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The following rectangle is composed of 8 congruent parts. Area of each part is
It is given that
Length =12 cm
Breadth =6 cm
We know that
Area of a rectangle = Length × Breadth
Substituting the values
Area of a rectangle =12×6=72 cm2
As there are 8 congruent parts
Area of each part =72/8=9 cm2
Therefore, the area of each part is 9 cm2.
Area of a right triangle is 54 cm2. If one of its legs is 12 cm long, its perimeter is
It is given that
Area of a right triangle =54 cm2
We know that
Area =1/2× base × height =1/2ab
Here
Perimeter =a+b+√a2+b2
P=a+√a2+4(A/a)2+2A/a
Substituting the values
P=12+√122+4(54/12)2+2(54/12)P=36 cm
Find the Area of parallelogram QPON shown in the figure below.
We know that
Area of a parallelogram = base × height
Substituting the values from the figure
Area of a parallelogram =6×8=48 cm2
Therefore, the area of parallelogram QPON is 48 cm2.
All the congruent triangles have equal area. Is this statement true?
Two triangles are said to be congruent if their corresponding sides and angles are equal.
We know that the Congruent triangle have equal area.
Therefore, all the congruent triangles have equal area.
A nursery school play ground is 160 m long and 80 m wide. In it 80 m×80 m is kept for swings and in the remaining portion, there is 1.5 m wide path parallel to its width and parallel to its remaining length as shown in Figure below. The remaining area is covered by grass. Find the area covered by grass.
Area of the school playground is 160 m×80 m=12800 m2
Area kept for swings =80 m×80 m=6400 m2
Area of path parallel to the width of playground
=80 m×1.5 m=120 m2
Area of path parallel to the remaining length of
playground
=80 m×1.5 m=120 m2.
Area common to both paths =1.5 m×1.5 m=2.25 m2.
[since it is taken twice for measurement it is to be subtracted from the area of paths]
Total area covered by both the paths
=(120+120−2.25)m2=237.75 m2.
Area covered by grass = Area of school playground – (Area
kept for swings + Area covered by paths)
=12800 m2−[6400+237.75]m2=(12800−6637.75)m2=6162.25 m2.
In Figure below, ABCD is a parallelogram, in which AB=8 cm,AD=6 cm and altitude AE=4 cm. Find the altitude corresponding to side AD.
Area of parallelogram ABCD=AB×AE=8×4 cm2 =32 cm2
Let altitude corresponding to AD be h. Then,
h×AD=32 or h×6=32 or h=326=163
Thus, altitude corresponding to AD is 163 cm.
A rectangular-shaped swimming pool with dimensions 30 m×20 m has 5 m wide cemented path along its length and 8 m wide path along its width (as shown in Figure below). Find the cost of cementing the path at the rate of Rs 200 per m2.
Area covered by swimming pool =30 m×20 m=600 m2.
Length of outer rectangle =(30+8+8)m=46 m
and its breadth =(20+5+5)m=30 m
So, the area of outer rectangle
=46 m×30 m=1380 m2.
Area of cemented path =
Area of outer rectangle – Area of swimming pool
=(1380−600)m2=780 m2.
Cost of cementing 1 m2 path =₹200
So, total cost of cementing the path
=₹780×200=₹156000
The Circumference of a circle is 33 cm. Find its area.
Let the radius of the circle be r. Then, 2πr=33
i.e., r=332π=332×722=214
Thus, radius is 214 cm
So, area of the circle =πr2=227⋅214⋅214=6938
Thus, area of the circle is 6938 cm2.
Rectangle ABCD is formed in a circle as shown in Figure below. If AE =8 cm and AD=5 cm, find the perimeter of the rectangle.
DE=EA+AD=(8+5)cm=13 cm
DE is the radius of the circle.
Also, DB is the radius of the circle.
Next, AC = DB [Since diagonals of a rectangle are equal in length]
Therefore, AC=13 cm.
From △ADC,DC2=AC2−AD2=132−52=169−25= 144=122
So, DC=12
Thus, length of DC is 12 cm.
Hence, perimeter of the rectangle ABCD
=2(12+5)cm=34 cm
Observe the shapes 1, 2, 3, and 4 in the figure below. Which of the following statements is not correct?
Trick: Count the number of squares around the shape.
Shape 1:
We know that
Perimeter =1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 +1+1 =22 units
Here
Area =18×1=18sq units
Shape 2:
We know that
Perimeter =1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 =18 units
Here
Area =18×1=18sq units
Shape 3:
We know that
Perimeter =1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 +1+1 =22 units
Area =16×1=16 sq units
Shape 4:
We know that
Perimeter =1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 +1+1 =22 units
Here
Area =18×1=18sq units
36 unit squares are joined to form a rectangle with the least perimeter. Perimeter of the rectangle is
It is given that
Area of rectangle =36 unit squares
We know that
36=6×6
It can be written as
36=2×3×2×3
So we get
36=22×3236=4×9
Here the sides of a rectangle are 4 cm and 9 cm
Perimeter =2(l+b)
Substituting the values
=2(4+9)=2(13)=26 units
Therefore, the perimeter of the rectangle is 26 units.
A wire is bent to form a square of side 22 cm. If the wire is rebent to form a circle, its radius is
It is given that
Side of a square =22 cm
Perimeter of square and circumference of circle are equal as the length of wire is same. Based on the question
Perimeter of square = Circumference of circle
4× side =2πr
We know that π=22/7
Substituting the values
4×22=2×22/7×r
By further calculation
r=(4×22×7)/(2×22)r=14 cm
Therefore, its radius is 14 cm.
Area of a rectangle and the area of a circle are equal. If the dimensions of the rectangle are 14cm × 11 cm, then radius of the circle is
It is given that
Dimensions of the rectangle =14×11 cm
From the question
Area of rectangle = Area of circle
1×b=πr2
Substituting the values
14×11=22/7×r2
By further calculation
r2=(14×11×7)/22r2=49r=7 cm
Therefore, the radius of the circle is 7 cm.
Area of the shaded portion in the figure below is
From the figure
Length of rectangle =5 cm
Breadth of rectangle =3+1=4 cm
We know that
Area of shaded portion =1/2× Area of rectangle =1/2×(l×b)
Substituting the values
=1/2×(5×4)=1/2×20
So we get
=10 cm2
Therefore, the area of the shaded portion is 10 cm2.
Area of parallelogram ABCD (Figure below) is not equal to
A parallelogram is defined as a quadrilateral in which both pairs of opposite sides are parallel and equal.
We know that
Area of parallelogram = base × height
From the figure
Area of parallelogram ABCD=AD×BE
As AD=BC
Area of parallelogram ABCD=BC×BE
Or
Area of parallelogram ABCD=DC×BF
Therefore, the area of the parallelogram is not equal to DE ×DC.
Area of triangle MNO of the figure below is
We know that
Area of triangle =1/2×b×h
From the figure
Area of triangle =1/2×NO×OQ
Therefore, the area of the triangle MNO is 1/2×NO×OQ.
Ratio of area of △MNO to the area of parallelogram MNOP in the same figure below is
We know that
Area of △MNO: Area of parallelogram MNOP=1/2×b×h:b×h
It can be written as
=1/2×NO×OQ:MP×OQ
So we get
=1/2×NO:NO[ As NO=MP]=1:2
Therefore, the ratio is 1:2
Ratio of areas of Δ MNO, ΔMOP and ΔMPQ in the figure below is
Using the figure
Area of ΔMNO=1/2×NO×MO
Substituting the values
=1/2×4×5=10 cm2
Area of △MOP=1/2×OP×MO
Substituting the values
=1/2×2×5=5 cm2
Area of △MPQ=1/2×PQ×MO
Substituting the values
=1/2×6×5=15 cm2
We know that
Required ratio =10:5:15=2:1:3
Therefore, the ratio of areas is 2:1:3.
In the figure below, EFGH is a parallelogram, and altitudes FK and FI are 8 cm and 4 cm respectively. If EF=10 cm, then area of EFGH is
A parallelogram is defined as a quadrilateral in which both pairs of opposite sides are parallel and equal.
In parallelogram EFGH,
EF=HG=10 cm
We know that
Area of parallelogram EFGH= Base × Corresponding height
Substituting the values
=10×4=40 cm2
Therefore, the area of EFGH is 40 cm2.
In the figure below, if PR=12 cm,QR=6 cm and PL=8 cm, then QM is
It is given that
PR=12 cmQR=6 cmPL=8 cm
In right angled triangle PLR, Using the Pythagoras theorem
PR2=PL2+LR2LR2=PR2−PL2
Substituting the values
LR2=144−64LR2=80LR=√80=4√5 cm
Here
LR=LQ+QRLQ=LR−QR
Substituting the values LQ=4√5−6 cm
Area of triangle PLR A1=1/2×LR×PL
Substituting the values LQ=4√5−6 cm
Area of triangle PLR A1=1/2×LR×PL
Substituting the values A1=1/2×4√5×8=16√5 cm2
Area of triangle PLQ A2=1/2×LQ×PL
Substituting the values A2=1/2×(4√5−6)×8A2=16√5−24 cm2
We know that
Area of triangle PLR = Area of triangle PLQ+ Area of triangle PQR
Substituting the values 16√5=16√5−24+ Area of triangle PQR
Area of triangle PQR=24 cm2
1/2×PR×QM=241/2×12×QM=24QM=4 cm
Therefore, QM is 4 cm.
In Figure below, ΔMNO is a right-angled triangle. Its legs are 6 cm and 8 cm long. Length of perpendicular NP on the side MO is
Given, MNO is a right-angled triangle
The legs are 6 cm and 8 cm long
We have to find the length of perpendicular NP on the side MO.
By using Pythagoras theorem,
MO2=MN2+MO2MO2=(6)2+(8)2MO2=36+64MO2=100
Taking square root,
MO=10 cm
Area of triangle =1/2× base × height
1/2×MN×NO=1/2×MO×NP6×8=10×NPNP=48/10NP=4.8 cm
Therefore, the value of NP is 4.8 cm
Area of a right-angled triangle is 30 cm2. If its smallest side is 5 cm, then its hypotenuse is
Given, area of a right-angled triangle is 30 cm2
The length of the smallest side is 5 cm
We have to find the length of the hypotenuse.
Area of triangle =1/2× base × height
30=1/2×5×& height
Height =30(2)/5
Height =6(2)
Height =12 cm
By using Pythagoras theorem,
( Hypotenuse )2=( height )2+( base )2
&( Hypotenuse )2=(12)2+(5)2
( Hypotenuse )2=144+25
( Hypotenuse )2=169
Taking square root,
Hypotenuse =13 cm
Therefore, the hypotenuse is 13 cm.
Length of tape required to cover the edges of a semicircular disc of radius 10 cm is
Perimeter of semicircle = circumference of semicircle + diameter
Circumference of semicircle =πr
Where, r is the radius of circle
=22/7×10=22(10)/7=220/7=31.4 cm Perimeter of semicircle =31.4+10(2)=31.4+20=51.4 cm
Therefore, the perimeter of semicircle is 51.4 cm
If p squares of each side 1 mm makes a square of side 1 cm, then p is equal to
Given, p squares of each side 1 mm makes a square of side 1 cm
We have to find the value of p.
Area of square =( side )2
Area of square with side 1 cm=(1)2=1 cm2
Area of square with side 1 mm=(1)2=1 mm2
Given, p× area of square with side 1 mm= area of square of side 1 cm
We know, 1 cm=10 mm
So, p×1 mm2(102)mm2
p=100/1
Therefore, p=100
If each side of a rhombus is doubled, how much will its area increase?
Given, each side of a rhombus is doubled.
We have to find the increase in its area.
Area of rhombus = base × corresponding height
Let the side of rhombus be b
Since the side is doubled, side =2b
Old Area =bh
New area =2bh
Therefore, the area has increased by 2 times.
If the sides of a parallelogram are increased to twice its original lengths, how much will the perimeter of the new parallelogram?
Given, the sides of a parallelogram are increased to twice its original lengths.
We have to find the increase in perimeter of the new parallelogram
Perimeter of parallelogram =2 (length + breadth)
Let the length be l
Let the breadth be b
So, perimeter =2(l+b)
Now, length =2l
Breadth =2b
New perimeter =2(2l+2b)
=2[2(l+b)]
Therefore, the new perimeter is increased by 2 times.
If the radius of a circle is increased to twice its original length, how much will the area of the circle increase?
Given, the radius of a circle is increased to twice its original length
We have to find the increase in area of the circle
Area of circle =πr2
Where, r is the radius of the circle
Now, r=2r
Area of circle =π(2r)2
=4πr2
Therefore, the area of the circle will be increased by 4 times.
What will be the area of the largest square that can be cut out of a circle of radius 10 cm?
Given, radius of circle is 10 cm
We have to find the area of the largest square that can be cut out of the circle.
The diagonal of the largest square will be equal to the diameter of the circle.
Let the side of the square be b.
Area of square =( side )2=b2
Considering right-angled triangle DAB,
By using Pythagoras theorem,
BD2=AD2+AB2(20)2=b2+b22b2=400b2=200
Area of square =200 cm2
Therefore, area of the largest square is 200 cm2
What is the radius of the largest circle that can be cut out of the rectangle measuring 10 cm in length and 8 cm in breadth?
Given, the dimension of the rectangle is 10 cm in length and 8 cm in breadth
We have to find the radius of the largest circle that can be cut out of the rectangle.
From the figure, we observe that the diameter of the largest circle will be equal to the breadth of the rectangle.
So, diameter of circle =8 cm
Radius = diameter /2
=8/2
=4 cm
Therefore, the radius is 4 cm
The area of a square is 100 cm2. The circumference (in cm ) of the largest circle cut of it is
Given, area of square is 100 cm2
We have to find the circumference of the largest circle cut out of the square.
Area of square =( side )2
side 2=100
Taking square root,
side =10 cm
The diameter of the circle will be equal to the side of the square.
Diameter of circle =10 cm
Radius =10/2=5 cm
Circumference of circle =2πr
Where, r is the radius of circle
=2π(5)=10πcm
Therefore, the circumference of the circle is 10πcm.
The area of a semicircle of radius 4r is
Given, radius of semicircle is 4r
We have to find the area of the semicircle
Area of semicircle =πr2/2
Where, r is the radius of the circle
Given, r=4r
Area =π(4r)2/2
=16πr2/2
=8πr2
Therefore, the area of the semicircle is 8πr2 square units.
A hedge boundary needs to be planted around a rectangular lawn of size 72 m × 18 m. If 3 shrubs can be planted in a metre of hedge, how many shrubs will be planted in all?
Given, a hedge boundary needs to be planted around a rectangular lawn
of size 72 m×18 m.
3 shrubs can be planted in a metre of hedge.
We have to find the number of shrubs that can be planted in all.
Perimeter of rectangle =2 (length × breadth )
Length of rectangular lawn =72 m
Breadth of rectangular lawn =18 m
Perimeter of rectangular lawn =2(72+18)
=2(90)
=180 m
In one meter 3 shrubs can be planted.
Number of shrubs that can be planted in rectangular lawn =3×
perimeter of lawn
=3(180)
=540 shrubs
Therefore, the number of shrubs that can be planted is 540.
In the Figure below, area of △AFB is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, find the altitude of the parallelogram to the base AB of length 10 cm. What is the area of △DAO, where O is the mid point of DC?
Given, the area of △AFB is equal to the area of parallelogram ABCD.
The length of altitude EF is 16 cm
We have to find the altitude of the parallelogram to the base AB of length 10 cm.
Area of triangle =1/2× base × height
Area of parallelogram = base × corresponding height
Area of △AFB= area of parallelogram ABCD
1/2×AB×EF=AB×EG1/2×10×16=10×EG10×8=10×EGEG=8 cm
Therefore, the corresponding altitude to the base AB is 8 cm.
We have to find the area of △DAO, where O is the midpoint of DC.
Since O is the midpoint of DC
DC=DO+CODO=CO=5 cm
Area of △DAO=1/2×DO× height
=1/2×5×8=5(4)=20 cm2
The ratio of the area of ΔWXY to the area of ΔWZY is 3:4 (Fig. 9.33). If the area of ΔWXZ is 56 cm2 and WY=8 cm, find the lengths of XY and YZ.
Given, the ratio of the area of △WXY to the area of △WZY is 3:4
The area of △WXZ is 56 cm2 and WY=8 cm.
We have to find the lengths of XY and YZ.
Area of triangle =1/2× base × height
Area of △WXY : area of △WZY=3:4
1/2×XY×WY:1/2×YZ×WY=3:4
XY:YZ=3:4
Area of △WXZ=1/2×XZ×WY
56=1/2×XZ×8
56=4×XZ
XZ=56/4
XZ=14 cm
From the figure
XZ=XY+YZYZ=XZ−XYYZ=14−XYNow,XY/YZ=3/4XY/(14−XY)=3/4XY(4)=3(14−XY)4XY=3(14)−3XY4XY+3XY=3(14)7XY=3(14)XY=3(14)/7XY=3(2)=6 cmYZ=14−6=8 cm
In the figure below, find the area of the parallelogram ABCD if the area of the shaded triangle is 9 cm2.
Given, area of the shaded triangle is 9 cm2
We have to find the area of the parallelogram ABCD.
Area of triangle =1/2× base × height
9=1/2×3×h
9(2)=3×h
h=9(2)/3
=3(2)
h=6 cm
Area of parallelogram = base × corresponding height
=BE×h
=(3+4)×6
=7(6)
=42 cm2
Therefore, the area of parallelogram is 42 cm2.
In the figure below, ABCD is a square with AB=15 cm. Find the area of the square BDFE.
Given, ABCD is a square
AB=15 cm
We have to find the area of the square BDFE.
From the figure,
BD is the diagonal of the square ABCD.
Diagonal of square = side ×√2
=15√2 cm
So, the length of the side of square BDFE =15√2 cm
Area of square =( side )2
=(15√2)2=(15)2×2=225(2)=450 cm2
Therefore, the area of square BDFE is 450 cm2.
Altitudes MN and MO of parallelogram MGHK are 8 cm and 4 cm long respectively (Figure below). One side GH is 6 cm long. Find the perimeter of MGHK.
Given, MGHK is a parallelogram
The altitudes MN and MO of parallelogram are 8 cm and 4 cm
The length of one side GH is 6 cm.
We have to find the perimeter of MGHK.
Area of parallelogram = base × corresponding height
=GH×MN=6×8=48 cm2
Now, taking base HK and altitude MO
Area of parallelogram =HK×MO
48=HK×4
HK=48/4
HK=12 cm
Perimeter of parallelogram =2 (length + breadth )
Perimeter of parallelogram MGHK=2(GH+HK)
=2(6+12)=2(18)=36 cm
In the figure below, area of △PQR is 20 cm2 and area of △PQS is 44 cm2. Find the length RS, if PQ is perpendicular to QS and QR is 5 cm.
Given, the area of △PQR is 20 cm2
The area of △PQS is 44 cm2
QR=5 cm
We have to find the length RS, if PQ is perpendicular to QS.
Area of triangle =1/2× base × height
Area of △PQR=1/2×QR×PQ
2O=1/2×5×PQPQ=2O(2)/5=4(2)PQ=8 cm
Area of △PQS=1/2×QS×PQ
44=1/2×QS×8QS=44(2)/8=44/4QS=11 cm
PQ is perpendicular to QS.
So, QS =QR+RS11=5+RSRS=11−5RS=6 cm
Find the areas of the shaded region in the figure below.
Consider r as the radius of smaller circle and R as the radius of bigger circle
From the figure, we can see,
r=7/2=3.5 cmR=7/2+7=21/2=10.5 cm
We know that
Area of shaded region = Area of bigger circle – Area of smaller circle
=πR2−πr2
Taking π as common
=π(R2−r2)
Substituting the values
=π[(10.5)2−(3.5)2]
By further calculation
=π(110.25−12.25)
Sowe get
=22/7×98=308 cm2
Find the areas of the shaded region shown in the figure below.
Given
Diameter of the complete circle =14 cm
We know that
Radius =14/2=7 cm
Here
Area of the complete circle =πr2
Substituting the values
=22/7×7×7=154 cm2
It is given that
Diameter of small circle =7/4 cm
Radius =7/(4×2)=7/8 cm
Here
Area of two small circles =2×πr2
Substituting the values
=2×22/7×7/8×7/8=77/16=4.8 cm2
We know that
Area of shaded region = Area of complete circle – Area of two small circles
Substituting the values
=154−4.8149.2 cm2
Therefore, the area of the shaded region is 149.2 cm2.
A circle with radius 16 cm is cut into four equal parts and rearranged to form another shape as shown in the figure below.
Does the perimeter change? If it does change, by how much does it increase or decrease?
We know that
Curved part of the new shape is 1/4 th the perimeter of the circle and the radius is the straight line
It is given that
Radius of the original circle =16 cm
Perimeter =2πr
Substituting the values
=2π×16=32π
Perimeter of curved part of new shape =1/4×32π=8π
Here
Perimeter of new shape =(4× Perimeter of curved part )+(2× radius )
Substituting the values
=(4×8π)+(2×16)=32π+32
Perimeter is increased by 32 cm.
Ishika has designed a small oval race track for her remote control car. Her design is shown in the figure below. What is the total distance around the track? Round your answer to the nearest whole cm.
We know that
Total distance around the track = Length of two parallel strips + Length of two semi circles
Given
Radius r=16 cm
Substituting the values
=(2×52)+(2×π×16)
By further calculation
=104+(2×3.14×16)
Sowe get
=104+100.5009=205 cm
Therefore, the total distance around the track is 205 cm.
The dimensions of a plot are 200 m×150 m. A builder builds 3 roads which are 3 m wide along the length on either side and one in the middle. On either side of the middle road he builds houses to sell. How much area did he get for building the houses?
It is given that
Dimension of a plot =200 m×150 m
Width of road =3 m
We know that
Total area available for houses = Area of total plot – Area of 3 roads
Substituting the values
=(200×150)−3×(3×200)
By further calculation
=30000−1800=28200 m2
Therefore, the area he got for building the houses is 28200 m2.
How much distance, in metres, a wheel of 25 cm radius will cover if it rotates 350 times?
It is given that
Radius of wheel =25 cm=25/100 m=1/4 m[ As 1 cm=1/100 m]
We know that
Distance traveled in one rotation =2πr
Substituting the values
=2×22/7×1/4=11/7=1.57 m
Therefore,
Distance traveled in 350 rotations =1.57×350
=550 m
Therefore, the wheel covers a distance of 550 m.
Calculate the area of the shaded region in the figure below, where all of the short line segments are at right angles to each other and 1 cm long.
Given, the figure represents a large rectangle in which all of the short line segments are at right angles to each other and 1 cm long.
We have to find the area of the shaded region.
Considering larger square,
side =9 cm
Area of square =( side )2
=(9)2=81 cm2
Given, short line segments are at right angles to each other and 1 cm long
By joining all the short lines using dotted lines, we obtain 41 identical squares.
Area of square =( side )2
Considering one square,
Side =1 cm
Area of square =(1)2=1 cm2
Area of 41 identical squares =41(1)=41 cm2
Area of shaded region = area of larger square – area of 41 identical squares
=81−41=40 cm2
Therefore, the area of the shaded region is 40 cm2.
ABCD is a given rectangle with length as 80 cm and breadth as 60 cm.P,Q,R,S are the mid points of sides AB,BC,CD,DA respectively. A circular rangoli of radius 10 cm is drawn at the centre as shown in the figure below. Find the area of shaded portion.
Given, ABCD is a rectangle with length 80 cm and breadth 60 cm.
P,Q,R and S are the midpoints of AB,BC,CD, and DA respectively.
A circular rangoli of radius 10 cm is drawn at the centre.
We have to find the area of the shaded portion.
Area of rectangle ABCD= length × breadth
=80(60)
=4800 cm2
The triangles SAP, PBQ, RCQ and RDS are congruent
S is the midpoint of DA
So, AS=60/2=30 cm
P is the midpoint of AB
So, AP=80/2=40 cm
Area of triangle =1/2× base × height
Area of triangle SAP =1/2×AP×AS
=1/2×40×30
=20(30)
=600 cm2
Area of 4 congruent triangles =4(600)
=2400 cm2
Area of circle =πr2
Area of circular rangoli =(3.14)(10)2
=3.14(100)=314 cm2
Area of shaded portion = area of rectangle PQRS – area of circular rangoli
=2400−314=2086 cm2
Therefore, the area of the shaded portion is 2086 cm2.
4 squares each of side 10 cm have been cut from each corner of a rectangular sheet of paper of size 100 cm × 80 cm. From the remaining piece of paper, an isosceles right triangle is removed whose equal sides are each of 10 cm length. Find the area of the remaining part of the paper.
Given 4 squares each side 10 cm is cut from each corner of a rectangle sheet of paper of size 100 cm×80 cm.
An isosceles triangle is removed from the remaining piece of paper whose equal sides are 10 cm length.
We have to find the remaining part of the paper.
Area of rectangular sheet = length × breadth
=100(80)=8000 cm2
Area of one square =( side )2
=(10)2=100 cm2
So, area of 4 squares =4(100)=400 cm2
Area of triangle =1/2× base × height
Area of isosceles triangle =1/2×10×10
=10(5)=50 cm2
Area of the remaining part of the sheet = area of rectangular sheet – area of 4 squares – area of isosceles triangle
=8000−400−50=7600−50=7550 cm2
Therefore, the area of the remaining part is 7550 cm2.
A dinner plate is in the form of a circle. A circular region encloses a beautiful design as shown in the figure below. The inner circumference is 352 mm and outer is 396 mm. Find the width of circular design.
Given, a dinner plate is in the form of a circle.
A circular region encloses a beautiful design.
The inner circumference is 352 mm and the outer is 396 mm.
We have to find the width of the circular design.
Circumference =2πr
Outer circumference =2πR
396=2(22/7)R396=(44/7)RR=396(7)/44R=9(7)R=63 mm
Inner circumference =2πr
352=2(22/7)r352=(44/7)rr=352(7)/44r=8(7)r=56 mm
Width of circular ring =R−r
=63−56=7 mm
Therefore, the width of the design is 7 mm.
The moon is about 384000 km from earth and its path around the earth is nearly circular. Find the length of path described by moon in one complete revolution. (Take π=3.14 )
Given, the moon is about 384000 km from earth and its path around the earth is nearly circular.
We have to find the length of the path described by the moon in one complete revolution.
The length of the path by moon in one complete revolution = circumference
Circumference =2πr
Given, radius =384000 km
=2(3.14)(384000)
=2(314)(3840)
=2411520 km
Therefore, the circumference is 2411520 km.
A photograph of Billiard/Snooker table has dimensions as 110 th of its actual size as shown in the figure below. The portion excluding six holes each of diameter 0.5 cm needs to be polished at rate of ₹200 per m2. Find the cost of polishing.
Given, a photograph of a billiard/snooker table has dimensions as 1/10 th of its actual size.
The portion excluding six holes each of diameter 0.5 cm needs to be polished.
The rate of polishing is ₹200 per m2.
We have to find the total cost of polishing.
Original length =25(10)=250 cm
Original breadth =10(10)=100 cm
Area of table = length × breadth
=250(100)
=25000 cm2
Area of circle =πr2
Given, diameter =0.5 cm
Radius =0.5/2=0.25 cm
=(22/7)(0.25)2
=22(0.0625)/7
=0.1964 cm2
Area of 6 holes =6(0.1964)
=1.179 cm2
Area to be polished = area of table – area of 6 holes
=25000−1.179=24998.82 cm2
We know, 1 m=100 cm
Area to be polished in m2=24998.82/10000 =2.4999 m2
Cost of polishing 1 m2=₹200
Cost of polishing table =200(2.4999) =₹499.98
Therefore, the total cost of polishing is ₹499.98.
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