Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 15 Perimeter And Area Of Plane Figures are provided here with simple step-by-step explanations. These solutions for Perimeter And Area Of Plane Figures are extremely popular among Class 10 students for Maths Perimeter And Area Of Plane Figures Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 683:

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 683:

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 683:

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 683:

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 683:

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 683:

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 683:

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 683:

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 683:

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 683:

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 683:

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 683:

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 683:

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 683:

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 684:

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 684:

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 684:

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 684:

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 684:

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 ​the isosceles right-angled triangle = 2a+2a
                                                                                =2×10+1.41×10=20+14.1=34.1 cm

Page No 684:

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 691:

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 691:

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 692:

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 692:

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 692:

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 692:

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 ₹172.80.

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

 

Page No 692:

Answer:

The length and breadth of floor of a rectangular hall is 24 m and 18 m respectively.
The area of rectangular floor is 24×18=432 m2.
The length and breadth of carpet is 2.5 m and 80 cm or 0.8 m respectively.
The area of carpet is 2.5×0.8=2 m2.

So, the number of such carpets required to cover the floor of rectangular hall is = the area of floorthe area of carpet=4322=216 carpets.

Page No 692:

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 692:

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 692:

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 692:

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 m2 of the path = Rs 75
∴ Cost of gravelling 1254 m2 of the path = 1254×75
                                                                   = Rs 94050

Page No 692:

Answer:

(i) 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.

(ii) 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 692:

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 692:

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 692:

Answer:

It is given that the dimensions of rectangular park is 50 m × 40 m.
∴ Area of the rectangular park = 50 × 40 = 2000 m2
Area of the grass surrounding the pond = 1184 m2
Now,
Area of the rectangular pond
= Area of the rectangular park − Area of the grass surrounding the rectangular pond
= 2000 − 1184
= 816 m2
Let the uniform width of the surrounding grass be x.
∴ Length of the rectangular pond = (50 − 2x) m
Breadth of the rectangular pond = (40 − 2x) m
Now,
Area of rectangular pond = 816 m2
∴ (50 − 2x) × (40 − 2x) = 816
⇒ 2000 − 80− 100x + 4x2 = 816
⇒ 4x2 − 180x + 2000 − 816 = 0
⇒ 4x2 − 180+ 1184 = 0
⇒ x2 − 45+ 296 = 0
 x2 − 37x − 8x + 296 = 0
⇒ x(− 37) − 8(− 37) = 0
⇒ (x − 8)(x − 37) =  0
⇒ x − 8 = 0 or x − 37 = 0
 x = 8 or x = 37
For x = 37,
Length of rectangular pond = 50 − 2 × 37 = −24 m, which is not possible
So, x ≠ 37
Therefore, x = 8.
When x = 8,
Length of the rectangular pond = 50 − 2 × 8 = 50 − 16 = 34 m
Breadth of the rectangular pond = 40 − 2 × 8 = 40 − 16 = 24 m.

Page No 692:

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 693:

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 693:

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 693:

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 693:

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
Thus, the perimeter of the square is 45.24 cm.

Page No 693:

Answer:

We know that 1 hectare = 10000 m2.
So, Area of square field = 8 hectares = 80000 m2.
12diagonal2=80000diagonal2=160000diagonal=160000=400 m.
Now, distance to be travelled = 400 m
speed is given to be 4 km/h = 4×1000 m60 min=400060m/min=2003m/min
Therefore, time taken=distancespeed=4002003=400×3200=6 minutes.

Page No 693:

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 693:

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 693:

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 693:

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 693:

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 693:

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 694:

Answer:

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

Page No 694:

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 694:

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 694:

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 694:

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 694:

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 694:

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 694:

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 694:

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 694:

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 697:

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 697:

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 697:

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 697:

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 697:

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 697:

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 697:

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 697:

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 697:

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 697:

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 697:

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 697:

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 698:

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 698:

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 698:

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 698:

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 698:

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 698:

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 698:

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 698:

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 701:

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 701:

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 701:

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 701:

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 701:

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 701:

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 701:

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 701:

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 701:

Answer:

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

Page No 701:

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 701:

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 701:

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 702:

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 702:

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 702:

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 702:

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 702:

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 702:

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 702:

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 702:

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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