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Page No 772:

Question 1:

Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.

Answer:

Given: base=24 cm, correponding height=14.5 cm
Area of a triangle =12×base×coresponding height

                       =12×24×14.5=174 cm2

Page No 772:

Question 2:

Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Find the height corresponding to the longest side.

Answer:

Let the sides of the triangle be ​a = 20 cm, b = 34 cm and c = 42 cm.
Let s be the semi-perimeter of the triangle.
s=12(a+b+c)s=12(20+34+42)s=48 cm

Area of the triangle =​s(s-a)(s-b)(s-c)48(48-20)(48-34)(48-42)48×28×14×6112896336 cm2

Length of the longest side is 42 cm.

Area of a triangle =12×b×h
336=12×42×h 672=42h 67242=h h=16 cm

The height corresponding to the longest side is 16 cm.

Page No 772:

Question 3:

Find the area of the triangle whose sides are 18 cm, 24 cm and 30 cm. Also, find the height corresponding to the smallest side.

Answer:

Let the sides of triangle be ​a = 18 cm, b = 24 cm and c = 30 cm.
Let s be the semi-perimeter of the triangle.
s=12(a+b+c)s=12(18+24+30)s=36 cm


Area of a triangle =​s(s-a)(s-b)(s-c)=36(36-18)(36-24)(36-30)=36×18×12×6=46656=216 cm2
The smallest side is 18 cm long. This is the base.

Now, area of a triangle =12×b×h
216=12×18×h 216=9h 2169=h h=24 cm

The height corresponding to the smallest side is 24 cm.

Page No 772:

Question 4:

The sides of a triangle are in the ratio 5 : 12 : 13, and its perimeter is 150 m. Find the area of the triangle.

Answer:

Let the sides of a triangle be 5x m ,12x m and 13x m.

Since, perimeter is the sum of all the sides,
5x+12x+13x=15030x =150or, x = 15030=5

The  lengths of the sides are:
a=5×5=25 mb=12×5=60 mc=13×5=65 mSemiperimeter (s) of the triangle = Perimeter2=25+60+652=1502=75 mArea of triangle = ss-as-bs-c=7575-2575-6075-65=75×50×15×10= 562500=750 m2

Page No 772:

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 field. Also, find the cost of ploughing the field at ₹40 per 100 m2.

Answer:

Let the sides of the triangular field be 25x, 17x and 12x.As, perimeter=540 m25x+17x+12x=54054x=540x=54054x=10So, the sides are 250 m, 170 m and 120 m.Now, semi-perimeter, s=250+170+1202=5402=270 mSo, area of the field=270270-250270-170270-120=270×20×100×150=33×10×2×10×102×3×5×10=32×103=9000 m2Also, the cost of ploughing the field=9000×40100=3,600

Page No 772:

Question 6:

The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm. Find the area of the triangle.

Answer:

The perimeter of a right-angled triangle = 40 cm
Therefore , a+b+c= 40 cm
Hypotenuse = 17 cm
Therefore, c = 17 cm
a+b+c= 40 cm
 a+b+17 = 40
 a+b = 23
 b = 23 -a...........(i)
Now, using Pythagoras' theorem, we have:
 a2+b2=c2 a2+(23-a)2=172 a2+529-46a+a2=289 2a2-46a+529-289=0 2a2-46a+240=0 a2-23a+120=0 (a-15)(a-8)=0 a=15 or a= 8

Substituting the value of a=15, in equation(i) we get:
b = 23-a
 = 23 - 15
= 8 cm

If we had chosen a=8 cm, then, b=23-8=15 cm

In any case,
​  Area of a triangle =  12×base×height                                    =12×8×15                                    =60 cm2

Page No 772:

Question 7:

The difference between the sides at right angle in a right-angled triangle is 7 cm. The area of the triangle is 60 cm2. Find its perimeter.

Answer:

Given:
Area of the triangle = 60 cm2
Let the sides of the triangle be a, b and c, where a is the height, b is the base and c is hypotenuse of the triangle.
a-b=7cm
a = 7 + b.......(1)
Area of triangle =12×b×h
 60=12×b×(7+b) 120=7b+b2 b2+7b-120=0 (b+15)(b-8)=0 b=-15 or  8

Side of a triangle cannot be negative.
Therefore, b = 8 cm.

Substituting the value of b = 8 cm, in equation (1):
a = 7+8 = 15 cm

Now,  a = 15 cm, b = 8 cm

Now, in the given right triangle, we have to find third side.
 (Hyp)2=(First side)2+(Second side)2 Hyp2=82+152 Hyp2=64+225 Hyp2=289 Hyp=17cm
So, the  third side is 17 cm.

Perimeter of a triangle = a+b+c.
Therefore, required perimeter of the triangle =15+8+17=40 cm.

Page No 772:

Question 8:

The lengths of the two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangle is 24 cm2, find the perimeter of the triangle.

Answer:

Given:
Area of triangle = 24 cm2
Let the sides be a and b, where a is the height and b is the base of triangle.

a-b=2 cm
a = 2 + b.......(1)

Area of triangle =12×b×h

 24=12×b×(2+b) 48=b+12b2 48=2b+b2 b2+2b-48=0(b+8)(b-6)=0 b=-8 or 6
Side of a triangle cannot be negative.

Therefore, b = 6 cm.

Substituting the value of b = 6 cm in equation(1), we get:
a = 2+6 = 8 cm

Now,  a = 8 cm, b = 6 cm

In the given right triangle we have to find third side. Using the relation

 (Hyp)2=(Oneside)2+(Otherside)2 Hyp2=82+62 Hyp2=64+36 Hyp2=100 Hyp=10 cm

So, the third side is 10 cm.

So, perimeter of the triangle = a b + c
= 8+6+10
​=24 cm

Page No 772:

Question 9:

Each side of an equilateral triangle is 10 cm. Find

(i) the area of the triangle and (ii) the height of the triangle.

Answer:

i The area of the equilateral triangle=34×side2=34×102=34×100=253 cm2or 25×1.732=43.3 cm2So, the area of the triangle is 253 cm2 or 43.3 cm2.ii As, area of the equilateral  triangle=253 cm212×Base×Height=25312×10×Height=2535×Height=253Height=2535=53or height=5×1.732=8.66 m The height of the triangle is 53 cm or 8.66 cm.

Page No 772:

Question 10:

The height of an equilateral triangle is 6 cm. Find its area. [Take 3 = 1.73]

Answer:

Let the side of the equilateral triangle be x cm.

As, the area of an equilateral triangle=34side2=x234Also, the area of the triangle=12×Base×Height=12×x×6=3xSo, x234=3xx34=3x=123x=123×33x=1233x=43 cmNow, area of the equilateral triangle=3x=3×43=123=12×1.73=20.76 cm2

Page No 772:

Question 11:

If the area of an equilateral triangle is 363cm2, find its perimeter.

Answer:

Area of equilateral triangle = 363 cm2

Area of equilateral triangle = 34×a2, where a is the length of the side.
 363=34×a2 144=a2 a=12 cm

Perimeter of a triangle = 3a
                              =3×12=36 cm

Page No 772:

Question 12:

If the area of an equilateral triangle is 813cm2, find its height.

Answer:


Area of the equilateral triangle = 813 cm2
Area of an equilateral triangle =34×a2, where a is the length of the side.
 813=34×a2 324=a2 a=18 cm
Height of triangle = 32×a

                         =32×18=93 cm

Page No 772:

Question 13:

The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.

Answer:

Base = 48 cm
Hypotenuse =50 cm
First we will find the height of the triangle; let the height be 'p'.
 Hypotenuse2=base2+p2 502=482+p2 p2=502-482 p2=(50-48)(50+48) p2=2×98 p2=196 p=14 cm


Area of the triangle=12×base×height

                      =12×48×14=336 cm2

Page No 772:

Question 14:

The hypotenuse of a right-angled triangle measures 65 cm and its base is 60 cm. Find the length of perpendicular and the area of the triangle.

Answer:

Hypotenuse = 65 cm
​Base = 60 cm
In a right-angled triangle,
Hypotenuse2=Base2+Perpendicular2 652=602+ perpendicular2 652-602=perpendicular2 perpendicular2=(65-60)(65+60) perpendicular2=5×125perpendicular2=625 perpendicular=25 cm​

Area of triangle = 12×base×perpendicular

                         = 12×60×25= 750 cm2



Page No 773:

Question 15:

Find the area of a right-angled triangle, the radius of whose circumcircle measure 8 cm and the altitude drawn to the hypotenuse measures 6 cm.

Answer:


Given: Radius = 8 cm
​Height = 6 cm
Area=?
In a right-angled triangle, the centre of the circumcircle is the mid-point of the hypotenuse.

Hypotenuse=2×(radius of circumcircle) for a right triangle                     = 2×8                     =16 cmSo, hypotenuse= 16 cm
Now, base = 16 cm and height = 6 cm

Area of the triangle =12×base×height

                                  =12×16×6=48 cm2

Page No 773:

Question 16:

Find the length of the hypotenuse of an isosceles right-angled triangle whose area is 200 cm2. Also, find its perimeter. [Given: 2 = 1.41]

Answer:

In a right isosceles triangle, base=height=a

Therefore,
  Area of the triangle = 12×base×height=12×a×a=12a2

Further, given that area of isosceles right triangle = 200 cm2

12a2=200a2=400or, a=400 =20 cm

In an isosceles right triangle, two sides are equal ('a') and the third side is the hypotenuse, i.e. 'c

Therefore, c = a2+a2

                   = 2a2= a2= 20×1.41= 28.2 cm

Perimeter of the triangle = a+a+c
​                                       =20+20+28.2

                                        = 68.2 cm
 
The length of the hypotenuse is 28.2 cm and the perimeter of the triangle is 68.2 cm.

Page No 773:

Question 17:

The base of an isosceles triangle measures 80 cm and its area is 360 cm2. Find the perimeter of the triangle.

Answer:

Given  :
Base = 80 cm
Area = 360 cm2
Area of an isosceles triangle = 14b4a2-b2

    360=14×804a2-802 360=204a2-6400 18=2a2-1600 9=a2-1600
Squaring both the sides, we get:
 81=a2-1600 a2=1681 a=41 cm
Perimeter =2a+b
 =241+80=82+80=162 cm
So, the perimeter of the triangle is 162 cm.
         

Page No 773:

Question 18:

Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measures 12 cm. Find the area of the triangle.

Answer:

Let the height of the triangle be h cm.
Each of the equal sides measures a = h+2 cm and b = 12 cm (base).

Now,
Area of the triangle = Area of the isosceles triangle
12×base×height=14×b4a2-b2

12×12×h=14×12×4(h+2)2-1446h=34h2+16h+16-1442h=4h2+16h+16-144On squaring both the sides, we get:4h2=4h2+16h+16-14416h-128=0h=8

Area of the triangle = 12×b×h
                                =12×12×8 = 48 cm2

Page No 773:

Question 19:

Find the area and perimeter of an isosceles right-angled triangle, each of whose equal sides measures 10 cm. [Given: 2 = 1.41]

Answer:

Let:
Length of each of the equal sides of the isosceles right-angled triangle = a = 10 cm
And,
Base = Height = a
Area of isosceles right-angled triangle=12×Base×Height=12×10×10=50 cm2

The hypotenuse of an isosceles right-angled triangle can be obtained using Pythagoras' theorem.

If h denotes the hypotenuse, we have:

h2=a2+a2h=2a2h=2a h=102 cm

∴ Perimeter of â€‹the isosceles right-angled triangle = 2a+2a
                                                                                =2×10+1.41×10=20+14.1=34.1 cm

Page No 773:

Question 20:

In the given figure, ∆ABC is an equilateral triangle the length of whose side is equal to 10 cm, and ∆DBC is right-angled at D and BD = 8 cm. Find the area of the shaded region. Take 3 = 1.732.

Answer:

Given:
Side of equilateral triangle ABC = 10 cm
BD = 8 cm
Area of equilateralABC=34a2 (where a = 10 cm)
Area of equilateral ABC=34×102= 253= 25×1.732= 43.30 cm2

In the right BDC, we have:BC2=BD2+CD2102=82+CD2CD2=102-82CD2=36CD=6

Area of triangle BCD = 12×b×h

                                      =12×8×6=24 cm2

Area of the shaded region = Area of ABC - Area ofBDC
                                           = 43.30 - 24
                                           = 19.3 cm2



Page No 780:

Question 1:

The perimeter of a rectangular plot of land is 80 m and its breadth is 16 m. Find the length and area of the plot.

Answer:

As, perimeter=80 m2length+breadth=802length+16=802×length+32=802×length=80-32length=482 length=24 mNow, the area of the plot=length×breadth=24×16=384 m2

So, the length of the plot is 24 m and its area is 384 m2.

Page No 780:

Question 2:

The length of a rectangular park is twice its breadth, and its perimeter measures 840 m. Find the area of the park.

Answer:

Let the breadth of the rectangular park be b.
∴ Length of the rectangular park=l=2b
Perimeter = 840 m

840=2 l+b840=2 2b+b840=23b840=6bb=140 mThus, we have:l=2b =2×140 =280 m

Area=l×b        =280×140        =39200 m2   



Page No 781:

Question 3:

One side of a rectangle is 12 cm long and its diagonal measures 37 cm. Find the other side and the area of the rectangle.

Answer:

One side of the rectangle = 12 cm
Diagonal of the rectangle = 37 cm

The diagonal of a rectangle forms the hypotenuse of a right-angled triangle. The other two sides of the triangle are the length and the breadth of the rectangle.

Now, using Pythagoras' theorem, we have:

one side2  +other side2 =hypotenuse2
 122+other side2=372 144+other side2=1369 other side2=1369-144 other side2=1225 other side=1225 other side = 35 cm

Thus, we have:
Length  = 35 cm
Breadth = 12 cm
Area of the rectangle=35×12=420 cm2

Page No 781:

Question 4:

The area of a rectangular plot is 462 m2 and its length is 28 m. Find the perimeter of the plot.

Answer:

Area of the rectangular plot = 462 m2
Length (l) = 28 m
Area of a rectangle = Length (l) x Breadth (b)462=28 x bb=16.5 m

Perimeter of the plot = 2l+b
                                  =228+16.5=2×44.5=89 m

Page No 781:

Question 5:

A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3. The area of the lawn is 3375 m2. Find the cost of fencing the lawn at Rs 65 per metre.

Answer:

Let the length and breadth of the rectangular lawn be 5x m and 3x m, respectively.

Given:
Area of the rectangular lawn = 3375 m2

3375=5x×3x3375=15x2337515=x2225=x2x=15 
 
Thus, we have:

 l=5x=5×15 = 75 mb=3x=3×15 = 45 m


 Perimeter of the rectangular lawn=2(l+b)=2(75+45)=2(120)=240 m

Cost of fencing 1 m lawn = Rs 65
∴ Cost of fencing 240 m lawn = 240×65=Rs 15,600

Page No 781:

Question 6:

A room is 16 m long and 13.5 m broad. Find the cost of covering its floor with 75-m-wide carpet at ₹60 per metre.

Answer:

As, the area of the floor=length×breadth=16×13.5=216 m2And, the width of the carpet=75 mSo, the length of the carpet required=Area of the floorWidth of the carpet=21675=2.88 mNow, the cost of the carpet required=2.88×60=172.80

Hence, the cost of covering the floor with carpet is â‚¹172.80.

Disclaimer: The answer given in the textbook is incorrect. The same has been rectified above.

 

Page No 781:

Question 7:

The floor of a rectangular hall is 24 m and breadth 80 cm, will be required to cover the floor of the hall?

Answer:

Given:
Length = 24 m
Breadth = 18 m
Thus, we have:
Area of the rectangular hall=24×18                                                =432 m2

Length of each carpet = 2.5 m
Breadth of each carpet = 80 cm = 0.80 m

Area of one carpet = 2.5×0.8=2 m2

Number of carpets required = Area of the hallArea of the carpet=4322=216

Therefore, 216 carpets will be required to cover the floor of the hall.

Page No 781:

Question 8:

A 36 m-long, 15 broad verandah is to be paved with stones, each measuring 6 dm by 5 cm. How many stones will be required?

Answer:

Area of the verandah=Length×Breadth=36×15=540 m2
Length of the stone = 6 dm = 0.6 m
Breadth of the stone = 5 dm = 0.5 m
Area of one stone=0.6×0.5=0.3 m2

 Number of stones required=Area of the verendahArea of the stone=5400.3=1800

Thus, 1800 stones will be required to pave the verandah.

Page No 781:

Question 9:

The area of a rectangle is 192 cm2 and its perimeter is 56 cm. Find the dimensions of the rectangle.

Answer:

Area of the rectangle = 192 cm2
Perimeter of the rectangle = 56 cm

Perimeter=2(length+breadth)56=2l+bl+b=28l=28-b

Area=length×breadth192=28-bxb192=28b-b2b2-28b+192=0b-16b-12=0b=16 or 12

Thus, we have;
l=28-b l=28-12 l=16

We will take length as 16 cm and breath as 12 cm because length is greater than breadth by convention.

Page No 781:

Question 10:

A rectangular park 35 m long 18 m wide is to be covered with grass, leaving 2.5 m uncovered all around it. Find the area to be laid with grass.

Answer:

The field is planted with grass, with 2.5 m uncovered on its sides.

The field is shown in the given figure.



Thus, we have;
Length of the area planted with grass = 35-(2.5+2.5)=35-5=30 m

Width of the area planted with grass = 18-(2.5+2.5)=18-5=13 m

Area of the rectangular region planted with grass = 30×13=390 m2

Page No 781:

Question 11:

A rectangular plot measures 125 m by 78 m. It has gravel path 3 m wide all around on the outside. Find the area of the path and the cost of gravelling it at Rs 75 per m2.

Answer:

The plot with the gravel path is shown in the figure.


Area of the rectangular plot = l×b
Area of the rectangular plot = 125×78=9750 m2
Length of the park including the path = 125 + 6 = 131 m
Breadth of the park including the path = 78 + 6 = 84 m
Area of the plot including the path
=131×84=11004 m2

Area of the path = 11004-9750
                           = 1254 m2
Cost of gravelling 1 m2of the path = Rs 75
∴ Cost of gravelling 1254 m2of the path = 1254×75
                                                                   = Rs 94050

Page No 781:

Question 12:

A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide. If the area of the path is 420 m2., find the width of the path.

Answer:

Area of the rectangular field = 54×35=1890 m2

Let the width of the path be x m. The path is shown in the following diagram:


Length of the park excluding the path = (54 - 2x) m
Breadth of the park excluding the path = (35 - 2x ) m

Thus, we have:
Area of the path = 420 m2
420=54×35- (54-2x)(35-2x)  420=1890-1890-70x-108x+4x2420=-4x2+178x4x2-178x+420=02x2-89x+210=02x2-84x-5x+210=02x(x-42)-5(x-42)=0(x-42)(2x-5) = 0x-42=0 or 2x-5=0 x=42 or x=2.5

The width of the path cannot be more than the breadth of the rectangular field.
∴ x = 2.5 m

​Thus, the path is 2.5 m wide.

Page No 781:

Question 13:

The length and the breadth of a rectangular garden are in the ratio 9 : 5. A path 3.5 m wide, running all around inside it has an area of 1911 m2. Find the dimensions of the garden.

Answer:

Let the length and breadth of the garden be 9x m and 5x m, respectively,
Now,
Area of the garden = (9x×5x)=45x2
Length of the garden excluding the path = (9x- 7)
Breadth of the garden excluding the path = (5x-7)
Area of the path = 45x2-(9x-7)(5x-7)
 1911=45x2-45x2-63x-35x+49 1911=45x2-45x2+63x+35x-49 1911=98x-49 1960=98x x=196098 x=20
Thus, we have:
Length = 9x=20×9=180 m
Breadth = 5x=5×20=100 m

Page No 781:

Question 14:

A room 4.9 m long and 3.5 m broad is covered with carpet, leaving an uncovered margin of 25 cm all around the room. If the breadth of the carpet is 80 cm, find its cost at Rs 80 per meter.

Answer:

Width of the room left uncovered = 0.25 m
Now,
Length of the room to be carpeted = 4.9-0.25+0.25=4.9-0.5=4.4 m
Breadth of the room be carpeted = 3.5-0.25+0.25=3.5-0.5=3 m

Area to be carpeted = 4.4×3=13.2 m2

Breadth of the carpet = 80 cm = 0.8 m
We know:
Area of the room = Area of the carpet

Length of the carpet =Area of the roomBreadth of the carpet=13.20.8=16.5 m

Cost of 1 m carpet = Rs 80
Cost of 16.5 m carpet = 80×16.5=Rs 1,320

Page No 781:

Question 15:

A carpet is laid on the floor of a room 8 m by 5 m. There is a border of constant width all around the carpet. If the area of the border is 12 m2, find its width.

Answer:

Let the width of the border be x m.
The length and breadth of the carpet are 8 m and 5 m, respectively.
Area of the carpet = 8×5=40 m2
Length of the carpet without border = (8-2x)
Breadth of carpet without border = (5-2x)
Area of the border = 12 m2
Area of the carpet without border = (8-2x)(5-2x)
Thus, we have:12=40-(8-2x)(5-2x)12=40-(40-26x+4x2)12=26x-4x2

 26x-4x2=12 4x2-26x+12=0 2x2-13x+6=0

(2x-1)(x-6)=02x-1=0 and x-6=0x=12 and x=6

Because the border cannot be wider than the entire carpet, the width of the carpet is 12 m, i.e., 50 cm.

Page No 781:

Question 16:

A 80 m by 64 m rectangular lawn has two roads, each 5 m wide, running through its   middle, one parallel to its length and the other parallel to its breadth. Find the cost of gravelling the roads at Rs 40 per m2.

Answer:

The length and breadth of the lawn are 80 m and 64 m, respectively.
The layout of the roads is shown in the figure below:

Area of the road ABCD = 80×5=400 m2
Area of the road PQRS = 64×5=320 m2
Clearly, the area EFGH is common in both the roads.
Area EFGH = 5×5=25 m2
Area of the roads = 400+320-25
                             = 695 m2
Given:
Cost of gravelling 1 m2 area = Rs 40
∴ Cost of gravelling 695 m2 area = 695×40
                                                       = Rs 27,800

Page No 781:

Question 17:

The dimensions of a room are 14 m × 10 m × 6.5 m. There are two doors and 4 windows in the room. Each door measures 2.5 m × 1.2 m and each window measures 1.5 × 1 m. Find the cost of painting the four walls of the room at Rs 35 per m2.

Answer:

The room has four walls to be painted.

Area of these walls=2(l×h)+2(b×h)=2×14×6.5+2×10×6.5=312 m2

Now,
Area of the two doors = 2×2.5×1.2=6 m2

Area of the four windows = 4×1.5×1=6 m2

The walls have to be painted; the doors and windows are not to be painted.

∴ Total area to be painted=312-6+6=300 m2
Cost for painting 1 m2 = Rs 35
Cost for painting 300 m2 = 300×35=Rs 10,500



Page No 782:

Question 18:

The cost of painting the four walls of a room 12 m long at ₹30 per m2 is â‚¹7560 and the cost of covering the floor with mat at ₹25 per m2 is â‚¹2700. Find the dimensions of the room.

Answer:

As, the rate of covering the floor=25 per m2And, the cost of covering the floor=2700So, the area of the floor=270025length×breadth=10812×breadth=108breadth=10812 breadth=9 mAlso,As, the rate of painting the four walls=30 per m2And, the cost of painting the four walls=7560So, the area of the four walls=7560302length+breadthheight=252212+9height=252221height=25242×height=252height=25242 height=6 m

So, the dimensions of the room are 12 m×9 m×6 m.

Page No 782:

Question 19:

Find the area and perimeter of a square plot of land whose diagonal is 24 m long. Take 2=1.41

Answer:

Area of the square = 12×Diagonal2

                               =12×24×24
                               =288 m2
Now, let the side of the square be x m.
Thus, we have:
Area=Side2288=x2x=122x=16.92

Perimeter = 4×Side
                =4×16.92=67.68 m

Thus, the perimeter of the square plot is 67.68 m.

Page No 782:

Question 20:

Find the length of the diagonal of a square of area 128 cm2. Also find the perimeter of the square, correct to two decimal places.

Answer:

Area of the square = 128 cm2
Area=12d2     (where d is a diagonal of the square)128=12d2d2=256d=16 cm

Now,
Area = Side2
128=Side2Side=11.31 cm

Perimeter = 4(Side)
                =411.31=45.24 cm

Page No 782:

Question 22:

The cost of harvesting a square field at ₹900 per hectare is ₹8100. Find the cost of putting a fence around it at ₹18 per metre.

Answer:

As, the rate of the harvesting=900 per hectareAnd, the cost of harvesting=8100So, the area of the square field=8100900=9 hectarethe area=90000 m2               As, 1 hectare=10000 m2side2=90000side=90000So, side=300 mNow, perimeter of the field=4×side=4×300=1200 mSince, the rate of putting the fence=18 per mSo, the cost of putting the fence=1200×18=21,600

Page No 782:

Question 23:

The cost of fencing a square lawn at Rs 14 per metre is Rs 28000. Find the cost of mowing the lawn at Rs 54 per 100 m2.

Answer:

Cost of fencing the lawn = Rs 28000
Let l be the length of each side of the lawn. Then, the perimeter is 4l.
We know:
Cost=Rate×Perimeter28000=14×4l

28000=56lOr,l=2800056=500 m

Area of the square lawn=500×500=250000 m2

Cost of mowing 100 m2 of the lawn = Rs 54
Cost of mowing 1 m2 of the lawn=Rs 54100


∴ Cost of mowing 250000 m2 of the lawn=250000×54100=Rs 135000

Page No 782:

Question 24:

In the given figure, ABCD is a quadrilateral in which diagonal BD = 24 cm, AL  BD and CM  BD such that AL = 9 cm and CM = 12 cm. Calculate the area of the quadrilateral.

 

Answer:


We have,BD=24 cm, AL=9 cm, CM=12 cm, ALBD and CMBDArea of the quadrilateral=arABD+arBCD=12×BD×AL+12×BD×CM=12×24×9+12×24×12=108+144=252 cm2

So, the area of the quadrilateral ABCD is 252 cm2.

Page No 782:

Question 25:

Find the area of the quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm. Also find the perimeter of the quadrilateral. Take 3 = 1.73.

Answer:

BDC is an equilateral triangle with side a= 26 cm.
Area of BDC = 34a2
                      =34×262=1.734×676=292.37 cm2

By using Pythagoras' theorem in the right-angled triangle DAB, we get:
AD2+AB2=BD2242+AB2=262AB2=262-242AB2=676-576AB2=100AB=10 cm

Area of ABD=12×b×h
                      =12×10×24=120 cm2

Area of the quadrilateral = Area of BCD + Area of ABD
                                         =292.37+120 
                                         = 412.37 cm2

Perimeter of the quadrilateral = AB + BC + CD + AD
                                                 = 24 + 10 + 26 + 26
                                                 = 86 cm

Page No 782:

Question 26:

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ∠ACB = 90° and AC = 15 cm.

Answer:

In the right-angled ACB:

AB2=BC2+AC2172=BC2+152172-152=BC264=BC2BC=8 cm

Perimeter = AB+BC+CD+AD
                 = 17+8+12+9
                 = 46 cm

Area of ABC=12(b×h)

                     =12(8×15)
                     = 60 cm2
 In ADC:AC2=AD2+CD2So, ADC is a right-angled triangle at D.

Area of ADC=12×b×h
                      =12×9×12=54 cm2


∴ Area of the quadrilateral = Area of ABC + Area of ADC
                                             = 60 + 54
                                             = 114 cm2

Page No 782:

Question 27:

Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.

Answer:

Area of ABD = s(s-a)(s-b)(s-c)
s=12(a+b+c)s=42+20+342s=48 cm

Area of ABD = 48(48-42)(48-20)(48-34)

                        =48×6×28×14=112896=336 cm2

Area of BDC = s(s-a)(s-b)(s-c)
s=12(a+b+c)s=21+20+292s=35 cm

Area of BDC = 35(35-29)(35-20)(35-21)

                        =35×6×15×14=44100=210 cm2

∴ Area of quadrilateral ABCD = Area of ABD + Area of BDC
​                                                  = 336 + 210
                                                  = 546 cm2

Page No 782:

Question 28:

Find the area of a parallelogram with base equal to 25 cm and the corresponding height measuring 16.8 cm.

Answer:

Given:
Base = 25 cm
Height = 16.8 cm
∴ Area of the parallelogram=Base×Height=25 cm×16.8 cm=420 cm2



Page No 783:

Question 29:

The adjacent sides of a parallelogram are 32 cm and 24 cm. If the distance between the longer sides is 17.4 cm, find hte distance between the shorter sides.

Answer:

Longer side = 32 cm
Shorter side = 24 cm
Let the distance between the shorter sides be x cm.
Area of a parallelogram = Longer side × Distance between the longer sides
                                     = Shorter side×Distance between the shorter sides
or, 32×17.4=24×x

or, x=32×17.424=23.2 cm

∴ Distance between the shorter sides = 23.2 cm

Page No 783:

Question 30:

The area of a parallelogram is 392 m2. If its altitude is twice the corresponding base, determine the base and the altitude.

Answer:

Area of the parallelogram = 392 m2
Let the base of the parallelogram be b m.
Given:
Height of the parallelogram is twice the base.
∴ Height = 2b m
​Area of a parallelogram = Base x Height
392 = b ×2b392=2b23922=b2196 = b2b=14
∴ Base = 14 m
Altitude = 2×Base=2×14=28 m

Page No 783:

Question 31:

The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.

Answer:

Parallelogram ABCD is made up of congruent ABC and ADC.

Area of triangle ABCs(s-a)(s-b)(s-c)   (Here, s is the semiperimeter.)

Thus, we have:

s=a+b+c2s=34+20+422s=48 cm

Area of ABC=48(48-34)(48-20)(48-42)=48×14×28×6=336 cm2

Now,
Area of the parallelogram = 2×Area of ABC
                                           =2×336=672 cm2

Page No 783:

Question 32:

Find the area of the rhombus, the lengths of whose diagonals are 30 cm and 16 cm. Also find the perimeter of the rhombus.

Answer:

Area of the rhombus = 12×d1×d2, where d1 and d2 are the lengths of the diagonals.

                                  =12×30×16=240 cm2

Side of the rhombus = 12d12+d22

                                 =12302 +162=121156=12×34=17 cm

Perimeter of the rhombus = 4a
                                          = 4×17
                                          = 68 cm

Page No 783:

Question 33:

The perimeter of a rhombus is 60 cm. If one of its diagonals is 18 cm long, find (i) the length of the other diagonal, and (ii) the area of the rhombus.

Answer:

Perimeter of a rhombus = 4a    (Here, a is the side of the rhombus.)

60 = 4aa= 15 cm

(i) Given:
One of the diagonals is 18 cm long.
d1=18 cm

Thus, we have:
 Side=12d12+d2215=12182+d2230=182+d22Squaring both sides, we get:900=182+d22900=324+d22d22=576d2=24 cm

∴ Length of the other diagonal = 24 cm

(ii) Area of the rhombus=12d1×d2
                                      =12×18×24=216 cm2
​

Page No 783:

Question 34:

The area of a rhombus is 480 cm2, and one of its diagonals measures 48 cm. Find (i) the length of the other diagonal, (ii) the length of each of its sides, and (iii) its perimeter.

Answer:

i) Area of a rhombus=12×d1×d2, where d1 and d2 are the lengths of the diagonals.
480=12×48×d2d2=480×248d2=20 cm

∴ Length of the other diagonal = 20 cm
   
(ii) Side = 12d12+d22

             =12482+202=122304+400=122704=12×52=26 cm

∴ Length of the side of the rhombus = 26 cm

(iii) Perimeter of the rhombus = 4×Side
                                                 =4×26=104 cm

Page No 783:

Question 35:

The parallel sides of a trapezium are 12 cm and 9 cm and the distance between them is 8 cm. Find the area of the trapezium.

Answer:

Area of the trapezium=12×sum of the parallel sides×distance between the parallel sides=12×12+9×8=21×4=84 cm2

So, the area of the trapezium is 84 cm2.

Page No 783:

Question 36:

The shape of the cross section of a canal is a trapezium. If the canal is 10 m wide at the top, 6 m wide at the bottom and the area of its cross section is 640 m2, find the depth of the canal.

Answer:


Area of the canal = 640 m2
Area of trapezium = 12×Sum of parallel sides×Distance between them
640=12×10+6×h128016=h h=80 m

Therefore, the depth of the canal is 80 m.

Page No 783:

Question 37:

Find the area of a trapezium whose parallel sides are 11 ma and 25 m long, and the nonparallel sides are 15 m and 13 m long.

Answer:



Draw DEBC and DL perpendicular to AB.
The opposite sides of quadrilateral DEBC are parallel. Hence, DEBC is a parallelogram.
DE = BC = 13 m
Also,
AE=(AB-EB)=(AB-DC)=(25 - 11)=14 m
For DAE:
Let:
AE = a =14 m
DE = b = 13 m
DA = c =15 m

Thus, we have:

s=a+b+c2s= 14+13+152=21 m

Area of DAE =s(s-a)(s-b)(s-c)
                      =21×(21-14)×(21-13)×(21-15)=21×7×8×6=7056=84 m2

Area of DAE=12×AE×DL
84=12×14×DL84×214=DLDL=12 m

Area of trapezium = 12×Sum of parallel sides×Distance between them
                              =12×(11+25)×12=12×36×12=216 m2



Page No 786:

Question 1:

The length of a rectangular hall is 5 m more than its breadth. If the area of the hall is 750 m2, then its length is
(a) 15 m
(b) 20 m
(c) 25 m
(d) 30 m

Answer:

(d) 30 m

Let the length of the rectangle be x m.
∴ Breadth of the rectangle=(x-5) m

Area=x(x-5)=x2-5xx2-5x=750 
x2-5x-750=0x2-30x+25x-750=0x(x-30)+25(x-30)=0(x+25)(x-30)=0 x+25 = 0 and x-30 = 0 x=-25 and x=30

Length cannot be negative.
∴ Length = = 30 m

Page No 786:

Question 2:

The length of a rectangular field is 23 m more than its breadth. If the perimeter of the field is 206 m, then its area is
(a) 2420 m2
(b) 2520 m2
(c) 2480 m2
(d) 2620 m2

Answer:

(b) 2520 m2

Let the breadth of the field be x m.
∴ Length = (x + 23) m

Now,
Perimeter=2(Length+Breadth)=2(x+x+23)=(4x+46) m
Thus, we have:
4x+46=2064x=206-46=160x=1604=40
∴ Breadth = x = 40 m
Length = + 23 =40+23=63 m
Area=Length×Breadth=63×40=2520 m2

Page No 786:

Question 3:

The length of a rectangular field is 12 m and the length of its diagonal is 15 m. The area of the field is
(a) 108 m2
(b) 180 m2
(c) 303m2
(d) 1215m2

Answer:

(a) 108 m2

Length of the rectangular field = 12 m
Diagonal = 15 m


Diagonal2=Length2+Breadth2
Breadth=Diagonal2-Length2=152-122=225-144=9 m

∴ Area of the field=Length×Breadth=12×9=108 m2

Page No 786:

Question 4:

Choose the correct answer of the following question:

The cost of carpeting a room 15 m long with a carpet 75 cm wide, at ₹70 per metre, is ₹8400. The width of the room is

(a) 9 m                    (b) 8 m                    (c) 6 m                    (d) 12 m

Answer:

We have,Width of the carpet=75 cm=0.75 m and length of the room=15 mLength of the carpet=Cost of carpetingRate of carpeting=840070=120 mNow, area of the carpet required for carpeting=120×0.75 m2area of the floor=120×0.75 m215×breadth=120×0.75breadth=120×0.7515So, breadth=6 m

Hence, the correct answer is option (c).

Page No 786:

Question 5:

The length of a rectangle is thrice its breadth and the length of its diagonal is 810cm.. The perimeter of the rectangle is
(a) 1510 cm
(b) 1610cm
(c) 2410cm
(d) 64 cm

Answer:

(d) 64 cm

Let the breadth of the rectangle be x cm.
∴ Length of the rectangle = 3x cm
We know:
Diagonal=(Length)2+(Breadth)2810=x2+(3x)2810=x2+9x2810=x10x=8

Now,
Breadth of the rectangle = x = 8 cm
Length of the rectangle = 3x = 24 cm
Perimeter of the rectangle=2(Length+Breadth)=2(8+24)=64 cm

Page No 786:

Question 6:

On increasing the length of a rectangle by 20% and decreasing its breadth by 20%, what is the change in its area?
(a) 20% increase
(b) 20% decrease
(c) No change
(d) 4% decrease

Answer:

(d) 4% decrease

Let:
Length=x
breadth=y
Area=xy

Now,
New length=x+20%x=x+15x=65x

New breadth=y-20%y=y-15y=45y

New area=65x×45y=2425xy

Difference in the areas=xy-2425xy=125xy

Difference in percentage=125xyxy×100%=4%

Page No 786:

Question 7:

A rectangular ground 80 m × 50 m has a path 1 m wide outside around it. The area of the path is
(a) 264 m2
(b) 284 m2
(c) 400 m2
(d) 464 m2

Answer:

(a) 264 m2

Length of the ground including the path=80+2=82 m
Breadth of the ground including the path=50+2=52 m

Total area (including the path)=Length×Breadth=82×52=4264 m2

Area of the field=80×50=4000 m2

Area of the path=4264-4000=264 m2

Page No 786:

Question 8:

The length of the diagonal of a square is 102cm. Its area is
(a) 200 cm2
(b) 100 cm2
(c) 150 cm2
(d) 1002 cm2

Answer:

(b) 100 cm2

A diagonal of a square forms the hypotenuse of a right-angled triangle with base and height equal to side a.

Diagonal2=a2+a2Diagonal2=2a2a=12Diagonal=12×102=10 cm

∴ Area of the square=a2=10×10=100 cm2

Page No 786:

Question 9:

The area of a square field is 6050 m2. The length of its diagonal is
(a) 135 m
(b) 120 m
(c) 112 m
(d) 110 m

Answer:

(d) 110 m

Let the diagonal of the square field be d m.

In case of a square field, d2=2a2, where a is the side of the square field.
Now,
Area of a square field=a2
d2=2a2d2=2×Area of the square fieldd=2×Area of the square field

∴ d=2×6050=12100=110

Page No 786:

Question 10:

The length of a square field is 0.5 hectare. The length of its diagonal is
(a) 150 m
(b) 1002
(c) 100 m
(d) 502m

Answer:

(c) 100 m

Disclaimer :- The length cannot be in hectare So we used is as area of the square.
Area of the square field =0.5×10000=5000 m2



The diagonal divides the square into two isosceles right-angled triangles.

Using Pythagoras' theorem, we have:

Diagonal2=a2+a2=2a2
Area of a square = a2

∴ Diagonal​=2 area=2×5000=10000=100 m

Page No 786:

Question 11:

The area of an equilateral triangle is 43cm2. Its perimeter is
(a) 9 cm
(b) 12 cm
(c) 123 cm
(d) 63 cm

Answer:

(b) 12 cm

Area of an equilateral triangle =34a2 (where a is the length of the side)
Thus, we have:
43 = 34a2
a2=16a=4 cm

Perimeter of the equilateral triangle = 3a =3×4=12 cm

Page No 786:

Question 12:

Each side of an equilateral triangle is 8 cm. Its area is
(a) 24 cm2
(b) 243cm2
(c) 163cm2
(d) 83cm2 

Answer:

(c) 163cm2

Let the side of the equilateral triangle be a.
Given:
a = 8 cm
Now,
Area of the equilateral triangle=34a2=34×8×8=163 cm2



Page No 787:

Question 13:

Choose the correct answer of the following question:

Each side of an equilateral triangle is 63 cm. The altitude of the triangle is

(a) 8 cm                    (b) 9 cm                    (c) 33 cm                    (d) 6 cm

Answer:

As, area of an equilateral triangle=34×side212×Base×Height=34×63212×63×Height=34×36×333×Height=273Height=27333 Height=9 cm

Hence, the correct option is (b).

Page No 787:

Question 14:

Choose the correct answer of the following question:

The height of an equilateral triangle is 33 cm. Its area is

(a) 63 cm2                    (b) 27 cm2                    (c) 93 cm2                    (d) 273 cm2

Answer:

Let the side of the equilateral triangle be x.As, area of the equilateral triangle=34×side212×Base×Height=34×x212×x×33=x2343x32=x234332=x34x=33×423x=6 cmNow, the area of the triangle=34×62=34×36=93 cm2

Hence, the correct answer is option (c).

Page No 787:

Question 15:

Choose the correct answer of the following question:


The base and height of a triangle are in the ratio 3:4 and its area is 216 cm2. The height of the triangle is

(a) 18 cm                    (b) 24 cm                    (c) 21 cm                    (d) 28 cm

Answer:

Let the base of the triangle be 3x and its height be 4x.As, the area of the triangle=216 cm212×Base×Height=21612×3x×4x=2166x2=216x2=2166x2=36x=36x=6 cmSo, the height of the triangle=4×6=24 cm

Hence, the correct answer is option (b).

Page No 787:

Question 16:

Choose the correct answer of the following question:

The lengths of the sides of a triangular field are 20 m, 21 m and 29 m. The cost of cultivating the field at ₹9 per m2 is

(a) ₹2610                    (b) ₹3780                    (c) ₹1890                    (d) ₹1800

Answer:

As, the sides of the triangle are 20 m, 21 m and 29 mSo, the semi-perimeter=20+21+292=35 mNow, the area of the triangular field=3535-2035-2135-29=35×15×14×6=7×5×5×3×7×2×2×3=2×3×5×7=210 m2 The cost of cultivating the field=210×9=1890

Hence, the correct answer is option (c).

Page No 787:

Question 17:

The side of a square is equal to the side of an equilateral triangle. The ratio of their areas is
(a) 4 : 3
(b) 2:3
(c) 4:3
(d) none of these

Answer:

(c) 4:3

Let:
Length of the side of the square = Length of the side of the equilateral triangle = a unit

Now,

Area of the square=a×a=a2 unit2

Area of the equilateral triangle=34a2 unit2

 Ratio of areas=Area of the squareArea of the equilateral triangle=a234a2=43

Page No 787:

Question 18:

The side of an equilateral triangle is equal to the radius of a circle whose area is 154 cm2. The area of the triangle is
(a) 49 cm2
(b) 4934cm2
(c) 734cm2
(d) 77 cm2

Answer:

(b) 4934 cm2

Area of a circle=πr2
154 = πr2r=154×722=7×7=7 cm

The radius of the circle is equal to the side of the equilateral triangle.  

 r = a (Here, a is the side of the equilateral triangle.)

a=7 cm

∴ Area of the equilateral triangle=34a2=34×7×7=4934 cm2

Page No 787:

Question 19:

Choose the correct answer of the following question:

The area of a rhombus is 480 cm2 and the length of one of its diagonals is 20 cm. The length of each side of the rhombus is

(a) 24 cm                    (b) 30 cm                    (c) 26 cm                    (d) 28 cm

Answer:


We have,BD=20 cmBO=BD2=202=10 cmAs, area of the rhombus ABCD=480 cm212×AC×BD=48012×AC×20=480AC×10=480AC=48010AC=48 cmAO=AC2=482=24 cmNow, in AOB,Using Pythagoras theorem,AB2=AO2+BO2=242+102=576+100AB2=676AB=676 AB=26 cm

Hence, the correct answer is option (c).

Page No 787:

Question 20:

Choose the correct answer of the following question:

One side of a rhombus is 20 cm long and one of its diagonals measures 24 cm. The area of the rhombus is

(a) 192 cm2                    (b) 480 cm2                    (c) 240 cm2                    (d) 384 cm2

Answer:



WE have,AB=BC=CD=DA=20 cm and BD=24 cmAlso, BO=BD2=242=12 cmIn AOB,Using Pythagoras theoremAO2=AB2-BO2=202-122=400-144AO2=256AO=256AO=16 cmAC=2AO=2×16=32 cmNow, the area of the rhombus ABCD=12×AC×BD=12×32×24=384 cm2

Hence, the correct answer is option (d).



Page No 790:

Question 1:

In the given figure ABCD is a quadrilateral in which ∠ABC = 90°, ∠BDC = 90°, AC = 17 cm, BC = 15 cm, BD = 12 cm and CD = 9 cm. The area of quad. ABCD is


(a) 102 cm2
(b) 114 cm2
(c) 95 cm2
(d) 57 cm2

Answer:

(b) 114 sq cm

Using Pythagoras' theorem in ABC, we get:
AC2=AB2+BC2AB=AC2-BC2=172-152=8 cm

Area of ABC=12×AB×BC=12×8×15=60 cm2

Area of BCD=12×BD×CD=12×12×9=54 cm2

∴ Area of quadrilateral ABCD=ArABC+ArBCD=54+60=114 cm2

Page No 790:

Question 2:

In the given figure ABCD is a trapezium in which AB = 40 m, BC = 15 m, CD = 28 m, AD = 9 m and CEAB. Area of trap. ABCD is


(a) 306 m2
(b) 316 m2
(c) 296 m2
(d) 284 m2

Answer:

(a) 306 m2

In the given figure, AECD is a rectangle.

Length AE = Length CD = 28 m

Now,
BE=AB-AE=40-28=12 m

Also,
AD = CE = 9 m
Area of trapezium=12×Sum of parallel sides×Distance between them=12×(DC+AB)×CE=12×(28+40)×9=12×68×9=306 m2

In the given figure, if DA is perpendicular to AE, then it can be solved, otherwise it cannot be solved.

Page No 790:

Question 3:

The sides of a triangle are in the ratio 12 : 14 : 25 and its perimeter is 25.5 cm. The largest side of the triangle is
(a) 7 cm
(b) 14 cm
(c) 12.5 cm
(d) 18 cm

Answer:

(c) 12.5 cm

Let the sides of the triangle be 12x cm, 14x cm and 25x cm.
Thus, we have:
Perimeter=12x+14x+25x25.5 =51xx = 25.551=0.5

∴ Greatest side of the triangle=25x=25×0.5=12.5 cm

Page No 790:

Question 4:

The parallel sides of a trapezium are 9.7 cm and 6.3 cm, and the distance between them is 6.5 cm. The area of the trapezium is
(a) 104 cm2
(b) 78 cm2
(c) 52 cm2
(d) 65 cm2

Answer:

(c) 52 cm2

Area of trapezium=12Sum of parallel sides×Distance between them=12×9.7+6.3×6.5=8×6.5=52.0 cm2

Page No 790:

Question 5:

Find the area of an equilateral triangle having each side of length 10 cm.

Answer:

Given:
Side of the equilateral triangle = 10 cm
Thus, we have:
 Area of the equilateral triangle=34side2=34×10×10=25×1.732=43.3 cm2

Page No 790:

Question 6:

Find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.

Answer:

Area of an isosceles triangle:
=14b4a2-b2  (Where a is the length of the equal sides and b is the base)=14×244132-242=64×169-576=6676-576=6100=6×10=60 cm2

Page No 790:

Question 7:

The longer side of a rectangular hall is 24 m and the length of its diagonal is 26 m. Find the area of the hall.

Answer:

Let the rectangle ABCD represent the hall. 

Using the Pythagorean theorem in the right-angled triangle ABC, we have:
Diagonal2=Length2+Breadth2

Breadth=Diagonal2-Length2=262-242=676-576=100=10 m

∴ Area of the hall=Length×Breadth=24×10=240 m2

Page No 790:

Question 8:

The length of the diagonal of a square is 24 cm. Find its area.

Answer:

The diagonal of a square forms the hypotenuse of an isosceles right triangle. The other two sides are the sides of the square of length a cm.

Using Pythagoras' theorem, we have:

Diagonal2=a2+a2=2a2Diagonal=2a 

 Diagonal of the square=2a24=2a

 a=242
Area of the square=Side2=2422=24×242=288 cm2

Page No 790:

Question 9:

Find the area of a rhombus whose diagonals are 48 cm and 20 cm long.

Answer:

 Area of the rhombus=12Product of diagonals=1248×20=480 cm2

Page No 790:

Question 10:

Find the area of a triangle whose sides are 42 cm, 34 cm and 20 cm.

Answer:

To find the area of the triangle, we will first find the semiperimeter of the triangle.

Thus, we have:

s=12a+b+c=1242+34+20=12×96=48 cm

Now,

Area of the triangle=ss-as-bs-c=4848-4248-3448-20=48×6×14×28=112896=336 cm2

Page No 790:

Question 11:

A lawn is in the form of a rectangle whose sides are in the ratio 5 : 3 and its area is 3375 m2. Find the cost of fencing the lawn at Rs 20 per metre.

Answer:

Let the length and breadth of the lawn be 5x m and 3x m, respectively.

Now,

Area of the lawn=5x×3x=5x2
15 x2=3375x=337515x=225=15

Length = 5x=5×15=75 mBreadth =3x=3×15=45 m

∴ Perimeter of the lawn=2Length+Breadth=275+45=2×120=240 m
Total cost of fencing the lawn at Rs 20 per metre =240×20=Rs 4800



Page No 791:

Question 12:

Find the area of a rhombus each side of which measures 20 cm and one of whose diagonals is 24 cm.

Answer:

Given:
Sides are 20 cm each and one diagonal is of 24 cm.
The diagonal divides the rhombus into two congruent triangles, as shown in the figure below.


We will now use Hero's formula to find the area of triangle ABC.
First, we will find the semiperimeter.
s=12a+b+c=1220+20+24=642=32 m

Area of ABC=ss-as-bs-c=3232-2032-2032-24=32×12×12×8=36864=192 cm2

Now,
Area of the rhombus =2×Area of triangle ABC=192×2=384 cm2

Page No 791:

Question 13:

Find the area of a trapezium whose parallel sides are 11 cm and 25 cm long and non-parallel sides are 15 cm and 13 cm.

Answer:

We will divide the trapezium into a triangle and a parallelogram.

Difference in the lengths of parallel sides=25-11=14 cm
We can represent this in the following figure:

Trapezium ABCD is divided into parallelogram AECD and triangle CEB.

  1. Consider triangle CEB.
In triangle CEB, we have:
EB=25-11=14 cm

Using Hero's theorem, we will first evaluate the semiperimeter of triangle CEB and then evaluate its area.

Semiperimeter, s=12a+b+c=1215+13+14=422=21 cm

Area of triangle CEB=ss-as-bs-c=2121-1521-1321-14=21×6×8×7=7056=84 cm2

Also,

Area of triangle CEB=12Base×Height

Height of triangle CEB=Area×2Base=84×214=12 cm
  1. Consider parallelogram AECD.
​​Area of parallelogram AECD=Height×Base=AE×CF=12×11=132 cm2

Area of trapezium ABCD=ArBEC+Arparallelogram AECD=132+84=216 cm2

Page No 791:

Question 14:

The adjacent sides of a ∥gm ABCD measure 34 cm and 20 cm and the diagonal AC is 42 cm long. Find the area of the ∥gm.

Answer:

The diagonal of a parallelogram divides it into two congruent triangles. Also, the area of the parallelogram is the sum of the areas of the triangles.

We will now use Hero's formula to calculate the area of triangle ABC.

Semiperimeter, s=1234+20+42=1296=48 cm

Area of ABC=ss-as-bs-c=4848-4248-3448-20=48×6×14×28=112896=336 cm2

Area of the parallelogram=2×AreaABC=2×336=672 cm2

Page No 791:

Question 15:

The cost of fencing a square lawn at Rs 14 per metre is Rs 2800. Find the cost of mowing the lawn at Rs 54 per 100 cm2.

Answer:

Given:
Cost of fencing = Rs 2800
Rate of fencing = Rs 14

Now,
Perimeter=Total costRate=280014=200 m

Because the lawn is square, its perimeter is 4 a, where a is the side of the square).
4a=200a=2004 =50 m

Area of the lawn = Side2=502=2500 m2

Cost for mowing the lawn per 100 m2= Rs 54

Cost for mowing the lawn per 1 m2= Rs 54100

Total cost for mowing the lawn  per 2500 m2=54100×2500=Rs 1350

Page No 791:

Question 16:

Find the area of quad. ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diag. BD = 20 cm.

Answer:

Quadrilateral ABCD is divided into triangles ABD and BCD.
We will now use Hero's formula.
For ABD:
Semiperimeter, s=1242+20+34=962=48 cm

Area of ABD=ss-as-bs-c=4848-4248-3448-20=48×6×14×28=112896=336 cm2

For BCD:
s=1220+21+29=702=35 cm

Area ofBCD=ss-as-bs-c=3535-2035-2135-29=35×15×14×6=44100=210 cm2

Thus, we have:
Area of quadrilateral ABCD=ArABD+ArBDC=336+210=546 cm2

Page No 791:

Question 17:

A parallelogram and a rhombus are equal in area. The diagonals of the rhombus measure 120 m add 44 m. If one of the sides of the ∥gm is 66 m long, find its corresponding altitude.

Answer:

Area of the rhombus=12Product of diagonals=12120×44=2640 m2

Area of the parallelogram=Base×Height=66×Height

Given:
The area of the rhombus is equal to the area of the parallelogram.

Thus, we have:66×Height=2640Height=264066=40 m

∴ Corresponding height of the parallelogram = 40 m

Page No 791:

Question 18:

The diagonals of a rhombus are 48 cm and 20 cm long. Find the perimeter of the rhombus.

Answer:

Diagonals of a rhombus perpendicularly bisect each other. The statement can help us find a side of the rhombus. Consider the following figure.


ABCD is the rhombus and AC and BD are the diagonals. The diagonals intersect at point O.

We know:
DOC =90°
DO=OB=12DB=12×48=24 cm
Similarly,
AO=OC=12AC=12×20=10 cm

Using Pythagoras' theorem in the right-angled triangle DOC, we get:

DC2=DO2+OC2=242+102=576+100=676=26 cm

DC is a side of the rhombus. 
We know that in a rhombus, all sides are equal.
∴ Perimeter of ABCD=26×4=104 cm

Page No 791:

Question 19:

The adjacent sides of a parallelogram are 36 cm and 27 cm in length. If the distance between the shorter sides is 12 cm, find the distance between the longer sides.

Answer:



Area of a parallelogram=Base×Height AB×DE = BC×DFDE = BC×DFAB=27×1236=9 cm

∴ Distance between the longer sides = 9 cm

Page No 791:

Question 20:

In a four-sides field, the length of the longer diagonal is 128 m. The lengths of perpendiculars from the opposite vertices upon this diagonal are 22.7 m and 17.3 m. Find the area of the field.

Answer:

The field, which is represented as ABCD, is given below.

The area of the field is the sum of the areas of triangles ABC and ADC.

Area of the triangle ABC=12AC×BF=12128×22.7=1452.8 m2

Area of the triangle ADC=12AC×DE=12128×17.3=1107.2 m2

Area of the field = Sum of the areas of both the triangles=1452.8+1107.2=2560 m2



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