Page No 196:
Table of Contents
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 180o.
∴∴ x + 72o + 58o = 180o
⇒ x + 130o = 180o
⇒ x = 180o −- 130o
⇒ x = 50o
The measure of the third angle of the triangle is 50o.
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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 = 180o
∴ 1x + 3x + 5x = 180o
⇒ 9x = 180o
⇒ x = 20o
1x = 20o
3x = 60o
5x = 100o
The measures of the angles are 20o, 60o and 100o.
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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 90o.
It is given that one of the acute angled of the right angled triangle is 50o.
We know that the sum of the measures of all the angles of a triangle is 180o.
Now, let the third angle be x.
Therefore, we have:
90o + 50o + x = 180o
⇒ 140o + x = 180o
⇒ x = 180o −- 140o
⇒ x = 40o
The third acute angle is 40o.
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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 = 110o and ∠B = ∠C
Now, the sum of the measures of all the angles of a traingle is 180o .
∠A + ∠B + ∠C = 180o
⇒ 110o + ∠B + ∠B = 180o
⇒ 110o + 2∠B = 180o
⇒ 2∠B = 180o −- 110o
⇒ 2∠B = 70o
⇒ ∠B = 70o / 2
⇒ ∠B = 35o
∴ ∠C = 35o
The measures of the three angles:
∠A = 110o, ∠B = 35o, ∠C = 35o
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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 = 180o
⇒ ∠B +∠C + ∠B + ∠C = 180o
⇒ 2∠B + 2∠C = 180o
⇒ 2(∠B +∠C) = 180o
⇒ ∠B + ∠C = 180/2
⇒ ∠B + ∠C = 90o
∴∴ ∠A = 90o
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 130o, which is greater than 90o.
(ii) It is an acute angle triangle as all the angles in it are less than 90o.
(iii) It is a right angle triangle as one angle is 90o.
(iv) It is an obtuse angle triangle as one angle is 92o, which is greater than 90o.
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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 60o.
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 60o.
(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 90o each, then the sum of two angles of a triangle will be 180o, which is not possible.
(ii) No
For example, let the two angles be 120o and 150o. Then, their sum will be 270o, which cannot form a triangle.
(iii) Yes
For example, let the two angles be 50o and 60o, which on adding, gives 110o. They can easily form a triangle whose third angle is 180o −- 110o = 70o.
(iv) No
For example, let the two angles be 70o and 80o, which on adding, gives 150o. They cannot form a triangle whose third angle is 180o −- 150o = 30o, which is less than 60o.
(v) No
For example, let the two angles be 50o and 40o, which on adding, gives 90o . Thus, they cannot form a triangle whose third angle is 180o −- 90o = 90o, which is greater than 60o.
(vi) Yes
Sum of all angles = 60o + 60o + 60o = 180o
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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 180o.
(iii) The sides of a scalene triangle are of different lengths.
(iv) Each angle of an equilateral triangle measures 60o.
(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.
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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° = 160o
This is not equal to the sum of all the angles of a triangle.
(b) Sum = 50° + 70° + 60° = 180o
Hence, it is possible to construct a triangle with these angles.
(c) Sum = 40° + 80° + 65° = 185o
This is not equal to the sum of all the angles of a triangle.
(d) Sum = 72° + 28° + 90° = 190o
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) 80o
Let the measures of the given angles be (2x)o, (3x)o and (4x)o.
∴∴ (2x)o + (3x)o + (4x)o = 180o
⇒ (9x)o = 180o
⇒ x = 180 / 9
⇒ x = 20o
∴∴ 2x = 40o, 3x = 60o, 4x = 80o
Hence, the measures of the angles of the triangle are 40o, 60o, 80o.
Thus, the largest angle is 80o.
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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 90o degrees.
Let the two angles be x and y, such that x + y = 90o .
Let the third angle be z.
Now, we know that the sum of all the angles of a triangle is 180o.
x + y + z = 180o
⇒ 90o + z = 180o
⇒ z = 180o −- 90o
= 90o
The third angle is 90o.
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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 = 70o
The triangle is an isosceles triangle.
We know that the angles opposite to the equal sides of an isosceles triangle are equal.
∴∴ ∠B = 70o
We need to find the vertical angle ∠C.
Now, sum of all the angles of a triangle is 180o.
∠A + ∠B + ∠C = 180o
⇒ 70o + 70o + ∠C = 180o
⇒ 140o + ∠C = 180o
⇒ ∠C = 180o −- 140o
⇒ ∠C = 40o
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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 = 180o
⇒ 40o + 2∠B = 180o
⇒ 2∠B = 140o
⇒ ∠B = 70o
Bisectors of an angle divide the angle into two equal angles.
So, in ∆BOC:
∠OBC = 35o and ∠OCB = 35o
∠BOC + ∠OBC + ∠OCB = 180o
⇒ ∠BOC + 35o + 35o = 180o
⇒ ∠BOC = 180o – 70o
⇒ ∠BOC = 110o
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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 + 30o + 25o = 180o
x = 180o −- 55o
x = 125o
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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 60o.
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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.
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