We have a square ABCD such that its diagonals AC and BD intersect at O.
(i) To prove that the diagonals are equal, i.e. AC = BD
In ΔABC and ΔBAD, we have
AB =BA
[Common] 
BC = AD [Opposite sides oj the square ABCD]
∠ABC = ∠BAD [All angles of a square are equal to 90^ο]
∴ ABC ≅ ΔBAD [SAS criteria]
⇒ Their corresponding parts are equal.
⇒  AC = BD ...(1)
(ii) To prove that '0' is the mid-point of AC and BD.
∵ AD II BC and AC is a transversal. [∵ Opposite sides of a square are parallel]
∴ ∠1 = ∠3 [Interior alternate angles]
Similarly, ∠2 = ∠4 [Interior alternate angles]
Now, in ΔOAD and ΔOCB, we have
AD = CB [Opposite sides oj the square ABCD]
∠1 = ∠3 [Proved]
∠2 = ∠4 [Proved]
∴ ΔOAD == ΔOCB [ASA criteria]
∴ Their corresponding parts are equal.
⇒ OA = OC and OD = OB
⇒ O is the mid-point of AC and BD, i.e. the diagonals AC and BD bisect each other at O. ...(2)
(iii) To prove that AC ⏊ BD.
In ΔOBA and ΔODA, we have
OB = OD [Proved]
BA = DA [Opposite sides of the square]
OA = OA [Common]
∴ ΔOBA ≅ ΔODA [SSS criteria]
⇒ Their corresponding parts are equal.
⇒ ∠AOB = ∠AOD
But ∠AOB and ∠AOD form a linear pair.
∴ ∠AOB + ∠AOD = 180^ο
⇒ ∠AOB = ∠AOD = 90^ο
⇒ AC ⏊ BD ... (3)
From (1), (2) and (3), we get AC and BD are equal and bisect each other at right angles.

Proof: In \(\triangle\)ABF and \(\triangle\)ACE,
AB =AC
\(\angle\)A=\(\angle\)A
AF =AE
\(\therefore\) \(\triangle\)ABF = \(\triangle\)ACE (by SAS cong.)
\(\therefore\) By c.p.c.t BF = CE.
Alternative Method:
AB = AC\(\Rightarrow \frac { AB }{ 2 } =\frac { AC }{ 2 } \)
\(\Rightarrow \) AE=AF,
sice E and F are the mid-points of AB and AC
In \(\triangle\)ABF and \(\triangle\)ACE,
AB = AC (Given) 
\(\angle\)A = \(\angle\)A (Common) 
AF = AE (Proved) 
\(\therefore \triangle ABF\cong \triangle ACE\) (By SAS cong.) 
\(\therefore\) BF = CE. (By c.p.c.t) 

(a) 114.89 cm²  
(b) \(40\sqrt { 3 } \)cm²  
(c) \(24\sqrt { 3 } \)cm²  
(d) 330 m²

'a' = a, 'b' =a, 'c' =a
\(s=\frac { 'a'+'b'+'c' }{ 2 } =\frac { a+a+a }{ 2 } =\frac { 3a }{ 2 } \)
\(\therefore \) Area of the signal board
\(=\sqrt { s(s-'a')(s-'b')(s-'c') } \)
\(\\ =\sqrt { \frac { 3a }{ 2 } \left( \frac { 3a }{ 2 } -a \right) \left( \frac { 3a }{ 2 } -a \right) \left( \frac { 3a }{ 2 } -a \right) } \)
\(\\ =\sqrt { \frac { 3a }{ 2 } \left( \frac { a }{ 2 } \right) \left( \frac { a }{ 2 } \right) \left( \frac { a }{ 2 } \right) } =\sqrt { \frac { { 3a }^{ 4 } }{ 16 } } =\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\)
Perimeter = 180 cm
\(\Rightarrow 'a'+'b'+'c'\ =\ 180\)
\(\\ \Rightarrow a+a+a=180\)
\(\\ \Rightarrow 3a=180\)
\(\\ \Rightarrow a=\frac { 180 }{ 3 } \)
\(\Rightarrow \)  a = 60 cm
\(\therefore \)  Area of the signal board
 \(=\frac { \sqrt { 3 } }{ 4 } \)  a² =\( \frac { \sqrt { 3 } }{ 4 } \) (60)² = \(900\sqrt { 3 } \) cm²
Alternatively,
\(s=\frac { 3a }{ 2 } =\frac { 3 }{ 2 } (60)=90\)cm
Area of the signal board
\(=\sqrt { s(s-'a')(s-'b')(s-'c') } \)
\(=\sqrt { 90(90-60)(90-60)(90-60) } \)
\(\\ =\sqrt { 90(30)(30)(30) } =900\sqrt { 3 } \) cm². 

\((2a-3b)^3\)
\({ (2a-3b) }^{ 3 }={ (2a) }^{ 3 }-{ (3b) }^{ 3 }-2(2a)(3b)(2a-3b)\) | Using Identity VII
\(=8{ a }^{ 3 }-27{ b }^{ 3 }-18ab(2a-3b)\)
\(=8{ a }^{ 3 }-27{ b }^{ 3 }-36{ a }^{ 2 }b+54{ ab }^{ 2 }\)

\(\frac { 3 }{ 13 } \)= 0.230769230769...=\(0.\overline { 230769 } \)

AB = 10 cm

54 cm

Given ABCD is a parallelogram.
\(\therefore \)  AB = CD and AD = BC
and E is the mid-point of side BC, i.e. BE = CE
To prove AF = 2 AB
Proof  Now, in \(\triangle DEC\) and \(\triangle FEB\), we have
\(\angle DEC=\angle BEF\) [vertically opposite angles]
BE = CE [given]
and \(\angle DCE=\angle FBE\)  [since, \(DC\parallel AF\) and BC is transversal, so alternate interior angles]
\(\therefore \triangle DEC\cong \triangle FEB\)  [by ASA congruence rule]
So, DC = BF     [by CPCT]   .... (i)
Also, we have DC = AB [given]    ... (ii)
From Eqs.(i) and (ii), we get
AB = BF
\(\therefore AF=2AB\)

First prove that ΔADB ≅ ΔADC by SAS congruence rule. Then show that ㄥABD=ㄥACD by CPCT Ans: 1:1 Alternate MehodSince, AB=AC
∴ ㄥABC = ㄥACB
[-:angles opposite to equal sides are equal]
Similarly,  ㄥDBC =ㄥDCB   [∵ DB = DC]
Now,  ㄥABC-ㄥDBC=ㄥACB-ㄥDCB
⇒  ㄥABD = ㄥACD
⇒ \(\frac { \angle ABD }{ \angle ACD } =1\Rightarrow \angle ABD:\angle ACD=1:\)

For finding area of the park, we have
2s = 50 m + 80 m + 120 m = 250 m.
i.e., s = 125 m
Now, s – a = (125 – 120) m = 5 m,
s – b = (125 – 80) m = 45 m,
s – c = (125 – 50) m = 75 m.
Therefore, area of the park = \(\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{125 \times 5 \times 45 \times 75} \mathrm{~m}^{2}\)
\(=375 \sqrt{15} \mathrm{~m}^{2}\)
Also, perimeter of the park = AB + BC + CA = 250 m
Therefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate)
= 247 m
And so the cost of fencing = Rs.20 x 247 = RS. 4940

Since we do not know what 0 3 . is , let us call it ‘x’ and so
x = 0.3333
Now here is where the trick comes in. Look at
10 x = 10 x (0.333...) = 3.333
Now, 3.3333 = 3 + x, since x = 0.3333
Therefore, 10 x = 3 + x
Solving for x, we get
\(9 x=3, \text { i.e., } x=\frac{1}{3}\)

FE IIBC, ED II AB
\(\therefore\)BDEF is a parallelogram

\(\because \)   ABD=\(90°\),\(\therefore \)   FED = \(90°\)

\(\angle \)OAB+\(\angle \)OBA=\(\frac { 1 }{ 2 } \)\(​​\angle \)A+\(\frac { 1 }{ 2 } \)\(\angle \)B
=\(\frac { 1 }{ 2 } \)(\(\angle \)A+\(\angle \)B)=(\(180°\))=\(90°\)
\(\therefore \) \(\angle \)AOB=\(90°\)

In a parallelogram, opposite sides and opposite angles are equal.

Use factor theorem

\(x+1=0\ \ Rightarrow \ x=-1\)
\(\therefore\) Remainder=\((-1)^11+101=-1+101\)
=100

Remainder theorem

\(p(x)=(p-x)^3+14=(17-x)^3+14\)
\(=(17)^3-x^3-3(17)^2(x)+3.17.x^2+14\)
\(\because\)  Degree=3

(i) In quadrilateral APCQ,
AP || QC (Since AB || CD)  .....................(1)
\(\mathrm{AP}=\frac{1}{2} \mathrm{AB}, \quad \mathrm{CQ}=\frac{1}{2} \mathrm{CD}\)
Also, AB = CD 
So, AP = QC  ...................(2)
Therefore, APCQ is a parallelogram [From (1) and (2)]
(ii) Similarly, quadrilateral DPBQ is a parallelogram, because
DQ || PB and DQ = PB
(iii) In quadrilateral PSQR,
SP || QR (SP is a part of DP and QR is a part of QB)
Similarly, SQ || PR
So, PSQR is a parallelogram.

Let p(x) = x³ – 23x² + 142x – 120
We shall now look for all the factors of –120. Some of these are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.
By trial, we find that p(1) = 0. So x – 1 is a factor of p(x).
Now we see that x³ – 23x² + 142x – 120 = x³ – x² – 22x² + 22x + 120x – 120
= x²(x –1) – 22x(x – 1) + 120(x – 1) 
= (x – 1) (x² – 22x + 120) [Taking (x – 1) common]
We could have also got this by dividing p(x) by x – 1.
Now x² – 22x + 120 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
x² – 22x + 120 = x² – 12x – 10x + 120
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
So, x³ – 23x² – 142x – 120 = (x – 1)(x – 10)(x – 12)

Draw XOX' and mark O as 0. Take OA = 1 unit.
Draw AB ⊥ OX and cut off AB = 1 unit.

Now, OAB is a right triangle.
∴ OA²+ AB² = OB²
or 1² + 1²= OB²
or 2 = OB²
⇒ OB = \(\sqrt{2}\)
With centre O and OB as radius, draw an arc intersecting OX at C.
Thus, OC = \(\sqrt{2}\) on the real line XOX'.

We are given that AB = BC and have to prove that
DE = EF.
Let us join A to E intersecting m at G.
Let trapezium ACFD is divided into two triangles, namely ΔACF and ΔAFD.
In ΔACF, it is given that B is the mid-point of AC(AB = BC) and BG II CF (Since m || n)
So, G is the mid-point of AF (By the converse of midpoint theorem)
Now in ΔAFD, we can apply the sam argument as G is the mid-point of AF, GE IIAD so E is the mid-point of DF
i.e., DE = EF
In otherwords l, m and n cut off equal intercepts on q also.

Let the sides of wall are a=122m, b=22m and c=120 m.
Then, semi-perimeter, \(s=\frac{a+b+c}{2}=\frac{122+22+120}{2}=\frac{264}{2}=132m\)
Now , area of the wall=\(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{132(132-122)(132-22)(132-120)}\)
\(=\sqrt{132\times10\times110\times12}=\sqrt{11\times12\times10\times11\times10\times12}\)
\(=11\times 12\times10=1320m^2\)
Given, earning on 1 m² per year=Rs5000
ஃ Earning on 1320m² per year=1320 x 5000=Rs6600000
Now, earning in 12 months =Rs 6600000
ஃ   Earning  in months\(=\frac{6600000\times3}{12}=Rs1650000\)
Hence, the rent paid by the company for 3 months is Rs 1650000.

Given In right angled \(\triangle ABC,\angle C=90^{ 0 }\) M is the mid-point of AB AM=MB and DM =CM
To Prove (i) \(\triangle AMC,\cong \triangle BMD\)
(ii) \(\angle DBC\) is a right angle
(iii) \(\triangle DBC=\triangle ACB\) 
(iv) \(CM=\frac { 1 }{ 2 } AB\)
Proof (i) In \(\triangle AMC,\triangle BMD\)
AM=BM [given]
mc=dm [given]
and \(\angle AMC=\angle DMB\) [vertically opposite angles]
\(\therefore \triangle AMC=\triangle BMD\) [by SAS congruence rule]
(ii) From part (i)
\(\triangle AMC=\triangle BMD\)
Then AC=DB
and \(\triangle ACB=\triangle BDM\)
which are alternate angles.
\(\therefore BD||CA\) and BC is a transversal
Then \(\angle ACB+\angle DBC=180^{ 0 }\)
\(\Rightarrow 90^{ 0 }+\angle DBC=180^{ 0 }[\therefore \angle C=90^{ 0 }]\)
\(\Rightarrow \angle DBC=90^{ 0 }\)
Hence \(\angle DBC\) is a right angle
(iii) In \(\triangle DBC,\triangle ACB\)
DB=AC[Proved above]
BC=BC[common side]
and \(\angle DBC=\angle BCA\)[each 90^0.]
\(\therefore \angle DBC\equiv \angle BCA\) [by sas congruence rule]
(iv) From part (iii)
\(\triangle DBC\equiv ACB\) [by CPCT]
\(\Rightarrow CM+MD=AB[\therefore CD=CM+MD]\)
\(\Rightarrow CM+CM=AB[\therefore DM=CM]\)
\(\Rightarrow 2CM=AB\)
\(\therefore CM=\frac { 1 }{ 2 } AB\)
Hence proved

We know that “If two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then the two triangles are congruent.” – This is ASA Congruence Rule
So, Reason is correct
Now, In triangles ABC and PQR,∠A = ∠P, ∠C = ∠R and AC = PR.
∴ By ASA congruence criteria, Δ ABC ≅ Δ PQR
So, Assertion is also correct.
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

We know that “If two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then triangles are congruent.” - This is ASA Congruence Rule.
So, Reason is correct.
In △BDL and △CDM, we have
BD = CD (D is midpoint)
DL = DM    (Given)
and ∠BLD = ∠CMD    (90° each)
∴ △BDL ≅ △CDM (RHS criterion)
⇒ BL = CM (CPCT)
So, Assertion is also correct 
Correct option is (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A).