How Many Squares - Class 5 Maths - Extra Questions

Find the area of the shaded portion in the following figures.
1794585_62ed521eb6db432f9b80a62ed0c5c6a7.png

Area of shaded portion = area of trapezium ABCH + area of trapezium CDEF
1676354_1794585_ans_5fd4da1537294fe18ccacfd8b55cab40.png


Find the area
1794551_1bb213a4fac9433b8501deb4349bfbe7.png

Area of the given figure= Area of DCF + Area of EDG+ Area ot trapezium FCBH+ Area of EGA+ Area of AHB
Therefore,
$$  \Rightarrow $$ area of DCF = $$ \frac{1}{2} \times base\times height $$ = $$ \frac{1}{2} \times 100 \times 100 $$ = 10000/2 = $$ 5000 m^{2} $$


Convert the following
$$50cm^{ 2 } in MM^{ 2 }$$

$$1c{m^2} = 100m{m^2}\\$$

$$\\50c{m^2} = 5,000m{m^2}$$'


Fill in the blanks: 
The area of a closed plane figure is measured in .......

The area of a closed plane figure is measured in sq. units. 


Find the area of the field shown along side. All dimension are in metres
565098_a92807554f2849aea561e1e9085015d5.png

Area of $$ABCDE =$$ Area of $$\triangle ABH +$$ area of trap $$BCFH +$$ area of $$\triangle CDF +$$ Area of $$\triangle AED$$
Now, area of $$ABH$$
$$= \times AH\times HB$$
$$= \dfrac {1}{2} \times 25\times 25$$
$$= \dfrac {625}{2}m^{2} = 312.5 m^{2}$$
Area of trap $$BCFH = \dfrac {1}{2}\times (HB + FC)\times HF$$
$$= \dfrac {1}{2} (25 + 50)\times 55 m^{2}$$
$$= \dfrac {75\times 55}{2}m^{2} = 2062.5\ m^{2}$$
Area of $$\triangle CDF = \times FC \times DF$$
$$= \dfrac {1}{2}\times 50\times 50 m^{2} = 1250\ m^{2}$$
Area of $$\triangle AED = \times AD \times EG$$
$$= \dfrac {1}{2} \times 130\times 60$$
$$= 3900 m^{2}$$
Thus, area of $$ABCDE = 312.5 m^{2} + 2062.5 m^{2} + 1250 m^{2} + 3900 m^{2}$$
$$= 7525 m^{2}$$


Find the area enclosed by the given figure
565380_a8817236955d47dfb6c912b1c200aa2a.png


From the fig. it is clear that $$CDEF$$ is a trapezium and $$ABCF$$ is a square.

Area of trapezium $$= \dfrac{a+b}{2}\times$$ height

Where $$a=ED$$ and $$b=FC$$

Area of figure $$ABCDEF =$$ Area of $$CDEF +$$ Area of square $$ABCF$$

                                         $$=\dfrac { 1 }{ 2 } \times \left( 7+18 \right) \times 8+\left( 18\times 18 \right) $$

                                        $$= 100 + 324$$

                                        $$= 424$$ $${ cm }^{ 2 }$$

941582_565380_ans_bdb6e8645a9f4df698682a6ab546cb82.png


Diagram of the adjacent picture frame has outer dimensions $$28\ cm \times 24\ cm$$ and inner dimensions $$20\ cm\times 16\ cm$$. Find the area of the shaded part of a frame, if the width of each section is the same
565388_fd085e08994a4caa9ad04e75ff3fe004.png

Given:
 outer dimensions $$28\ cm \times 24\ cm$$ and inner dimensions $$20\ cm\times 16\ cm$$

Area of shaded region =$$ (area EFJI - area EFGH) $$+ 2 $$\times $$ area $$\triangle AIH$$

= $$\left( 20\times 20-16\times 20 \right) +2\times \dfrac { 1 }{ 2 } \times 4\times 4$$

= $$\left( 400-320 \right) +16=80+16=96{ cm }^{ 2 }$$

946742_565388_ans_4bd2d115ff8b43ac90aa6e1cfccae8f1.png


Find the area enclosed by the given figure
565382_7f6c74b92e184cb0bb2e4fbef81383be.png


Area of figure $$ABCDEF =$$ ar. trap $$ADEF +$$ ar. rec $$ABCD$$

                                         $$=\dfrac { 1 }{ 2 } \left( 6+15 \right) \times 8+20\times 15$$

                                        $$= 84 + 300 = 384$$ $${ cm }^{ 2 }$$

941585_565382_ans_170c224ea8be40d5a2de3a5377a0957a.png


Find the area of each of the following fields all dimensions are in meters 
1318348_f72caea06d774f7f921f906e602d7b29.JPG


$$(1)$$ Area of the $$Δ DEI = 1/2$$ $$×$$ Base $$×$$ Height
$$⇒ 1/2 × 60 × 40$$ 
$$= 1200{m^2}$$
$$(2)$$ Area of the Trapezium $$EFGI = 1/2 ×$$ $$($$Sum of the two parallel sides$$)$$ $$×$$ Height
$$⇒ 1/2 × (60 + 50) × 70$$
$$⇒ 1/2 × 110 × 70$$
$$= 3850{m^2}$$
$$(3)$$ Area of the $$Δ AFG = 1/2$$ $$×$$ Base $$×$$ Height
$$⇒ 1/2 × 50 × 50$$
$$= 1250{m^2}$$
$$(4)$$ Area of the Trapezium$$ ABCH = 1/2$$ $$×($$Sum of the two parallel sides$$)$$ $$×$$ Height
$$⇒ 1/2 × (40 + 30) × 80$$
$$⇒ 1/2 × 70 × 80$$
$$= 2800 sq m$$
$$(5)$$ Area of the$$ Δ CDH =$$ $$1/2 ×$$ Base $$×$$ Height 
The height of this figure is not clear in the image. I think it is $$80 m$$ and I have assumed it $$80 m.$$
$$⇒ 1/2 × 40 × 80$$
$$= 1600{m^2}$$
Total area $$=$$ Area of $$Δ DEI +$$ Area of the Trapezium $$EFGI +$$ Area of $$Δ AFG +$$ Area of Trapezium $$ABCH +$$ Area of $$Δ CDH$$
$$⇒ 1200 + 3850 + 1250 + 2800 + 1600$$
$$= 10700$$ $${m^2}$$
So$$,$$ the area of the figure $$ABCDEF$$ is $$10700{m^2}$$


On a map drawn to a scale of $$1 : 250000$$, a triangular plot of land has the following measurements : $$AB = 3 cm, BC = 4 cm$$ and $$\angle ABC = 90^{\circ}$$. Calculate
(i) the actual length of $$AB$$ in km.
(ii) the area of the plot in sq. km:

$$AC=\sqrt{3^2+4^2}=5cm$$
Actual length of AC$$=\cfrac{5\times250000}{100\times10000}km$$
                                 $$=1.25km$$
Actual length of AB $$=\cfrac{3\times250000}{100\times1000}km$$
                                  $$=0.75km$$
Actual length of BC$$=\sqrt{1.25^2-0.75^2}$$
                                 $$=1km$$
Area $$=\cfrac{1}{2}\times1\times0.75$$
         $$=0.375 km^2$$

1210382_1244348_ans_020d35185c3c43708e94818dddcfcbde.png


Rajni says $$1\ sq.m=100^{2}sq.cm$$. Do you agree? Explain

Yes it is true that $$1\;{\rm{sq}}\;{\rm{m}} = {100^2}\;{\rm{sq}}{\rm{.}}\;{\rm{cm}}$$.

It is known that $$1\;{\rm{m}} = 100\;{\rm{cm}}$$

Squaring both sides, we get,

$${\left( {1\;{\rm{m}}} \right)^2} = {\left( {100\;{\rm{cm}}} \right)^2}$$

$$1\;{{\rm{m}}^2} = {100^2}\;{\rm{c}}{{\rm{m}}^2}$$



The diagram shows a field $$ABCD$$. 
a) Calculate the area of the field $$ABCD$$.
b) Calculate the perimeter of the field $$ABCD$$.
1137601_e1895b920dd54f44aff7818bbb155931.PNG

Area of ABCD=Area of $$\triangle ABC$$ + Area of $$\triangle ADC$$

$$=\dfrac{1}{2}100\times 70+\dfrac{100\times 110 \sin 40}{2}$$

$$=3500+3535.3$$

$$=7035.3$$ sq. units

Perimeter of ABCD$$=AB+BC+CD+AD$$

$$=70+122.066+110+72.436$$

$$=374.502$$


Find the area of each of the following fields in the figure. All dimensions are in meters
565392_5564d94790394a1591f81f68e4e62448.png


Area of $$\triangle DEI$$ = $$\dfrac { 1 }{ 2 } \times EI\times DI=\dfrac { 1 }{ 2 } \times 60\times 40=1200{ m }^{ 2 }$$

Area of $$\triangle $$$$DHC $$= $$\dfrac { 1 }{ 2 } \times HC\times DH=\dfrac { 1 }{ 2 } \times 40\times 80=1600{ m }^{ 2 }$$

Area of trapezium $$ ABCH$$ $$=$$ $$\dfrac { 1 }{ 2 } \left( CH+AB \right) \times HA$$
 $$=$$ $$\dfrac { 1 }{ 2 } \left( 40+30 \right) \times 70$$
$$=2450{ m }^{ 2 }$$

Area of $$\triangle $$$$AGF $$= $$\dfrac { 1 }{ 2 } \times AG\times GH=\dfrac { 1 }{ 2 } \times 50\times 50=1250{ m }^{ 2 }$$

Area of trapezium $$ EFGI$$$$=$$ $$\dfrac { 1 }{ 2 } \left( EI+FG \right) \times IG=\dfrac { 1 }{ 2 } \times \left( 60+50 \right) \times 70=3850{ cm }^{ 2 }$$

946747_565392_ans_a649d2e5d13e4bb4972844e4a300195f.png


Ravi completed the shape like this. What is the area occupied by the blue triangle?
695914_991fea2f58db41f7a7c817023f186a6a.png

Area of blue triangle$$=\cfrac { 1 }{ 2 } \times 4\times 2=2$$ sq. units.


Find the area of the polygon ABCDE whose measurementrs in meter are given in the following figure.
1247110_96ee6ebabce94418ac152c7ce1e170e1.png

$$ \rightarrow $$ There are three triangles comprising the pentagon
& area of triangle by heron's formula = $$ \sqrt{s(s-a)(s-b)(s-c)}$$
$$ \triangle _{1}$$
a = b = 8 , c = 12 
$$ S = \frac{8+8+12}{2} = 14$$
$$ \triangle _{1} = \sqrt{14(6)(6)(2)}$$
$$ = 6 \times 2\sqrt{7}$$
$$ \triangle _{2}$$
a = b = 12, b = 10
S = 17
$$ \triangle _{2} \sqrt{17(5)(5)(7)}$$
$$ = 5 \sqrt{119}$$
$$ \triangle _{3}$$
a = 10, b = 10, c = 12
S = 16
$$ \triangle _{3} = \sqrt{16(6)(6)(4)}$$
$$ = 6 \times 4 \times 2 = 48$$
$$ \Rightarrow $$ Total area $$= 48 + 12 \sqrt{7}+5\sqrt{119} m^{2} $$

1167314_1247110_ans_966ee6616cfb43ad8c5b874f6baaa6b6.jpg


Find the area of each of the following figure, all dimensions are in meters 
1318349_06c8c4ef0ecc4b10a1a93008aa28d657.JPG

$$(1)$$ Area of the $$Δ DEI = 1/2$$ $$×$$ Base $$×$$ Height
$$⇒ 1/2 × 60 × 40$$ 
$$= 1200{m^2}$$
$$(2)$$ Area of the Trapezium $$EFGI = 1/2 ×$$ $$($$Sum of the two parallel sides$$)$$ $$×$$ Height
$$⇒ 1/2 × (60 + 50) × 70$$
$$⇒ 1/2 × 110 × 70$$
$$= 3850{m^2}$$
$$(3)$$ Area of the $$Δ AFG = 1/2$$ $$×$$ Base $$×$$ Height
$$⇒ 1/2 × 50 × 50$$
$$= 1250{m^2}$$
$$(4)$$ Area of the Trapezium$$ ABCH = 1/2$$ $$×($$Sum of the two parallel sides$$)$$ $$×$$ Height
$$⇒ 1/2 × (40 + 30) × 80$$
$$⇒ 1/2 × 70 × 80$$
$$= 2800 sq m$$
$$(5)$$ Area of the$$ Δ CDH =$$ $$1/2 ×$$ Base $$×$$ Height 
The height of this figure is not clear in the image. I think it is $$80 m$$ and I have assumed it $$80 m.$$
$$⇒ 1/2 × 40 × 80$$
$$= 1600{m^2}$$
Total area $$=$$ Area of $$Δ DEI +$$ Area of the Trapezium $$EFGI +$$ Area of $$Δ AFG +$$ Area of Trapezium $$ABCH +$$ Area of $$Δ CDH$$
$$⇒ 1200 + 3850 + 1250 + 2800 + 1600$$
$$= 10700$$ $${m^2}$$
So$$,$$ the area of the figure $$ABCDEF$$ is $$10700{m^2}$$


Find the area of the following combined figure.
1402623_ad2c8577a6264abc95f133e89b9cdcaa.png

Area of first rectangle $$15\times 4=60cm^2$$
Area of second rectangle $$(17-4)\times 4\\13\times 4=52cm^2$$
Area of third rectangle $$(15-4)\times 4\\11\times 4=44cm^2$$
Area of fourth rectangle $$12\times 4=48cm^2$$
So the area of combined figure is $$60+52+44+48=204cm^2$$


Find the area of the following figure

1367942_f0c98d1fd430495cb3145b8479c630e3.png

Area of $$ABCDEA=$$Area of $$\triangle{AHB}+$$Area of trapezium $$HBCF+$$Area of $$\triangle{FCD}+$$Area of $$\triangle{DGE}+$$Area of $$\triangle{AGE}$$
$$=\dfrac{1}{2}\times HA\times HB+\dfrac{1}{2}\times FH\left(FC+HB\right)+\dfrac{1}{2}\times FD\times FC+\dfrac{1}{2}\times DG\times EG+\dfrac{1}{2}\times GA\times EG$$
$$=\dfrac{1}{2}\times 50\times 50+\dfrac{1}{2}\times 110\left(100+50\right)+\dfrac{1}{2}\times 100\times 100+\dfrac{1}{2}\times 180\times 120+\dfrac{1}{2}\times 80\times 120$$
$$=\dfrac{1}{2}\left[2500+16500+10000+21600+9600\right]$$
$$=\dfrac{60200}{2}=30100$$sq.units
$$\therefore$$Area of $$ABCDEA=30,100$$sq.units


Find the enclosed area
1404508_7884a5db6e7e4bc786a4a053e7631fd3.png


Area of rectangle is $$10\times 50=500cm^2$$
Area of $$2$$ trapezium is $$2\times \dfrac 12(20+10)\times 10=2\times 150=300cm^2$$
Area of whole figure is $$500+300=800cm^2$$


Find the area of the shaded region given in below figue . 
1404501_b74b0a8cad0f46d7aa48bca22e1b4031.PNG


1306398_1404501_ans_8babca17b91d4bdda78671bbb5a88449.jpeg


Find the area of the following figure.
1396474_68a3b26fd3ef4211922d4f5db22b84da.png

Area of total figure =Area of rectangle + Area of Triangle
Area of rectangle $$=l\times b$$
$$=6\times 10=60\ cm^2$$
Base of triangle $$DF=CF-CD=10-6=4cm$$
Area of triangle :
$$=\dfrac{1}{2}(b\times h)\\=\dfrac{1}{2}\times4\times4\\=4\times2=8\ cm^2$$

Area of total figure =$$60+8=68\ cm^2$$


In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and $$\angle BOD={ 90 }^{ 0 }$$. Find the area of the shaded region.
1392148_446ad4cc3d2340098cdd799a6de3c3a6.png

Area of the shaded region $$=$$ area of the circle $$-$$area of the triangle ABC$$-$$area of the quadrant COD

Here, in triangle ABC $$\angle CAB=90^o$$    (angle in a semicircle)
Hence triangle ABC is right-angled at A

Then applying Pythagoras theorem,
$$BC^2=AC^2+AB^2$$
$$BC^2=24^2+7^2$$
$$BC^2=625$$
$$BC=25\ cm$$

Therefore diameter of circle $$=25 cm$$
i.e., Radius$$=\dfrac{25}{2}=12.5\ cm$$

Then area of circle $$=\pi{r^2}=490.625\ cm^2$$

Area of triangleABC$$=\dfrac{1}{2}\times{base}\times{height}$$
$$=84\ cm^2$$

Area of quadrant $$=\dfrac{1}{4}\times{\pi{r^2}}$$
$$=122.65625\ cm^2$$

Therefore area of shaded region $$=490.625-84-122.65625=283.96875 \ cm^2$$


A sheet of paper measures $$50 \,cm$$ by $$30 \,cm$$. A strip $$6 \,cm$$ wide is cut from all around. Find the area of the remaining sheet and also the area of the strip cut out.

Area of sheet $$= l\times b$$
$$=50\times 30=1500$$ sq.cm
A strip $$6$$ cm wide is cut from all around.So
Length of remaining sheet$$=50-6-6=38$$ cm
Breadth of remaining sheet$$=30-6-6=18$$ cm
Area of remaining sheet$$=38\times 18=684$$ sq.cm
Area of strip cut out$$=1500-684=816$$ sq.cm


An agriculture field has a shape of an irregular polygon as shown in the figure. It's dimensions are given. Calculate it's area.
Given:$$AB=220\ m,\,AL=20\ m,\,AM=30\ m,\,BN=160\ m,\,OB=40\ m,\,TM=OS=60\ m,\,LP=20\ m,\,NQ=60\ m,\,MN=30~ m$$

1377555_bdd907f3fc5b498ebe35fd8ac9383fd7.png

To find the area of the field we need to find the sum of area of its parts.
Area of $$\triangle{ALP}=\dfrac{1}{2}\times\,AL\times\,LP$$
$$=\dfrac{1}{2}\times 20\times\,20=200\text{ m}^2$$
Area of trapezium $$PLNQ=\dfrac{1}{2}\times\,LN\times\left(LP+NQ\right)$$
$$=\dfrac{1}{2}\times 40\times\left(20+60\right)$$
$$=\dfrac{1}{2}\times 40\times 80=1600\text{ m}^2$$
Area of ractangle $$QNBR=BN\times NQ=160\times 60=9600\text{ m}^2$$
Area of $$\triangle{AMT}=\dfrac{1}{2}\times AM\times MT=\dfrac{1}{2}\times 30\times 60=900\text{ m}^2$$
Area of rectangle $$TMOS=TM\times\,OM=60\times 150=9000\text{ m}^2$$
Area of $$\triangle{SOB}=\dfrac{1}{2}\times OS\times OB=\dfrac{1}{2}\times 60\times 80=1200\text{ m}^2$$
$$\therefore\,$$Area of the field$$=200+1600+9600+900+9000+1200=22,500\text{ m}^2$$


In the figure, square ABCDABCD Find the area of shaded region. 
1405765_d026a16a9516416dbd0e698b66f173af.png

The radius of sector $$A-PCQ=20\sqrt 2cm$$
Area of sector is $$\dfrac 14\pi (20\sqrt 2)^2\\\dfrac 14\pi(800)=200\pi$$
The radius of sector $$C-BXD=20cm$$
Area of sector is $$\dfrac 14\pi (20)^2\\\dfrac 14\pi(400)=100\pi$$
Area of shaded region is $$200\pi-100\pi=100\pi$$


The boundary of the shaded region in the given figure consists of  three semicircles the smaller being equal. If the diameter of the larger is 28 cm find 
(i) the length of the boundary
(ii) the area of the shaded region.
1449975_9d9ec18a56074dfc81c2c23bc32558f9.PNG

$$\textbf{Hint: We will solve this question by Mensuration}$$

$$\textbf{Step-1: Finding the length of the boundary}$$
                $$\text{The total length of the boundary is:}$$
                $$l=$$$$\text{Circumference of large semicircle+2(Circumference of small semicircles)}$$
                $$\text{Let the radii of the large and small semicircles be}$$ $$r_1,r_2$$
                $$r_1=14\: cm,r_2=7\: cm$$
                $$l=\pi(14)+2(\pi\times 7)$$
                $$l=\dfrac{22}{7} \times 14+\dfrac{22}{7} \times 2 \times 7$$
                $$l=44+44=88\: cm$$
$$\textbf{Step-2: Finding the area of the shaded region}$$
                $$Area=$$ $$\text{Area of large semicircle}$$
                          $$=\dfrac{1}{2} \pi r_1^2$$ 
                          $$=\dfrac{1}{2}\times \dfrac{22}{7} \times 14 \times 14$$
                          $$=308\: cm^2$$

$$\textbf{Final Answer: Length of the boundary is 88 cm. Area of the shaded region is}$$ $$\mathbf{308\: cm^2.}$$


Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.
1599213_593c007af11b4112a2155c20ffc7a712.png


1870957_1599213_ans_d2ffcdbd018e430fac964d1c2e83de0e.jpg


Each square is having side=1 sq cm.
The area of the smallest stamp is:-
1457995_098ce2ff2d96453ebe77cf9d3c132b89.PNG

Smallest stamp is E having 4 small squares each having side 1 sq. cms, thus the area is 4 square cms.


In given figure ABC is quadrant of a radius of 14 cm and semicircle is drawn BC as diameter. Find the area of the shaded region.
1596249_f743a9b84d5847f9b14167669bf4fc1a.png

i) Area of part II
Area of Segment $$ABC-$$ ARea of $$\triangle ABC$$
$$=\dfrac {\theta}{360}\times \pi r^2 -\dfrac {1}{2}\times AC\times AB$$
$$=\left\{\dfrac {90}{360}\times \dfrac {22}{7}\times (14)^2-\dfrac {1}{2}\times 14\times 14\right\}$$
$$=\dfrac {1}{4}\times \dfrac {22}{7}\times 196-98$$
$$=56\ sq.cm$$
$$=56\ sq.cm$$.
ii) Area of shade region part III:
$$=\dfrac {1}{2}\pi r^2 -56$$
$$=\dfrac {1}{2}\times \dfrac {22}{7}\times \left(\dfrac {14\sqrt 2}{2}\right)^2 -56$$
$$=\dfrac {1}{2}\times \dfrac {22}{7}\times 98-56$$
$$=154-56$$
$$=98\ sq.cm$$


Find the area of shaded region.
1405376_c8bd848c36e94de38154e6af7f4c0026.png

The diameter of semicircles is side of square $$2a$$
$$\implies a+2a+a=42\\a=\dfrac{42}4=10.5cm$$
Area of square is $$(2a)^2=21^2=441cm^2$$
Area of semicircles is $$4\times \dfrac{\pi r^2}{2}\\2\times \dfrac{22}{7}\times 10.5\times 10.5\\44\times 1.5\times 10.5=693cm^2$$


Each square has side of 1 sq. cm
What is the occupied area of the largest stamp?
1460098_4489e07057cf4fb7a3ebc12b68fed2f6.PNG

Largest stamp is A and its occupied area is 18 squares of 1 square cm each. So, the occupied area is 18 square cms. 


Write true or false for each of the following using the figures shown above:

(a) area of fig. (i) is the same as the area of fig. (ii).
(b) Perimeter of fig. (i) is the same as fig. (ii).
(c) If fig. (i) is divided into squares of unit length, then its area is $$13$$ unit squares.
(d) Perimeter of fig. (ii) is 18 units. 

1793713_d88dff3cee7a467eabd30182dcda89b8.png

(a) False, 
In given figure, numbers of small identical blocks in $$(i)$$ are $$16$$ and in $$(ii)$$ are $$12$$ . Hence, area of $$(i)$$ is greater than $$(ii)$$

(b) False. 
Because, total edges of small blocks in closed shape $$(i)$$ are $$16$$ and in closed shape $$(ii)$$ are $$18$$.
So, perimeter of $$(ii)$$ is greater than $$(i)$$

(c) False 
Area of $$1$$ squares $$=1\times 1= 1$$ unit squares 
Number of squares $$=16$$ 
Hence, total area $$=16\times 1=16$$ unit squares 

(d) True 
Because, total edges of small blocks in closed shape $$(ii)$$ are $$18$$
So, perimeter of $$(ii)$$ is $$18$$ units

1673949_1793713_ans_4d6568fa79e04947b63ab2d7bd2f5da4.png


Find the area of the following fields. All dimensions are in meters.
1794537_224c04c43b734f74943f1a577da824b9.png

Area of the given figure
= Area of EFH + Area of rectangle EDCI + Area of Trapezium FHJG + Area of trapezium ICBK+ Area of GJA + Area of KBA


Split the following shapes into rectangles and find their areas. (The measures are given in centimetres). 
1801822_6cbe169896c54eb586fb41dc88865c26.png

(a) The given figure is split into 2 rectangles
Length of Part $$I = 12 cm $$
Breadth of Part $$I = 2 cm$$ 
Area of Part $$I$$ = $$Length \times Breadth$$ $$= (12 \times 2) cm^2 = 24 cm^2$$ 

Length of Part $$II = 8 cm$$ 
Breadth of Part $$II = 2 cm $$
Area of Part $$II = Length \times Breadth  = (8 \times 2) cm^2 = 16 cm^2$$
$$\therefore$$ Total area = Area of Part l + Area of Part || $$= (24 +16) cm^2 = 40 cm^2$$

(b)The given figure is divided into $$5$$ parts  
Here length of all the rectangles $$= 7 cm $$
and breadth of as the rectangles $$=7 cm$$ 
Area $$= l \times b = 7 \times 7 = 49 cm^2$$
Total number of rectangles $$= 5$$ 
$$\therefore$$ Total area $$= 5 \times 49 cm^2 = 245 cm^2$$ 

(c) The given figure is divided into $$2$$ rectangles 
Length of Part $$I$$ $$= 4 cm$$
Breadth of Part $$I$$ $$=(5-2-2) cm=1cm$$ 
Area $$= l \times b = 4 \times 1 cm^2= 4 cm^2$$ 
Length Part $$II$$ $$= 5 cm$$
Breadth Part $$II$$ $$= 1 cm$$ 
Area $$= l \times b = 5 \times 1 cm^2= 5cm^2$$ 
$$\therefore Total\,\, area = 4 cm^2 + 5 cm^2 = 9 cm^2$$ 

1775521_1801822_ans_564ef9161ced4e229b573d16484ef270.png


.
1795291_77889724b7164671956049dd035fc282.PNG


1) Area of trapezium ABCD=
$${ A }_{ 1 }=\frac { 1 }{ 2 } \times \left( sum\quad of\quad parallel\quad sides \right) \times height$$
$$\therefore { A }_{ 1 }=\frac { 1 }{ 2 } \times \left( AB+CD \right) \times EF$$
$$\therefore { A }_{ 1 }=\frac { 1 }{ 2 } \times \left( 12+9 \right) \times 4$$
$$\therefore { A }_{ 1 }=\frac { 1 }{ 2 } \times 21\times 4$$
$$\therefore { A }_{ 1 }=42$$ sq.cm.

2) Area of trapezium ECGH=
$$\therefore { A }_{ 2 }=\frac { 1 }{ 2 } \times \left( sum\quad of\quad parallel\quad sides \right) \times height$$
$$\therefore { A }_{ 2 }=\frac { 1 }{ 2 } \times \left( EH+CG \right) \times EC$$
$$\therefore { A }_{ 2 }=\frac { 1 }{ 2 } \times \left( EH+\left( BG-BC \right)  \right) \times EC$$
$$\therefore { A }_{ 2 }=\frac { 1 }{ 2 } \times \left( 8+\left( 16-4 \right)  \right) \times 3$$
$$\therefore { A }_{ 2 }=\frac { 1 }{ 2 } \times \left( 8+12 \right) \times 3$$
$$\therefore { A }_{ 2 }=\frac { 1 }{ 2 } \times 20\times 3$$
$$\therefore { A }_{ 2 }=30$$ sq.cm.

Thus, area of the shaded region=
$$A={ A }_{ 1 }+{ A }_{ 2 }$$
$$\therefore A=42+30$$
$$\therefore A=72$$ sq.cm.

1885893_1795291_ans_8e7c1d2458d34e9e89660daf1766d6e7.png


For the figure, given above, find the area of shaded region: (All measurements are in $$cm$$)

1831340_3e4026b9833f4f348777c57f4d58c1ea.png

Area of shaded portion = $$2 * 20 + 2 * 8 + 2 * (12 + 2)$$
By further calculation,
$$= 40 + 16 + 28$$
$$= 84 cm^{2}$$.


There is a pentagonal shaped park as shown in the following figure: For finding its area Jaspreet and Rahul divided it in two different ways. Find the area of this park using both ways. Can you suggest some other say of finding its area? 
1809982_008ec8fffa1442fc8954473ee8c54086.PNG

(i) Jaspreet's figure is divided into two trapeziums equal in area: $$DEAM$$ and $$DCBM$$

Now,
Area of trapezium $$DEAM=\dfrac { 1 }{ 2 } \times (AE+DM)\times AM$$
$$=\dfrac { 1 }{ 2 } \times (32+18)\times 9=\cfrac{50\times 9}2=225{ m }^{ 2 }$$

$$\therefore$$ Total area $$=225\times 2=450{ m }^{ 2 }$$

(ii) Rahul's figure is divided into shapes of a square and an isosceles triangle.

Area of square $$ABCE={(side)}^{2}={(18)}^{2}=324{ m }^{ 2 }$$
And, area of isosceles $$\triangle EDC=\cfrac { 1 }{ 2 } \times EC\times DC=\cfrac { 1 }{ 2 } \times  18\times 14=126{ m }^{ 2 }$$

$$\therefore$$ Total area $$=324+126=450{ m }^{ 2 }$$


The third way to find out the area of the given figure is to divide it into three triangles as shown:
Here $$DL\bot ED$$ and $$DL=14m$$
Area of $$\triangle DEC=\cfrac { 1 }{ 2 } \times  EC\times LD=\cfrac { 1 }{ 2 } \times 18\times 14=126{ m }^{ 2 }$$
Area of $$\triangle AEB=\cfrac { 1 }{ 2 } \times AB\times AE=\cfrac { 1 }{ 2 } \times 18\times 18=162{ m }^{ 2 }$$
Area of $$\triangle BEC=\cfrac { 1 }{ 2 } \times  18\times 18=162{ m }^{ 2 }$$
Hence, area of pentagon $$ABCDE$$ = Area $$\triangle DEC$$ + Area of $$\triangle AEB$$ + Area of $$\triangle BEC$$
$$=(126+162+162){ m }^{ 2 }=450{ m }^{ 2 }$$

1730394_1809982_ans_36c72034dc20456db509f2002d945ae1.png


The figure (ii) given below shows a field with the measurements given in metre.Find the area of the field .
1812356_e2c83f9b37264a18ac5fe870f5c18fa4.png


1977522_1812356_ans_86bae3e776b8473280ba0252459b3b4d.jpeg


By splitting the following figures into rectangle, find their areas (The measures are given in centimetres). 
1802842_9227505a2e004068b7ea2b7c2393f3f0.PNG

(a) Area of the figure,
 $$ = (3 \times 1 + 3 \times 1 + 3 \times 1) sq \ m$$
$$ = (3 + 3 + 3) sq \ m$$
$$ = 9 sq \ m$$
(b) Area of the figure,
 $$ = (3 \times 3 + 1 \times 2 + 3 \times 3 + 4 \times 2) sq \ m$$
$$ = (9 + 2 + 9 + 8) sq \ m$$
$$ = 28 sq \ m$$
2004633_1802842_ans_b37035362d514bf683e734b6a6409bec.png


Find the area of the shaded portion in the following figures all measurements are given in cm.
1810695_1c29d798424543d992419ad6a8665ba8.png

(i) Given,
Outer length $$=30\ cm$$
Outer breadth $$=10\ cm$$

Length of each rectangle of the corner $$(l)=\dfrac{30-18}2=6\ cm$$
and breadth $$(b)=\dfrac{10-6}2=\dfrac42=2cm$$

So
Area of 4 corner $$=(6\times 2)\times 4=48{ cm }^{ 2 }$$
And area of inner rectangle $$=18\times 6=108{ cm }^{ 2 }$$

Therefore
Area of shaded portion $$=108+48=156{ cm }^{ 2 }$$

(ii) Area of rectangle  $$I=4\times 2=8{ cm }^{ 2 }$$
Area of rectangle  $$II=4\times 1=4{ cm }^{ 2 }$$
Area of rectangle  $$III=6\times 1=6{ cm }^{ 2 }$$
And area of square  $$IV=1\times 1=1{ cm }^{ 2 }$$
Total area of shaded portion $$=8+4+6+1=19{ cm }^{ 2 }$$


Calculate the area enclosed by the given shapes. All measurements are in cm.
1809812_1c9fdb7f382742a486cb1d6d433422c9.PNG

(i) Its seen that,
Total area of the figure = Area of trapezium $$ABCD$$ + Area of rectangle $$GAFE$$

Now, Area of trapezium $$ABCD$$
$$=\cfrac12 \times $$ (Sum of parallel sides) $$\times$$ (height)
$$=\cfrac12 \times (AB+CD)\times (AF+FD)$$
$$=\cfrac12 \times (5+3)\times (5+4)$$
$$=\cfrac12 \times 8\times 9=36{ cm }^{ 2 }$$

Area of rectangle $$GAFE=Length\times Breadth$$
                                         $$=2\times 5=10{ cm }^{ 2 }$$

Hence,
Total area of the figure $$=36+10\ { cm }^{ 2 }=46{\ cm }^{ 2 }$$

(ii) Its seen that,
Area of given figure = Area of rect. $$ABCD$$ + Area of parallelogram $$BIHJ$$ + Area of rectangle $$EFGH$$

Now, Area of rectangle $$ABCD=Length \times Breadth$$
                                                 $$=AD\times CD$$
                                                 $$=9\times 2$$
                                                 $$=18cm^2$$
Area of rectangle $$EFGH=Length\times Breadth$$
                                         $$=(EJ+JH)\times EF$$
                                         $$=(7+2)\times 2=7\times 2=18{ cm }^{ 2 }$$

Area of parallelogram $$BIHJ=2\times 5=10{ cm }^{ 2 }$$
(Since distance between $$BI$$ And $$HJ=9-2-2=5cm$$)

Hence
Total area of the figure $$=(18+18+10){ cm }^{ 2 }=46{ cm }^{ 2 }$$

1730278_1809812_ans_4a38d51aa7e742379614dc1caee8627d.png


Irfan and Ishan were painting a wall. Irfan painted the diamond-shaped portion. Ishan painted the rest. What area did each paint?
1314847_c1079416659746928565835f1676c252.PNG

$$\begin{array}{l} Area\, \, of\, wall=8\times 2.4=19.2{ m^{ 2 } } \\ Ishan\, \, area\, \, of\, \, pa{ { int } }=4\times \frac { 1 }{ 2 } \times 1.2\times { 4^{ 2 } }=4\times 2.4=9.6{ m^{ 2 } } \\ \therefore Ishan\, area\, \, of\, \, pa{ { int } }=19.2-9.6=9.6\, { m^{ 2 } } \end{array}$$


Radha made a picture of an aeroplane with coloured paper as shown in fig $$8.15$$. Find the total area of the paper used.
1259691_95e649f1a28840b2a1dc4b5f3ce7ea11.JPG


Consider the problem

Semi perimeter of region $$I$$ (triangle)
$$\begin{array}{l} s=\dfrac { { 5+5+1 } }{ 2 }  \\ =\dfrac { { 11 } }{ 2 }  \\ =5.5cm \end{array}$$

By using Hero's formula, then the area of  region $$I$$ (triangle)
$$\begin{array}{l} =\sqrt { s\left( { s-5 } \right) \left( { s-5 } \right) \left( { s-1 } \right)  }  \\ =\sqrt { 5.5\left( { 5.5-5 } \right) \left( { 5.5-5 } \right) \left( { 5.5-1 } \right)  }  \\ =\sqrt { 5.5\times 0.5\times 0.5\times 4.5 }  \\ =2.5c{ m^{ 2 } }\left( { approx } \right) ..........\left( 1 \right)  \end{array}$$

And,

Area of region $$II$$ (rectangle)
Area of rectangle $$ = length \times Breadth$$

so,
$$\begin{array}{l} 1\times 6.5 \\ =6.5c{ m^{ 2 } }..........\left( 2 \right)  \end{array}$$

Now applying Pythagoras theorem on $$\Delta ABC$$ 

$$A{B^2} = A{C^2} + B{C^2}$$
$$\begin{array}{l} \Rightarrow A{ B^{ 2 } }={ 6^{ 2 } }+{ \left( { 1.5 } \right) ^{ 2 } } \\ \Rightarrow A{ B^{ 2 } }=36+2.25=38.25 \\ \Rightarrow AB=6.2c{ m^{ 2 } }\left( { approx } \right)  \end{array}$$

Semi perimeter of region $$IV$$ $$\Delta ABC$$

$$\begin{array}{l} s=\dfrac { { 6+1.5+6.2 } }{ 2 }  \\ =\dfrac { { 13.7 } }{ 2 }  \\ =6.85c{ m^{ 2 } } \end{array}$$

Using Hero's formula, Area of region $$IV$$ $$\Delta ABC$$

$$\begin{array}{l} =\sqrt { s\left( { s-6 } \right) \left( { s-1.5 } \right) \left( { s-6.184 } \right)  }  \\ =\sqrt { 6.85\left( { 6.85-6 } \right) \left( { 6.85-1.5 } \right) \left( { 6.85-6.2 } \right)  }  \\ =\sqrt { 6.85\times 0.85\times 5.35\times 0.65 }  \\ =4.5c{ m^{ 2 } }..............\left( 3 \right)  \end{array}$$

Area of region $$V$$ triangle $$ = Area\,of\,region\,IV\left( {\Delta ABC} \right) = 4.5c{m^2}.......\left( 4 \right)$$

Here it seems Region $$III$$ is a trapezium and we do not know the height of this trapezium, so we have to draw this trapezium separately so we can find its area.
So, Draw $$ST \bot PQ\,and\,RU \bot PQ$$. doing so, we have $$PT = 0.5cm\,and\,UQ = 0.5cm$$

Now, we can use Pythagoras theorem on $$\Delta SPT$$ to find the height of trapezium $$PQRS$$.

$$S{P^2} = S{T^2} + P{T^2}$$
$$\begin{array}{l} \Rightarrow { 1^{ 2 } }=S{ T^{ 2 } }+{ \left( { 0.5 } \right) ^{ 2 } } \\ \Rightarrow S{ T^{ 2 } }=1-0.25=0.75 \\ \Rightarrow ST=0.87cm\left( { approx } \right)  \end{array}$$

So, Area of region $$III$$ (Trapezium) 
$$\begin{array}{l} =\frac { 1 }{ 2 } \left( { SR+PQ } \right) \times ST \\ =\frac { 1 }{ 2 } \left( { 1+2 } \right) \times 0.87 \\ =1.305\, c{ m^{ 2 } }.........\left( 5 \right)  \end{array}$$

Now, Adding $$(1)$$,$$(2)$$,$$(3)$$,$$(4)$$ and $$(5)$$ to find the total area of aeroplane picture.

$$\begin{array}{l} =2.5+6.5+4.5+4.5+1.305 \\ =19.305c{ m^{ 2 } }\left( { approx } \right)  \end{array}$$

1194133_1259691_ans_9f4704d6613b4d6e9cc33979bd50e5e8.PNG


Find the area of the shaded portion.
1367577_9cd9e14e040c4b2ea9e29df2b0fef75d.PNG

$$\textbf{Step 1: Find the area of shaded portion. }$$
                $$\text{Length of the rectangle(l) = 25 m}$$
                $$\text{Breadth of the rectangle(b) = 16 m}$$
                $$\text{Area of rectangle}$$ $$=l\times b$$
                                                 $$=25\times 16\ m^2$$
                                                 $$=400\ m^2$$

                $$\text{Area of each unshaded corner}$$ $$=\dfrac{1}{4}\pi r^2$$

                                                                        $$=\dfrac{1}{4}\times 3.14\times (2.1)^2\ m^2$$

                                                                        $$=\dfrac{1}{4}\times 3.14\times 4.41\ m^2$$

                                                                        $$=3.46\ m^2$$
                $$\text{Therefore, area of the 4 corners}$$ $$= 3.46\times 4\ m^2=13.84\ m^2$$
                $$\text{Thus, area of the shaded portion}$$ $$=(400-13.84)\ m^2$$
                                                                              $$=386.16\ m^2$$
              
$$\textbf{Hence, the area of the shaded portion is 386.16 }\mathbf{m^2}.$$


.
1795279_c14cef03705a4f019d691a2592678109.png

1) Area of $$\triangle EGD=\frac { 1 }{ 2 } \times GD\times EG$$
$$\therefore \triangle EGD=\frac { 1 }{ 2 } \times \left( GF+FD \right) \times EG$$
$$\therefore \triangle EGD=\frac { 1 }{ 2 } \times \left( 180 \right) \times 120$$
$$\therefore \triangle EGD=10800$$ sq.units

2) Area of $$\triangle EGA=\frac { 1 }{ 2 } \times GA\times EG$$
$$\therefore \triangle EGA=\frac { 1 }{ 2 } \times \left( GH+HA \right) \times EG$$
$$\therefore \triangle EGA=\frac { 1 }{ 2 } \times \left( 30+50 \right) \times 120$$
$$\therefore \triangle EGA=4800$$ sq.units

3) Area of $$\triangle AHB=\frac { 1 }{ 2 } \times AH\times HB$$
$$\therefore \triangle AHB=\frac { 1 }{ 2 } \times 50\times 50$$
$$\therefore \triangle AHB=1250$$ sq.units

4) Area of $$\Box CFHB=\frac { 1 }{ 2 } \times \left( sum\quad of\quad parallel\quad sides \right) \times height$$
$$\therefore \Box CFHB=\frac { 1 }{ 2 } \times \left( CF+HB \right) \times FH$$
$$\therefore \Box CFHB=\frac { 1 }{ 2 } \times \left( 100+50 \right) \times \left( 80+30 \right) $$
$$\therefore \Box CFHB=\frac { 1 }{ 2 } \times 150\times 110$$
$$\therefore \Box CFHB=8250$$ sq.units

5) Area of $$\triangle DFC=\frac { 1 }{ 2 } \times DF\times FC$$
$$\therefore \triangle DFC=\frac { 1 }{ 2 } \times 100\times 100$$
$$\therefore \triangle DFC=5000$$ sq.units

Thus, total area of polygon is,
$$A=10800+4800+1250+8250+5000$$
$$\therefore A=30100$$ sq.units


.
1795294_566a956f61474161b61644e9e0f90d14.PNG

1) Area of the triangle is calculated by Heron's formula.
$$a=5 cm$$, $$b=5 cm$$ and $$c=6 cm$$

$$\therefore s=\frac { a+b+c }{ 2 } $$

$$\therefore s=\frac { 5+5+6 }{ 2 } $$

$$\therefore s=8\quad cm$$

Thus, area of the two triangles is given by,
$${ A }_{ 1 }=2\times \sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right)  } $$

$$\therefore { A }_{ 1 }=2\times \sqrt { 8\times \left( 8-5 \right) \times \left( 8-5 \right) \times \left( 8-6 \right)  } $$

$$\therefore { A }_{ 1 }=2\times \sqrt { 8\times 3\times 3\times 2 } $$

$$\therefore { A }_{ 1 }=2\times \sqrt { 16\times 9 } $$

$$\therefore { A }_{ 1 }=2\times 4\times 3$$

$$\therefore { A }_{ 1 }=24\quad { cm }^{ 2 }$$

2) Radius of the semi-circle is $$r=3 cm$$

Thus, area of the two semi-circles is given by,
$${ A }_{ 2 }=2\times \frac { \pi { r }^{ 2 } }{ 2 } $$

$$\therefore { A }_{ 2 }=2\times \frac { 22 }{ 7 } \times \frac { { 3 }^{ 2 } }{ 2 } $$

$$\therefore { A }_{ 2 }=\frac { 22 }{ 7 } \times 9$$

$$\therefore { A }_{ 2 }=28.286\quad { cm }^{ 2 }$$

3) Area of the square is given by,
$${ A }_{ 3 }=6\times 6$$

$$\therefore { A }_{ 3 }=36\quad { cm }^{ 2 }$$

Thus, total area of polygon is given by,
$$A=24+28.286+36$$

$$\therefore A=88.286\quad { cm }^{ 2 }$$


.
1795295_797c66e35275414a9b04174cff286400.PNG

1) Volume of the prism is,
$$V=Area\quad of\quad base\times height$$

$$\therefore V=\left( \frac { 1 }{ 2 } \times DF\times DE \right) \times AD$$

$$\therefore V=\left( \frac { 1 }{ 2 } \times \frac { x }{ 2 } \times x \right) \times 2x$$

$$\therefore V=\left( \frac { { x }^{ 2 } }{ 4 }  \right) \times 2x$$

$$\therefore V=\frac { { x }^{ 3 } }{ 2 } $$

2) Refer to the figure (b) above.
Volume of the cuboid is given by,
$$V=l\times b\times h$$

$$\therefore V=3y\times 2y\times y$$

$$\therefore V=6{ y }^{ 3 }$$

3) 

Given ABCDEF is regular hexagon. Thus, each interior angle is $${ 12 }0^{ 0 }$$

1) In $$\triangle ABG$$, $$sin30=\frac { BG }{ AB } $$

$$\therefore \frac { 1 }{ 2 } =\frac { BG }{ 2p } $$

$$\therefore BG=p$$

Similarly, $$cos30=\frac { AG }{ AB } $$

$$\therefore \frac { \sqrt { 3 }  }{ 2 } =\frac { AG }{ 2p } $$

$$\therefore AG=\sqrt { 3 } p$$

Now, $$AC=2\times AG$$

$$\therefore AC=2\times \sqrt { 3 } p$$

Now, area of triangle ABC is given by,
$$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \times AC\times BG$$

$$\therefore A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \times 2\times \sqrt { 3 } p\times p$$

$$\therefore A\left( \triangle ABC \right) =\sqrt { 3 } { p }^{ 2 }$$

2) $$\triangle ABC\cong \triangle FED$$

$$\therefore A\left( \triangle FED \right) =\sqrt { 3 } { p }^{ 2 }$$

3) Area of rectangle ACDF is given by,
$$A\left( \Box ACDF \right) =AC\times AF$$

$$\therefore A\left( \Box ACDF \right) =2\times \sqrt { 3 } p\times 2p$$

$$\therefore A\left( \Box ACDF \right) =4\sqrt { 3 } { p }^{ 2 }$$

Thus, total area of polygon is given by,
$$A=\sqrt { 3 } { p }^{ 2 }+\sqrt { 3 } { p }^{ 2 }+4\sqrt { 3 } { p }^{ 2 }$$

$$\therefore A=6\sqrt { 3 } { p }^{ 2 }$$

1892530_1795295_ans_5c977257f0db47e79bae359da75fcf05.png


 
1485475_555d19d4fd8747ff86fc7ccbbcf2cbc4.jpg


2071432_1485475_ans_d1e8576e449944c5b75e5b7184020520.jpg


.
1795293_1af04dabd4ce47ea9ea004154df96814.PNG

1) Diameter of the semi-circle is $$d=12 cm$$
Thus, radius of semi-circle is $$r=6 cm$$

Thus, area of semi-circle is given by,
$${ A }_{ 1 }=\frac { \pi { r }^{ 2 } }{ 2 } \quad { cm }^{ 2 }$$

$$\therefore { A }_{ 1 }=\frac { 22 }{ 7 } \times \frac { { 6 }^{ 2 } }{ 2 } $$

$$\therefore { A }_{ 1 }=\frac { 22 }{ 7 } \times \frac { 36 }{ 2 } $$

$$\therefore { A }_{ 1 }=56.57\quad { cm }^{ 2 }$$        (1)

2) Area of square is given by,
$${ A }_{ 2 }=12\times 12$$

$$\therefore { A }_{ 2 }=144\quad { cm }^{ 2 }$$         (2)

3) Radius of quarter circle is $$r=6 cm$$
Thus, area of quarter circle is given by,

$${ A }_{ 3 }=\frac { \pi { r }^{ 2 } }{ 4 } $$

$$\therefore { A }_{ 3 }=\frac { 22 }{ 7 } \times \frac { { 6 }^{ 2 } }{ 4 } $$

$$\therefore { A }_{ 3 }=\frac { 22 }{ 7 } \times \frac { 36 }{ 4 } $$

$$\therefore { A }_{ 3 }=28.29\quad { cm }^{ 2 }$$         (3)

Thus, total area of polygon is given by,
$$A={ A }_{ 1 }+{ A }_{ 2 }+{ A }_{ 3 }$$

$$\therefore A=56.57+144+28.29$$

$$\therefore A=228.86\quad { cm }^{ 2 }$$


Class 5 Maths Extra Questions