Rs Aggarwal 2019 Solutions for Class 9 Math 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 Math 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 2019 Book of Class 9 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 533:

Question 1:

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

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:

Question 2:

The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.

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:

Question 3:

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

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:

Question 4:

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.

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:

Question 5:

Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.

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:

Question 6:

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 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:

Question 7:

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 ₹ 5 per m2.

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:

Question 8:

Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m.

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:

Question 9:

Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.

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:

Question 10:

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

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:

Question 11:

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find  the area of the triangle.

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:

Question 12:

The perimeter of triangle is 50 cm. One side of the triangle is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side. Find the area of the triangle.

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:

Question 13:

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m, 15 m. The advertisements yield an earning of Rs 2000 per m2 a year. A company hired one of its walls for 6 months. How much rent did it pay?

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:

Question 14:

The perimeter of an isosceles triangle is 42 cm and its base is 112 times each of the equal sides. Find (i) the length of each side of the triangle, (ii) the area of the triangle, and (iii) the height of the triangle.

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:

Question 15:

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

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:

Question 16:

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

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:

Question 17:

Each side of an equilateral triangle measures 8 cm. Find (i) the area of the triangle, correct to 2 places of decimal and (ii) the height of the triangle, correct to 2 places of decimal. Take 3=1.732.

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:

Question 18:

The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take 3=1.732.

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:

Question 19:

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

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:

Question 20:

Find the area of the shaded region in the figure given below.

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:

Question 21:

The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle. Find its area. (Given, 6=2.45).

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:

Question 22:

Find the perimeter and area of a quadrilateral ABCD in which BC = 12 cm, CD = 9 cm, BD = 15 cm, DA = 17 cm and ∠ABD = 90°.

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:

Question 23:

Find the perimeter and area of the quadrilateral ABCD in which AB = 21 cm, BAC = 90°, AC = 20 cm, CD = 42 cm and AD = 34 cm.

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:

Question 24:

Find the area of the quadrilateral ABCD in which BCD is an equilateral triangle, each of whose sides is 26 cm, AD = 24 cm and ∠BAD = 90°. Also, find the perimeter of the quadrilateral. (Given: 3 = 1.73.)

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:

Question 25:

Find the area of a parallelogram ABCD in which AB = 28 cm, BC = 26 cm and diagonal AC = 30 cm.

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:

Question 26:

Find the area of a parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC = 16 cm. [Given: 3=1.73]

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:

Question 27:

In the given figure ABCD is a quadrilateral in which diagonal BD = 64 cm, AL BD and CMBD such that AL = 16.8 cm and CM = 13.2 cm. Calculate the area of quadrilateral ABCD.

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:

Question 28:

The area of a trapezium is 475 cm2 and its height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.

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:

Question 29:

In the given figure, a ∆ABC has been given in which AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram DBCE of the same area as that of ∆ABC is constructed. Find the height DL of the parallelogram.

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:

Question 30:

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs 5 to plough 1 m2 of the field, find the total cost of ploughing the field.

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:

Question 31:

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, a minimum of 3-m-wide space should be left in the front and back each and 2 m wide space on each of the other sides. Find the largest area where house can be constructed.

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:

Question 32:

A rhombus-shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs 5 per cm2. Find the cost of painting.

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:

Question 33:

The difference between the semiperimeter and the sides of a ∆ABC are 8 cm, 7 cm and 5 cm respectively. Find the area of the triangle.

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:

Question 34:

A floral design on a floor is made up of 16 tiles, each triangular in shape having sides 16 cm, 12 cm and 20 cm. Find the cost of polishing the tiles at Re 1 per sq cm.

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:

Question 35:

An umbrella is made by stitching 12 triangular pieces of cloth, each measuring (50 cm × 20 cm × 50 cm). Find the area of the cloth used in it.

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:

Question 36:

In the given figure, ABCD is a square with diagonal 44 cm. How much paper of each shade is needed to make a kite given in the figure?

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:

Question 37:

A rectangular lawn, 75 m by 60 m, has two roads, each road 4 m wide, running through the middle of the lawn, one parallel to length and the other parallel to breadth, as shown in the figure. Find the cost of gravelling the roads at Rs 50 per m2.

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:

Question 38:

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:

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:

Question 39:

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

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:

Question 40:

The difference between the lengths of the parallel sides of a trapezium is 8 cm, the perpendicular distance between these sides is 24 cm and the area of the trapezium is 312 cm2. Find the length of each of the parallel sides.

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:

Question 41:

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

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:

Question 42:

A parallelogram and a square have the same area. If the sides of the square measure 40 m and altitude of the parallelogram measures 25 m, find the length of the corresponding base of the parallelogram.

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:

Question 43:

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

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:

Question 44:

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:

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:

Question 1:

In a ABC it is given that base = 12 cm and height = 5 cm. Its area is
(a) 60 cm2
(b) 30 cm2
(c) 153 cm2
(d) 45 cm2

Answer:

(b) 30 cm2

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

Page No 540:

Question 2:

The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is
(a) 96 cm2
(b) 120 cm2
(c) 144 cm2
(d) 160 cm2

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:

Question 3:

Each side of an equilateral triangle measure 8 cm. The area of the triangle is
(a) 83 cm2
(b) 163 cm2
(c) 323 cm2
(d) 48 cm2

Answer:

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

Page No 540:

Question 4:

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is
(a) 165 cm2
(b) 85 cm2
(c) 163 cm2
(d) 83 cm2

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:

Question 5:

The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is
(a) 8 cm
(b) 30 cm
(c) 4 cm
(d) 11 cm

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:

Question 6:

Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is
(a) 510 cm2
(b) 50 cm2
(c) 103 cm2
(d)
75 cm2

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:

Question 7:

Each side of an equilateral triangle is 10 cm long. The height of the triangle is
(a) 103 cm
(b) 53 cm
(c) 102 cm
(d)
5 cm

Answer:

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

Page No 541:

Question 8:

The height of an equilateral triangle is 6 cm. Its area is
(a) 123 cm2
(b) 63 cm2
(c) 122 cm2
(d)
18 cm2

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:

Question 9:

The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is
(a) 480 m2
(b) 320 m2
(c) 384 m2
(d) 360 m2

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:

Question 10:

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 cm. The area of the triangle is
(a) 375 cm2
(b) 750 cm2
(c) 250 cm2
(d)
500 cm2

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:

Question 11:

The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is
(a) 24 cm
(b) 18 cm
(c) 30 cm
(d)
12 cm

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:

Question 12:

The base of an isosceles triangle is 16 cm and its area is 48 cm2. The perimeter of the triangle is
(a) 41 cm
(b) 36 cm
(c) 48 cm
(d)
324 cm

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:

Question 13:

The area of an equilateral triangle is 363 cm2. Its perimeter is
(a) 36 cm
(b) 123 cm
(c) 24 cm
(d) 30
cm

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:

Question 14:

Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is
(a) 156 cm2
(b) 78 cm2
(c) 60 cm2
(d) 120
cm2

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:

Question 15:

The base of a right triangle. is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is
(a) 168 cm2
(b) 252 cm2
(c) 336 cm2
(d) 504
cm2

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:

Question 16:

The area of an equilateral triangle is 813 cm2. Its height is
(a) 93 cm
(b) 63 cm
(c) 183 cm
(d) 9
cm

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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