Important Questions: The Triangle and Its Properties

∠EFD+∠FED = x  
(Exterior angle property of a triangle) 
⇒ 28^o+42^o = ∠x or ∠x = 70^o

∠1 =∠2 and  ∠2 =∠3  [Alternate angles]
So, ∠1 =∠3  and  ∠1+∠4+∠5 = 180^o  [Angle sum property]
Also, ∠8 = ∠6

In ΔCED, CE=ED 
∴ ∠EDC = ∠ECD  
[Angles opposite to equal sides are equal] ⇒ ∠ECD = 28^o 
 Also,  ∠ECD = ∠BCA (Vertically opposite angles) ⇒ ∠BCA = 28^o 
In ΔBCA, y = 62^o+28^o 
Exterior angle property] ⇒ y = 90^o

Since ABC is an equilateral triangle. 
∴ ∠CAB =∠ABC =∠BCA = 60^o  
And ∠DBA = ∠DAB = (60^o−x) 
[∵ DA = DB] 
In ΔDAB, ∠DAB∠DAB+∠ADB =180^o   
⇒ 2(60^o−x)+88^o = 180^o 
⇒ 2(60^o−x) = 92^o
⇒ 60^o−x = 46^o ⇒ x = 14^o

In ΔAEB, 

 ∠A=∠DAE+∠BAD ⇒ ∠A=60^o+90^o=15^o 
And,  AE=AB ⇒ ∠ABE=∠AEB  
[Angles opposite to equal sides are equal]
Now, ∠A+∠ABE+∠AEB=180^o   (Angle sum property) 
⇒ 2∠AEB=180^o −150^o = 30^o ⇒ ∠AEB = 15^o  
Now, ∠E=60^o ⇒ ∠DEF=60^o−15^o = 45^o 
∴ In ΔEFD, ∠DEF+∠EDF+∠EFD
= 180^o ⇒ 45^o+60^o+y = 180^o 
⇒ y = 180^o−(45^o+60^o) = 75^o

∠UXV = y (Vertically opposite angles) 
∴ y = 45^o In ΔXYZ y+x+63^o = 180^o  
(Angle sum property) 
⇒ 45^o+x+63^o = 180^o
⇒ x = 180^o−(45^o+63^o) 
⇒  x= 180^o−108^o = 72^o

We have, ABCD, CEFG and CIHJ are all squares. 

 So, ∠1+∠2+x=90^o ???
(i) 36^o+∠1+x=90^o q ??.
(ii) x+∠2+27^o=90^o ??.
(iii) Adding (ii) and (iii),
we get 36^o+x+27^o+(∠A+∠2+x)=180^o 
⇒ 63^o+x+90^o=180^o 
(From (i)) ⇒ x=180^o−153^o = 27^o

In ΔFGC, ∠CBF = 60^o
(Angle of equilateral triangle)  
∴  x+60^o+92^o = 180^o      
⇒ x = 180^o−152^o=  28^o        
Now, In ΔBCF, ∠CBF = 60^o            
∠FCB = 180^o−92^o  (Linear pair)            
⇒ ∠FCB=88^o          
∴ ∠BFC+88^o+60^o=180^o                       
(Angle sum property)      
⇒ ∠BFC=180^o−148^o = 32^o
And ∠AFE = 90^o ⇒ y+32^o = 90^o
⇒ y = 90^o−32^o = 58^o
∴ y−2x = 58^o−2×28^o = 58^o−56^o = 2^o

∠FCA = ∠BFD  
(Corresponding angles) 
⇒ x = 51^o 
Now, in ΔABC y = 51^o+83^o (Exterior angle property) 
⇒ y = 134^o So, x+y = 51^o+134^o = 185^o

It is given that, 
AB+BC = 10cm   ?..(i) 
BC+CA = 12cm    ?..(ii) 
CA+AB = 16cm      ?..(iii)
Adding (i), (ii) and (iii); we get 
2(AB+BC+CA) = 10+12+16 
⇒ AB+BC+CA = 19cm

∠CDB+∠BDE = 90^o  
(Angle of a rectangle) 
⇒ 48^o+∠BDE = 90^o 
⇒ ∠BDE = 90^o−48^o = 42^o 
In ΔBED ∠EBD+∠BDE+∠BED = 180^o  
(Angle sum property) 
⇒ 86^o+42^o+∠BED = 180^o 
⇒ ∠BED = 180^o−(86^o+42^o) = 52^o

∆MDC is an equilateral triangle. therefore all the angles will be 60°.

angle MCB=BCD-MCD

=90-60 = 30°

NOW,

In ∆BCD,

BC=CD

Therefore angle XBC = 45°.

when angle sum is applied in ∆CBX, X=105°

∠KLO = ∠MLN 
∴ ∠MLN = 70^o in ∠LMN,∠MLN+∠LNM+∠LMN = 180^o 
(Angle sum property) 
⇒ 70^o+∠LNM+50^o=180^o
 ⇒ ∠LNM = 180^o−(70^o+50^o) = 60^o

In ΔPRQ,  PR² = PQ²+QR² 

(By Pythagoras theorem) 
(26)² =(24)²+QR² or QR² 
= 676−576 = 100 
⇒ QR = √100 ⇒ QR = 10
∴ The distance of the foot of the ladder from the wall is 10 m.

Let KB is original height of the tree. In ΔABC, 

AC² = AB²+BC² = 52+122 =25+144=169
∴ AC = √169 = 13m 
KB = KA+AB = (13+5)m = 18m
∴ Original height of the tree is 18m.

Since diagonals of a rhombus bisect each other at 90^o . 
Given: BD = 42 cm and AC = 56 cm 

∴ BK = 1/2 BD = 42/2 = 21cm 
AK=1/2 AC = 56/2 = 28 cm  
In ΔKAB,AB² = AK²+BK² = (28)²+(21)²=784+441=1225 
∴ AB=√1225 = 35cm 
∴  Perimeter of the field ABCD = 4×35 = 140 cm;

Let AB = length of ladder, AC = height of window 
 
 In ΔABC, (AB)² = (AC)²+(BC)²
⇒ (34)²  =(16)²+BC² 
or  
BC² = (34)²−(16)² ⇒ BC² = 1156−256 = 900 
∴ BC = √900 = 30m

ABCD is a rectangle, 

In ΔACB, AC² = AB²+BC²  (By Pythagoras theorem)
(35)² = (28)²+BC² or  BC² = (35)²−(28)²
⇒ BC² = 1225−784 ⇒ BC² = 441
∴ BC==21cm
∴  Perimeter of rectangle 
= 2×(28+21)cm = 2×(49)cm = 98cm