RS Agarwal Solution | Class 6th | Chapter-16 | Triangles | Edugrown

Page No 196:

Exercise 16A

Question 1:

Take three noncollinear points A, B and C on a page of your notebook. Join AB, BC and CA. What figure do you get?
Name: (i) the side opposite to ∠C
           (ii) the angle opposite to the side BC
           (iii) the vertex opposite to the side CA
           (iv) the side opposite to the vertex B
Figure

ANSWER:

We get a triangle by joining the three non-collinear points A, B, and C.
(i) The side opposite to ∠C is AB.
(ii) The angle opposite to the side BC is ∠A.
(iii) The vertex opposite to the side CA is B.
(iv) The side opposite to the vertex B is AC.

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Question 2:

The measures of two angles of a triangle are 72° and 58°. Find the measure of the third angle.

ANSWER:

The measures of two angles of a triangle are 72° and 58°. 
Let the third angle be x. 
Now, the sum of the measures of all the angles of a triangle is 180^o​.
 ∴∴    x + 72^o + 58^o = 180^o
    ⇒ x + 130^o = 180^o                   
    ⇒ x = 180^o​ −- 130^o
    ⇒ x = 50^o
​The measure of the third angle of the triangle is 50^o​.

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Question 3:

The angles of a triangle are in the ratio 1 : 3 : 5. Find the measure of each of the angles.

ANSWER:

The angles of a triangle are in the ratio 1:3:5. 
Let the measures of the angles of the triangle be (1x), (3x) and (5x)
Sum of the measures of the angles of the triangle = 180^o
      ∴ 1x + 3x + 5x = 180^o
        ⇒ 9x = 180^o
        ⇒ x = 20^o
 1x = 20^o
3x = 60^o
​5x = 100^o
The measures of the angles are 20^o, 60^o and 100^o. 

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Question 4:

One of the acute angles of a right triangle is 50°. Find the other acute angle.

ANSWER:

In a right angle triangle, one of the angles is 90^o.
It is given that one of the acute angled of the right angled triangle is 50^o.
We know that the sum of the measures of all the angles of a triangle is 180^o.
Now, let the third angle be x.
​Therefore, we have:
            90^o​ + 50^o + x = 180^o
   ⇒        140^o + x = 180^o
​   ⇒                    x = 180^o −- 140^o
  ⇒                     x =  40^o
​ The third acute angle is 40^o​.

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Question 5:

One of the angles of a triangle is 110° and the other two angles are equal. What is the measure of each of these equal angles?

ANSWER:

Given:
∠A = 110^o and ​∠B = ∠C
Now, the sum of the measures of all the angles of a traingle is 180^o .
              ∠A + ∠B + ∠C = 180^o
      ⇒    110^o + ​∠B + ∠B = 180^o
​      ⇒    110^o  + 2​∠B = 180^o
​       ⇒                2​∠B = 180^o −- 110^o
       ⇒                2∠B =  70^o
      ⇒                  ∠B = 70^o / 2
      ⇒                  ∠B = 35^o

      ∴ ​∠C = 35^o
The measures of the three angles:
∠A = 110^o, ∠B = 35^o, ​∠C = 35^o

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Question 6:

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

ANSWER:

Given:
 ∠A = ∠B + ∠C
​We know:
       ∠A + ∠B + ∠C = 180^o
    ⇒ ∠B +∠C + ∠B + ∠C = 180^o
​    ⇒ 2∠B + 2∠C = 180^o
​    ⇒ 2(∠B +∠C) = 180^o
​    ⇒ ∠B + ∠C = 180/2
    ⇒ ​∠B + ∠C = 90^o
∴∴ ∠A = 90^o
This shows that the triangle is a right angled triangle.

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Question 7:

In a ∆ABC, if 3∠A = 4 ∠B = 6 ∠C, calculate the angles.

ANSWER:

Let 3∠A = 4 ∠B = 6 ∠C = x
Then, we have:
 ∠A = x3, ∠B = x4, ∠C = x6But, ∠A + ∠B + ∠C = 180°∴ x3 + x4 + x6 = 180°or 4x + 3x + 2×12 = 180°or 9x = 180° × 12 = 2160°or x = 240° ∴ ∠A = 2403 = 80°, ∠B = 2404 = 60°, ∠C = 2406 = 40°∠A = x3, ∠B = x4, ∠C = x6But, ∠A + ∠B + ∠C = 180°∴ x3 + x4 + x6 = 180°or 4x + 3x + 2×12 = 180°or 9x = 180° × 12 = 2160°or x = 240° ∴ ∠A = 2403 = 80°, ∠B = 2404 = 60°, ∠C = 2406 = 40°

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Question 8:

Look at the figures given below. State for each triangle whether it is acute, right or obtuse.
Figure

ANSWER:

(i) It is an obtuse angle triangle as one angle is 130^o, which is greater than 90^o.

(ii) It is an acute angle triangle as all the angles in it are less than 90^o.

(iii) It is a right angle triangle as one angle is 90^o.

(iv) It is an obtuse angle triangle as one angle is 92^o, which is greater than 90^o.

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Question 9:

In the given figure some triangles have been given. State for each triangle whether it is scalene, isosceles or equilateral.
Figure

ANSWER:

Equilateral Triangle: A triangle whose all three sides are equal in length and each of the three angles measures 60^o.
Isosceles Triangle: A triangle whose two sides are equal in length and the angles opposite them are equal to each other.
Scalene Triangle: A triangle whose all three sides and angles are unequal in measure.

(i) Isosceles
   AC = CB = 2 cm
(ii) Isosceles
   DE = EF = 2.4 cm
(iii) Scalene
   All the sides are unequal.
(iv) Equilateral
    XY = YZ = ZX = 3 cm
(v) Equilateral
     All three angles are 60^o.
(vi) Isosceles
     Two angles are equal in measure.
(vii) Scalene
    All the angles are unequal.

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Question 10:

Draw a ∆ABC. Take a point D on BC. Join AD. How many triangles do you get? Name them.
Figure

ANSWER:

In ∆ABC, if we take a point D on BC, then we get three triangles, namely ∆ADB, ∆ADC and ∆ABC.

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Question 11:

Can a triangle have
(i) two right angles?
(ii) two obtuse angles?
(iii) two acute angles?
(iv) each angle more than 60°?
(v) each angle less than 60°?
(vi) each angle equal to 60°?

ANSWER:

(i) No
     If the two angles are 90^o each, then the sum of two angles of a triangle will be 180^o​, which is not possible.
(ii) No
      For example, let the two angles be 120^o and 150^o. Then, their sum will be 270^o​, which cannot form a triangle.
(iii) Yes
       For example, let the two angles be 50^o and 60^o​, which on adding, gives 110^o. They can easily form a triangle whose third angle is 180^o −- 110^o = 70^o​.
(iv) No
      For example, let the two angles be 70^o​ and 80^o, which on adding, gives 150^o. They cannot form a triangle whose third angle is 180^o​ −- 150^o = 30^o, which is less than 60^o.
(v) No
      For example, let the two angles be 50^o and 40^o, which on adding, gives 90^o . Thus, they cannot form a triangle whose third angle is 180^o −- 90^o = 90^o​, which is greater than 60^o.
(vi) Yes
      Sum of all angles = 60^o + 60^o + 60^o​ = 180^o

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Question 12:

Fill in the blanks.
(i) A triangle has …… sides, …… angles and …… vertices.
(ii) The sum of the angles of a triangle is …… .
(iii) The sides of a scalene triangle are of ……. lengths.
(iv) Each angle of an equilateral triangle measures …… .
(v) The angles opposite to equal sides of an isosceles triangle are ……. .
(vi) The sum of the lengths of the sides of a triangle is called its ………. .

ANSWER:

(i) A triangle has 3 sides 3 angles and 3 vertices.
(ii) The sum of the angles of a triangle is 180^o. 
(iii) The sides of a scalene triangle are of different lengths.
(iv) Each angle of an equilateral triangle measures 60^o.
(v) The angles opposite to equal sides of an isosceles triangle are equal.
(vi) The sum of the lengths of the sides of a triangle is called its perimeter.

Page No 197:

Exercise 16B

Question 1:

How many parts does a triangle have?
(a) 2
(b) 3
(c) 6
(d) 9

ANSWER:

Correct option: (c)
A triangle has 6 parts: three sides and three angles.

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Question 2:

With the angles given below, in which case the construction of triangle is possible?
(a) 30°, 60°, 70°
(b) 50°, 70°, 60°
(c) 40°, 80°, 65°
(d) 72°, 28°, 90°

ANSWER:

Correct option: (b) 
(a) Sum = 30° + 60° + 70° = 160^o
     This is not equal to the sum of all the angles of a triangle. 
(b) Sum = 50° + 70° + 60° = 180^o
     Hence, it is possible to construct a triangle with these angles.
(c) Sum = 40° + 80° + 65° = 185^o
      This is not equal to the sum of all the angles of a triangle.
(d) Sum = 72° + 28° + 90° = 190^o​
     This is not equal to the sum of all the angles of a triangle.

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Question 3:

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle is
(a) 60°
(b) 80°
(c) 76°
(d) 84°

ANSWER:

(b) 80^o
Let the measures of the given angles be (2x)^o, (3x)^o​ and (4x)^o.
∴∴ (2x)^o + (3x)^o + (4x)^o = 180^o
  ⇒ (9x)^o = 180^o
​  ⇒ x = 180 / 9
  ⇒ x = 20^o
∴∴ ​2x =  40^o, 3x = 60^o, 4x = 80^o

Hence, the measures of the angles of the triangle are 40^o​, 60^o, 80^o.
Thus, the largest angle is 80^o.

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Question 4:

The two angles of a triangle are complementary. The third angle is
(a) 60°
(b) 45°
(c) 36°
(d) 90°

ANSWER:

Correct option: (d)
The measure of two angles are complimentary if their sum is 90^o degrees. 
Let the two angles be x and y, such that x + y = 90^o .
Let the third angle be z.
Now, we know that the sum of all the angles of a triangle is 180^o​.
   x + y + z​ = 180^o
 ⇒ 90^o + z = 180^o
 ⇒ z  = 180^o −- 90^o 
        = 90^o
​The third angle is 90^o.

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Question 5:

One of the base angles of an isosceles triangle is 70°. The vertical angle is
(a) 60°
(b) 80°
(c) 40°
(d) 35°

ANSWER:

Correct option: (c)
Let ∠A = 70^o
The triangle is an isosceles triangle.
We know that the angles opposite to the equal sides of an isosceles triangle are equal.
∴∴ ​∠B = 70^o
​We need to find the vertical angle ​∠C.
Now, sum of all the angles of a triangle is 180^o.
    ∠A + ∠B + ∠C = 180^o
⇒​ 70^o + 70^o​ + ∠C = 180^o
⇒ 140^o + ​∠C = 180^o
⇒ ∠C = 180^o −- 140^o
⇒ ∠C = 40^o

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Question 6:

A triangle having sides of different lengths is called
(a) an isosceles triangle
(b) an equilateral triangle
(c) a scalene triangle
(d) a right triangle

ANSWER:

Correct option: (c)
A triangle having sides of different lengths is called a scalene triangle.

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Question 7:

In an isosceles ∆ABC, the bisectors of ∠B and ∠C meet at a point O. If ∠A = 40°, then ∠BOC =
?
(a) 110°
(b) 70°
(c) 130°
(d) 150°

ANSWER:

Correct option: (a)

In the isosceles ABC, ​the bisectors of ∠B and ∠C meet at point O.
Since the triangle is isosceles, the angles opposite to the equal sides are equal.
∠B = ∠C
∴∴ ∠A + ∠B + ∠C = 180^o
  ⇒  40^o + 2∠B = 180^o
  ⇒ 2∠B = 140^o
  ⇒ ∠B = 70^o
Bisectors of an angle divide the angle into two equal angles.
So, in  ∆BOC:
∠OBC = 35^o and ∠OCB = 35^o
∠BOC + ∠OBC + ∠OCB = 180​^o
  ⇒ ∠BOC + 35^o + 35^o = 180^o
  ⇒ ∠BOC = 180^o​ – 70^o
  ⇒ ∠BOC = 110^o

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Question 8:

The sides of a triangle are in the ratio 3 : 2 : 5 and its perimeter is 30 cm. The length of the longest side is
(a) 20 cm
(b) 15 cm
(c) 10 cm
(d) 12 cm

ANSWER:

Correct option: (b)
The sides of a triangle are in the ratio 3:2:5.
Let the lengths of the sides of the triangle be (3x), (2x), (5x).
We know:
 Sum of the lengths of the sides of a triangle = Perimeter
   (3x) + (2x) + (5x) = 30
  ⇒ 10x = 30
  ⇒ x =  30 
            10
  ⇒ x = 3
First side = 3x = 9 cm
Second side = 2x = 6 cm
Third side = 5x = 15 cm
The length of the longest side is 15 cm.

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Question 9:

Two angles of a triangle measure 30° and 25° respectively. The measure of the third angle is
(a) 35°
(b) 45°
(c) 65°
(d) 125°

ANSWER:

Correct option: (d)
Two angles of a triangle measure 30° and 25°, respectively.
  Let the third angle be x.
   x + 30^o + 25^o = 180^o
​                    x = 180^o −- 55^o
                     x = 125^o

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Question 10:

Each angle of an equilateral triangle measures
(a) 30°
(b) 45°
(c) 60°
(d) 80°

ANSWER:

Correct option: (c)
Each angle of an equilateral triangle measures 60^o.

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Question 11:

In the adjoining figure, the point P lies
(a) in the interior of ∆ABC
(b) in the exterior of ∆ABC
(c) on ∆ABC
(d) outside ∆ABC
Figure

ANSWER:

Correct option: (c)
Point P lies on ∆ABC.