the right angled triangle ADE.
AD² = AE² + DE²
⇒ 12² = 6² + DE²
⇒  144 = 36 + DE²
⇒ DE² = 144 - 36
⇒ DE² = 108
⇒ DE = ${\displaystyle \sqrt{36\times 3}}$
⇒ DE = ${\displaystyle 6\sqrt{3}}$

Area( ${\displaystyle \square \text{ABCD})=\text{Area}(\Delta \text{ADE})+\text{Area}(\square \text{DEFC})+\text{Area}(\Delta \text{CFB})}$

⇒ Area( ${\displaystyle \square \text{ABCD})=\frac{1}{2}\times 6\times 6\sqrt{3}+18\times 6\sqrt{3}+\frac{1}{2}\times 6\times 6\sqrt{3}}$

⇒ Area( ${\displaystyle \square \text{ABCD})=6\times 6\sqrt{3}+18\times 6\sqrt{3}}$

⇒ Area( ${\displaystyle \square \text{ABCD})=144\sqrt{3}}$= 249.41 cm²

Question 28

The perimeter of a rectangular board is 70 cm. Taking its length as x cm, find its width in terms of x.
If the area of the rectangular board is 300 cm²; find its dimensions.

Sol:

Let b be the breadth of the rectangle. then its perimeter
2( x + b ) = 70
x + b = 35
b = 35 - x

Again
${\displaystyle x}$ x b = 300
x( 35 - x ) = 300
x² - 35x + 300 = 0
( x - 15 )( x - 20 ) = 0
x = 15, 20.
Hence its length is 20 cm and width is 15 cm.

Question 29

The area of a rectangular is 640 m². Taking its length as x cm; find in terms of x, the width of the rectangle. If the perimeter of the rectangle is 104 m; find its dimensions.

Sol:

Let b be the width of the rectangle.
${\displaystyle x}$ x b = 640

b = ${\displaystyle \frac{640}{x}}$

Again perimeter of the rectangle is 104 m.
Hence,
${\displaystyle 2(x+\frac{640}{x})}$ = 104

x² - 52x + 640 = 0
( x - 32 )( x - 20 ) = 0
x = 32, 20.

Hence
length = 32 m
width = 20 m.

Question 30

The length of a rectangle is twice the side of a square and its width is 6 cm greater than the side of the square. If the area of the rectangle is three times the area of the square; find the dimensions of each.

SOL:

Let a be the length of the sides of the square.

According to the question,
2a x ( a + 6 ) = 3a²
2a² + 12a = 3a²
a = 12

Hence sides of the square are 12 cm each and
Length of the rectangle = 2a = 24 cm.
Width of the rectangle = a + 6 = 18 cm.

Question 31

ABCD is a square with each side 12 cm. P is a point on BC such that area of  ΔABP: area of trapezium APCD = 1: 5. Find the length of CP.

Sol:

The figure is shown below:

${\displaystyle \frac{\text{Area of ΔABP}}{\text{Area of trapezium APCD}}=\frac{1}{5}}$

⇒ ${\displaystyle \frac{\frac{1}{2}\times 12\times (12-\text{CP})}{\frac{1}{2}\times (12+\text{CP})\times 12}=\frac{1}{5}}$

⇒ 60 - 5CP = 12 + CP
⇒  6CP = 48
⇒ CP = 8 cm.

Question 32

A rectangular plot of land measures 45 m x 30 m. A boundary wall of height 2.4 m is built all around the plot at a distance of 1 m from the plot. Find the area of the inner surface of the boundary wall.

Sol:

Length of the wall = 45 + 2 = 47 m

Breath of the wall = 30 + 2 = 32 m

Hence area of the inner surface of the wall is given by

A = ( 47 x 2 x 2.4 ) + ( 32 x 2 x 2.4 )
= 225.6 + 153.6
= 379.2 m²

Question 33

A wire when bent in the form of a square encloses an area = 576 cm². Find the largest area enclosed by the same wire when bent to form;
(i) an equilateral triangle.
(ii) A rectangle whose adjacent sides differ by 4 cm.

Sol:

Let a be the length of each side.
a² = 576
a = 24 cm
4a = 96 cm
Hence length of the wire = 96 cm

(i) For the equilateral triangle,
side = ${\displaystyle \frac{96}{3}}$ = 32 cm

Area = ${\displaystyle \frac{\sqrt{3}}{4}}$( side )²

        = ${\displaystyle \frac{\sqrt{3}}{4}}$ x 32²

        = 256√3 sq.cm

(ii) Let the adjacent side of the rectangle be x and y cm.
Since the perimeter is 96 cm, we have,
2( x + y ) = 96
Hence,
x + y = 48
x - y = 4
x = 26
y = 22
Hence area of the rectangle is = 26 x 22 = 572 sq.cm

Question 34

The area of a parallelogram is y cm² and its height is h cm. The base of another parallelogram is x cm more than the base of the first parallelogram and its area is twice the area of the first. Find, in terms of y, h, and x, the expression for the height of the second parallelogram.

Sol:

Let 'y' and 'h' be the area and the height of the first parallelogram respectively.

Let 'height' be the height of the second parallelogram

base of the first parallelogram = ${\displaystyle \frac{y}{h}}$cm

the base of the second parallelogram = ${\displaystyle (\frac{y}{h}+x)}$ cm

${\displaystyle (\frac{y}{h}+x)}$ x height = 2y

height = ${\displaystyle \frac{2hy}{y+hx}}$

Question 35

The distance between parallel sides of a trapezium is 15 cm and the length of the line segment joining the mid-points of its non-parallel sides is 26 cm. Find the area of the trapezium.

Sol:

EF = ${\displaystyle \frac{1}{2}}$ x ( AD + BC ) = 26 cm

Area of the trapezium
= ${\displaystyle \frac{1}{2}}$ x ( AD + BC ) x h

= 26 x 15
= 390 cm².              

Question 36

The diagonal of a rectangular plot is 34 m and its perimeter is 92 m. Find its area.

Sol:

Let a and b be the sides of the rectangle
Since the perimeter is 92 m, we have,
2( a + b ) = 92
⇒ a + b = 46 m                      ...(1)
Also, given that diagonal of a trapezium is 34 m.
⇒ a² + b² = 34²                     ....(2)

We know that
( a + b )² - a² - b² = 2ab

From equations (1) and (2), we have,
46² - 34² = 2ab
⇒ 2ab = 960
⇒ ab = ${\displaystyle \frac{960}{2}}$
⇒ ab = 480m².