Let us consider the ΔPMN such that ∠M = 90^о 
Since, ∠M + ∠N + ∠P = 180^о
[Sum of angles of a triangle]
∵ ∠M = 90^о [∵ PM ⊥ ℓ ]
⇒ ∠N< ∠M
⇒  PM < PN ... (1)
Similarly, PM < PN₁ ,.. (2)
PM < PN₂ ...(3)
From (1), (2) and (3). we have PM is the smallest line segment drawn from P on the line l . Thus, the perpendicular segment is the shortest line segment drawn on a line from a point not on it.

Let the given field be in the shape of a trapezium ABCD in which AB=25 m, CD= 10 m, BC = 13 m and AD = 14 m.
From D, draw DE 11 BC meeting AB at E. Also, draw DF \(\bot \) AB.
\(\therefore \) DE = BC =13 m
AE = AB - EB = AB - DC
= 25 - 10 = 15 m

For AED
a = 14 m, b = 13 m, c=  15 m
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 14+13+15 }{ 2 } =\frac { 42 }{ 2 } =21\) m
\(\therefore \)    Area of the \(\Delta \)AED  \(=\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 21(21-14)(21-13)(21-15) }\)
\( \\ =\sqrt { 21(7)(8)(6) } =\sqrt { \left( 7\times 3 \right) (7)\left( 4\times 2 \right) \left( 2\times 3 \right) } \)
\(=7\times 3\times 2\times 2=84\) m²
\(\Rightarrow \frac { 1 }{ 2 } \times \)AE\(\times \)DE = 84
\(\Rightarrow \ \frac { 1 }{ 2 } \times \)15\(\times \)DF = 84
\(\Rightarrow \) DF = \(\frac { 84\times 2 }{ 15 } \)
\(\Rightarrow \) DF=  \(\frac { 56 }{ 5 } \) m = 11.2 m
\(\Rightarrow\) Height of the trapezium is 11.2 m.
\(\therefore \) Area of parallelogram EBCD = Base \(\times \) Height
= EB\(\times \) DF = 10 \(\times \)\(\frac { 56 }{ 5 } \) = 112 m²
\(\therefore \) Area of the field = Area of AED + Area of parallelogram EBCD = 84 m² + 112 m² = 196 m².                 

Area of paper of shade I \(2\times \left( \frac { 1 }{ 2 } \times 16\times 16 \right) =256\)cm²
Similarly, Area of paper of shade II = 256 cm²
For area of paper of shade III
a = 8 cm, b = 6 cm, c = 6cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 8+6+6 }{ 2 } =10\) cm
\(\therefore \) Area of paper of shade III = \(\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 10(10-8)(10-6)(10-6) } \)
 \(=\sqrt { (10)(2)(4)(4) } =8\sqrt { 5 } \) 
= 17.89 cm²

\(\therefore \)Ray QP stands on line ST
\(\therefore \) \(\angle \)PQS +\(\angle \)PQR = \(180^{ 0 }\)
|Linear Pair Axiom
Ray RP stands on line ST 
\(\therefore \) \(\angle \)PRQ +\(\angle \)PRT= \(180^{ 0 }\)
| Linear Pair Axiom
From (1) and (2) we obtain
\(\angle \)PQS+\(\angle \)PQR =\(\angle \)PRQ +\(\angle \)PRT
\(\Rightarrow \)\(\angle \) PQS =\(\angle \)PRT
 \(\therefore \) \(\angle \)PQR=\(\angle \)PRQ(Given)

(i) In ABC, we have
AB = AC [Given]
∴ ∠C = ∠B
[Angle opposite to equal sides are equal]
⇒ \(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠B
or ∠OCB = ∠OBC
⇒ OB = OC 
[Sides opposite to equal angles are equal]
(ii) In ΔABO and ΔACO, we have
AB =AC [Given]
OB = OC [Proved
∠OBA = ∠OCA  [∵ \(\frac{1}{2}\)∠B-\(\frac{1}{2}\)∠C]
∴ Using SAS criteria,
ΔABO =ΔACO
⇒ ∠OAB = ∠OAC [c.p.c.t.]
⇒ AO bisects LA. 

Let AB and CD two lines intersecting at O

This leads to two pairs of vertically  opposite angles,namely
(i)  \(\angle \)AOC and \(\angle \)BOD
(II) \(\angle \)AOD and \(\angle \)BOC
We are to prove that
(i)   \(\angle \)AOC and \(\angle \)BOD
(II) \(\angle \)AOD and \(\angle \)BOC
\(\therefore \) Ray OA stands on line CD 
Therefore (i)   \(\angle \)AOC and \(\angle \)AOD=\(180^{ 0 }\)
|Linear Pair Axiom
From (1) and (2) 
\(\angle \)AOC and \(\angle \)AOD= \(\angle \)AOD and \(\angle \)BOD \(\angle \)AOC and \(\angle \)BOD
\(\Rightarrow \) Similarly we can prove that 
\(\angle \)AOD and \(\angle \)BOC

Let us join AC.
Now, in ΔABC  AB < BC
[∵ AB the smallest side of quadrilateral ABCD]
BC >AB
[Angle opposite to BC] < [Angle opposite to AB]
∠BAC > ∠BCA ... (1)
Again in ΔACD,
CD > AD
[∵ CD is the longest side ofthe quadrilateral ABCD]
∴ [Angle opposite to CD] > [Angle opposite to AD]
⇒ ∠CAD > ∠ACD ...(2)
Adding (1) and (2), we get
[∠BAC + CAD] > [∠BCA + ∠ACD]
⇒ ∠A > ∠C
Similarly, by joining BD, we have
∠B > ∠D

Given AB॥CD and O is the mid-point of AD.

∴ AO = OD
To prove (i) ΔAOB ≅ ΔDOC
(ii) O is also the mid-point of BC, ie OB = OC
Proof (i) In  ΔAOB and ΔDOC,
ㄥABO =ㄥDCO  [alternate angles as AB ‖ CD and BC is a transversal line]
ㄥAOB = ㄥDOC  [vertically opposite angles]
and OA = OD  [from Eq. (i)]
∴ ㄥAOB ≅ ΔDOC    [by AAS congruence rule]
(ii) Bty using corresponding part of congruent triangles, we get 
OB = OC
Hence, O is also the mid-point of BC.

Since AB = 9 m and BC = 40 m, Ð B = 90°, we have

\( \mathrm{AC} =\sqrt{9^{2}+40^{2}} \mathrm{~m} \)
\(=\sqrt{81+1600} \mathrm{~m} \)
\(=\sqrt{1681} \mathrm{~m}=41 \mathrm{~m} \)
Therefore, the first group has to clean the area of triangle ABC, which is right angled.
\(Area of \Delta \mathrm{ABC}=\frac{1}{2} \times base \times height \)
\(=\frac{1}{2} \times 40 \times 9 \mathrm{~m}^{2}=180 \mathrm{~m}^{2} \)
The second group has to clean the area of triangle ACD, which is scalene having sides 41 m, 15 m and 28 m.
Here, \(s=\frac{41+15+28}{2} \mathrm{~m}=42 \mathrm{~m}\)
Therefore, area of \(\Delta \mathrm{ACD}=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{42(42-41)(42-15)(42-28)} \mathrm{m}^{2} \)
\( =\sqrt{42 \times 1 \times 27 \times 14} \mathrm{~m}^{2}=126 \mathrm{~m}^{2} \)
So first group cleaned 180 m² which is (180 – 126) m², i.e., 54 m² more than the area cleaned by the second group.
Total area cleaned by all the students = (180 + 126) m² = 306 m².

Let ABCD be the field.
Perimeter = 400 m
So, each side = 400 m ÷ 4 = 100 m.
i.e. AB = AD = 100 m.
Let diagonal BD = 160 m.
Then semi-perimeter s of D ABD is given by
\(s=\frac{100+100+160}{2} \mathrm{~m}=180 \mathrm{~m}\)
Therefore, area of \(\Delta \mathrm{ABD}=\sqrt{180(180-100)(180-100)(180-160)}\)
\(=\sqrt{180 \times 80 \times 80 \times 20} \mathrm{~m}^{2}=4800 \mathrm{~m}^{2}\)
Therefore, each of them will get an area of 4800 m².

Here, we need to draw a line AB parallel to line PQ, through point M as shown in Fig. Now, AB || PQ and PQ || RS.
Therefore, AB || RS
Now, \(\angle\) QXM + \(\angle\) XMB = 180°
(AB || PQ, Interior angles on the same side of the transversal XM)
But \(\angle\) QXM = 135°
So, 135° + \(\angle\) XMB = 180°
Therefore, \(\angle\) XMB = 45° (1)
Now, \(\angle\) BMY = \(\angle\) MYR (AB || RS, Alternate angles)
Therefore, \(\angle\) BMY = 40° (2)
Adding (1) and (2), you get
\(\angle\) XMB + \(\angle\) BMY = 45° + 40°
That is, \(\angle\) XMY = 85°

(i) In  ΔABD and ΔACD. we have
AB = AC [Given]
AD = AD [Common]
BD = CD [Given]
∴ ΔABD ≅ ΔACD [SSS Criteria]
(ii) In ΔABP and ΔACP, we have
AB = AC [Given]
∴ AB = AC⇒ ∠B = ∠C
[∵ Angle opposite to equal sides are equal]
AP = AP [Common]
∴ ΔABP ≅ ACP [SAS Criteria]
(iii) Since, ΔABP ≅ ΔACP
∴ Their corresponding parts are congruent.
⇒ ∠BAP = ∠CAP
∴ AP is the bisector of ∠A  ...(1)
Again, in ΔBDP and ΔCDP, we have
BD = CD [Given]
∠DBP = ∠CDP [Angles opposite to equal sides]
DP = DP [Common]
⇒ ∠BDP = ∠CDP
∴ ∠BDP = ∠CDP [c.p.c.t.]
⇒  DP (or AP) is the bisector of ∠D.  ...(2)
From (1) and (2) AP is the bisector of ∠A as well as ∠D.
(ii) ∴ ΔABP ≅ ΔACP
∴ Their corresponding parts are equal.
⇒ ∠APB = ∠APC
But ∠APB + ∠APC = 180^० [Linear pair]
∴ ∠APB = ∠APC = 90^० 
⇒ AP ⊥ BC
⇒  AP is the perpendicular bisector of BC

∵ POQ is a straight line. [Given]
∴ ∠POS + ∠ROS + ∠ROQ = 180^о
But OR ⊥ PQ
∴ ∠ROQ = 90^о
∴ ∠POS + ∠ROS + 90^о = 180^o
⇒ ∠POS + ∠ROS = 90^о
Now, we have ∠ROS + ∠ROQ = ∠QOS ...(1)
⇒ ∠ROS + 90^о = ∠QOS ...(2)
From (1) and (2), we have
∠ROS + [∠POS + ∠ROS] = ∠QOS
⇒ 2∠ROS + ∠POS = ∠QOS
⇒ 2∠ROS = [∠QOS - ∠POS]
∴ ∠ROS = \(\frac{1}{2}\)[∠QOS - ∠POS]

Let the sides of triangle be a = 26 cm, b = 28 cm and c = 30 cm. Let s be the semi-perimeter of the triangle
Then, 
\(s=\frac{a+b+c}{2}=\frac{26+28+30}{2}=\frac{84}{2}=42cm\)
and area of triangle \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{42\times(42-26)(42-28)(42-30)}\)
\(=\sqrt{42\times16\times14\times12}\)
\(=\sqrt{7\times6\times16\times7\times2\times6\times2}=336cm^2\)
Let h be the height of the parallelogram.
Then, area of parallelogram = base x height = 28 x b
∵ area of parallelogram =Area of triangle
ஃ  \(28\times b=336\Rightarrow b=\frac{336}{28}\Rightarrow b=12cm\)
Hence, the height of the parallelogram is 12 cm.​​​​​​​

Let given sides of the wall be a = 15m, b = 11m and c = 6 m.
ஃ  Semi-perimeter, \(s=\frac{a+b+c}{2}=\frac{15+11+6}{2}=\frac{32}{2}=16m\)
Now, area of the wall= \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{16(16-15)(16-11)(16-6)}\)
\(=\sqrt{16\times1\times5\times10}=\sqrt{4\times4\times5\times5\times2}=5\times4\sqrt2\)
\(=20\sqrt{2}m^2\)
Hence, the area painted in colour is \(=20\sqrt{2}m^2\)