Let us join DB.
We have, DN || CB, DM || AB, and ∠B = 90°
∴ DMBN is a rectangle.
∴ DN = MB and DM = NB
The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.
∴ ∠CDB = 90°
⇒ ∠2 + ∠3 = 90° … (1)
In ΔCDM,
∠1 + ∠2 + ∠DMC = 180°
⇒ ∠1 + ∠2 = 90° … (2)
In ΔDMB,
∠3 + ∠DMB + ∠4 = 180°
⇒ ∠3 + ∠4 = 90° … (3)
From equation (1) and (2), we obtain
∠1 = ∠3
From equation (1) and (3), we obtain
∠2 = ∠4
In ΔDCM and ΔBDM,
∠1 = ∠3 (Proved above)
∠2 = ∠4 (Proved above)
∴ ΔDCM ∼ ΔBDM (AA similarity criterion)
=> (BM)/(DM) = (DM)/(MC)
=> (DN)/(DM) = (DM)/(MC) (BM = DN)
⇒ DM² = DN x MC

Prove \(\triangle PNK\) and \(\triangle LMK\), similar then, \(\frac { NK }{ MK } =\frac { PN }{ LM } \)
\(\frac { c }{ b+c } =\frac { x }{ a } \Rightarrow x=\frac { ac }{ b+c } \)

In \(\triangle ABC, DE\parallel BC\)
Let EC = x cm
\(\Rightarrow \frac { AD }{ DB } =\frac { AE }{ EC } \)

[by basic proportionality theorem]
\(\Rightarrow \frac { 3 }{ 4 } =\frac { 6 }{ x } \Rightarrow \)  x = 8 cm
\(\therefore \)  EC = 8 cm

Let the three numbers in A.P. be a - d, a, a + d
\(\therefore\)    Sum is   a - d + a + a + d =15
\(\Rightarrow\)  3a = 15
\(\Rightarrow\)  a = 5
Also, product is ( a - d ) (a) ( a + d ) = 80
( 5 - d ) (5) ( 5 + d ) = 80
5 ( 25 - d² ) = 8
\(\Rightarrow\)      25 - d² = \({80 \over 5}\)
\(\Rightarrow\)     25 - d2 = 16
\(\Rightarrow\)          d² = 9
\(\Rightarrow\)    d = \(\pm\) 3 
\(\therefore\) Three numbes in A.P. are 5, -3, 5, 5 + 3
i.e.,   2, 5, 8
or  5 + 3,  5, 5 - 3
i.e.,  8, 5, 2

Here, S_q = 63q - 3q²             ....(i)
Also,  S_q-1 = 63 ( q - 1 ) - 3 ( q - 1 )²
= 63q - 63 - 3 ( q2 + 1 - 2q )
\(\Rightarrow\)   500 + 100d + 50 + 50 - d = 500 - 100d + 50 + 5 + d - 594
\(\Rightarrow\)   198d = - 594
\(\Rightarrow\)    d = - 3
Thus, the three digits are 5 - (  - 3 ), 5 and 5 - 3, (i.e., 8, 5 and 2 )
Hence, the required number is 852.

Given;     a₄ = 18
\(\Rightarrow\)  a + 3d = 18    ...... (i)
and   a₁₅ - a₉ = 30
\(\Rightarrow\)   (15 - 9)d = 30
\(\Rightarrow\)     6d = 30
\(\Rightarrow\)    d = 5
Putting the value of d in (i), we have
  a + 3d = 18
\(\Rightarrow\)  a + 3 x 5 = 18
\(\Rightarrow\)      a + 15 = 18
\(\Rightarrow\)     a = 3
Required AP is 3, 8, 13,......

Let BQ =aunits, DQ = b units

∵ PQ || AB ∴ ㄥ1 = ㄥ2
and ㄥADB = ㄥPDQ
∴ ΔADB ~ ΔPDQ
Similarily ΔADB ~ΔPDQ
∴ \({AB\over PQ}={BD\over DQ}\)
\({x\over z}={a+b\over b}\)
\({x\over z}={a\over b}+1⇒{x\over z}-1={a\over b}\ \ \ ...(i)\)
Also ΔBCD ~ ΔBPQ
∴ \({BD\over BQ}={CD\over PQ}⇒{a+b\over a}={y\over z}\)
\(1+{b\over a}={y\over z}⇒{a\over b}={y\over z}-1\)
⇒ \({b\over a}={y-z\over z}⇒{a\over b}={z\over y-z}\ \ \ (ii)\)
From (i) and (ii)
\({x\over z}-1={z\over y-z}⇒{x\over z}={z\over y-z}+1\)
\({x\over z}={z+y-z\over y-z}\)
\({z\over z}={y\over y-z}⇒{z\over x}={y-z\over y}\)
\({z\over x}=1-{z\over y}\)
\(z\left(1\over x\right)=z\left({1\over z}-{1\over y}\right)⇒{1\over x}={1\over z}-{1\over y}\)
\(⇒{1\over x}+{1\over y}={1\over z}\)

Given: \(\triangle\)ABC~ \(\triangle\)PQR
To Prove: \(\frac { ar(\triangle ABC) }{ ar(\triangle PQR) } ={ \left( \frac { AB }{ PQ } \right) }^{ 2 }\)
\(={ \left( \frac { BC }{ QR } \right) }^{ 2 }={ \left( \frac { AC }{ PR } \right) }^{ 2 }\)
Construction: Draw AD\(\bot\)BC and PE\(\bot\)QR.
Proof: \(\triangle\)ABC ~ \(\triangle\)PQR

\(\frac { AB }{ PQ } =\frac { BC }{ QR } =\frac { AC }{ PR } \)
(Corresponding sides of similar triangles)   ...(i)
\(\angle\)B = \(\angle\)Q
In \(\triangle\)ADB and \(\triangle\)PEQ
\(\angle\)B=\(\angle\)Q
\(\angle\)ADB=\(\angle\)PEQ    [each 90⁰]
\(\triangle\)ADB ~ \(\triangle\)PEQ  (AA similarity)
\(\Rightarrow \quad \frac { AD }{ PE } =\frac { AB }{ QQ } \)
(corresponding sides of similar triangles)    ...(ii)
From eq.(i) and eq.(ii),
\(\Rightarrow \frac { AB }{ PQ } =\frac { BC }{ QR } =\frac { AC }{ PR } =\frac { AD }{ PE } ...(iii)\)
Now \(\frac { ar(\triangle ABC) }{ ar(\triangle PQR) } =\frac { \frac { 1 }{ 2 } \times BC\times AD }{ \frac { 1 }{ 2 } \times QR\times PE } \)
\(=\left( \frac { BC }{ QR } \right) \times \left( \frac { AD }{ PE } \right) \)
\(\frac { ar(\triangle ABC) }{ ar(\triangle PQR) } ={ \left( \frac { BC }{ QR } \right) }^{ 2 }...(iv)\)
From eq (iii) and eq (iv),
\(\frac { ar(\triangle ABC) }{ ar(\triangle PQR) } ={ \left( \frac { AB }{ PQ } \right) }^{ 2 }\)
\(={ \left( \frac { BC }{ QR } \right) }^{ 2 }={ \left( \frac { AC }{ PR } \right) }^{ 2 }\)

Ist instalment = Rs 1000
IInd instalment = Rs 1000 + Rs 100 = IIIrd instalment = Rs 1100 + Rs 100 = Rs 1200 and so on
Let number of instalments = n
\(\therefore\)  1000 + 1100 + 1200 + .... up to n terms = 118000
\(\Rightarrow\)  \({n\over2}[2\times1000+(n-1)100]=118000\)
\(\Rightarrow\)  n( 100n + 1900 ) = 236000
\(\Rightarrow\)  100n² + 1900n - 2360 = 0
\(\Rightarrow\)  n² + 19n - 2360 = 0
\(\Rightarrow\)  ( n + 59 ) ( n - 40 ) = 0.
\(\Rightarrow\) n = - 59 ( rejected ) or n = 40
\(\therefore\)   Total no. of instalment = 40th instalment
\(\therefore\)    a₄₀ = a + 39d
= 1000 + 39 x 100
= Rs 4900
Loan repaid i 30 instalments = \({30\over2}[2\times1000+(30-1)\times100]\)
\(={30\over2}(2000+2900)\)
= 15 x 4900 = Rs 73500
\(\therefore\)    Amount to be repaid after 30th instalment = 118000 - 73500 = Rs 44500 

Here     a = 1, d = 1 and n = 31
S_n = \({n\over2}[2a+(n-1)d]\)
\(={31\over2}[2\times1+(31-1)]\)
\(={31\over2}[2+30]={31\over2}\times 32\)
= 496
\(\therefore\)    Piggy bank amount =  Rs 496
Amount spent = Rs 204
Amount left = Rs 100
Total pocket money = Rs 800

Let AB be the lamp-post and ED be the position of girl after 4s.
Given, height of the girl, ED = 100 cm
and height of the lamp-post, AB = 5 m = 500 cm
Distance of the girl from lamp-post after 4 s
= 1.9 x 4 = 7.6 m = 760 cm
[\(\because\)  distance = speed x time]
i.e. BD = 760 cm
Let DC = x cm
In \(\triangle CDE\) and \(\triangle CBA\),
\(\angle DCE=\angle BCA\) [common angle]
\(\angle CDE=\angle CBA\)  [each 90°]
\(\therefore \triangle CDE\sim \triangle CBA\) [by AA similarity criterion]
So, \(\frac { CD }{ CB } =\frac { DE }{ BA } \Rightarrow \frac { x }{ x+760 } =\frac { 100 }{ 500 } \)
\(\Rightarrow\)  5x = x + 760
\(\Rightarrow\)  4x = 760
\(\Rightarrow\)  x = 190 cm
Hence, the length of her shadow 4s is 190 cm.

\(\frac{16}{9}\)

Let h be the height of the tower.

[\(\because\)  80 cm = 0.8 m]
Since, it is clear that both the triangles will be similar.
\(\therefore \frac{1}{0.8}=\frac{h}{30} \quad \Rightarrow h = \frac{30}{0.8}\) = 37.5 m

Principal = Rs 10000
Rate of interest = 85 per annum
Since the interest is compounded yearly.
Therefore,   Amount(A) = \(10000\left(1+\frac{8}{100}\right)\)
Now, the sequence becomes  \(10000\left( 1+\frac{8}{100} \right),\)
\(10000\left( 1+\frac{8}{100} \right)^{2},10000\left( 1+\frac{8}{100} \right)^{3}\) , ...
Here, a₂ - a₁ \(\neq\) a₃ - a₂
Therefore, the given amount of money in the account every year does not form an A.P.

8

(i) (c) Let CD = h m be the height of the tower BE = 20 m be the height of Vijay's house and GF be the height of Ajay's house.

\(\begin{array}{rlrl} \triangle A C D & \sim\triangle A B E \\ \end{array}\)
\(\therefore\) \(\begin{array}{rlrl} \frac{A C}{A B} & =\frac{C D}{E B} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \frac{50}{10} & =\frac{h}{20} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & h & =100 \mathrm{~m} \end{array}\)
(ii) (d) Given, AB = 12 m, let  AC = h
In similar \(\Delta\) ABE and \(\Delta\)ACD,
\(\begin{aligned} \frac{A B}{A C} =\frac{B E}{C D} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{12}{h} & =\frac{20}{100} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & h =\frac{12 \times 100}{20}=12 \times 5=60 \mathrm{~m} \end{aligned}\)
(iii) (b) Let height of Ajay's house be GF = h₁
Since, \(\Delta\)HFG - \(\Delta\)HCD
\(\begin{array}{ll} \therefore & \frac{H F}{H C}=\frac{F G}{C D} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{20}{50}=\frac{h_1}{100} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & h_1=\frac{20 \times 100}{50}=40 \mathrm{~m} \end{array}\)
(a) Given, HC = 40 m
Let length of the shadow of Ajay's house be HF = l m
Since, \(\Delta\)HFG \(\sim\) \(\Delta\)HCD
\(\begin{aligned} & \therefore & \frac{H F}{H C} & =\frac{F G}{C D} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow & \frac{l}{40} & =\frac{40}{100} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & l & =\frac{40 \times 40}{100}=16 \mathrm{~m} \end{aligned}\)
(v) (d) Given, AC = 40 m
Let length of the shadow of Vijay's house be AB = l m
Since,  \(\triangle A B E \sim \triangle A C D\)
\(\begin{array}{rlrl} \therefore & & \frac{A B}{A C} & =\frac{E B}{C D} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \frac{l}{40} & =\frac{20}{100} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & l l & =\frac{20 \times 40}{100}=8 \mathrm{~m} \end{array}\)

(i) (c)
(ii) (c)
(iii) (d)
(iv) (b)
(v) (c)

Here the savings form an A.P. i.e., Rs 2.75, Rs 3, Rs 3.25, ...
So, a = 2.75, d = 3 - 2.75 = 0.25
(i) (b): Amount saved by Anuj on 14^th day
= t₁₄ = a + 13d = 2.75 + 13(0.25) = ₹ 6
(ii) (d): Total amount saved by Anuj in 8 days
\(=S_{8}=\frac{8}{2}[2(2.75)+7(0.25)]=₹ 29\)
(iii) (a): Amount saved by Anuj on 30^th day
= t₃₀ = a + 29d = 2.75 + 29(0.25) = ₹ 10
(iv) (b): Number of days in June = 30
\(\therefore S_{30}=\frac{30}{2}[2(2.75)+29(0.25)]=₹ 191.25\)
(v) (d): Let on nth day, he saves 10 times as he saves on 1^st day.
t_n = 10(2.75) \(\Rightarrow\) a + (n - 1)d = 27.5 \(\Rightarrow\) n = 100.