Steps of construction:
i) Construct a triangle ABC.
ii) Mark an exterior angle outside the triangle ABC,and name the point as E.

iii) Now, ACE is the exterior angle.
iv) Draw a bisecter of ㄥACE

 We know that each angle of equilateral triangle is 60^o,So have to draw an angle 60^oand bisect it.
Construction:
i) Draw any line OP.
ii) With 0 as centre and any suitable radius, draw an arc to meet OP at R.
iii) With R as centre and same radius draw an arc to meet the previous arc at S.
iv) Join OS and Produce it to Q, then ㄥPOQ=60^o
v) With R as centre and any suitable radius (not necessarily)equal to radius of step 1 (but >\(\frac { 1 }{ 2 } \)RS),draw an arc. Also, with 5 as centre and radius draw another arc to meet the previous arc at Y
vi) Join OY and produced it, the OY is the required bisector of ㄥPOQ(i.e,ㄥPOY=30^o)

\(\angle QPR=75^o\) (Vertically opposite angles)
Again, \(\angle PQR+\angle QPR=105^o\) (Exterior angle)
\(\Rightarrow \angle PQR+75^o=105^o\)
\(\Rightarrow PQR=30^o\)

In \(\triangle\)ABC, AC >AB
\(\therefore\) \(\angle\) ABC > \(\angle\)ACB
(Angles opposite to larger side is greater)
\(\therefore\) \(\angle\)ABC + \(\angle\)1 > \(\angle\)ACB + \(\angle\)1
(Adding \(\angle 1\)on both sides) Y.
\(\therefore\) \(\angle\)ABC + \(\angle\)1 > \(\angle\)ACB + \(\angle\)2
(AD bisects \(\angle\)A, \(\angle\)1 = \(\angle\)2) 
\(\therefore\) \(\angle\)ADC > \(\angle\)ADB.
(Exterior angle property of triangle)

∠APR=ㄥDRP
or ㄥ1=ㄥ2
But these are alternate interior angles
SP II RQ, SR II PQ
PQRS is a parallelogram
∠APR+ㄥBPR=180°,(linear pair)
⇒ \(\frac{1}{2}\) ∠APR+\(\frac{1}{2}\)ㄥBPR=\(\frac{1}{2}\)x180°
⇒ ∠1+ㄥ3=90°
⇒ ∠SPQ=90°

PQRS is a rectangle

Since opposite angles of a parallelogram are equal
3x - 2 = 63 - 2x
⇒ x=13°
Angles of a parallelogram :
(39 - 2)°, (180- 37)°, (63 - 26)°, (180- 37)°
i.e., 37°,143°,37°,143°.

9x²+6xy+y² = (3x)²+2\(\times\)(3x)\(\times\)y+y²
=(3x+y)²
[\(\because\) a²+2ab+b²=(a+b)²]

AB = CD(Given)
\(\Rightarrow\)AB + BC = BC + CD
\(\Rightarrow\)AC = BD
Euclid's axiom used: If equals are added to equals, the wholes are equal.

PQ = QR
QX = QY

If equals are subtracted from equals, the remainders are also equal.
We have PQ - QX = QR - QY
PX = RY

Euclid's axioms
(i) Things which are equal to the same thing are equal to one another.
(ii) If equals are added to equals, the wholes are equal.

DE, EF

(a) 28    
(b) 26

(x-1)(x-3)(x-4)

2

3

34

7/6,4/3,3/2,5/3 and 11/6

AB = BC (given)
BX = BY (given)
If equals are subtracted from equals, then remains are also equal.
AB - BX = BC - BY
\(\Rightarrow\) AX = CY

Given: The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles.
To Prove: Quadrilateral ABCD is a square.
Proof: In \(\Delta\)OAD and \(\Delta\)OCB,
OA = OC             I Given
OD = OB       I Given
\(\angle AOD=\angle COB\)    |Vertically Opposite Angles
 \(\therefore \Delta OAD\cong \Delta OCB\)    I SAS Congruence Rule

\(\therefore \)   AD = CB           I C.P.C.T.
 \(\angle ODA=\angle OBC\)  |C.P.C.T
\(\angle BDA=\angle DBC\)
.\(\therefore \)   AD II BC
Now, \(\because\) AD = CB and AD II CB
\(\therefore \)  Quadrilateral ABCD is a II gm.      I A quadrilateral is a parallelogram if a pair of opposite sides are parallel and  equal.
In \(\therefore \) \(\Delta\) AOB and \(\Delta\)AOD,
AO = AO                          I Common
OB = OD          I Given
 \(\angle \)AOB = \(\angle \)AOD               |Each=\(90°\)
\(\therefore \)  \(\Delta\)AOB \(\cong \)  \(\Delta\) AOD         I SAS Congruence Rule
AB = AD                 I C.P.C.T.
Now, \(\because\) ABCD is a parallelogram and
AB=AD
 \(\because\) ABCD is a rhombus.
Again, in \(\Delta\) ABC and \(\Delta\) BAD,
AC = BD            I Given
BC = AD            I \(\because\)  ABCD is a rhombus
AB =  BA                     I Common
\(\therefore \) \(\Delta\) ABC \(\cong \) \(\Delta\) BAD       I SSS Congruence Rule
\(\therefore \)    \(\angle \) ABC = \(\angle\) BAD    I C.P.C.T.
\(\because\)    AD II BC                 I Opp. sides of IIgm ABCD and transversal AB intersects them.
\(\therefore \) \(\angle ABC+\angle BAD=180°\)      I Sum of consecutive interior angles on the  same side of a transversal is 180°
ABC = \(\angle\)BAD = \(90°\)
Similarly,\(\angle\)BCD =\(\angle\)ADC =\(90°\)
 \(\therefore \)  ABCD is a square.

Coefficient of x² = 0

\(\frac { 2 }{ 11 } \)= 0.1818...= \(0.\overline { 18 } \)

PQ II DB, SR II DB
 
\(\therefore\) PQ II SR
Similarly, PS IIQR
\(\therefore\) PQRS is a parallelogram.

Two equilateral triangles of the same side length are congruent

Required angle = \(360^{ 0 },\) - \(60^{ 0 },\)=\(300^{ 0 },\)

Required angle=\(180^{ 0 }\)-(\(180^{ 0 }\)+\(9^{ 0 }\)) =\(9^{ 0 }\)​​​​​​​

Steps of Construction
1. Draw a line segment AB = 5 cm.
2. Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point P.
3. Taking P as centre and with the same radius as before, draw an arc intersecting the previously draw arc, say at a point E.
4. Draw the ray AC passing through E.
5. From ray AC, cut off AD = 6 cm. Then, \(\angle DAB\) is the required angle of 60° such that AD = 6 cm.
6. Now, taking A and D as centres and radius 1 more than \(\frac { 1 }{ 2 } \) AD, draw arcs on both sides of the line segment AD (to intersect each other).
 

Let in \(\triangle \)ABC \(\angle A=\angle B+\angle C\)
We know that \(\angle A=\angle B+\angle C=180^{ 0 }\)
The sum of the three angles of a triangle is \(180^{ 0 }\) 
\(\Rightarrow \angle A+\angle A=180^{ 0 }\)
\(\Rightarrow 2\angle A=180^{ 0 }\)
\(\Rightarrow \angle A=\frac { 180^{ 0 } }{ 2 } =90^{ 0 }\)
Hence Triangle ABC is a right-angled triangle.

Number of terms = 3
Terms: 3x², -5x, 7