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RD Chapter 11- Co-ordinate Geometry Ex-VSAQS Interview Questions Answers

Question 1 : Define a triangle.-

Answer 1 : A figure bounded by three lines segments in a plane is called a triangle.

Question 2 : Write the sum of the angles of an obtuse triangle..

Answer 2 : The sum of angles of an obtuse triangle is 180°.

Question 3 : In ∆ABC, if ∠B = 60°, ∠C = 80° and the bisectors of angles ∠ABC and ∠ACB meet at a point O, then find the measure of ∠BOC.

Answer 3 :

In ∆ABC, ∠B = 60°, ∠C = 80°
OB and OC are the bisectors of ∠B and ∠C
∵ ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 60° + 80° = 180°
⇒ ∠A + 140° = 180°
∴ ∠A = 180°- 140° = 40°
 
= 90° + – x 40° = 90° + 20° = 110°

Question 4 : If the angles of a triangle are in the ratio 2:1:3. Then find the measure of smallest angle.

Answer 4 :

Sum of angles of a triangle = 180°
Ratio in the angles = 2 : 1 : 3
Let first angle = 2x
Second angle = x
and third angle = 3x
∴ 2x + x + 3x = 180° ⇒ 6x = 180°
∴ x == 30°
∴ First angle = 2x = 2 x 30° = 60°
Second angle = x = 30°
and third angle = 3x = 3 x 30° = 90°
Hence angles are 60°, 30°, 90°

Question 5 : State exterior angle theorem.

Answer 5 :

Given : In ∆ABC, side BC is produced to D
 
To prove : ∠ACD = ∠A + ∠B
Proof: In ∆ABC,
∠A + ∠B + ∠ACB = 180° …(i) (Sum of angles of a triangle)
and ∠ACD + ∠ACB = 180° …(ii) (Linear pair)
From (i) and (ii)
∠ACD + ∠ACB = ∠A + ∠B + ∠ACB
∠ACD = ∠A + ∠B
Hence proved.

Question 6 : The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.

Answer 6 :

In ∆ABC,
∠A + ∠C = ∠B
 
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
∴ ∠B + ∠A + ∠C = 180°
⇒ ∠B + ∠B = 180°
⇒ 2∠B = 180°
⇒ ∠B = = 90°
∴ Third angle = 90°

Question 7 : In the figure, if AB || CD, EF || BC, ∠BAC = 65° and ∠DHF = 35°, find ∠AGH.

Answer 7 :

Given : In figure, AB || CD, EF || BC ∠BAC = 65°, ∠DHF = 35°
 
∵ EF || BC
∴ ∠A = ∠ACH (Alternate angle)
∴ ∠ACH = 65°
∵∠GHC = ∠DHF
(Vertically opposite angles)
∴ ∠GHC = 35°
Now in ∆GCH,
Ext. ∠AGH = ∠GCH + ∠GHC
= 65° + 35° = 100°

Question 8 : In the figure, if AB || DE and BD || FG such that ∠FGH = 125° and ∠B = 55°, find x and y.

Answer 8 :

In the figure, AB || DF, BD || FG
 
∠FGH = 125° and ∠B = 55°
∠FGH + FGE = 180° (Linear pair)
⇒ 125° + y – 180°
⇒ y= 180°- 125° = 55°
∵ BA || FD and BD || FG
∠B = ∠F = 55°
Now in ∆EFG,
∠F + ∠FEG + ∠FGE = 180°
(Angles of a triangle)
⇒ 55° + x + 55° = 180°
⇒ x+ 110°= 180°
∴ x= 180°- 110° = 70°
Hence x = 70, y = 55°

Question 9 : If the angles A, B and C of ∆ABC satisfy the relation B – A = C – B, then find the measure of ∠B.

Answer 9 :

In ∆ABC,
∠A + ∠B + ∠C= 180° …(i)
and B – A = C – B
 
⇒ B + B = A + C ⇒ 2B = A + C
From (i),
B + 2B = 180° ⇒ 3B = 180°
∠B =  = 60°
Hence ∠B = 60°

Question 10 : In ∆ABC, if bisectors of ∠ABC and ∠ACB intersect at O at angle of 120°, then find the measure of ∠A.

Answer 10 :

In ∆ABC, bisectors of ∠B and ∠C intersect at O and ∠BOC = 120°
 
But ∠BOC = 90°+  
90°+  ∠A= 120°
⇒   ∠A= 120°-90° = 30°
∴ ∠A = 2 x 30° = 60°


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