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

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

In the given figure, ∠A can be calculated by using the concept of exterior angles.
Since ∠B and ∠C are both exterior angles measuring 120º, we can calculate the interior angles
at B and C by subtracting each exterior angle from 180°.
So, the interior angle at B and C is:
180° – 120° = 60°
Since the sum of the interior angles in a triangle is 180°, we can find ∠A as follows:
∠A = 180° – 60° – 60° = 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

Sum of interior opposite angles = Exterior angle

or 50° + x = 110°

or x = 60°

In given ΔABC, ∠C = 110°
Let ∠A = ∠B = x° (Angle opposite to equal sides of a triangle are equal)
x + x + 110° = 180°
⇒ 2x + 110° = 180°
⇒ 2x = 180° – 110°
⇒ 2x = 70°
⇒ x = 35°
Thus, ∠A = ∠B = 35°

(i) PQ = 5 cm, PR = 6 cm and QR = 7 cm
PQ ≠ PR ≠ QR
Thus, ∆PQR is a scalene triangle.
(ii) AB = 4 cm, AC = 4 cm
AB = AC
Thus, ∆ABC is an isosceles triangle.
(iii) MN = 3 cm, ML = 3 cm and NL = 3 cm
MN = ML = NL
Thus, ∆MNL is an equilateral triangle.

Using the exterior angle property:
Exterior Angle = Sum of Interior Opposite Angles
130^∘= 45^∘+ Other Angle
Other Angle = 130^∘− 45^∘= 85^∘

∠PRS = ∠RPQ + ∠PQR (By exterior angle property)
⇒ 115º = 45º + ∠PQR
⇒ ∠PQR = 115º - 45º ⇒ ∠PQR = 70º

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