Triangles and

Q 1 – What is the difference between a triangle and triangular region ?

The triangle is defined as a plane figure which is formed by three line segments which are non-parallel whereas a triangular region is the region which is inside a triangle.

Image result for Q 1 – What is the difference between a triangle and triangular region ?

Q 2 – If the angles of a triangle are in the ratio 1: 2: 3 determine three angles.  

Ratio in three angles of a triangle = 1 : 2 : 3

Let first angle = x

Then second angle = 2x

and third angle = 3x

∴ x+2x + 3x = 180º ( sum of angles of a triangle)

⇒6x = 180º ⇒ x 180º/6 =30º

∴ First angle = x=30º

Second angle = 2x = 2 × 30º =60º

and third angle = 3x = 3 × 30º=90º

∴ Angles are 30º,60º,90º

Q 3 – Two angles of a triangle are measures 75  and 35  find the measures of the third angle.

→ ∠ A =75

→ ∠ B = 35

→ ∠ c = x

As we know that Sum of all the angles of a triangle is 180°.So ,

→ ∠ A +∠ B +∠ c = 180

→ 75 + 35 + x =  180

→ 110 + x = 180

→ x = 180 – 110

→ x = 70

      

Q 4 – Each of the two equals angles of a triangle is twice the third angle. Find the angles of the triangles.    

As per the fact, the sum of all angles of triangle is 180°. Also, the two angles of isosceles triangle are equal.

Let the angle be x. So, the mathematical expression will be –

2x + 2x + x = 180°

5x = 180°

x = 36°

The two same angles of is osceles triangle = 2x

Two same angles = 2*36°

Two same angles = 72°

Angle of isosceles triangle = x

Angle of isosceles triangle = 36°

Thus, angles of is osceles triangle are 36°, 72° and 72°

Q 5 – One of the angles of a triangle is 130  and the other two angles are equals. What is the measures of each of those equal angles ?

Ans. Given one of the angles of a triangle is 130o

Also given that remaining two angles are equal

So let the second and third angle be x

We know that sum of all the angles of a triangle = 180o

130o + x + x = 180o

130o + 2x = 180o

2x = 180o – 130o

2x = 50o

x = 50/2

x = 25o

Therefore the two other angles are 25o each

Q 6 – The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10o. Find the three angles.

. Given that angles of a triangle are arranged in ascending order of magnitude

Also given that difference between two consecutive angles is 10o

Let the first angle be x

Second angle be x + 10o

Third angle be x + 10o + 10o

We know that sum of all the angles of a triangle = 180o

x + x + 10o + x + 10o +10o = 180o

3x + 30 = 180

3x = 180 – 30

3x = 150

x = 150/3

x = 50o

First angle is 50o

Second angle x + 10o = 50 + 10 = 60o

Third angle x + 10o +10o = 50 + 10 + 10 = 70o

Q 7 – If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.                    

Given that one angle of a triangle is equal to the sum of the other two

Let the measure of angles be x, y, z

Therefore we can write above statement as x = y + z

x + y + z = 180o

Substituting the above value we get

x + x = 180o

2x = 180o

x = 180/2

x = 90o

Q 8 – If one angle is 90o then the given triangle is a right angled triangle If one angle of a triangle is 60o and the other two angles are in the ratio 1: 2, find the angles.     

Given that one of the angles of the given triangle is 60o.

Also given that the other two angles of the triangle are in the ratio 1: 2.

Let one of the other two angles be x.

Therefore, the second one will be 2x.

We know that the sum of all the three angles of a triangle is equal to 180o.

60o + x + 2x = 180o

3x = 180o – 60o

3x = 120o

x = 120o/3

x = 40o

2x = 2 × 40o

2x = 80o

Hence, we can conclude that the required angles are 40o and 80o.

Q 9 – In △ABC, ∠A = 100 , AD bisects ∠A and AD  BC. Find ∠B.

Q 10 – In △ABC, if 3∠A = 4∠B = 6∠C, calculate the angles.      

. We know that for the given triangle, 3∠A = 6∠C

∠A = 2∠C ……. (i)

We also know that for the same triangle, 4∠B = 6∠C

∠B = (6/4) ∠C …… (ii)

We know that the sum of all three angles of a triangle is 180o.

Therefore, we can say that:

∠A + ∠B + ∠C = 180o (Angles of △ABC)…… (iii)

On putting the values of ∠A and ∠B in equation (iii), we get:

2∠C + (6/4) ∠C + ∠C = 180o

(18/4) ∠C = 180o

∠C = 40o

From equation (i), we have:

∠A = 2∠C = 2 × 40

∠A = 80o

From equation (ii), we have:

∠B = (6/4) ∠C = (6/4) × 40o

∠B = 60o

∠A = 80o, ∠B = 60o, ∠C = 40o

Therefore, the three angles of the given triangle are 80o, 60o, and 40o

Q 11 – In Fig., ∆ABC is right angled at A, Q and R are points on line BC and P is a point such that QP ∥ AC and RP ∥ AB. Find ∠P

 

In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.

∠QCA = ∠CQP (Alternate interior angles)

Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,

∠ABC = ∠PRQ (alternate interior angles).

We know that the sum of all three angles of a triangle is 180o.

Hence, for ∆ABC, we can say that:

∠ABC + ∠ACB + ∠BAC = 180o

∠ABC + ∠ACB + 90o = 180o (Right angled at A)

∠ABC + ∠ACB = 90o

Using the same logic for △PQR, we can say that:

∠PQR + ∠PRQ + ∠QPR = 180o

∠ABC + ∠ACB + ∠QPR = 180o (∠ACB = ∠PQR and ∠ABC = ∠PRQ)

Or,

90o+ ∠QPR =180o (∠ABC+ ∠ACB = 90o)

∠QPR = 90o

Q 12 – In the six cornered figure, AC, AD and AE are joined. Find ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA.    


 Therefore in ∆ABC, we have

∠CAB + ∠ABC + ∠BCA = 180° …….. (i)

In△ ACD, we have

∠DAC + ∠ACD + ∠CDA = 180° …….. (ii)

In △ADE, we have

∠EAD + ∠ADE + ∠DEA =180° ………. (iii)

In △AEF, we have

∠FAE + ∠AEF + ∠EFA = 180° ………. (iv)

Adding (i), (ii), (iii), (iv) we get

∠CAB + ∠ABC + ∠BCA + ∠DAC + ∠ACD + ∠CDA + ∠EAD +

∠ADE + ∠DEA + ∠FAE +

∠AEF +∠EFA = 720°

Therefore ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA =

720°

Q 13 – One of the exterior angles of a triangle is 80  and the interior opposite angles are on the ratio 3 : 5 find the angles of the triangle.        

Given: Exterior angle =80∘

Ratio of opposite interior angles =3:5

Let the interior angles be 3x and 5x

⇒80=3x+5x [Exterior angle is equal to the sum of interior opposite angles.]

⇒80=8x

⇒x=10

The two interior angles are 3×10=30∘, and 5×10=50∘

The other angle =180−(30+50)=100 [Angle sum Property]

Hence, all three angles of the given triangle are 30∘,50∘,100∘

Q 14 – Two poles of height 6m and 11m stands on a plane ground, if the distance between their feet is 12m. Find the distance between their tops. (Hint: Find the hypotenuse of a right triangle having the sides (11 – 6) m = 5 and 12 m)

Let AB and CD be the poles of height 6m and 11m.

Therefore, CP = 11 – 6 = 5m

From the figure, it can be observed that AP = 12m

By Pythagoras theorem for ΔAPC, we get,

AP2 = PC2 + AC2

(12m)2 + (5m)2 = (AC)2

AC2 = (144+25) m2 = 169 m2

AC = 13m

Therefore, the distance between their tops is 13 m.

Q 15 – In Fig. AD and CF are respectively perpendiculars to sides BC and AB of  ABC. If ∠FCD = 50  , Find ∠BAD.          


As we know the total of all three angle of a triangle = 180

 FCB

∠ FCB + ∠ CBF + ∠ BFC = 180  

50  +  ∠ CBF + 90  =180

∠ CBF = 180  – 50  – 90  = 40

Applying  same way for ABD,

∠ ABD + ∠ BDA + ∠ BAD = 180

∠ BAD =         180 – 90  – 40 = 50

Q 16 – An exterior angle of a triangle is 110  and one of the interior opposite angles is 30 . Find the other two angles of the triangles.

In the diagram, PQR is a Triangle.

∴ ∠a + 110° = 180° ❪Linear pair❫

⇒ ∠a = 180° – 110°

⇒ ∠a = 70°

Hence, ∠a = 70°.

∴ ∠P + ∠Q + ∠R = 180°

❪Sum of ∠s of a Triangle = 180°❫

⇒ ∠P + 30° + ∠a = 180°

⇒ ∠P + 30° + 70° = 180°

⇒ ∠P + 100° = 180°

⇒ ∠P = 180° – 100°

⇒ ∠P = 80°

Hence, ∠P = 80°.

Therefore, the other two angles of the triangle are 70° and 80°.


Q 17 – In a ABC, AD is the altitude from A such that AD = 12 cm, BD = 9 cm and DC = 16 cm. Examine if ABC is right angled at A.

Image result for Q 17 – In a ABC, AD is the altitude from A such that AD = 12 cm, BD = 9 cm and DC = 16 cm. Examine if ABC is right angled at A.

Consider △ADC,

∠ADC = 90o (AD is an altitude on BC) Using the Pythagoras theorem, we get

122 + 162 = AC2
AC2 = 144 + 256
= 400
AC = 20 cm

Again consider △ADB,

∠ADB = 90o (AD is an altitude on BC) Using the Pythagoras theorem, we get

122 + 92 = AB2
AB2 = 144 + 81 = 225
AB = 15 cm
Consider △ABC,

BC2 = 252 = 625
AB2 + AC2 = 152 + 202 = 625
AB2 + AC2 = BC2

Because it satisfies the Pythagoras theorem, therefore △ABC is right angled at A.

Q 18 – One of the exterior angles of a triangle is 80  and the interior opposite angles are equal to each other. What is the measure of these two angles?       

Let us assume that A and B are the two interior opposite angles.

We know that ∠A is equal to ∠B.

We also know that the sum of interior opposite angles is equal to the exterior angle.

Therefore from the figure we have,

∠A + ∠B = 80

∠A +∠A = 80 (because ∠A = ∠B)

2∠A = 80

∠A = 40/2 =40

∠A= ∠B = 40

Thus, each of the required angles is of 40.

Q 19 – A student when asked to measure two exterior angles of ABC observed that the exterior angles at A and B are of 103  and 74 respectively. Is this possible?  Why or why not?            

We know that sum of internal and external angle is equal to 180

Internal angle at A + External angle at A = 180

Internal angle at A + 103 =180

Internal angle at A = 77

Internal angle at B + External angle at B = 180

Internal angle at B + 74 = 180

Internal angle at B = 106

Sum of internal angles at A and B = 77 + 106 = 183

It means that the sum of internal angles at A and B is greater than 180, which cannot be possible.

Back