Given, AD and CE are altitudes which intersect each other at the point P.
In ΔAEP and ΔCDP,
∠AEP = ∠CDP           [Each 90°]
and ∠APE = ∠CPD    [Vertically opposite angles]
\(\therefore\) ΔAEP ∼ ΔCDP       [by AA similarity criterion]

No, because in \(\Delta\)ABC, \(\angle\)BAC is given but the measure of included side AC is not given.

No, the two triangles are not similar.

In Δ POQ, DE || OQ    [given]
\(\therefore \frac{P E}{E Q}=\frac{P D}{D O}\) ...(i)
[by basic proportionality theorem] 
In ΔPOR,    DF || OR     [given]
\(\therefore \frac{P F}{F R}=\frac{P D}{D O}\) ...(ii)
[by basic proportionality theorem]
From Eqs. (i) and (ii), we get 
\(\frac{P E}{E Q}=\frac{P F}{F R}\)
In ΔPQR, we have \(\frac{P E}{E Q}=\frac{P F}{F R}\)
∴  EF || QR
[by Converse of basic proportionally theorem]
Hence proved.

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
\( \frac{P E}{E Q}=\frac{4}{4.5}=\frac{8}{9} \)
\(\frac{P F}{F R}=\frac{8}{9} \)
\(\frac{P F}{F R}=\frac{P E}{E Q} \)
Therefore, EF is parallel to QR

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
\(\frac{A D}{D B}=\frac{A E}{E C}\)
\(\frac{1.5}{3}=\frac{1}{x}\)
\(x=\frac{3 \times 1}{1.5}\)
x =  2
∴ EC = 2 cm

Given, \(\angle ABC={ 90 }^{ ° }\) and \(\angle AMP={ 90 }^{ ° }\)
In \(\triangle ABC\) and \(\triangle AMP\)
\(\angle ABC=\angle AMP\)  [each 90°]
and \(\angle BAC=\angle MAP\) [common angle]
\(\therefore \triangle ABC\sim \triangle AMP\) [by AA similarity criterion]

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
\( \frac{P E}{E Q}=\frac{3.9}{3}=1.3 \)
\(\frac{P F}{F R}=\frac{3.6}{2.4}=1.5\)
Hence, \(\frac{P E}{E Q} \neq \frac{P F}{F R}\)
Therefore, EF is not parallel to QR

(i) (a) Pair of the equilateral triangles are similar figures.
(b) Pair of the squares are similar figures.
(ii) (a) A triangle and a quadrilateral form a pair of non-similar figures.
(b) A square and a circle form a pair of non-similar figures.

116 cm, 462 cm²

Here, \({PQ\over PR}=sin30^0\)
\(⇒\ {PQ\over 10}={1\over 2}⇒\ PQ=5cm\)
And \({QR\over PR}=cos30^0\)
\(⇒\ {QR\over 10}={\sqrt3\over 2}⇒ QR=5\sqrt3cm\)

∴ Area of ΔPQR\(={1\over 2}\times PQ\times QR\)
\(={2\over 2}\times5\times5\sqrt3\)
\(={25\over 2}\times1.73={43.25\over 2}\)
=21.625cm²

15 cm

Four times

12.5 cm

33 cm

20 cm

2000

190.14 m²

17.5 cm

192.5 cm²

14 cm²

Radius of the wheel = r = 49cm
Distance covered in one = 2πr
\(=2\times{22\over 7}\times49\)
2 x 22 x 7cm
Total distance covered = 2002m
∴ Number of revolutions=\(Total\ distance\ covered\over Distance\ covedred\ in\ one\ revolution\)
\(={2002\times100\over 2\times22\times7}=750\)

Let r be the radius of cycle wheel
\(r={21\over2}cm\)
Distance travelled in one revolution = 2πr
\(=2\times{22\over 7}\times{21\over 2}=66cm\)
Total distance covered = 1.98km
=1.98 x 100 x 100
=198000cm
∴ Number of revolutions=\(Total\ distance\ covered\over distance\ covered\ in\ one\ revolution\)
\(={198000\over66}=3000\)

Here, area of shaded region
=Area of circle-Area of \(\Delta AOB\)
\(={2\over7}\times7\times7-{1\over2}\times7\times7\)
\(=154-{49\over2}\)
=129.5 cm²

Area of an equilateral ΔABC=\(\frac { \sqrt { 3 } }{ 4 } \times (2)^{ 2 }=\sqrt { 3 } \)
∵ Central angle of each sector = 60^o
∴ Area of 3 sectors=\(3\times \frac { { \pi r }^{ 2 }\theta }{ { 360 }^{ o } } \)
\(=\frac { 3\times \pi \times 1\times 1\times { 60 }^{ o } }{ { 360 }^{ o } } =\frac { \pi }{ 2 } \)
∴ Area of shaded region = Area of ΔABC Area of 3 sectors
=\(\sqrt { 3 } -\frac { \pi }{ 2 } \)

Area of shaded region = Area of outer circle-Area of inner circle
\(=\pi R^2-\pi r^2=\pi(R^2-r^2)\)

Diameter AB=28m
Radius (r₁)=\(={28\over 2}=14m\)
Diameter AO=14m
Radius (r₂)=\(={14\over 2}=7m\)
radius(r₃)=7m
Perimeter of the shaded region=πr₁+(πr₂+πr₃)=π​​​​​​​[r₁+r₂+r₃]
\(={22\over 7}[14+7+7]={22\over 7}\times28=88m\)

Area of sector OCDE \(=\frac{60^{\circ}}{360^{\circ}} \pi r^{2}\)
\(\begin{array}{l} =\frac{1}{6} \times \frac{22}{7} \times 6 \times 6 \\ =\frac{132}{7} \mathrm{~cm}^{2} \end{array}\)
Area of triangle OAB \(=\frac{\sqrt{3}}{4}(12)^{2}=\frac{\sqrt{3} \times 12 \times 12}{4}=36 \sqrt{3} \mathrm{~cm}^{2}\) 
Area of circle \(=\pi r^{2}=\frac{22}{7} \times 6 \times 6=\frac{792}{7} \mathrm{~cm}^{2}\)
Area of shaded region = Area of ΔOAB + Area of circle − Area of sector OCDE
\(\begin{array}{l} =36 \sqrt{3}+\frac{792}{7}-\frac{132}{7} \\ =\left(36 \sqrt{3}+\frac{660}{7}\right) c m^{2} \end{array}\)

ABCD is a square
Given: side of the square =14cm
∴ Area of the square = (side)² (14)² = 196 cm²
Radius of the semicircle APD=\({1\over 2}\)(side of square)=\({1\over 2}\times14=7cm\)
Area of the semicircle APD=\({1\over 2}\pi r^2={1\over 2}\times{22\over 7}\times7\times7=11\times7=77cm^2\)
Similarly, area of the semicircle BPC= 77cm² 
Total area of both the semicircles=77 + 77 - 154 cm².
Area of the shaded region = Area of square - area of both semicircles = 196 - 154 -42cm²

Area of the circle=1256cm²
⇒ A=πr²
A.T.Q 1256=πr²
\({1256\over3.14}=r^2\)
\(⇒\sqrt{1256\over 3.14}=r\)
​​​​​
⇒ 20cm = radius
⇒ 40cm = diameter
Here, Diagonals of rhombus = diameter of the circle
∴ Area of rhombus = \({1\over 2}d_1\times d_2\)
\(={1\over 2}\times40\times40=800cm^2\) 

Radius of wheel=70cm
Distance covered in 1 revolution=2πr
\(=2\times{22\over 7}\times70=440cm=4.4m\)
\(Speed=66km/h={66000\over 60}m/minute\)
=1100m/minute
∴ No. of revolutions\(={1100\over 4.4}=250\)

Let radius of circle be r cm
Its area = πr² sq units
Side of an equilateral triangle = diameter of the circle = 2r
Area of the equilateral triangle
\(={\sqrt3\over 4}\times(2r)^2\)
\(={\sqrt3\over 4}\times4r^2=\sqrt3r^2\)
\({Area\ of\ the\ circle\over Area\ of\ the\ triangle }={\pi r^2\over \sqrt3 r^2}={\pi\over\sqrt3}\)

Let radius of the protractor be r cm

⇒ Perimeter=\(\left[{1\over2}\times2\pi r+2r\right]\)
66cm=[πr+2r]
⇒ \(66=\left({22\over 7}+2\right)r=\left(22+14\over7\right)r\)
\(=\left(36\over 7\right)r\)
\(⇒\ \ {7\times66\over 36}=r⇒ r={77\over 6}cm\)

Inner radius \(= {220\times 7\over 22\times 2}=35 cm\) and Width of the track = 15 m
Outer radius = 35 + 15 = 50 m
Area of the track \(=\ \pi(50^2-35^2)={22\over 7}(50-35)(50+35)={22\over 7}\times 1585\ m^2=4007.14\ sq.m\)

BO = OD = r = 60 cm
\(\theta\) = 360° - BOD = 360° - 90° = 270°
(i) Area of the table top (sector)
= \(\frac { \theta }{ 360° } { \pi r }^{ 2 }=\frac { 270° }{ 360° } \times 3.14\times 60\times 60\)
= \(45\times 60\times \frac { 314 }{ 100 } \times 60+120\)
= (ii) Perimeter (table top) = \(\frac { \theta }{ 360° } { 2\pi r }+2r\)
= \(\frac { 270° }{ 360° } \times 2\times \frac { 314 }{ 100 } \times 60+120\)
= \(\frac { 270° }{ 360° } \times 2\times \frac { 90\times 314 }{ 100 } +120=\frac { 2826+1200 }{ 10 } =402.6 \ cm\)

Length of the rectangle AB = 60 m 
Breadth of the rectangle = BC = 28 m 
Diameter of the shaded portion = 28 m 
⇒ Radius of the shaded portion = \(28\over2\)m = 14m
Grass portion = shaded portion 
⇒ Area of the plot without grass = area of the rectangle ABCD - area of the shaded portion 
= \(\left[ 60\times 28-\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times { (14) }^{ 2 } \right] { m }^{ 2 }\) = (1680 - 308)m² = 1372 m²

Time taken by minute hand to make one circle = 60 minutes.
\(\therefore\) Angle described in 60 minutes = \(360^o\)
Angle described in 5 minutes [i.e., from 7.00 a.m. to 7.05 a.m.] = \({360^o\over 60^o}\times 5=30^o\)
Radius of circle = length of minute hand = \(\sqrt {21}cm\)
Area swept = \({\theta \over 360^o}\times r^2={30^oover 360^o}\times {22\over 7}\times \sqrt{21}\times \sqrt {21}={1\over 12}\times {22\over 7}\times 21 = {11\over 2}\ cm^2=5.5 cm^2\)

Perimeter = length of major arc + 2r
\(={270^o\over 306^o} \times 2 \times + 2r= {3\over 2}\times {22\over 7}\times 42+2\times 42+2\times 42 = 198 + 84=284\ cm\)

length of arc AB = 5\(\pi\) cm
Let AOB = \(\theta\)
Now   i = \(\frac { \theta }{ 360° } \)x 2\(\pi\)r
5\(\pi\) = \(\frac { \theta }{ 180° } \)\(\pi\)r ⇒ \({900\over r}=\theta\)
Now, Area of sector = \(\frac { \theta }{ 360° } \)x 2\(\pi\)r²
20\(\pi\) =  \(\frac { r }{ 360° } \)x 2\(\pi\)r² ⇒ 20 = \(900°\over360°\)r ⇒ r = 8 cm

Let length of the pendulum he I cm.
Length of the arc = 8.8 cm
⇒ \(\frac { \theta \pi i }{ 180 } =8.8cm\Rightarrow \frac { 30°\times \pi \times i }{ 180° } =8.8\)
⇒ \(\pi i=8.8\times 6\Rightarrow \frac { 22 }{ 7 } \times i=52.8cm\)
⇒ \(i=\frac { 52.8\times 7 }{ 22 } cm\quad =\quad 16.8\quad cm\)

Distance covered in 5000 revolutions = 11 km.
Distance covered in 1 revolution
Distance covered in 1 revolution \(={11000\over 5000}m={11\over 5}m\)
Distance covered in 1 revolution = circumference of the wheel
\(\Rightarrow\)\(2\ \pi r ={11\over 5} \Rightarrow 2\times {22\over 7}\times r = {11\over 5}\)
\(\Rightarrow\)  \(r \Rightarrow {11\over 5}\times 7\times {1\over 2\times 22}={7\over 20}m\)
\(\therefore\) Diameter \(=2\times r=2\times {7\over 20}={7\over 10}\times 100 cm=70cm\)

Area of sector OAPB \(={x\over 360^o}\times \pi r^2\)
According to the question,
Ares of sector \(OAPB = {5\over 18}\)of the area of the circle.
\(\Rightarrow\)  \({x\over 360^o} \times \pi r^2={15\over 18}\pi r^2\Rightarrow {x\over 360^o}={5\over 18} \Rightarrow x=100^o\).

Given, sector angle, \(\theta\)= 80°
and distance or radius, r = 16.5 km
\(\begin{aligned} \therefore \text { Area of sector } & =\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{80^{\circ}}{360^{\circ}} \times 3.14 \times(16.5)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{2 \times 3.14 \times 272.25}{9} \end{aligned}\)
\(\begin{aligned} =\frac{1709.73}{9} \end{aligned}\)
= 189.97 km²
which is the required area of the sea over which the ships are warned.

Given, umbrella to be a flat circle. So, the central angle of an umbrella is 360°.
Since, umbrella has 8 ribs.
\(\therefore\) Angle between two ribs.
\(=\frac{360^{\circ}}{8}=45^{\circ}\)
Area between two ribs = Area of one sector of the umbrella
\(=\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(45)^2[\because r=45, \text { given }]\)
\(\begin{aligned} & =\frac{22}{7 \times 8}(45)^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22275}{28} \mathrm{~cm}^2 \end{aligned}\)

Given, radius of a circle, AO = 10 cm and \(\angle\)AOC = 90°
Area of \(\triangle A O C=\frac{1}{2} \times O A \times O C=\frac{1}{2} \times 10 \times 10=50 \mathrm{~cm}^2\)
\(\begin{aligned} \text { Area of sector } O A E C O & =\frac{\theta}{360^{\circ}} \times \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(10)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{314}{4}=78.5 \mathrm{~cm}^2 \end{aligned}\)

(i) Area of minor segment AECDA
= Area of sector OAECO - Area of \(\Delta\)AOC
= 78.5 - 50 = 28.5 cm²
(ii) Area of major sector OAFGCO
= Area of circle - Area of sector OAECO
= 3.14 \(\times\)(10)² - 78.5
= 314 - 78.5 = 235.5 cm²

Let, Radius of the circle = r
\(\therefore\)  Circumference of the circle = \(2\pi r\)
A.T.Q \(2\pi r\) = 22 cm
\(\Rightarrow\) \(2\times {22\over 7}\times r=22\ \ \ \Rightarrow\ \ \ \ r={{22\times 7\over 2\times 22}}={7\over 2}cm\)
Area of quadrant of the circle \(={\pi r^2\theta\over 360^o}={22\over 7}\times {7\times 7\over 2\times 2}\times {90^o\over 360^o}={22\times 7\over 2\times 2\times 4}={77\over 8}cm^2\)