2D Quiz

Hint

It is given that
Length \(=12 \mathrm{~cm}\)
Breadth \(=6 \mathrm{~cm}\)
We know that
Area of a rectangle \(=\) Length \(\times\) Breadth
Substituting the values
Area of a rectangle \(=12 \times 6=72 \mathrm{~cm}^2\)
As there are 8 congruent parts
Area of each part \(=72 / 8=9 \mathrm{~cm}^2\)
Therefore, the area of each part is \(9 \mathrm{~cm}^2\).

Hint

It is given that
Area of a right triangle \(=54 \mathrm{~cm}^2\)
We know that
Area \(=1 / 2 \times\) base \(\times\) height \(=1 / 2 a b\)
Here
Perimeter \(=a+b+\sqrt{a^2}+b^2\)
\(
P=a+\sqrt{a^2}+4(A / a)^2+2 A / a
\)
Substituting the values
\(
\begin{aligned}
& P=12+\sqrt{12^2}+4(54 / 12)^2+2(54 / 12) \\
& P=36 \mathrm{~cm}
\end{aligned}
\)

Hint

We know that
Area of a parallelogram \(=\) base \(\times\) height
Substituting the values from the figure
Area of a parallelogram \(=6 \times 8=48 \mathrm{~cm}^2\)
Therefore, the area of parallelogram QPON is \(48 \mathrm{~cm}^2\).

Hint

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.

Hint

Area of the school playground is \(160 \mathrm{~m} \times 80 \mathrm{~m}=12800 \mathrm{~m}^2\)
Area kept for swings \(=80 \mathrm{~m} \times 80 \mathrm{~m}=6400 \mathrm{~m}^2\)
Area of path parallel to the width of playground
\(
=80 \mathrm{~m} \times 1.5 \mathrm{~m}=120 \mathrm{~m}^2
\)
Area of path parallel to the remaining length of
playground
\(
=80 \mathrm{~m} \times 1.5 \mathrm{~m}=120 \mathrm{~m}^2 .
\)
Area common to both paths \(=1.5 \mathrm{~m} \times 1.5 \mathrm{~m}=2.25 \mathrm{~m}^2\).
[since it is taken twice for measurement it is to be subtracted from the area of paths]
Total area covered by both the paths
\(
\begin{aligned}
& =(120+120-2.25) \mathrm{m}^2 \\
& =237.75 \mathrm{~m}^2 .
\end{aligned}
\)
Area covered by grass = Area of school playground \(–\) (Area
kept for swings + Area covered by paths)
\(
\begin{aligned}
& =12800 \mathrm{~m}^2-[6400+237.75] \mathrm{m}^2 \\
& =(12800-6637.75) \mathrm{m}^2 \\
& =6162.25 \mathrm{~m}^2 .
\end{aligned}
\)

Hint

Area of parallelogram \(\mathrm{ABCD}=\mathrm{AB} \times \mathrm{AE}=8 \times 4 \mathrm{~cm}^2\) \(=32 \mathrm{~cm}^2\)
Let altitude corresponding to \(\mathrm{AD}\) be \(h\). Then,
\(
\begin{array}{ll}
& h \times \mathrm{AD}=32 \\
\text { or } & h \times 6=32 \\
\text { or } & h=\frac{32}{6}=\frac{16}{3}
\end{array}
\)
Thus, altitude corresponding to \(\mathrm{AD}\) is \(\frac{16}{3} \mathrm{~cm}\).

Hint

Area covered by swimming pool \(=30 \mathrm{~m} \times 20 \mathrm{~m}=600 \mathrm{~m}^2\).
Length of outer rectangle \(=(30+8+8) \mathrm{m}=46 \mathrm{~m}\)
and its breadth \(=(20+5+5) m=30 \mathrm{~m}\)
So, the area of outer rectangle
\(
=46 \mathrm{~m} \times 30 \mathrm{~m}=1380 \mathrm{~m}^2 \text {. }
\)
Area of cemented path \(=\)
Area of outer rectangle – Area of swimming pool
\(
=(1380-600) \mathrm{m}^2=780 \mathrm{~m}^2 .
\)
Cost of cementing \(1 \mathrm{~m}^2\) path \(=₹ 200\)
So, total cost of cementing the path
\(
\begin{aligned}
& =₹ 780 \times 200 \\
& =₹ 156000
\end{aligned}
\)

Hint

Let the radius of the circle be \(r\). Then, \(\quad 2 \pi r=33\)
i.e., \(\quad r=\frac{33}{2 \pi}=\frac{33}{2} \times \frac{7}{22}=\frac{21}{4}\)
Thus, radius is \(\frac{21}{4} \mathrm{~cm}\)
So, area of the circle \(=\pi r^2=\frac{22}{7} \cdot \frac{21}{4} \cdot \frac{21}{4}=\frac{693}{8}\)
Thus, area of the circle is \(\frac{693}{8} \mathrm{~cm}^2\).

Hint

\(
\mathrm{DE}=\mathrm{EA}+\mathrm{AD}=(8+5) \mathrm{cm}=13 \mathrm{~cm}
\)
\(\mathrm{DE}\) 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, \(\mathrm{AC}=13 \mathrm{~cm}\).
From \(\triangle \mathrm{ADC}, \mathrm{DC}^2=\mathrm{AC}^2-\mathrm{AD}^2=13^2-5^2=169-25=\) \(144=12^2\)
So, \(\mathrm{DC}=12\)
Thus, length of DC is \(12 \mathrm{~cm}\).
Hence, perimeter of the rectangle ABCD
\(=2(12+5) \mathrm{cm}=34 \mathrm{~cm}\)

Hint

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 \(=18 \times 1=18 \mathrm{sq}\) 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 \(=18 \times 1=18 \mathrm{sq}\) 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 \(=16 \times 1=16\) 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 \(=18 \times 1=18 \mathrm{sq}\) units

Hint

It is given that
Area of rectangle \(=36\) unit squares
We know that
\(
36=6 \times 6
\)
It can be written as
\(
36=2 \times 3 \times 2 \times 3
\)
So we get
\(
\begin{aligned}
& 36=2^2 \times 3^2 \\
& 36=4 \times 9
\end{aligned}
\)
Here the sides of a rectangle are \(4 \mathrm{~cm}\) and \(9 \mathrm{~cm}\)
Perimeter \(=2(l+b)\)
Substituting the values
\(
\begin{aligned}
& =2(4+9) \\
& =2(13) \\
& =26 \text { units }
\end{aligned}
\)
Therefore, the perimeter of the rectangle is 26 units.

Hint

It is given that
Side of a square \(=22 \mathrm{~cm}\)
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
\(4 \times\) side \(=2 \pi r\)
We know that \(\pi=22 / 7\)
Substituting the values
\(
4 \times 22=2 \times 22 / 7 \times r
\)
By further calculation
\(
\begin{aligned}
& r=(4 \times 22 \times 7) /(2 \times 22) \\
& r=14 \mathrm{~cm}
\end{aligned}
\)
Therefore, its radius is \(14 \mathrm{~cm}\).

Hint

It is given that
Dimensions of the rectangle \(=14 \times 11 \mathrm{~cm}\)
From the question
Area of rectangle \(=\) Area of circle
\(1 \times b=\pi r^2\)
Substituting the values
\(
14 \times 11=22 / 7 \times r^2
\)
By further calculation
\(
\begin{aligned}
& r^2=(14 \times 11 \times 7) / 22 \\
& r^2=49 \\
& r=7 \mathrm{~cm}
\end{aligned}
\)
Therefore, the radius of the circle is \(7 \mathrm{~cm}\).

Hint

From the figure
Length of rectangle \(=5 \mathrm{~cm}\)
Breadth of rectangle \(=3+1=4 \mathrm{~cm}\)
We know that
Area of shaded portion \(=1 / 2 \times\) Area of rectangle \(=1 / 2 \times(l \times b)\)
Substituting the values
\(
\begin{aligned}
& =1 / 2 \times(5 \times 4) \\
& =1 / 2 \times 20
\end{aligned}
\)
So we get
\(
=10 \mathrm{~cm}^2
\)
Therefore, the area of the shaded portion is \(10 \mathrm{~cm}^2\).

Hint

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 \(\times\) height
From the figure
Area of parallelogram \(A B C D=A D \times B E\)
As \(A D=B C\)
Area of parallelogram \(A B C D=B C \times B E\)
Or
Area of parallelogram \(A B C D=D C \times B F\)
Therefore, the area of the parallelogram is not equal to DE \(\times D C\).

Hint

We know that
Area of triangle \(=1 / 2 \times b \times h\)
From the figure
Area of triangle \(=1 / 2 \times \mathrm{NO} \times \mathrm{OQ}\)
Therefore, the area of the triangle MNO is \(1 / 2 \times N O \times O Q\).

Hint

We know that
Area of \(\triangle M N O:\) Area of parallelogram \(M N O P=1 / 2 \times b \times h: b \times h\)
It can be written as
\(
=1 / 2 \times N O \times O Q: M P \times O Q
\)
So we get
\(
\begin{aligned}
& =1 / 2 \times N O: N O[\text { As } N O=M P] \\
& =1: 2
\end{aligned}
\)
Therefore, the ratio is \(1: 2\)

Hint

Using the figure
Area of \(\Delta \mathrm{MNO}=1 / 2 \times \mathrm{NO} \times \mathrm{MO}\)
Substituting the values
\(
\begin{aligned}
& =1 / 2 \times 4 \times 5 \\
& =10 \mathrm{~cm}^2
\end{aligned}
\)
Area of \(\triangle M O P=1 / 2 \times O P \times M O\)
Substituting the values
\(
\begin{aligned}
& =1 / 2 \times 2 \times 5 \\
& =5 \mathrm{~cm}^2
\end{aligned}
\)
Area of \(\triangle M P Q=1 / 2 \times P Q \times M O\)
Substituting the values
\(
\begin{aligned}
& =1 / 2 \times 6 \times 5 \\
& =15 \mathrm{~cm}^2
\end{aligned}
\)
We know that
Required ratio \(=10: 5: 15=2: 1: 3\)
Therefore, the ratio of areas is \(2: 1: 3\).

Hint

A parallelogram is defined as a quadrilateral in which both pairs of opposite sides are parallel and equal.
In parallelogram EFGH,
\(
\mathrm{EF}=\mathrm{HG}=10 \mathrm{~cm}
\)
We know that
Area of parallelogram \(\mathrm{EFGH}=\) Base \(\times\) Corresponding height
Substituting the values
\(
\begin{aligned}
& =10 \times 4 \\
& =40 \mathrm{~cm}^2
\end{aligned}
\)
Therefore, the area of EFGH is \(40 \mathrm{~cm}^2\).

Hint

It is given that
\(
\begin{aligned}
& P R=12 \mathrm{~cm} \\
& \mathrm{QR}=6 \mathrm{~cm} \\
& \mathrm{PL}=8 \mathrm{~cm}
\end{aligned}
\)
In right angled triangle PLR, Using the Pythagoras theorem
\(
\begin{aligned}
& P R^2=P L^2+L R^2 \\
& L R^2=P R^2-P L^2
\end{aligned}
\)
Substituting the values
\(
\begin{aligned}
& \mathrm{LR}^2=144-64 \\
& \mathrm{LR}^2=80 \\
& \mathrm{LR}=\sqrt{ } 80=4 \sqrt{5} \mathrm{~cm}
\end{aligned}
\)
Here
\(
\begin{aligned}
& \mathrm{LR}=\mathrm{LQ}+\mathrm{QR} \\
& \mathrm{LQ}=\mathrm{LR}-\mathrm{QR}
\end{aligned}
\)
Substituting the values \(L Q=4 \sqrt{5}-6 \mathrm{~cm}\)
Area of triangle PLR \(\mathrm{A} 1=1 / 2 \times \mathrm{LR} \times \mathrm{PL}\)
Substituting the values \(L Q=4 \sqrt{5}-6 \mathrm{~cm}\)
Area of triangle PLR \(\mathrm{A} 1=1 / 2 \times \mathrm{LR} \times \mathrm{PL}\)
Substituting the values \(\mathrm{A} 1=1 / 2 \times 4 \sqrt{5} \times 8=16 \sqrt{5} \mathrm{~cm}^2\)
Area of triangle PLQ \(\mathrm{A} 2=1 / 2 \times \mathrm{LQ} \times \mathrm{PL}\)
Substituting the values \(\begin{aligned}& A_2=1 / 2 \times(4 \sqrt{5}-6) \times 8 \\& A 2=16 \sqrt{5}-24 \mathrm{~cm}^2\end{aligned}\)
We know that
Area of triangle PLR = Area of triangle \(\mathrm{PLQ}+\) Area of triangle PQR
Substituting the values \(16 \sqrt{5}=16 \sqrt{5}-24+\) Area of triangle PQR
Area of triangle \(\mathrm{PQR}=24 \mathrm{~cm}^2\)
\(
\begin{aligned}
& 1 / 2 \times P R \times Q M=24 \\
& 1 / 2 \times 12 \times Q M=24 \\
& Q M=4 \mathrm{~cm}
\end{aligned}
\)
Therefore, \(\mathrm{QM}\) is \(4 \mathrm{~cm}\).

Hint

Given, MNO is a right-angled triangle
The legs are \(6 \mathrm{~cm}\) and \(8 \mathrm{~cm}\) long
We have to find the length of perpendicular NP on the side MO.
By using Pythagoras theorem,
\(
\begin{aligned}
& M O^2=M N^2+M O^2 \\
& M O^2=(6)^2+(8)^2 \\
& M O^2=36+64 \\
& M O^2=100
\end{aligned}
\)
Taking square root,
\(
M O=10 \mathrm{~cm}
\)
Area of triangle \(=1 / 2 \times\) base \(\times\) height
\(
\begin{aligned}
& 1 / 2 \times M N \times N O=1 / 2 \times M O \times N P \\
& 6 \times 8=10 \times N P \\
& N P=48 / 10 \\
& N P=4.8 \mathrm{~cm}
\end{aligned}
\)
Therefore, the value of NP is \(4.8 \mathrm{~cm}\)

Hint

Given, area of a right-angled triangle is \(30 \mathrm{~cm}^2\)
The length of the smallest side is \(5 \mathrm{~cm}\)
We have to find the length of the hypotenuse.
Area of triangle \(=1 / 2 \times\) base \(\times\) height
\(30=1 / 2 \times 5 \times \&\) height
Height \(=30(2) / 5\)
Height \(=6(2)\)
Height \(=12 \mathrm{~cm}\)
By using Pythagoras theorem,
\((\text { Hypotenuse })^2=(\text { height })^2+(\text { base })^2\)
\(\&(\text { Hypotenuse })^2=(12)^2+(5)^2\)
\((\text { Hypotenuse })^2=144+25\)
\((\text { Hypotenuse })^2=169\)
Taking square root,
Hypotenuse \(=13 \mathrm{~cm}\)
Therefore, the hypotenuse is \(13 \mathrm{~cm}\).

Hint

Perimeter of semicircle \(=\) circumference of semicircle \(+\) diameter
Circumference of semicircle \(=\pi r\)
Where, \(r\) is the radius of circle
\(
\begin{aligned}
& =22 / 7 \times 10 \\
& =22(10) / 7 \\
& =220 / 7 \\
& =31.4 \mathrm{~cm} \\
& \text { Perimeter of semicircle }=31.4+10(2) \\
& =31.4+20 \\
& =51.4 \mathrm{~cm}
\end{aligned}
\)
Therefore, the perimeter of semicircle is \(51.4 \mathrm{~cm}\)

Hint

Given, \(p\) squares of each side \(1 \mathrm{~mm}\) makes a square of side \(1 \mathrm{~cm}\)
We have to find the value of \(p\).
Area of square \(=(\text { side })^2\)
Area of square with side \(1 \mathrm{~cm}=(1)^2=1 \mathrm{~cm}^2\)
Area of square with side \(1 \mathrm{~mm}=(1)^2=1 \mathrm{~mm}^2\)
Given, \(p \times\) area of square with side \(1 \mathrm{~mm}=\) area of square of side \(1 \mathrm{~cm}\)
We know, \(1 \mathrm{~cm}=10 \mathrm{~mm}\)
So, \(\mathrm{p} \times 1 \mathrm{~mm}^2\left(10^2\right) \mathrm{mm}^2\)
\(p=100 / 1\)
Therefore, \(p=100\)

Hint

Given, each side of a rhombus is doubled.
We have to find the increase in its area.
Area of rhombus \(=\) base \(\times\) corresponding height
Let the side of rhombus be \(b\)
Since the side is doubled, side \(=2 b\)
Old Area \(=b h\)
New area \(=2 b h\)
Therefore, the area has increased by 2 times.

Hint

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 \(=2 b\)
New perimeter \(=2(2 {l}+2 {b})\)
\(=2[2(l+b)]\)
Therefore, the new perimeter is increased by 2 times.

Hint

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 \(=\pi r^2\)
Where, \(r\) is the radius of the circle
Now, \(r=2 r\)
Area of circle \(=\pi(2 r)^2\)
\(=4 \pi r^2\)
Therefore, the area of the circle will be increased by 4 times.

Hint

Given, radius of circle is \(10 \mathrm{~cm}\)
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 \(=(\text { side })^2=b^2\)
Considering right-angled triangle DAB,
By using Pythagoras theorem,
\(
\begin{aligned}
& B D^2=A D^2+A B^2 \\
& (20)^2=b^2+b^2 \\
& 2 b^2=400 \\
& b^2=200 \\
&
\end{aligned}
\)
Area of square \(=200 \mathrm{~cm}^2\)
Therefore, area of the largest square is \(200 \mathrm{~cm}^2\)

Hint

Given, the dimension of the rectangle is \(10 \mathrm{~cm}\) in length and \(8 \mathrm{~cm}\) 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 \(=8 \mathrm{~cm}\)
Radius \(=\) diameter \(/ 2\)
\(=8 / 2\)
\(=4 \mathrm{~cm}\)
Therefore, the radius is \(4 \mathrm{~cm}\)

Hint

Given, area of square is \(100 \mathrm{~cm}^2\)
We have to find the circumference of the largest circle cut out of the square.
Area of square \(=(\text { side })^2\)
side \(^2=100\)
Taking square root,
side \(=10 \mathrm{~cm}\)
The diameter of the circle will be equal to the side of the square.
Diameter of circle \(=10 \mathrm{~cm}\)
Radius \(=10 / 2=5 \mathrm{~cm}\)
Circumference of circle \(=2 \pi r\)
Where, \(r\) is the radius of circle
\(
\begin{aligned}
& =2 \pi(5) \\
& =10 \pi \mathrm{cm}
\end{aligned}
\)
Therefore, the circumference of the circle is \(10 \pi \mathrm{cm}\).

Hint

Given, radius of semicircle is \(4 r\)
We have to find the area of the semicircle
Area of semicircle \(=\pi r^2 / 2\)
Where, \(r\) is the radius of the circle
Given, \(r=4 r\)
Area \(=\pi(4 r)^2 / 2\)
\(=16 \pi r^2 / 2\)
\(=8 \pi r^2\)
Therefore, the area of the semicircle is \(8 \pi r^2\) square units.

Hint

Given, a hedge boundary needs to be planted around a rectangular lawn
of size \(72 \mathrm{~m} \times 18 \mathrm{~m}\).
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 \(\times\) breadth \()\)
Length of rectangular lawn \(=72 \mathrm{~m}\)
Breadth of rectangular lawn \(=18 \mathrm{~m}\)
Perimeter of rectangular lawn \(=2(72+18)\)
\(=2(90)\)
\(=180 \mathrm{~m}\)
In one meter 3 shrubs can be planted.
Number of shrubs that can be planted in rectangular lawn \(=3 \times\)
perimeter of lawn
\(=3(180)\)
\(=540\) shrubs
Therefore, the number of shrubs that can be planted is 540.

Hint

Given, the area of \(\triangle A F B\) is equal to the area of parallelogram \(A B C D\).
The length of altitude EF is \(16 \mathrm{~cm}\)
We have to find the altitude of the parallelogram to the base \(A B\) of length \(10 \mathrm{~cm}\).
Area of triangle \(=1 / 2 \times\) base \(\times\) height
Area of parallelogram \(=\) base \(\times\) corresponding height
Area of \(\triangle \mathrm{AFB}=\) area of parallelogram \(\mathrm{ABCD}\)
\(
\begin{aligned}
& 1 / 2 \times A B \times E F=A B \times E G \\
& 1 / 2 \times 10 \times 16=10 \times E G \\
& 10 \times 8=10 \times E G \\
& E G=8 \mathrm{~cm}
\end{aligned}
\)
Therefore, the corresponding altitude to the base \(A B\) is \(8 \mathrm{~cm}\).
We have to find the area of \(\triangle D A O\), where \(O\) is the midpoint of \(D C\).
Since \(O\) is the midpoint of \(D C\)
\(
\begin{aligned}
& \mathrm{DC}=\mathrm{DO}+\mathrm{CO} \\
& \mathrm{DO}=\mathrm{CO}=5 \mathrm{~cm}
\end{aligned}
\)
Area of \(\triangle D A O=1 / 2 \times D O \times\) height
\(
\begin{aligned}
& =1 / 2 \times 5 \times 8 \\
& =5(4) \\
& =20 \mathrm{~cm}^2
\end{aligned}
\)

Hint

Given, the ratio of the area of \(\triangle W X Y\) to the area of \(\triangle W Z Y\) is \(3: 4\)
The area of \(\triangle W X Z\) is \(56 \mathrm{~cm}^2\) and \(W Y=8 \mathrm{~cm}\).
We have to find the lengths of \(X Y\) and \(Y Z\).
Area of triangle \(=1 / 2 \times\) base \(\times\) height
Area of \(\triangle W X Y\) : area of \(\triangle W Z Y=3: 4\)
\(1 / 2 \times X Y \times W Y: 1 / 2 \times Y Z \times W Y=3: 4\)
\(X Y: Y Z=3: 4\)
Area of \(\triangle W X Z=1 / 2 \times X Z \times W Y\)
\(56=1 / 2 \times X Z \times 8\)
\(56=4 \times X Z\)
\(X Z=56 / 4\)
\(X Z=14 \mathrm{~cm}\)
From the figure
\(
\begin{aligned}
& X Z=X Y+Y Z \\
& Y Z=X Z-X Y \\
& Y Z=14-X Y \\
& N o w, X Y / Y Z=3 / 4 \\
& X Y /(14-X Y)=3 / 4 \\
& X Y(4)=3(14-X Y) \\
& 4 X Y=3(14)-3 X Y \\
& 4 X Y+3 X Y=3(14) \\
& 7 X Y=3(14) \\
& X Y=3(14) / 7 \\
& X Y=3(2)=6 \mathrm{~cm} \\
& Y Z=14-6=8 \mathrm{~cm}
\end{aligned}
\)

Hint

Given, area of the shaded triangle is \(9 \mathrm{~cm}^2\)
We have to find the area of the parallelogram ABCD.
Area of triangle \(=1 / 2 \times\) base \(\times\) height
\(9=1 / 2 \times 3 \times h\)
\(9(2)=3 \times h\)
\(h=9(2) / 3\)
\(=3(2)\)
\(\mathrm{h}=6 \mathrm{~cm}\)
Area of parallelogram \(=\) base \(\times\) corresponding height
\(=\mathrm{BE} \times \mathrm{h}\)
\(=(3+4) \times 6\)
\(=7(6)\)
\(=42 \mathrm{~cm}^2\)
Therefore, the area of parallelogram is \(42 \mathrm{~cm}^2\).

Hint

Given, \(\mathrm{ABCD}\) is a square
\(
A B=15 \mathrm{~cm}
\)
We have to find the area of the square BDFE.
From the figure,
\(\mathrm{BD}\) is the diagonal of the square \(\mathrm{ABCD}\).
Diagonal of square \(=\) side \(\times \sqrt{2}\)
\(=15 \sqrt{ } 2 \mathrm{~cm}\)
So, the length of the side of square BDFE \(=15 \sqrt{ } 2 \mathrm{~cm}\)
Area of square \(=(\text { side })^2\)
\(
\begin{aligned}
& =(15 \sqrt{ } 2)^2 \\
& =(15)^2 \times 2 \\
& =225(2) \\
& =450 \mathrm{~cm}^2
\end{aligned}
\)
Therefore, the area of square BDFE is \(450 \mathrm{~cm}^2\).

Hint

Given, MGHK is a parallelogram
The altitudes MN and MO of parallelogram are \(8 \mathrm{~cm}\) and \(4 \mathrm{~cm}\)
The length of one side \(\mathrm{GH}\) is \(6 \mathrm{~cm}\).
We have to find the perimeter of MGHK.
Area of parallelogram \(=\) base \(\times\) corresponding height
\(
\begin{aligned}
& =\mathrm{GH} \times \mathrm{MN} \\
& =6 \times 8 \\
& =48 \mathrm{~cm}^2
\end{aligned}
\)
Now, taking base \(\mathrm{HK}\) and altitude \(\mathrm{MO}\)
Area of parallelogram \(=\mathrm{HK} \times M O\)
\(
48=\mathrm{HK} \times 4
\)
\(
\mathrm{HK}=48 / 4
\)
\(
\mathrm{HK}=12 \mathrm{~cm}
\)
Perimeter of parallelogram \(=2\) (length \(+\) breadth \()\)
Perimeter of parallelogram \(\mathrm{MGHK}=2(\mathrm{GH}+\mathrm{HK})\)
\(
\begin{aligned}
& =2(6+12) \\
& =2(18) \\
& =36 \mathrm{~cm}
\end{aligned}
\)

Hint

Given, the area of \(\triangle \mathrm{PQR}\) is \(20 \mathrm{~cm}^2\)
The area of \(\triangle \mathrm{PQS}\) is \(44 \mathrm{~cm}^2\)
\(
\mathrm{QR}=5 \mathrm{~cm}
\)
We have to find the length RS, if \(P Q\) is perpendicular to QS.
Area of triangle \(=1 / 2 \times\) base \(\times\) height
Area of \(\triangle \mathrm{PQR}=1 / 2 \times \mathrm{QR} \times \mathrm{PQ}\)
\(
\begin{aligned}
& 2 O=1 / 2 \times 5 \times P Q \\
& P Q=2 O(2) / 5 \\
& =4(2) \\
& P Q=8 \mathrm{~cm}
\end{aligned}
\)
Area of \(\triangle \mathrm{PQS}=1 / 2 \times \mathrm{QS} \times \mathrm{PQ}\)
\(
\begin{aligned}
& 44=1 / 2 \times Q S \times 8 \\
& Q S=44(2) / 8 \\
& =44 / 4 \\
& Q S=11 \mathrm{~cm}
\end{aligned}
\)
\(P Q\) is perpendicular to \(Q S\).
\(
\begin{aligned}
& \text { So, QS }=Q R+R S \\
& 11=5+R S \\
& R S=11-5 \\
& R S=6 \mathrm{~cm}
\end{aligned}
\)

Hint

Consider \(r\) as the radius of smaller circle and \(R\) as the radius of bigger circle
From the figure, we can see,
\(
\begin{aligned}
& r=7 / 2=3.5 \mathrm{~cm} \\
& \mathrm{R}=7 / 2+7=21 / 2=10.5 \mathrm{~cm}
\end{aligned}
\)
We know that
Area of shaded region = Area of bigger circle – Area of smaller circle
\(
=\pi R^2-\pi r^2
\)
Taking \(\pi\) as common
\(
=\pi\left(R^2-r^2\right)
\)
Substituting the values
\(
=\pi\left[(10.5)^2-(3.5)^2\right]
\)
By further calculation
\(
=\pi(110.25-12.25)
\)
Sowe get
\(
\begin{aligned}
& =22 / 7 \times 98 \\
& =308 \mathrm{~cm}^2
\end{aligned}
\)

Hint

Given
Diameter of the complete circle \(=14 \mathrm{~cm}\)
We know that
\({\text { Radius }}=14 / 2=7 \mathrm{~cm}\)
Here
Area of the complete circle \(=\pi r^2\)
Substituting the values
\(
\begin{aligned}
& =22 / 7 \times 7 \times 7 \\
& =154 \mathrm{~cm}^2
\end{aligned}
\)
It is given that
Diameter of small circle \(=7 / 4 \mathrm{~cm}\)
Radius \(=7 /(4 \times 2)=7 / 8 \mathrm{~cm}\)
Here
Area of two small circles \(=2 \times \pi r^2\)
Substituting the values
\(
\begin{aligned}
& =2 \times 22 / 7 \times 7 / 8 \times 7 / 8 \\
& =77 / 16=4.8 \mathrm{~cm}^2
\end{aligned}
\)
We know that
Area of shaded region \(=\) Area of complete circle \(–\) Area of two small circles
Substituting the values
\(
\begin{aligned}
& =154-4.8 \\
& 149.2 \mathrm{~cm}^2
\end{aligned}
\)
Therefore, the area of the shaded region is \(149.2 \mathrm{~cm}^2\).

Hint

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 \(=16 \mathrm{~cm}\)
Perimeter \(=2 \pi r\)
Substituting the values
\(
\begin{aligned}
& =2 \pi \times 16 \\
& =32 \pi
\end{aligned}
\)
Perimeter of curved part of new shape \(=1 / 4 \times 32 \pi=8 \pi\)
Here
Perimeter of new shape \(=(4 \times\) Perimeter of curved part \()+(2 \times\) radius \()\)
Substituting the values
\(
\begin{aligned}
& =(4 \times 8 \pi)+(2 \times 16) \\
& =32 \pi+32
\end{aligned}
\)
Perimeter is increased by \(32 \mathrm{~cm}\).

Hint

We know that
Total distance around the track = Length of two parallel strips + Length of two semi circles
Given
Radius \(r=16 \mathrm{~cm}\)
Substituting the values
\(
=(2 \times 52)+(2 \times \pi \times 16)
\)
By further calculation
\(
=104+(2 \times 3.14 \times 16)
\)
Sowe get
\(
\begin{aligned}
& =104+100.5009 \\
& =205 \mathrm{~cm}
\end{aligned}
\)
Therefore, the total distance around the track is \(205 \mathrm{~cm}\).

Hint

It is given that
Dimension of a plot \(=200 \mathrm{~m} \times 150 \mathrm{~m}\)
Width of road \(=3 \mathrm{~m}\)
We know that
Total area available for houses \(=\) Area of total plot \(–\) Area of 3 roads
Substituting the values
\(
=(200 \times 150)-3 \times(3 \times 200)
\)
By further calculation
\(
\begin{aligned}
& =30000-1800 \\
& =28200 \mathrm{~m}^2
\end{aligned}
\)
Therefore, the area he got for building the houses is \(28200 \mathrm{~m}^2\).

Hint

It is given that
Radius of wheel \(=25 \mathrm{~cm}=25 / 100 \mathrm{~m}=1 / 4 \mathrm{~m}[\) As \(1 \mathrm{~cm}=1 / 100 \mathrm{~m}]\)
We know that
Distance traveled in one rotation \(=2 \pi r\)
Substituting the values
\(
\begin{aligned}
& =2 \times 22 / 7 \times 1 / 4 \\
& =11 / 7=1.57 \mathrm{~m}
\end{aligned}
\)
Therefore,
Distance traveled in 350 rotations \(=1.57 \times 350\)
\(
=550 \mathrm{~m}
\)
Therefore, the wheel covers a distance of \(550 \mathrm{~m}\).

Hint

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,
\(
\text { side }=9 \mathrm{~cm}
\)
Area of square \(=(\text { side })^2\)
\(
\begin{aligned}
& =(9)^2 \\
& =81 \mathrm{~cm}^2
\end{aligned}
\)
Given, short line segments are at right angles to each other and \(1 \mathrm{~cm}\) long
By joining all the short lines using dotted lines, we obtain 41 identical squares.
Area of square \(=(\text { side })^2\)
Considering one square,
Side \(=1 \mathrm{~cm}\)
Area of square \(=(1)^2=1 \mathrm{~cm}^2\)
Area of 41 identical squares \(=41(1)=41 \mathrm{~cm}^2\)
Area of shaded region \(=\) area of larger square \(–\) area of 41 identical squares
\(
\begin{aligned}
& =81-41 \\
& =40 \mathrm{~cm}^2
\end{aligned}
\)
Therefore, the area of the shaded region is \(40 \mathrm{~cm}^2\).

Hint

Given, \(A B C D\) is a rectangle with length \(80 \mathrm{~cm}\) and breadth \(60 \mathrm{~cm}\).
\(P, Q, R\) and \(S\) are the midpoints of \(A B, B C, C D\), and DA respectively.
A circular rangoli of radius \(10 \mathrm{~cm}\) is drawn at the centre.
We have to find the area of the shaded portion.
Area of rectangle \(\mathrm{ABCD}=\) length \(\times\) breadth
\(=80(60)\)
\(=4800 \mathrm{~cm}^2\)
The triangles SAP, PBQ, RCQ and RDS are congruent
\(S\) is the midpoint of DA
So, \(A S=60 / 2=30 \mathrm{~cm}\)
\(P\) is the midpoint of \(A B\)
So, \(A P=80 / 2=40 \mathrm{~cm}\)
Area of triangle \(=1 / 2 \times\) base \(\times\) height
Area of triangle SAP \(=1 / 2 \times \mathrm{AP} \times \mathrm{AS}\)
\(=1 / 2 \times 40 \times 30\)
\(=20(30)\)
\(=600 \mathrm{~cm}^2\)
Area of 4 congruent triangles \(=4(600)\)
\(=2400 \mathrm{~cm}^2\)
Area of circle \(=\pi r^2\)
Area of circular rangoli \(=(3.14)(10)^2\)
\(
\begin{aligned}
& =3.14(100) \\
& =314 \mathrm{~cm}^2
\end{aligned}
\)
Area of shaded portion = area of rectangle PQRS – area of circular rangoli
\(
\begin{aligned}
& =2400-314 \\
& =2086 \mathrm{~cm}^2
\end{aligned}
\)
Therefore, the area of the shaded portion is \(2086 \mathrm{~cm}^2\).

Hint

Given 4 squares each side \(10 \mathrm{~cm}\) is cut from each corner of a rectangle sheet of paper of size \(100 \mathrm{~cm} \times 80 \mathrm{~cm}\).
An isosceles triangle is removed from the remaining piece of paper whose equal sides are \(10 \mathrm{~cm}\) length.
We have to find the remaining part of the paper.
Area of rectangular sheet \(=\) length \(\times\) breadth
\(
\begin{aligned}
& =100(80) \\
& =8000 \mathrm{~cm}^2
\end{aligned}
\)
Area of one square \(=(\text { side })^2\)
\(
\begin{aligned}
& =(10)^2 \\
& =100 \mathrm{~cm}^2
\end{aligned}
\)
So, area of 4 squares \(=4(100)=400 \mathrm{~cm}^2\)
Area of triangle \(=1 / 2 \times\) base \(\times\) height
Area of isosceles triangle \(=1 / 2 \times 10 \times 10\)
\(
\begin{aligned}
& =10(5) \\
& =50 \mathrm{~cm}^2
\end{aligned}
\)
Area of the remaining part of the sheet \(=\) area of rectangular sheet \(–\) area of 4 squares – area of isosceles triangle
\(
\begin{aligned}
& =8000-400-50 \\
& =7600-50 \\
& =7550 \mathrm{~cm}^2
\end{aligned}
\)
Therefore, the area of the remaining part is \(7550 \mathrm{~cm}^2\).

Hint

Given, a dinner plate is in the form of a circle.
A circular region encloses a beautiful design.
The inner circumference is \(352 \mathrm{~mm}\) and the outer is \(396 \mathrm{~mm}\).
We have to find the width of the circular design.
Circumference \(=2 \pi r\)
Outer circumference \(=2 \pi R\)
\(
\begin{aligned}
& 396=2(22 / 7) R \\
& 396=(44 / 7) R \\
& R=396(7) / 44 \\
& R=9(7) \\
& R=63 \mathrm{~mm}
\end{aligned}
\)
Inner circumference \(=2 \pi r\)
\(
\begin{aligned}
& 352=2(22 / 7) r \\
& 352=(44 / 7) r \\
& r=352(7) / 44 \\
& r=8(7) \\
& r=56 \mathrm{~mm}
\end{aligned}
\)
Width of circular ring \(=\mathrm{R}-\mathrm{r}\)
\(
\begin{aligned}
& =63-56 \\
& =7 \mathrm{~mm}
\end{aligned}
\)
Therefore, the width of the design is \(7 \mathrm{~mm}\).

Hint

Given, the moon is about \(384000 \mathrm{~km}\) 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 \(=2 \pi r\)
Given, radius \(=384000 \mathrm{~km}\)
\(=2(3.14)(384000)\)
\(=2(314)(3840)\)
\(=2411520 \mathrm{~km}\)
Therefore, the circumference is \(2411520 \mathrm{~km}\).

Hint

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 \(0.5 \mathrm{~cm}\) needs to be polished.
The rate of polishing is \(₹ 200\) per \(\mathrm{m}^2\).
We have to find the total cost of polishing.
Original length \(=25(10)=250 \mathrm{~cm}\)
Original breadth \(=10(10)=100 \mathrm{~cm}\)
Area of table \(=\) length \(\times\) breadth
\(=250(100)\)
\(=25000 \mathrm{~cm}^2\)
Area of circle \(=\pi r^2\)
Given, diameter \(=0.5 \mathrm{~cm}\)
Radius \(=0.5 / 2=0.25 \mathrm{~cm}\)
\(=(22 / 7)(0.25)^2\)
\(=22(0.0625) / 7\)
\(=0.1964 \mathrm{~cm}^2\)
Area of 6 holes \(=6(0.1964)\)
\(=1.179 \mathrm{~cm}^2\)
Area to be polished = area of table \(–\) area of 6 holes
\(
\begin{aligned}
& =25000-1.179 \\
& =24998.82 \mathrm{~cm}^2
\end{aligned}
\)
We know, \(1 \mathrm{~m}=100 \mathrm{~cm}\)
Area to be polished in \(\mathrm{m}^2=24998.82 / 10000\) \(=2.4999 \mathrm{~m}^2\)
Cost of polishing \(1 \mathrm{~m}^2=₹ 200\)
Cost of polishing table \(=200(2.4999)\) \(=₹ 499.98\)
Therefore, the total cost of polishing is \(₹ 499.98\).