Rs Aggarwal 2021 2022 Solutions for Class 9 Maths Chapter 14 Areas Of Triangles And Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Areas Of Triangles And Quadrilaterals are extremely popular among Class 9 students for Maths Areas Of Triangles And Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 9 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

Page No 533:

Answer:

We have:
Base = 24 cm
Height = 14.5 cm

Now,
Area of triangle=12×Base×Height=12×24×14.5=174 cm2

Page No 533:

Answer:

Let the height of the triangle be h m.
∴ Base = 3h m
Now,
Area of the triangle = Total CostRate=78358=13.5 ha =135000 m2
We have:
Area of triangle = 135000 m212×Base×Height =13500012×3h×h =135000h2=135000×23h2=90000h =300 m

Thus, we have:
Height = h = 300 m
Base = 3h = 900 m

Page No 533:

Answer:

Let: a=42 cm, b = 34 cm and c=20 cms= a+b+c2=42+34+202=48 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=48(48-42)(48-34)(48-20)=48×6×14×28=4×2×6×6×7×2×7×4=4×2×6×7=336 cm2

We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
Area of triangle = 336 cm212×Base×Height = 336Height = 336×242=16 cm

Page No 533:

Answer:

Let: a=18 cm, b = 24 cm and c=30 cms= a+b+c2=18+24+302=36 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=36(36-18)(36-24)(36-30)=36×18×12×6=12×3×6×3×12×6=12×3×6=216 cm2

We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:
Area of triangle = 216 cm212×Base×Height = 216Height = 216×218=24 cm

Page No 533:

Answer:

Let: a=91 m, b = 98 m and c=105 ms= a+b+c2=91+98+1052=147 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=147(147-91)(147-98)(147-105)=147×56×49×42=7×3×7×2×2×2×7×7×7×7×3×2=7×7×7×2×3×2=4116 m2

We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
Area of triangle = 4116 m212×Base×Height = 4116Height = 4116×2105=78.4 m

Page No 533:

Answer:

Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5×5 = 25 m
12×5 = 60 m
13×5 = 65 m

Now,
Let: a=25 m, b = 60 m and c=65 ms= 1502=75 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=75(75-25)(75-60)(75-65)=75×50×15×10=15×5×5×10×15×10=15×5×10=750 m2

Page No 533:

Answer:

Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25×10 = 250 m
17×10 = 170 m
12×10 = 120 m

Now,
Let: a=250 m, b =170 m and c=120 ms= 5402=270 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=270(270-250)(270-170)(270-120)=270×20×100×150=30×3×3×20×20×5×30×5=30×3×20×5=9000 m2

Cost of ploughing 1 m2 field = Rs 5
Cost of ploughing 9000 m2 field = 5×9000=Rs 45000.

Page No 533:

Answer:

(i) Let: a=85 m and b = 154 m Given:Perimeter = 324 mor, a+b+c =324c=324-85-154=85 ms= 3242=162 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=162(162-85)(162-154)(162-85)=162×77×8×77=1296×77×77=36×77×77×36=36×77=2772 m2


(ii) We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:
Area of triangle = 2772 m212×Base×Height = 2772Height = 2772×2154=36 m

Page No 533:

Answer:

We have: a=13 cm and b=20 cmArea of isosceles triangle=b44a2-b2                                         =204×4(13)2-202                                         =5×676-400                                         =5×276                                         =5×16.6                                         =83.06 cm2

Page No 533:

Answer:


Let PQR be an isosceles triangle and PXQR.
Now,
Area of triangle =360 cm2 12×QR×PX = 360h =72080=9 cmNow, QX = 12×80 = 40 cm and PX = 9 cm
Also,
PQ=QX2+PX2a=402+92=1600+81=1681=41 cm

∴ Perimeter = 80 + 41 + 41  = 162 cm

Page No 533:

Answer:

The ratio of the equal side to its base is 3 : 2.
Ratio of sides = 3 : 3 : 2.
Let the three sides of triangle be 3x, 3x, 2x.
The perimeter of isosceles triangle = 32 cm.
3x+3x+2x=32 cm8x=32x=4 cm
Therefore, the three side of triangle are 3x, 3x, 2x = 12 cm, 12 cm, 8 cm.
Let S be the semi-perimeter of the triangle. Then, S=1212+12+8=322=16
Area of the triangle will be
=SS-aS-bS-c=1616-1216-1216-8=16×4×4×8=4×48=4×4×22=322 cm2
Disclaimer: The answer does not match with the answer given in the book.



Page No 534:

Answer:

Let ABC be any triangle with perimeter 50 cm.
Let the smallest side of the triangle be x.
Then the other sides be x + 4 and 2x − 6.

Now,
x + x + 4 + 2x − 6 = 50        (∵ perimeter is 50 cm)
⇒ 4x − 2 = 50
⇒ 4x = 50 + 2
⇒ 4x = 52
x = 13

∴ The sides of the triangle are of length 13 cm, 17 cm and 20 cm.
∴ Semi-perimeter of the triangle is
s=13+17+202=502=25 cm

∴ By Heron's formula,
Area of ABC=ss-as-bs-c                           =2525-1325-1725-20                           =251285                           =2030 cm2

Hence, the area of the triangle is 2030 cm2.

Page No 534:

Answer:

The sides of the triangle are of length 13 m, 14 m and 15 m.
∴ Semi-perimeter of the triangle is
s=13+14+152=422=21 m

∴ By Heron's formula,
Area of =ss-as-bs-c                  =2121-1321-1421-15                  =21876                  =84 m2

Now,
The rent of advertisements per m2 per year = Rs 2000
The rent of the wall with area 84 m2 per year = Rs 2000 × 84
                                                                         = Rs 168000
The rent of the wall with area 84 m2 for 6 months = Rs 1680002
                                                                                = Rs 84000


Hence, the rent paid by the company is Rs 84000.

Page No 534:

Answer:

Let the equal sides of the isosceles triangle be a cm each.
∴ Base of the triangle, b = 32a cm
(i) Perimeter = 42 cm
or, a + a + 32a = 42
or, 2a +32a= 42

2a+32a = 427a2=42a=12 
 
So, equal sides of the triangle are 12 cm each.
Also,
Base = 32a = 32×12=18 cm
(ii)
Area of isosceles triangle = b44a2-b2=184×4(12)2-182        (a= 12 cm and b=18 cm)=4.5×576-324=4.5×252=4.5×15.87=71.42 cm2

(iii)
Area of triangle =71.42 cm212×Base×Height = 71.42Height = 71.42×218=7.94 cm

Page No 534:

Answer:

Area of equilateral triangle = 34×(Side)234×(Side)2 =363
(Side)2=144Side=12 cm

Thus, we have:
Perimeter = 3 × Side = 3 × 12 = 36 cm

Page No 534:

Answer:

Area of equilateral triangle = 34×(Side)234×(Side)2 =813
(Side)2=324Side=18 cm

Now, we have:
Height =32×Side=32×18=93 cm

Page No 534:

Answer:

Side of the equilateral triangle = 8 cm

(i)
Area of equilateral triangle = 34×(Side)2=34×(8)2 =1.732×644=27.71 cm2

(ii)
Height =32×Side=32×8=1.732×82=6.93 cm

Page No 534:

Answer:

Height of the equilateral triangle = 9 cm
Thus, we have:
Height =32×Side9=32×Side Side = 183=183×33=63 cm

Also,
Area of equilateral triangle = 34×(Side)2=34×(63)2 =10843=273=46.76 cm2

Page No 534:

Answer:


Let PQR be a right-angled triangle and PQQR.
Now,
PQ=PR2-QR2=502-482=2500-2304=196=14 cm

Area of triangle =12×QR×PQ =12×48×14=336 cm2

Page No 534:

Answer:

In right angled ∆ABD,
AB2 = AD2 + DB2          (Pythagoras Theorem)
AB2 = 122 + 162
AB2 = 144 + 256
AB2 = 400
AB = 20 cm

Area of ∆ADB = 12×DB×AD
                        = 12×16×12
                        = 96 cm2               ....(1)

In ∆ACB,
The sides of the triangle are of length 20 cm, 52 cm and 48 cm.
∴ Semi-perimeter of the triangle is
s=20+52+482=1202=60 cm

∴ By Heron's formula,
Area of ACB=ss-as-bs-c                          =6060-2060-5260-48                          =6040812                          =480 cm2          ...2


Now,
Area of the shaded region = Area of ∆ACB − Area of ∆ADB
                                          =
480 − 96
                                          = 384 cm2

Hence, the area of the shaded region in the given figure is 384 cm2.

Page No 534:

Answer:

In the given figure, ABCD is a quadrilateral with sides of length 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle.

 

Join AC.

In right angled ∆ABC,
AC2 = AB2 + BC2          (Pythagoras Theorem)
AC2 = 62 + 82
AC2 = 36 + 64
AC2 = 100
AC = 10 cm

Area of ∆ABC = 12×AB×BC
                        = 12×6×8
                        = 24 cm2               ....(1)

In ∆ACD,
The sides of the triangle are of length 10 cm, 12 cm and 14 cm.
∴ Semi-perimeter of the triangle is
s=10+12+142=362=18 cm

∴ By Heron's formula,
Area of ACD=ss-as-bs-c                          =1818-1018-1218-14                          =18864                          =246 cm2                          =242.45 cm2                          =58.8 cm2            ...2

Thus,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
                                             = (24 + 58.8) cm2
                                             = 82.8 cm2

Hence, the area of quadrilateral ABCD is 82.8 cm2.

Page No 534:

Answer:

We know that ABD is a right-angled triangle.
∴ AB2=AD2-DB2=172-152=289-225=64=8 cm
 Now,Area of triangle ABD=12×Base×Height = 12×AB×BD = 12×8×15 =60 cm2

Let:a=9 cm, b = 15 cm and c=12 cms= a+b+c2=9+15+122=18 cmBy Heron's formula, we have:Area of triangle DBC = s(s-a)(s-b)(s-c)=18(18-9)(18-15)(18-12)=18×9×3×6=6×3×3×3×3×6=6×3×3=54 cm2

Now,
Area of quadrilateral ABCD  = Area of ABD + Area of BCD
                                           = (60 + 54) cm2 =114 cm2
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 17 + 8 + 12 + 9 = 46 cm



Page No 535:

Answer:

In right angled ∆ABC,
BC2 = AB2 + AC2          (Pythagoras Theorem)
BC2 = 212 + 202
BC2 = 441 + 400
BC2 = 841
BC = 29 cm

Area of ∆ABC = 12×AB×AC
                        = 12×21×20
                        = 210 cm2               ....(1)

In ∆ACD,
The sides of the triangle are of length 20 cm, 34 cm and 42 cm.
∴ Semi-perimeter of the triangle is
s=20+34+422=962=48 cm

∴ By Heron's formula,
Area of ACD=ss-as-bs-c                          =4848-2048-3448-42                          =4828146                          =336 cm2            ...2

Thus,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
                                             = (210 + 336) cm2
                                             = 546 cm2

Also,
Perimeter of quadrilateral ABCD = (34 + 42 + 29 + 21) cm
                                                     = 126 cm

Hence, the perimeter and area of quadrilateral ABCD is 126 cm and 546 cm2, respectively.

Page No 535:

Answer:

We know that BAD is a right-angled triangle.

∴ AB=BD2-AD2=262-242=676-576=100=10 cm

 Now,Area of triangle BAD=12×Base×Height = 12×AB×AD = 12×10×24 =120 cm2

Also, we know that BDC is an equilateral triangle.
Area of equilateral triangle = 34×(Side)2=34×(26)2=34×676=1693 =292.37 cm2

Now,
Area of quadrilateral ABCD = Area of ABD + Area of BDC
                                         = (120 + 292.37) cm2 = 412.37 cm2
Perimeter of ABCD = AB + BC + CD + DA = 10 + 26+ 26 + 24 = 86 cm

Page No 535:

Answer:

Let: a=26 cm, b =30 cm and c=28 cms= a+b+c2=26+30+282=42 cmBy Heron's formula, we have:Area of triangle ABC = s(s-a)(s-b)(s-c)=42(42-26)(42-30)(42-28)=42×16×12×14=14×3×4×4×2×2×3×14=14×4×2×3=336 cm2

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = 2×336=672 cm2

Page No 535:

Answer:

Let: a=10 cm, b =16 cm and c=14 cms= a+b+c2=10+16+142=20 cmBy Heron's formula, we have:Area of triangle ABC = s(s-a)(s-b)(s-c)=20(20-10)(20-16)(20-14)=20×10×4×6=10×2×10×2×2×3×2=10×2×23=69.2 cm2

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = 2×69.2 cm2=138.4 cm2

Page No 535:

Answer:

Area of ABCD=Area of ABD+Area of BDC=12×BD×AL+12×BD×CM=12×BD(AL+CM)=12×64(16.8+13.2)=32×30=960 cm2

Page No 535:

Answer:

In the given figure, ABCD is a trapezium with parallel sides AB and CD.



Let the length of CD be x.
Then, the length of AB be x + 4.

Area of trapezium = 12×sum of parallel sides×height
475=12×x+x+4×19475×2=192x+4950=38x+7638x=950-7638x=874x=87438x=23

∴ The length of CD is 23 cm and the length of AB is 27 cm.

Hence, the lengths of two parallel sides is 23 cm and 27 cm.                                                                                                  

Page No 535:

Answer:

In ∆ABC,
The sides of the triangle are of length 7.5 cm, 6.5 cm and 7 cm.
∴ Semi-perimeter of the triangle is
s=7.5+6.5+72=212=10.5 cm

∴ By Heron's formula,
Area of ABC=ss-as-bs-c                          =10.510.5-7.510.5-6.510.5-7                          =10.5343.5                          =21 cm2            ...2

Now,
Area of parallelogram DBCE = Area of ∆ABC
                                               = 21 cm2

Also,
Area of parallelogram DBCE = base × height
21=BC×DL21=7×DLDL=217=3 cm

Hence, the height DL of the parallelogram is 3 cm.

Page No 535:

Answer:

In the given figure, ABCD is a trapezium having parallel sides 90 m and 30 m.



Draw DE perpendicular to AB, such that DE = BC.

In right angled ∆ADE,
AD2 = AE2 + ED2          (Pythagoras Theorem)
⇒ 1002 = (90 − 30)2 + ED2
⇒ 10000 = 3600 + ED2
ED2 = 10000 − 3600
ED2 = 6400
ED = 80 m

Thus, the height of the trapezium = 80 m        ...(1)

Now,
Area of trapezium = 12×sum of parallel sides×height
                              = 12×90+30×80
                              = 4800 m2

The cost to plough per m2 = Rs 5
The cost to plough 4800 m2 = Rs 5 × 4800
                                             = Rs 24000

Hence, the total cost of ploughing the field is Rs 24000.



Page No 536:

Answer:

Let ABCD be a rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front.



According to the laws, the length of the inner rectangle = 40 − 3 − 3 = 34 m and the breath of the inner rectangle = 15 − 2 − 2 = 11 m.

∴ Area of the inner rectangle PQRS = Length × Breath
                                                          = 34 × 11
                                                          = 374 m2

Hence, the largest area where house can be constructed is 374 m2.

Page No 536:

Answer:

Let the sides of rhombus be of length x cm.



Perimeter of rhombus = 4x
⇒ 40 = 4x
x = 10 cm

Now,
In ∆ABC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is
s=10+10+122=322=16 cm

∴ By Heron's formula,
Area of ABC=ss-as-bs-c                          =1616-1016-1016-12                          =16664                          =48 cm2            ...1

In ∆ADC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is
s=10+10+122=322=16 cm

∴ By Heron's formula,
Area of ADC=ss-as-bs-c                          =1616-1016-1016-12                          =16664                          =48 cm2            ...2


∴ Area of the rhombus = Area of ∆ABC + Area of ∆ADC
                                      = 48 + 48
                                      = 96 cm2

The cost to paint per cm2 = Rs 5
The cost to paint 96 cm2 = Rs 5 × 96
                                        = Rs 480
The cost to paint both sides of the sheet = Rs 2 × 480
                                                                = Rs 960

Hence, the total cost of painting is Rs 960.

Page No 536:

Answer:

Let the semi-perimeter of the triangle be s.
Let the sides of the triangle be a, b and c.
Given: sa = 8, sb = 7 and sc = 5      ....(1)

Adding all three equations, we get
3s − (a + b + c) = 8 + 7 + 5
⇒ 3s − (a + b + c) = 20
⇒ 3s − 2s = 20                        s=a+b+c2
s = 20 cm                  ...(2)

∴ By Heron's formula,
Area of =ss-as-bs-c                 =20875           from 1 and 2                 =2014 cm2 

Hence, the area of the triangle is 2014 cm2.

Page No 536:

Answer:

Area of one triangular-shaped tile can be found in the following manner:

Let: a=16 cm, b = 12 cm and c=20 cms= a+b+c2=16+12+202=24 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=24(24-16)(24-12)(24-20)=24×8×12×4=6×4×4×4×4×6=6×4×4=96 cm2

Now,
Area of 16 triangular-shaped tiles = 16×96=1536 cm2
Cost of polishing tiles of area 1 cm2 = Rs 1
Cost of polishing tiles of area 1536 cm2 = 1×1536 =Rs 1536

Page No 536:

Answer:

We know that the triangle is an isosceles triangle.
Thus, we can find out the area of one triangular piece of cloth.
Area of isosceles triangle = b44a2-b2=204×4(50)2-202       (a= 50 cm and b=20 cm)=5×10000-400=5×9600=5×406=2006 =490 cm2

Now,
Area of 1 triangular piece of cloth = 490 cm2
Area of 12 triangular pieces of cloth = 12×490 =5880 cm2

Page No 536:

Answer:

In the given figure, ABCD is a square with diagonal 44 cm.
AB = BC = CD = DA.        ....(1)

In right angled ∆ABC,
AC2 = AB2 + BC2          (Pythagoras Theorem)
⇒ 442 = 2AB2
⇒ 1936 = 2AB2
AB2 = 19362
AB2 = 968
AB = 222 cm      ...(2)

∴ Sides of square = AB = BC = CD = DA = 222 cm

Area of square ABCD = (side)2
                                    = (222)2
                                    = 968 cm2          ...(3)

Area of red portion = 9684=242 cm2
Area of yellow portion = 9682=484 cm2
Area of green portion = 9684=242 cm2

Now, in ∆AEF,
The sides of the triangle are of length 20 cm, 20 cm and 14 cm.
∴ Semi-perimeter of the triangle is
s=20+20+142=542=27 cm

∴ By Heron's formula,
Area of AEF=ss-as-bs-c                          =2727-2027-2027-14                          =277713                          =2139                          =131.04 cm2            ...4

Total area of the green portion = 242 + 131.04 = 373.04 cm2

Hence, the paper required of each shade to make a kite is red paper 242 cm2, yellow paper 484 cm2 and green paper 373.04 cm2.



Page No 537:

Answer:

Area of rectangle ABCD = Length × Breath
                                        = 75 × 4
                                        = 300 m2

Area of rectangle PQRS = Length × Breath
                                       = 60 × 4
                                       = 240 m2

Area of square EFGH = (side)2
                                    = (4)2
                                    = 16 m2

∴ Area of the footpath = Area of rectangle ABCD + Area of rectangle PQRS − Area of square EFGH
                                     = 300 + 240 − 16
                                     = 524 m2

The cost of gravelling the road per m2 = Rs 50
The cost of gravelling the roads 524 m2 = Rs 50 × 524
                                                                = Rs 26200

Hence, the total cost of gravelling the roads at Rs 50 per m2 is Rs 26200.

Page No 537:

Answer:

The top and the bottom of the canal are parallel to each other.
Let the height of the trapezium be h.

Area of trapezium = 12×sum of parallel sides×height
⇒ 640 = 12×10+6×h
⇒ 640 = 8×h
h = 6408
h = 80 m

Hence, the depth of the canal is 80 m.

Page No 537:

Answer:

In the given figure, ABCD is the trapezium.

 

Draw a line BE parallel to AD.

In ∆BCE,
The sides of the triangle are of length 15 m, 13 m and 14 m.
∴ Semi-perimeter of the triangle is
s=15+13+142=422=21 m

∴ By Heron's formula,
Area of BCE=ss-as-bs-c                          =2121-1521-1321-14                          =21687                          =84 m2            ...1

Also,
Area of ∆BCE = 12×Base×Height
84=12×14×Height84=7×HeightHeight=847Height=12 m

∴ Height of ∆BCE = Height of the parallelogram ABED = 12 m

Now,
Area of the parallelogram ABED = Base × Height
                                                     = 11 × 12
                                                     = 132 m2                     ...(2)

∴ Area of the trapezium = Area of the parallelogram ABED + Area of the triangle BCE
                                        = 132 + 84
                                        = 216 m2

Hence, the area of a trapezium is 216 m2.

Page No 537:

Answer:

Let the length of the parallel sides be x and x − 8.
The height of the trapezium = 24 cm

Area of trapezium = 12×sum of parallel sides×height
⇒ 312 = 12×x+x-8×24
⇒ 312 = 12(2x − 8)
⇒ 2x − 8 = 31212
⇒ 2x − 8 = 26
⇒ 2x = 26 + 8
⇒ 2x = 34
x = 17 cm

Hence, the lengths of the parallel sides are 17 cm and 9 cm.

Page No 537:

Answer:

Diagonals d1 and d2 of the rhombus measure 120 m and 44 m, respectively.

Base of the parallelogram = 66 m

Now,
Area of the rhombus = Area of the parallelogram
12×d1×d2=Base×Height12×120×44=66×Height60×44=66×Height2640=66×HeightHeight=264066Height=40 m

Hence, the measure of the altitude of the parallelogram is 40 m.

Page No 537:

Answer:

It is given that,
Sides of the square = 40 m
Altitude of the parallelogram = 25 m

Now,
Area of the parallelogram = Area of the square
Base×Height=side2Base×25=402Base×25=1600Base=160025Base=64 m

Hence, the length of the corresponding base of the parallelogram is 64 m.

Page No 537:

Answer:

It is given that,
The sides of rhombus = 20 cm.
One of the diagonal = 24 cm.

 

In ∆ABC,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is
s=20+20+242=642=32 cm

∴ By Heron's formula,
Area of ABC=ss-as-bs-c                          =3232-2032-2032-24                          =3212128                          =192 cm2            ...1

In ∆ACD,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is
s=20+20+242=642=32 cm

∴ By Heron's formula,
Area of ACD=ss-as-bs-c                          =3232-2032-2032-24                          =3212128                          =192 cm2            ...2


∴ Area of the rhombus = Area of ∆ABC + Area of ∆ACD
                                      = 192 + 192
                                      = 384 cm2

Hence, the area of a rhombus is 384 cm2.

Page No 537:

Answer:

It is given that,
Area of rhombus = 480 cm2.
One of the diagonal = 48 cm.

(i) Area of the rhombus = 12×d1×d2
480=12×48×d2480=24×d2d2=48024d2=20 cm

Hence, the length of the other diagonal is 20 cm.

(ii) We know that the diagonals of the rhombus bisect each other at right angles.

 

In right angled ∆ABO,
AB2 = AO2 + OB2          (Pythagoras Theorem)
AB2 = 242 + 102
AB2 = 576 + 100
AB2 = 676
AB = 26 cm

Hence, the length of each of the sides of the rhombus is 26 cm.

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

Hence, the perimeter of the rhombus is 104 cm.



Page No 540:

Answer:

(b) 30 cm2

Area of triangle = 12×Base×HeightArea of ABC=12×12×5=30 cm2

Page No 540:

Answer:

(a) 96 cm2

Let: a=20 cm, b = 16 cm and c=12 cms= a+b+c2=20+16+122=24 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=24(24-20)(24-16)(24-12)=24×4×8×12=6×4×4×4×4×6=6×4×4=96 cm2

Page No 540:

Answer:

(b) 163 cm2
Area of equilateral triangle = 34×(Side)2=34×(8)2=34×64=163 cm2

Page No 540:

Answer:

(b) 85 cm2
Area of isosceles triangle = b44a2-b2Here, a= 6 cm and b=8 cmThus, we have:84×4(6)2-82=84×144-64=84×80=84×45=85 cm2

Page No 540:

Answer:

(c) 4 cm
Height of isosceles triangle = 124a2-b2=12452-62       a=5 cm and b=6 cm=12×100-36=12×64=12×8=4 cm

Page No 540:

Answer:

(b) 50 cm2
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:
Area of triangle = 12×Base×Height=12×10×10=50 cm2



Page No 541:

Answer:

(b) 53 cm
Height of equilateral triangle=32×Side=32×10=53 cm

Page No 541:

Answer:

(a) 123 cm2
Height of equilateral triangle = 32×Side6=32×SideSide=123×33=123×3=43  cmNow,Area of equilateral triangle = 34×(Side)2=34×432=34×48=123 cm2

Page No 541:

Answer:

(c) 384 m2

Let: a=40 m, b = 24 m and c=32 ms= a+b+c2=40+24+322=48 mBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=48(48-40)(48-24)(48-32)=48×8×24×16=24×2×8×24×8×2=24×8×2=384 m2

Page No 541:

Answer:

(b) 750 cm2

Let the sides of the triangle be 5x cm, 12x cm and 13x cm.
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, the sides of the triangle are 5×5 cm, 12×5 cm and 13×5 cm, i.e., 25 cm, 60 cm and 65 cm.

Now,
Let: a=25 cm, b = 60 cm and c=65 cms= 1502=75 cmBy Heron's formula, we have:Area of triangle = s(s-a)(s-b)(s-c)=75(75-25)(75-60)(75-65)=75×50×15×10=15×5×5×10×15×10=15×5×10=750 cm2

Page No 541:

Answer:

(a) 24 cm

Let:a=30 cm, b = 24 cm and c=18 cms= a+b+c2=30+24+182=36 cmOn applying Heron's formula, we get:Area of triangle = s(s-a)(s-b)(s-c)=36(36-30)(36-24)(36-18)=36×6×12×18=12×3×12×6×3=12×3×6=216 cm2

The smallest side is 18 cm.
Hence, the altitude of the triangle corresponding to 18 cm is given by:
Area of triangle = 216 cm212×Base×Height = 216Height = 216×218=24 cm

Page No 541:

Answer:

(b) 36 cm

Let PQR be an isosceles triangle and PXQR.
Now,
Area of triangle =48 cm2 12×QR×PX = 48h =9616=6 cmAlso, QX = 12×24 = 12 cm and PX = 12 cm
PQ=QX2+PX2a=82+62=64+36=100=10 cm

∴ Perimeter = (10 + 10 + 16) cm = 36 cm

Page No 541:

Answer:

(a) 36 cm
Area of equilateral triangle = 34×(Side)234×(Side)2 =363

(Side)2=144Side=12 cm

Now,
Perimeter
 = 3 × Side = 3 × 12 = 36 cm

Page No 541:

Answer:

(c) 60 cm2
Area of isosceles triangle = b44a2-b2Here, a= 13 cm and b=24 cmThus, we have:244×4(13)2-242=6×676-576=6×100=6×10=60 cm2

Page No 541:

Answer:

(c) 336 cm2

Let PQR be a right-angled triangle and PQQR.
Now,
PQ=PR2-QR2=502-482=2500-2304=196=14 cm

Area of triangle =12×QR×PQ =12×48×14=336 cm2

Page No 541:

Answer:

(a) 93 cm
Area of equilateral triangle = 813 cm234×(Side)2=813(Side)2=81×4(Side)2=324Side=18 cmNow,Height = 32×Side=32×18=93 cm



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