Important Questions: The Triangle and Its Properties - Class 7 MCQ

∠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

Explanation: In an isosceles triangle, the angles opposite the equal sides are also equal. Since AB = AC in triangle ABC, the angles opposite these sides, which are ∠B and ∠C, must be equal. Additionally, AD being the altitude means it is also the angle bisector, further ensuring that ∠B = ∠C

∠A – 180° – [(180° – 120°) + (180° – 120°)] = 60°.

∠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

∠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

∠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

Sol: Let, ∠ABC = 3x and ∠ACB = 2x

We know, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ ∠ACD = ∠ABC + ∠ACB

∴ 100° = 3x + 2x ∴ 100° = 5x

∴ x = 100°/5 ∴ x = 20°

→ ∠ABC = 3x = 3 × 20° = 60° → ∠ACB = 2x = 2 × 20° = 40°