**Exercise 15.1**

** **

**Question :1. Take three non-collinear points A. B and C on a page of your notebook. Join AB, BC and CA. What figure do you get? Name the triangle. Also, name**

**(i) The side opposite to ∠B**

**(ii) The angle opposite to side AB**

**(iii) The vertex opposite to side BC**

**(iv) The side opposite to vertex B.**** **

**Solution 1:**

(i) The side opposite to ∠B = AC

(ii) The angle opposite to side AB = ∠C

(iii) The vertex opposite to side BC = A

(iv) The side opposite to vertex B = AC

** **

**Question :2. Take three collinear points A, B and C on a page of your note book. Join AB. BC and CA. Is the figure a triangle? If not, why?**** **

**Solution 2:**

The figure isn't a triangle at all. A triangle is a plane figure formed by three non-parallel line segments, according to description.

**Question :3. Distinguish between a triangle and its triangular region.**** **

**Solution 3:**

Triangle:

Three non- parallel line segments form a triangle, which is a plane figure.

Triangular region:

Its triangular area, on the other hand, includes the interior of the triangle as well as the triangle itself.** **

**Question :4. D is a point on side BC of a △CAD is joined. Name all the triangles that you can observe in the figure. How many are they?**

**Solution 4:**

The following three triangles can be seen in the diagram.

△ ABC △ ACD △ ADB

** **

**Question :5. A, B, C and D are four points, and no three points are collinear. AC and BD intersect at O. There are eight triangles that you can observe. Name all the triangles**

**Solution 5:**

Given A, B, C and D are four points, and no three points are collinear

△ ABC △ ACD △ DBC

△ ABD △ AOB △ BOC

△ COD △ AOD

** **

**Question :6. What is the difference between a triangle and triangular region? **

**Solution 6:**

Triangle:

Three non- parallel line segments form a triangle, which is a plane figure.

Triangular region:

Its triangular area, on the other hand, includes the interior of the triangle as well as the triangle itself.

**Question :7. Explain the following terms:**

**(i) Triangle**

**(ii) Parts or elements of a triangle**

**(iii) Scalene triangle**

**(iv) Isosceles triangle**

**(v) Equilateral triangle**

**(vi) Acute triangle**

**(vii) Right triangle**

**(viii) Obtuse triangle**

**(ix) Interior of a triangle**

**(x) Exterior of a triangle**** **

**Solution 7:**

(i) Triangle:

Three non- parallel line segments form a triangle, which is a plane figure.

(ii) Parts or elements of a triangle:

The three sides and the three angles of a triangle are together known as the parts or elements of that triangle.

(iii) Scalene triangle:

A scalene triangle is a triangle in which no two sides are equal.

(iv) Isosceles triangle:

An isosceles triangle is a triangle in which two sides are equal.

(v) Equilateral triangle:

An equilateral triangle is a triangle in which all three sides are equal.

(vi) Acute triangle:

An acute triangle is a triangle in which all the angles are less than 90^{o}.

(vii) Right triangle:

A right-angled triangle is a triangle in which one angle should be equal to 90^{o}.

(viii) Obtuse triangle:

An obtuse triangle is a triangle in which one angle is more than 90^{o}.

(ix) Interior of a triangle:

Many of the points that are enclosed inside a triangle make up the triangle's interior.

(x) Exterior of a triangle:

All such points that are not enclosed within the triangle make up the triangle's exterior.

** **

**Question :8. In Fig. 11, the length (in cm) of each side has been indicated along the side. State for each triangle angle whether it is scalene, isosceles or equilateral:**

**Solution 8:**

(i) It is a scalene triangle because no two sides are equal.

(ii) It is an isosceles triangle because two of its sides, viz. PQ and PR, are equal.

(iii) It is an equilateral triangle because all its three sides are equal.

(iv) It is a scalene triangle because no two sides are equal.

(v) It is an isosceles triangle because two of its sides are equal.

**Question :9. In Fig. 12, there are five triangles. The measures of some of their angles have been indicated. State for each triangle whether it is acute, right or obtuse.**

**Solution 9:**

(i) It is a right triangle because one of its angles is 90^{o}.

(ii) It is an obtuse triangle because one of its angles is 120^{o}, which is greater than 90^{o}

(iii) It is an acute triangle because all its angles are less than 90^{o}

(iv) It is a right triangle because one of its angle is 90^{o}.

(v) It is an obtuse triangle because one of its angle is 120^{o}, which is greater than 90^{o}.

** **

**Question :10. Fill in the blanks with the correct word/symbol to make it a true statement:**

**(i) A triangle has ……. sides.**

**(ii) A triangle has ……..vertices.**

**(iii) A triangle has ……..angles.**

**(iv) A triangle has ………parts.**

**(v) A triangle whose no two sides are equal is known as ………**

**(vi) A triangle whose two sides are equal is known as ………. **

**(vii) A triangle whose all the sides are equal is known as ………**

**(viii) A triangle whose one angle is a right angle is known as ……..**

**(ix) A triangle whose all the angles are of measure less than 90′ is known as ………**

**(x) A triangle whose one angle is more than 90′ is known as ……….**** **

**Solution 10:**

(i) A triangle has **three** sides.

(ii) A triangle has **three** vertices.

(iii) A triangle has **three** angles.

(iv) A triangle has **six **parts.

(v) A triangle whose no two sides are equal is known as **a scalene triangle.**

(vi) A triangle whose two sides are equal is known as **an isosceles triangle.**

(vii) A triangle whose all the sides are equal is known as **an equilateral triangle.**

(viii) A triangle whose one angle is a right angle is known as **a right triangle.**

(ix) A triangle whose all the angles are of measure less than 90′ is known as **an acute triangle.**

(x) A triangle whose one angle is more than 90′ is known as **an obtuse triangle.**

**Question :11. In each of the following, state if the statement is true (T) or false (F):**

**Solution 11:**

(i) A triangle has three sides. **True **

(ii) A triangle may have four vertices.

**Clarification:**

Any three non-parallel line segments can make up a triangle. **False**

(iii) Any three line-segments make up a triangle.

**Clarification:**

Any three non-parallel line segments can make up a triangle. **False**

(iv) The interior of a triangle includes its vertices.

**Clarification:**

The interior of a triangle is the region enclosed by the triangle and the vertices are not enclosed by the triangle. **False**

(v) The triangular region includes the vertices of the corresponding triangle. **True **

(vi) The vertices of a triangle are three collinear points.

**Clarification:**

The vertices of a triangle are three non-collinear points. **False**

(vii) An equilateral triangle is isosceles also. **True**

(viii) Every right triangle is scalene.

**Clarification:**

A right triangle can also be an isosceles triangle. **False**

(ix) Each acute triangle is equilateral.

**Clarification:**

Each acute triangle is not an equilateral triangle, but each equilateral triangle is an acute triangle. **False**

(x) No isosceles triangle is obtuse.

**Clarification:**

An isosceles triangle can be an obtuse triangle, a right triangle or an acute triangle. **False**

**Exercise 15.2**** **

**Question :1. Two angles of a triangle are of measures 150 ^{o} and 30^{o}. Find the measure of the third angle.**

**Solution 1:**

Given: Two angles of a triangle 150^{o} and 30^{o}

Let x be the required third angle

As we know the total of all three angle of a triangle = 180^{o}

So,150^{o} + 30^{o} + x = 180^{o}

135^{o} + x = 180^{o}

x = 180^{o} – 135^{o} = 45^{o}

Thus, the third angle is 45^{o}

** **

**Question :2. One of the angles of a triangle is 130 ^{o}, and the other two angles are equal. What is the measure of each of these equal angles?**

**Solution 2:**

Given: in a triangle one of the angle is 130^{o}, and the other two angles are equal.

So, let x be the second and third angle

As we know the total of all three angle of a triangle = 180^{o}

130^{o} + x + x = 180^{o}

130^{o} + 2x = 180^{o}

2x = 180^{o }– 130^{o}

2x = 50^{o}

x = 50/2 = 25^{o}

Thus, the two other angles are 25^{o} each

**Question :3. The three angles of a triangle are equal to one another. What is the measure of each of the angles?**** **

**Solution 3:**

Given: all three angles are equal

So, let x be each angle

As we know the total of all three angle of a triangle = 180^{o}

x + x + x = 180^{o}

3x = 180^{o}

x = 180/3 = 60^{o}

Thus, angle is 60^{o} each

** **

**Question :4. If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.**** **

**Solution 4:**

Given: The triangles’ angle are in the ratio of 1: 2: 3

So, we take angle first as x, angle second as 2x and angle third as 3x

As we know the total of all three angle of a triangle = 180^{o}

x + 2x + 3x = 180^{o}

6x = 180^{o}

x = 180/6 = 30^{o}

2x = 30^{o} × 2 = 60^{o}

3x = 30^{o} × 3 = 90^{o}

Thus, the angle first is 30^{o}, angle second is 60^{o} and angle third is 90^{o}.

**Question :5. The angles of a triangle are (x − 40) ^{o}, (x − 20)^{o} and (1/2 − 10)^{o}. Find the value of x.**

**Solution 5:**

Given: Angles of a triangle are (x − 40)^{o}, (x − 20)^{o} and ( 1/2− 10)^{o}.

As we know the total of all three angle of a triangle = 180^{o}

(x − 40)^{o} + (x − 20)^{o} + ( 1/2 − 10)^{o} = 180^{o}

x + x + 1/2 – 40^{o} – 20^{o }– 10^{o} = 180^{o}

x + x + 1/2 – 70^{o} = 180^{o}

5x/2 = 180^{o }+ 70^{o}

5x/2 = 250^{o}

x = 2/5 × 250^{o} = 100^{o}

So, the value of x is 100^{o}

**Question 6. The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10 ^{o}. Find the three angles.**

**Solution 6:**

Let x be the angle first; angle second be x + 10^{o}; angle third be x + 10^{o} + 10^{o}

As we know the total of all three angle of a triangle = 180^{o}

x + x + 10^{o} + x + 10^{o} +10^{o} = 180^{o}

3x + 30 = 180

3x = 180 – 30

3x = 150

x = 150/3 = 50^{o}

Angle first= 50^{o}

Angle second = x + 10^{o} = 50 + 10 = 60^{o}

Angle third = x + 10^{o} +10^{o} = 50 + 10 + 10 = 70^{o}

** **

**Question 7. Two angles of a triangle are equal and the third angle is greater than each of those angles by 30 ^{o}. Determine all the angles of the triangle**

**Solution 7:**

Let x be the first and second angle; according to the question third angle is greater than each of those angles by 30^{o}

Thus, the third angle is greater than the first and second by 30^{o} = x + 30^{o}

The angles first and the second are equal

As we know the total of all three angle of a triangle = 180^{o}

x + x + x + 30^{o} = 180^{o}

3x + 30 = 180

3x = 180 – 30

3x = 150

x = 150/3 = 50^{o}

Angle third = x + 30^{o} = 50^{o} + 30^{o} = 80^{o}

The angle first and the second is 50^{o} and the angle third is 80^{o}.

** **

**Question 8. If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.**** **

**Solution 8:**

Let the measure of angles be m, n, o

Thus, can write above statement as m = n + o

m + n + o = 180^{o}

Replacing the above value

m + m = 180^{o}

2m = 180^{o}

m = 180/2 = 90^{o}

The given triangle is a right-angled triangle because one angle is 90^{o}.

** **

**Question 9. If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.**** **

**Solution 9:**

Let the measure of angles be a, b and c

From the above statement we write as

a> b + c

b < a + c

c < a + b

Thus, triangle is an acute triangle

**Question 10. In each of the following, the measures of three angles are given. State in which cases the angles can possibly be those of a triangle:**

**(i) 63 ^{o}, 37^{o}, 80^{o}**

**(ii) 45 ^{o}, 61^{o}, 73^{o}**

**(iii) 59 ^{o}, 72^{o}, 61^{o}**

**(iv) 45 ^{o}, 45^{o}, 90^{o}**

**(v) 30 ^{o}, 20^{o}, 125^{o}**

**Solution 10:**

(i) 63^{o} + 37^{o }+ 80^{o} = 180^{o}

Angles form a triangle

(ii) 45^{o}+ 61^{o}+ 73^{o }= 179^{o}

179 ≠ 180^{o}

Thus, not a triangle

(iii) 59^{o}+ 72^{o}+ 61^{o} = 192^{o}

192 ≠ 180^{o}

Thus, not a triangle

(iv) 45^{o}+ 45^{o}+ 90^{o} = 180^{o}

Angles form a triangle

(v) 30^{o}, 20^{o}, 125^{o} =175^{o}

179 ≠ 180^{o}

Thus, not a triangle

** **

**Question 11. The angles of a triangle are in the ratio 3: 4: 5. Find the smallest angle.**** **

**Solution 11:**

Let 3x, 4x, 5x be the measure of the angles

As we know the total of all three angle of a triangle = 180^{o}

3x + 4x + 5x = 180^{o}

12x = 180^{o}

x = 180/12 = 15^{o}

Smallest angle = 3x

= 3 × 15^{o} = 45^{o}

Thus, smallest angle = 45^{o}

**Question 12. Two acute angles of a right triangle are equal. Find the two angles.**** **

**Solution 12:**

A Right triangle: whose one of the angle is a right angle

Let x, x, 90^{o} be the measure of angle

x + x + 90^{o}= 180^{o}

2x = 90^{o}

x = 90/2 = 45^{o}

The measure of two angles are 45^{o} and 45^{o}

** **

**Question 13. One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?**** **

**Solution 13:**

Let the measure of the angles be a, b, c

From the question we can write as

a > b + c;

b > a + c;

c > a + b

a or b or c > 90^{o} which is obtuse

Thus, triangle is an obtuse angle

**Question 14. In the six cornered figure, (fig. 20), AC, AD and AE are joined. Find ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA.**

**Solution 14:**

As we know the total of all three angle of a triangle = 180^{o}

Thus, in ∆ABC,

∠CAB + ∠ABC + ∠BCA = 180^{o }- (i)

In △ADE,

∠EAD + ∠ADE + ∠DEA =180^{o} - (ii)

In△ ACD,

∠DAC + ∠ACD + ∠CDA = 180^{o }- (iii)

In △AEF,

∠FAE + ∠AEF + ∠EFA = 180^{o} - (iv)

Adding (i), (ii), (iii), (iv);

∠CAB + ∠ABC + ∠BCA + ∠DAC + ∠ACD + ∠CDA + ∠EAD + ∠ADE + ∠DEA + ∠FAE + ∠AEF +∠EFA = 720^{o}

Thus, ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720^{o}

** **

**Question 15. Find x, y, z (whichever is required) from the figures (Fig. 21) given below:**

**Solution 15:**

(i) In △ABC and △ADE,

∠ ADE = ∠ ABC [corresponding angles]

x = 40^{o}

∠AED = ∠ACB [corresponding angles]

y = 30^{o}

As we know the total of all three angle of a triangle = 180^{o}

x + y + z = 180^{o} (Angles of △ADE)

By putting the values

40^{o} + 30^{o} + z = 180^{o}

z = 180^{o} – 70^{o}= 110^{o}

Thus, we can find that the angles of the given triangle are x=40^{o}, y=30^{o} and z=110^{o}

(ii) As we can see that in △ADC, ∠ADC is equal to 90^{o}.

(△ADC is a right triangle)

As we know the total of all three angle of a triangle = 180^{o}

45^{o} + 90^{o} + y = 180^{o} (Sum of the angles of △ADC)

135^{o} + y = 180^{o}

y = 180^{o} – 135^{o} = 45^{o}.

We can also say that in △ ABC, ∠ABC + ∠ACB + ∠BAC is equal to 180^{o}.

(Sum of the angles of △ABC)

40^{o} + y + (x + 45^{o}) = 180^{o}

40^{o} + 45^{o} + x + 45^{o} = 180^{o} (y = 45^{o})

x = 180^{o} –130^{o}

x = 50^{o}

Thus, the required angles are 45^{o} and 50^{o}.

(iii) As we know the total of all three angle of a triangle = 180^{o}

For △ABD:

∠ABD +∠ADB + ∠BAD = 180° (Sum of the angles of △ABD)

50^{o} + x + 50^{o} = 180^{o}

100^{o} + x = 180^{o}

x = 180^{o} – 100^{o} = 80^{o}

For △ ABC:

∠ABC + ∠ACB + ∠BAC = 180^{o} (Sum of the angles of △ABC)

50^{o} + z + (50^{o} + 30^{o}) = 180^{o}

50^{o} + z + 50^{o} + 30^{o} = 180^{o}

z = 180^{o} – 130^{o} =50^{o}

Using the values for △ADC:

∠ADC + ∠ACD + ∠DAC = 180^{o} (Sum of the angles of △ADC)

y + z + 30^{o} = 180^{o}

y + 50^{o} + 30^{o} = 180^{o} (z = 50^{o})

y = 180^{o} – 80^{o} = 100^{0}

Thus, the required angles are 80^{o}, 50^{o} and 100^{o}.

(iv) In △ABC and △ADE:

∠ADE = ∠ABC (Corresponding angles)

y = 50^{o}

, ∠AED = ∠ACB (Corresponding angles)

z = 40^{o}

As we know the total of all three angle of a triangle = 180^{o}

x + 50^{o} + 40^{o} = 180^{o} (Angles of △ADE)

x = 180^{o} – 90^{o} = 90^{o}

Thus, the required angles are 50^{o}, 40^{o} and 90^{o}.

** **

**Question 16. If one angle of a triangle is 60 ^{o} and the other two angles are in the ratio 1: 2, find the angles.**

**Solution 16:**

Let x be the one of the other two angles,

Thus, the 2x will be the second one,

As we know the total of all three angle of a triangle = 180^{o}

60^{o} + x + 2x = 180^{o}

3x = 180^{o} – 60^{o}

3x = 120^{o}

x = 120/3 = 40^{o}

2x = 2 × 40^{o}

2x = 80^{o}

Hence, The required angles are 40^{o} and 80^{o}.

**Question 17. If one angle of a triangle is 100 ^{o} and the other two angles are in the ratio 2: 3. Find the angles.**

**Solution 17:**

Let 2x be one of the other two angle, the 3x will be second angle.

As we know the total of all three angle of a triangle = 180^{o}

100^{o} + 2x + 3x = 180^{o}

5x = 180^{o} – 100^{o}

5x = 80^{o}

x = 80/5 = 16

2x = 2 ×16

2x = 32^{o}

3x = 3×16

3x = 48^{o}

Thus, the required angles are 32^{o} and 48^{o}.

** **

**Question 18. In △ABC, if 3∠A = 4∠B = 6∠C, calculate the angles.**** **

**Solution 18:**

According to the question the given triangle, 3∠A = 6∠C

∠A = 2∠C - (i)

According to the question the same triangle, 4∠B = 6∠C

∠B = (6/4)∠C- (ii)

As we know the total of all three angle of a triangle = 180^{o}

Thus, ∠A + ∠B + ∠C = 180^{o} (Angles of △ABC)- (iii)

By putting the values of ∠A and ∠B in equation (iii),

=2∠C + (6/4)∠C + ∠C = 180^{o}

=(18/4)∠C = 180^{o}

∠C = 40^{o}

From equation (i),

∠A = 2∠C = 2 × 40

∠A = 80^{o}

From equation (ii),

∠B = (6/4)∠C = (6/4)× 40^{o}

∠B = 60^{o}

Therefore, the three angles of the given triangle are ∠A = 80^{o}, ∠B = 60^{o}, ∠C = 40^{o}

**Question 19. Is it possible to have a triangle, in which?**

**(i) Two of the angles are right?**

**(ii) Two of the angles are obtuse?**

**(iii) Two of the angles are acute?**

**(iv) Each angle is less than 60 ^{o}?**

**(v) Each angle is greater than 60 ^{o}?**

**(vi) Each angle is equal to 60 ^{o}?**

**Solution 19:**

(i) No, even if a triangle has two right angles, the triangle's third angle must be zero, which is not true.

(ii) No, since the number of a triangle's three angles is always 180^{o}, as we know. If two obtuse angles are added together, their number exceeds 180^{o}, which is impossible in a triangle.

(iii) Yes, two acute angles are possible in right triangles and acute triangles.

(iv) No, because if each angle is less than 60^{o}, the number of all three angles is less than 180^{o}, which is impossible in a triangle.

(v) No, because if each angle is less than 60^{o}, the number of all three angles is less than 180^{o}, which is impossible in a triangle.

(vi) Yes, if each triangle angle is 60^{ o}, the total of all three angles would be 180^{ o}, which is possible in the case of a triangle.

** **

**Question 20. In△ ABC, ∠A = 100 ^{o}, AD bisects ∠A and AD ⊥ BC. Find ∠B**

**Solution 20:**

In the question given that in △ABC, ∠A = 100^{o}

Also, AD ⊥ BC

Consider △ABD

∠BAD = 100/2 [AD bisects ∠A]

∠BAD = 50^{o}

∠ADB = 90^{o} [AD perpendicular to BC]

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

Thus,

∠ABD + ∠BAD + ∠ADB = 180^{o} [Sum of angles of △ABD]

Or,

∠ABD + 50^{o} + 90^{o} = 180^{o}

∠ABD =180^{o} – 140^{o} = 40^{o}

**Question 21. In △ABC, ∠A = 50 ^{o}, ∠B = 70^{o} and bisector of ∠C meets AB in D. Find the angles of the triangles ADC and BDC**

**Solution 21:**

As we know the total of all three angle of a triangle = 180^{o}

For the given △ABC,

∠A + ∠B + ∠C = 180^{o} [Sum of angles of △ABC]

50^{o} + 70^{o} + ∠C = 180^{o}

∠C= 180^{o} –120^{o} = 60^{o}

∠ACD = ∠BCD =∠C2 [CD bisects ∠C and meets AB in D]

∠ACD = ∠BCD = 60/2 = 30^{o}

According to the same way for the given △ACD,

∠DAC + ∠ACD + ∠ADC = 180^{o}

50^{o} + 30^{o} + ∠ADC = 180^{o}

∠ADC = 180^{o}– 80^{o} = 100^{o}

According to the same way for the given △BCD,

∠DBC + ∠BCD + ∠BDC = 180^{o}

70^{o} + 30^{o} + ∠BDC = 180^{o}

∠BDC = 180^{o} – 100^{o} = 80^{o}

Thus,

For △ADC= ∠A = 50^{o}, ∠D = 100^{o} ∠C = 30^{o}

△BDC= ∠B = 70^{o}, ∠D = 80^{o} ∠C = 30^{o}

** **

**Question 22: In ∆ABC, ∠A = 60 ^{o}, ∠B = 80^{o}, and the bisectors of ∠B and ∠C, meet at O. Find**

**(i) ∠C**

**(ii) ∠BOC**** **

**Solution 22:**

(i) As we know the total of all three angle of a triangle = 180^{o}

For △ABC,

∠A + ∠B + ∠C = 180^{o} [Sum of angles of ∆ABC]

60^{o} + 80^{o} + ∠C= 180^{o}.

∠C = 180^{o} – 140^{o} = 40^{o}.

(ii)For △OBC,

∠OBC = ∠B/2 = 80/2 [OB bisects ∠B]

∠OBC = 40^{o}

∠OCB = ∠C/2 = 40/2 [OC bisects ∠C]

∠OCB = 20^{o}

If we apply the above logic to this triangle,

∠OCB + ∠OBC + ∠BOC = 180^{o} [Sum of angles of △OBC]

20^{o} + 40^{o} + ∠BOC = 180^{o}

∠BOC = 180^{o} – 60^{o} = 120^{o}** **

**Question 23. The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.**** **

**Solution 23:**

As we know the total of all three angle of a triangle = 180^{o}

For △ABC,

∠A + ∠B + ∠C = 180^{o}

∠A + 90^{o} + ∠C = 180^{o}

∠A + ∠C = 180^{o} – 90^{o}

∠A + ∠C = 90^{o}

For △OAC:

∠OAC = ∠A/2 [OA bisects ∠A]

∠OCA = ∠C/2 [OC bisects ∠C]

By applying same way to △OAC,

∠AOC + ∠OAC + ∠OCA = 180^{o} [Sum of angles of △AOC]

∠AOC + ∠A/2 +∠C/2 = 180^{o}

∠AOC + (∠A+∠C)/2 = 180^{o}

∠AOC + 90/2 = 180^{o}

∠AOC = 180^{o} – 45^{o} = 135^{o}** **

**Question 24. In △ABC, ∠A = 50 ^{o} and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.**

**Solution 24:**

∠ACD = ∠A + ∠B. [Exterior angle is equal to the sum of two opposite interior angles]

As we know the total of all three angle of a triangle = 180^{o}

∠ABC + ∠BCA + ∠CAB = 180^{o}

∠A + ∠B + ∠BCA = 180^{o}

∠BCA = 180^{o} – (∠A + ∠B)

But EC bisects ∠ACD

Therefore ∠ECA = ∠ACD/2

∠ECA = (∠A + ∠B)/2 [∠ACD = (∠A + ∠B)]

But EB bisects ∠ABC

∠EBC =∠ABC/2 = ∠B/2

∠EBC = ∠ECA + ∠BCA

∠EBC = (∠A + ∠B)/2 + 180^{o} – (∠A + ∠B)

If we use same steps for △EBC,

∠B/2 + (∠A + ∠B)/2 + 180^{o} – (∠A + ∠B) + ∠BEC = 180^{o}

∠BEC =∠A + ∠B – (∠A + ∠B)/2 – ∠B/2

∠BEC = ∠A/2

∠BEC = 50/2 = 25^{o}

** **

**Question 25. In ∆ABC, ∠B = 60°, ∠C = 40°, AL ⊥ BC and AD bisects ∠A such that L and D lie on side BC. Find ∠LAD**** **

**Solution 25:**

As we know the total of all three angle of a triangle = 180^{o}

△ABC, we can write

∠A + ∠B + ∠C = 180^{o}

∠A + 60^{o} + 40^{o} = 180^{o}

∠A = 80^{0}

And ∠DAC bisects ∠A

∠DAC = ∠A/2

∠DAC = 80^{o}/2

Apply same steps for the △ADC,

As we know the total of all three angle of a triangle = 180^{o}

∠ADC + ∠DCA + ∠DAC = 180^{o}

∠ADC + 40^{o} + 40^{o} = 180^{o}

∠ADC = 180^{o }– 80^{o} = 100^{0}

Exterior angle is equal to the sum of two interior opposite angles

Thus, ∠ADC = ∠ALD+ ∠LAD

But here AL perpendicular to BC

100^{o} = 90^{o} + ∠LAD

∠LAD = 10^{o}

**Question 26. Line segments AB and CD intersect at O such that AC ∥ DB. It ∠CAB = 35 ^{o} and ∠CDB = 55^{o}. Find ∠BOD.**

**Solution 26:**

AC parallel to BD and AB cuts AC and BD at A and B, respectively.

∠CAB = ∠DBA [Alternate interior angles]

∠DBA = 35^{o}

As we know the total of all three angle of a triangle = 180^{o}

For △OBD, we can say that:

∠DBO + ∠ODB + ∠BOD = 180°

35^{o} + 55^{o} + ∠BOD = 180^{o} (∠DBO = ∠DBA and ∠ODB = ∠CDB)

∠BOD = 180^{o} – 90^{o}

∠BOD = 90^{o}

** **

**Question 27. In Fig. 22, ∆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**

**Solution 27:**

AC parallel to QP and BR cuts AC and QP at C and Q, respectively.

∠QCA = ∠CQP [Alternate interior angles]

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

∠ABC = ∠PRQ [Alternate interior angles]

As we know the total of all three angle of a triangle = 180^{o}

For ∆ABC,

∠ABC + ∠ACB + ∠BAC = 180^{o}

∠ABC + ∠ACB + 90^{o} = 180^{o} [Right angled at A]

∠ABC + ∠ACB = 90^{o}

Applying same way for △PQR,

∠PQR + ∠PRQ + ∠QPR = 180^{o}

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

Or,

90^{o}+ ∠QPR =180^{o} (∠ABC+ ∠ACB = 90^{o})

∠QPR = 90^{o}

**Exercise 15.3 **

** **

**Question 1. In Fig. 35, ∠CBX is an exterior angle of ∆ABC at B. Name**

**(i) The interior adjacent angle**

**(ii) The interior opposite angles to exterior ∠CBX**

**Also, name the interior opposite angles to an exterior angle at A.**

**Solution 1:**

(i) The interior adjacent angle = ∠ABC

(ii) The interior opposite angles to exterior ∠CBX = ∠BAC and ∠ACB

Also the interior angles opposite to exterior =∠ABC and ∠ACB

** **

**Question 2. In the fig. 36, two of the angles are indicated. What are the measures of ∠ACX and ∠ACB?**

**Solution 2:**

△ABC, ∠A = 50^{o} and ∠B = 55^{o}

As we know the total of all three angle of a triangle = 180^{o}

∠A + ∠B + ∠C = 180^{o}

50^{o}+ 55^{o}+ ∠C = 180^{o}

∠C = 75^{o}

∠ACB = 75^{o}

∠ACX = 180^{o}− ∠ACB = 180^{o }− 75^{o} = 105^{o}

** **

**Question 3. In a triangle, an exterior angle at a vertex is 95 ^{o} and its one of the interior opposite angles is 55^{o}. Find all the angles of the triangle.**

**Solution 3:**

The sum of interior opposite angles is equal to the exterior angle.

∠ABC+ ∠BAC = ∠BCO

55^{o} + ∠BAC = 95^{o}

∠BAC= 95^{o}– 55^{o}

∠BAC = 40^{o}

As we know the total of all three angle of a triangle = 180^{o}

For the given △ABC,

∠ABC + ∠BAC + ∠BCA = 180^{o}

55^{o} + 40^{o} + ∠BCA = 180^{o}

∠BCA = 180^{o} –95^{o}

∠BCA = 85^{o}** **

**Question 4. One of the exterior angles of a triangle is 80 ^{o}, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?**

**Solution 4:**

Let us assume that P and Q are the two interior opposite angles.

We know that ∠P is equal to ∠Q

The sum of interior opposite angles is equal to the exterior angle.

Thus, from the figure,

∠P + ∠Q = 80^{o}

∠P +∠P = 80^{o } [because ∠P = ∠Q]

2∠P = 80^{o}

∠P = 80/2 =40^{o}

∠P= ∠Q = 40^{o}

Thus, each of the required angles is of 40^{o}.

**Question 5. The exterior angles, obtained on producing the base of a triangle both ways are 104 ^{o} and 136^{o}. Find all the angles of the triangle.**

**Solution 5:**

∠ABE and ∠ABC form a linear pair.

∠ABE + ∠ABC =180^{o}

∠ABC = 180^{o}– 136^{o}

∠ABC = 44^{o}

And, ∠ACD and ∠ACB form a linear pair.

∠ACD + ∠ACB = 180^{o}

∠ACB = 180^{o}– 104^{o}

∠ACB = 76^{o}

Sum of interior opposite angles is equal to the exterior angle.

Therefore, ∠BAC + ∠ABC = 104^{o}

∠BAC = 104^{o} – 44^{o} = 60^{o}

∠ACE = 76^{o} and ∠BAC = 60^{o}** **

**Question 6. In Fig. 37, the sides BC, CA and BA of a △ABC have been produced to D, E and F respectively. If ∠ACD = 105 ^{o} and ∠EAF = 45^{o}; find all the angles of the △ABC.**

**Solution 6:**

△ABC, ∠BAC and ∠EAF [vertically opposite angles]

Hence, ∠BAC = ∠EAF = 45^{o}

Considering the exterior angle property,

∠BAC + ∠ABC = ∠ACD = 105^{o}

By repositioning ∠ABC = 105^{o}– 45^{o} = 60^{o}

As we know the total of all three angle of a triangle = 180^{o}

∠ABC + ∠ACB +∠BAC = 180°

∠ACB = 75^{o}

Thus, the angles are 45^{o}, 60^{o} and 75^{o}.** **

**Question 7. In Fig. 38, AC perpendicular to CE and C ∠A: ∠B: ∠C= 3: 2: 1. Find the value of ∠ECD.**

**Solution 7:**

Let 3x, 2x and x be the angles of the triangle

As we know the total of all three angle of a triangle = 180^{o}

3x + 2x + x = 180^{o}

6x = 180^{o}

x = 30^{o}

Also, ∠ACB + ∠ACE + ∠ECD = 180^{o}

x + 90^{o} + ∠ECD = 180^{o} (∠ACE = 90^{o})

According to question x = 30^{o}

Thus, ∠ECD = 60^{o}

** **

**Question 8. A student when asked to measure two exterior angles of △ABC observed that the exterior angles at A and B are of 103 ^{o} and 74^{o} respectively. Is this possible? Why or why not?**

**Solution 8:**

As we know the total of all three angle of a triangle = 180^{o}

Internal angle at A + External angle at A = 180^{o}

Internal angle at A + 103^{o} =180^{o}

A = 77^{o}

Internal angle at B + External angle at B = 180^{o}

Internal angle at B + 74^{o} = 180^{o}

B = 106^{o}

Sum of internal angles at A and B

= 77^{o} + 106^{o} = 183^{o}

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

** **

**Question 9. In Fig.39, AD and CF are respectively perpendiculars to sides BC and AB of △ABC. If ∠FCD = 50 ^{o}, find ∠BAD**

**Solution 9:**

As we know the total of all three angle of a triangle = 180^{o}

△FCB,

∠FCB + ∠CBF + ∠BFC = 180^{o}

50^{o} + ∠CBF + 90^{o }= 180^{o}

∠CBF = 180^{o} – 50^{o} – 90^{o} = 40^{o}

Applying same way for △ABD,

∠ABD + ∠BDA + ∠BAD = 180^{o}

∠BAD = 180^{o} – 90^{o} – 40^{o} = 50^{o}

** **

**Question 10. In Fig.40, measures of some angles are indicated. Find the value of x.**

**Solution 10:**

As we know the total of all three angle of a triangle = 180^{o}

From the figure we have,

∠AED + 120^{o} = 180^{o} (Linear pair)

∠AED = 180^{o} – 120^{o} = 60^{o}

As we know the total of all three angle of a triangle = 180^{o}

Thus, for △ADE,

∠ADE + ∠AED + ∠DAE = 180^{o}

60^{o}+ ∠ADE + 30^{o} =180^{o}

∠ADE = 180^{o }– 60^{o }– 30^{o} = 90^{o}

According to the given figure,

∠FDC + 90^{o} = 180^{o} [Linear pair]

∠FDC = 180^{o }– 90^{o} = 90^{o}

Applying same way for △CDF,

∠CDF + ∠DCF + ∠DFC = 180^{o}

90^{o} + ∠DCF + 60^{o} = 180^{o}

∠DCF = 180^{o }– 60^{o }– 90^{o }= 30^{o}

∠DCF + x = 180^{o} [Linear pair]

30^{o} + x = 180^{o}

x = 180^{o} – 30^{o} = 150^{o}

**Question 11. In Fig. 41, ABC is a right triangle right angled at A. D lies on BA produced and DE perpendicular to BC intersecting AC at F. If ∠AFE = 130 ^{o}, find**

**(i) ∠BDE**

**(ii) ∠BCA**

**(iii) ∠ABC**

**Solution 11:**

(i) ∠BAF + ∠FAD = 180^{o} [Linear pair]

∠FAD = 180^{o }– ∠BAF = 180^{o }– 90^{o} = 90^{o}

According to the figure,

∠AFE = ∠ADF + ∠FAD [Exterior angle property]

∠ADF + 90^{o} = 130^{o}

∠ADF = 130^{o} − 90^{o} = 40^{o}

∠BDE = 40^{o}

(ii) As we know the total of all three angle of a triangle = 180^{o}

For △BDE,

∠BDE + ∠BED + ∠DBE = 180^{o}

∠DBE = 180^{o }– ∠BDE – ∠BED

∠DBE = 180^{o }– 40^{o }– 90^{o }= 50^{o}- Equation (i)

According to the figure,

∠FAD = ∠ABC + ∠ACB [Exterior angle property]

90^{o} = 50^{o} + ∠ACB

∠ACB = 90^{o }– 50^{o} = 40^{o}

(iii) By the equation ∠ABC = ∠DBE = 50^{o}

** **

**Question 12. ABC is a triangle in which ∠B = ∠C and ray AX bisects the exterior angle DAC. If ∠DAX = 70 ^{o}. Find ∠ACB.**

**Solution 12:**

In the question given information is that ABC is a triangle in which ∠B = ∠C

Also AX bisects the exterior angle DAC

∠CAX = ∠DAX [AX bisects ∠CAD]

∠CAX =70^{o }[given]

∠CAX +∠DAX + ∠CAB =180^{o}

70^{o}+ 70^{o} + ∠CAB =180^{o}

∠CAB =180^{o} –140^{o} =40^{o}

∠ACB + ∠CBA + ∠CAB = 180^{o} [Sum of the angles of ∆ABC]

∠ACB + ∠ACB+ 40^{o} = 180^{o} [∠C = ∠B]

2∠ACB = 180^{o }– 40^{o}

∠ACB = 140/2 = 70^{o}

** **

**Question 13. The side BC of △ABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC= 30 ^{o} and ∠ACD = 115^{o}, find ∠ALC**

**Solution 13:**

In the question given information is that ∠ABC= 30^{o} and ∠ACD = 115^{o}

According to the figure,

∠ACD and ∠ACL [linear pair]

∠ACD+ ∠ACB = 180^{o}

115^{o} + ∠ACB =180^{o}

∠ACB = 180^{o }– 115^{o} = 65^{o}

As we know the total of all three angle of a triangle = 180^{o}

Thus, for△ ABC,

∠ABC + ∠BAC + ∠ACB = 180^{o}

30^{o} + ∠BAC + 65^{o} = 180^{o}

∠BAC = 85^{o}

∠LAC = ∠BAC/2 = 85/2

Applying same way for △ALC,

∠ALC + ∠LAC + ∠ACL = 180^{o}

∠ALC + 85/2 + 65^{o} = 180^{o}

∠ACL = ∠ACB

∠ALC = 180^{o} – 85/2 – 65^{o}

∠ALC = 72 ½^{o}** **

**Question 14. D is a point on the side BC of △ABC. A line PDQ through D, meets side AC in P and AB produced at Q. If ∠A = 80 ^{o}, ∠ABC = 60^{o} and ∠PDC = 15^{o}, find**

**(i) ∠AQD**

**(ii) ∠APD**** **

**Solution 14:**

According to the figure

∠ABD and ∠QBD [form a linear pair]

∠ABC + ∠QBC =180^{o}

60^{o} + ∠QBC = 180^{o}

∠QBC = 120^{o}

∠PDC = ∠BDQ [Vertically opposite angles]

∠BDQ = 15^{o}

(i) △QBD:

∠QBD + ∠QDB + ∠BQD = 180^{o} [Sum of angles of △QBD]

120^{o}+ 15^{o} + ∠BQD = 180^{o}

∠BQD = 180^{o }– 135^{o}

∠BQD = 45^{o}

∠AQD = ∠BQD = 45^{o}

(ii) △AQP:

∠QAP + ∠AQP + ∠APQ = 180^{o} [Sum of angles of △AQP]

80^{o} + 45^{o} + ∠APQ = 180^{o}

∠APQ = 55^{o}

∠APD = ∠APQ = 55^{o}** **

**Question 15. Explain the concept of interior and exterior angles and in each of the figures given below. Find x and y (Fig. 42)**

**Solution 15:**

In a triangle if all the three angle elements inside the triangle then they are interior triangles.

The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

(i) ∠ACB + x = 180^{o} [Linear pair]

75^{o}+ x = 180^{o}

x = 105^{o}

As we know the total of all three angle of a triangle = 180^{o}

Thus, △ABC,

∠BAC+ ∠ABC +∠ACB = 180^{o}

40^{o}+ y +75^{o} = 180^{o}

y = 65^{o}

(ii) x + 80^{o }= 180^{o} [Linear pair]

x = 100^{o}

In △ABC,

As we know the total of all three angle of a triangle = 180^{o}

x + y + 30^{o} = 180^{o}

100^{o} + 30^{o} + y = 180^{o}

y = 50^{o}

(iii) As we know the total of all three angle of a triangle = 180^{o}

Thus, for △ACD,

30^{o} + 100^{o} + y = 180^{o}

y = 50^{o}

Again, according to the figure

∠ACB + 100° = 180°

∠ACB = 80^{o}

Applying same rule △ACB,

x + 45^{o} + 80^{o} = 180^{o}

x = 55^{o}

(iv) As we know the total of all three angle of a triangle = 180^{o}

Thus, for △DBC,

30^{o} + 50^{o} + ∠DBC = 180^{o}

∠DBC = 100^{o}

According to the figure

x + ∠DBC = 180^{o} [Linear pair]

x = 80^{o}

From the exterior angle property,

y = 30^{o} + 80^{o} = 110^{o}

**Question 16. Compute the value of x in each of the following figures:**

**Solution 16:**

(i) ∠ACD + ∠ACB = 180^{o} [linear pair]

By repositioning

∠ACB = 180^{o }– 112^{o} = 68^{o}

∠BAE + ∠BAC = 180^{o} [linear pair]

On rearranging we get,

∠BAC = 180^{o}– 120^{o} = 60^{o}

As we know the total of all three angle of a triangle = 180^{o}

Thus, for △ABC:

x + ∠BAC + ∠ACB = 180^{o}

x = 180^{o }– 60^{o }– 68^{o} = 52^{o}

x = 52^{o}

(ii) ∠ABC + 120^{o} = 180^{o} [linear pair]

∠ABC = 60^{o}

Again, According to the figure we

∠ACB+ 110^{o} = 180^{o} [linear pair]

∠ACB = 70^{o}

As we know the total of all three angle of a triangle = 180^{o}

Thus, consider △ABC,

x + ∠ABC + ∠ACB = 180^{o}

x = 50^{o}

(iii) ∠BAD = ∠ADC = 52^{o} [alternate angles]

As we know the total of all three angle of a triangle = 180^{o}

Thus, △DEC, we have

x + 40^{o }+ 52^{o }= 180^{o}

x = 88^{o}

(iv) We have a quadrilateral in the given figure, and we also know that the number of all the angles in a quadrilateral is 360^{o}.

Thus,

35^{o} + 45^{o} + 50^{o} + reflex ∠ADC = 360^{o}

By repositioning

Reflex ∠ADC = 230^{o}

230^{o} + x = 360^{o} [A complete angle]

x = 130^{o}

**Exercise 15.4**** **

**Question 1. In each of the following, there are three positive numbers. State if these numbers could possibly be the lengths of the sides of a triangle:**

**(i) 5, 7, 9**

**(ii) 2, 10, 15**

**(iii) 3, 4, 5**

**(iv) 2, 5, 7**

**(v) 5, 8, 20**** **

**Solution 1:**

(i) 5, 7, 9

Yes, since the sum of any two triangle sides is always greater than the third side, these numbers may be the lengths of the triangle's sides. 5 + 7 > 9, 5 + 9 > 7, 9 + 7 > 5

(ii) 2, 10, 15

No, these numbers can't be the lengths of a triangle's sides because the sum of any two triangle sides is always greater than the third, which isn't the case here.2 + 10 < 15

(iii) 3, 4, 5

Yes, Since the sum of any two triangle sides is always greater than the third side, these numbers may be the lengths of the triangle's sides, 3 + 4 > 5, 3 + 5 > 4, 4 + 5 > 3

(iv) 2, 5, 7

No, these numbers can't be the lengths of a triangle's sides because the sum of any two triangle sides is always greater than the third, which isn't the case here. Here, 2 + 5 = 7

(v) 5, 8, 20

No, these numbers can't be the lengths of the triangle's sides because the sum of any two triangle sides is always greater than the third, which isn't the case here. Here, 5 + 8 < 20

** **

**Question 2. In Fig. 46, P is the point on the side BC. Complete each of the following statements using symbol ‘=’,’ > ‘or ‘< ‘so as to make it true:**

**(i) AP… AB+ BP**

**(ii) AP… AC + PC**

**(iii) AP…. ½ (AB + AC + BC)**

** **

**Solution 2:**

(i) In △APB, AP < AB + BP, since the total of any two triangle sides is greater than the sum of the triangle's third side

(ii) In △APC, AP < AC + PC, since the total of any two triangle sides is greater than the sum of the triangle's third side

(iii) AP < ½ (AB + AC + BC)

In △ABP and △ACP, we can write as

AP < AB + BP- (i) (since the total of any two triangle sides is greater than the sum of the triangle's third side)

AP < AC + PC -(ii) (since the total of any two triangle sides is greater than the sum of the triangle's third side)

By adding (i) and (ii),

AP + AP < AB + BP + AC + PC

2AP < AB + AC + BC (BC = BP + PC)

AP < ½ (AB + AC + BC)

** **

**Question 3. P is a point in the interior of △ABC as shown in Fig. 47. State which of the following statements are true (T) or false (F):**

**(i) AP + PB < AB**

**(ii) AP + PC > AC**

**(iii) BP + PC = BC**

**Solution 3:**

(i) AP + PB < AB = False

**Clarification:**

While we know that the number of any two triangle sides is greater than the third side.

(ii) AP + PC > AC = True

**Clarification:**

While we know that the number of any two triangle sides is greater than the third side.

(iii) BP + PC = BC = False

**Clarification:**

While we know that the number of any two triangle sides is greater than the third side.

** **

**Question 4. O is a point in the exterior of △ABC. What symbol ‘>’,’<’ or ‘=’ will you see to complete the statement OA+OB….AB? Write two other similar statements and show that OA + OB + OC > ½ (AB + BC +CA)**** **

**Solution 4:**

While we know that the number of any two triangle sides is greater than the third side, in △OAB,

OA + OB > AB - (i)

In △OBC

OB + OC > BC - (ii)

In △OCA

OA + OC > CA - (iii)

By adding equations (i), (ii) and (iii)

OA + OB + OB + OC + OA + OC > AB + BC + CA

2(OA + OB + OC) > AB + BC + CA

OA + OB + OC > (AB + BC + CA)/2

Or

OA + OB + OC > ½ (AB + BC +CA)

Hence the proof.

** **

**Question 5. In △ABC, ∠A = 100 ^{o}, ∠B = 30^{o}, ∠C = 50^{o}. Name the smallest and the largest sides of the triangle.**

**Solution 5:**

We know that the smallest side is always opposite the smallest angle, which in this case is 30°, and that the smallest angle is always AC.

Furthermore, it is BC since the largest side is always opposite the largest angle, which in this case is 100°.

**Exercise 15.5**** **

**Question 1. State Pythagoras theorem and its converse.**** **

**Solution 1:**

The Pythagoras Theorem states that the hypotenuse square is always equal to the number of the squares of the other two sides in a right triangle.

If the square of one side of a triangle equals the number of the squares of the other two sides, the triangle is a right triangle, with the angle opposite the first side being a right angle.

** **

**Question 2. In right △ABC, the lengths of the legs are given. Find the length of the hypotenuse**

**(i) a = 6 cm, b = 8 cm**

**(ii) a = 8 cm, b = 15 cm**

**(iii) a = 3 cm, b = 4 cm**

**(iv) a = 2 cm, b =1.5 cm**** **

**Solution 2:**

(i) According to the Pythagoras theorem,

(Hypotenuse)^{2} = (Base)^{2} + (Height)^{2}

Let c be hypotenuse and a and b be other two legs of right-angled triangle

Then, c^{2 }= a^{2} + b^{2}

c^{2} = 6^{2} + 8^{2}

c^{2} = 36 + 64 = 100

c = 10 cm

(ii) Let c be hypotenuse and a and b be other two legs of right -angled triangle

Then, c^{2} = a^{2} + b^{2}

c^{2} = 8^{2} + 15^{2}

c^{2} = 64 + 225 = 289

c = 17cm

(iii) Let c be hypotenuse and a and b be other two legs of right -angled triangle

Then, c^{2} = a^{2} + b^{2}

c^{2} = 3^{2} + 4^{2}

c^{2 }= 9 + 16 = 25

c = 5 cm

(iv) Let c be hypotenuse and a and b be other two legs of right -angled triangle

Then, c^{2 }= a^{2} + b^{2}

c^{2 }= 2^{2 }+ 1.5^{2}

c^{2} = 4 + 2.25 = 6.25

c = 2.5 cm

** **

**Question 3. The hypotenuse of a triangle is 2.5 cm. If one of the sides is 1.5 cm. find the length of the other side.**** **

**Solution 3:**

Let c be hypotenuse and the other two sides be b and a

According to the Pythagoras theorem, we have

c^{2} = a^{2} + b^{2}

2.5^{2} = 1.5^{2} + b^{2}

b^{2} = 6.25 −2.25 = 4

b = 2 cm

Hence, the 2 cm is length of the other side.

** **

**Question 4. A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2 m away from the wall. Find the height of the wall to which the ladder reaches.**** **

**Solution 4:**

Let the height of the ladder reaches to the wall be h.

According to the Pythagoras theorem, we have

(Hypotenuse)^{2} = (Base)^{2} + (Height)^{2}

3.7^{2} = 1.2^{2} + h^{2}

h^{2} = 13.69 – 1.44 = 12.25

h = 3.5 m

Hence, 3.5 m is the height of the wall.

** **

**Question 5. If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is right-angled triangle.**** **

**Solution 5:**

In the given triangle, the largest side is 6 cm.

We know that in a right-angled triangle, the number of the smaller sides' squares should equal the larger side's square.

Thus, 3^{2} + 4^{2} = 9 + 16 = 25

But, 6^{2} = 36

3^{2} + 4^{2} = 25 which is not equal to 6^{2}

So, the given triangle is not a right-angled triangle.

**Question 6. The sides of certain triangles are given below. Determine which of them are right triangles.**

**(i) a = 7 cm, b = 24 cm and c= 25 cm**

**(ii) a = 9 cm, b = 16 cm and c = 18 cm**** **

**Solution 6:**

(i) We know that in a right-angled triangle, the number of the smaller sides' squares should equal the larger side's square.

Here, c is the larger side , which is 25 cm.

c^{2} = 625

a^{2}+ b^{2 }= 7^{2} + 24^{2}

= 49 + 576

= 625

= c^{2}

Thus, the given triangle is a right triangle.

(ii) We know that in a right-angled triangle, the number of the smaller sides' squares should equal the larger side's square.

Here, c is the larger side , which is 18 cm.

c^{2 }= 324

a^{2}+ b^{2 }= 9^{2 }+ 16^{2 }

= 81 + 256

= 337 which is not equal to c^{2}

Thus, the given triangle is not a right triangle.

** **

**Question 7. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m. Find the distance between their tops.**

**(Hint: Find the hypotenuse of a right triangle having the sides (11 – 6) m = 5 m and 12 m)**** **

**Solution 7:**

The distance between points A and B is the distance between the tops of the poles.

The points A, B, and C form a right triangle, with AB as the hypotenuse, as seen in the diagram.

By using the Pythagoras Theorem in △ABC,

(11−6)^{2 }+ 12^{2 }= AB^{2}

AB^{2} = 25 + 144

AB^{2} = 169

AB = 13

So, the distance between the tops of the poles is 13 m.

** **

**Question 8. A man goes 15 m due west and then 8 m due north. How far is he from the starting point?**** **

**Solution 8:**

If a man travels 15 meters west and then 8 meters north,

Let O represent the starting point and P represent the ending point.

The hypotenuse of the triangle is then OP.

We can find the distance OP using Pythagoras' theorem.

OP^{2} = 15^{2 }+ 8^{2}

OP^{2} = 225 + 64

OP^{2} = 289

OP = 17

Hence, the required distance is 17 m.** **

**Question 9. The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?**

**Solution 9:**

By using the Pythagoras theorem, the length of the ladder is = L^{2}

6^{2 }+ 8^{2 }= L^{2}

L^{2} = 36 + 64 =100

L = 10

Thus, the length of the ladder is 10 m.

When ladder is shifted,

Let the height of the ladder after it is shifted be H m.

By using the Pythagoras theorem, we can find the height of the ladder after it is shifted.

8^{2 }+ H^{2 }= 10^{2}

H^{2 }= 100 – 64 = 36

H = 6

Thus, the height of the ladder is 6 m.

** **

**Question 10. A ladder 50 dm long when set against the wall of a house just reaches a window at a height of 48 dm. How far is the lower end of the ladder from the base of the wall?**** **

**Solution 10:**

In the question given that length of a ladder is 50dm

Let the distance of the lower end of the ladder from the wall be x dm.

By using the Pythagoras theorem,

x^{2} + 48^{2 }= 50^{2}

x^{2} = 50^{2 }− 48^{2 }

= 2500 – 2304

= 196

H = 14 dm

Hence, the distance of the lower end of the ladder from the wall is 14 dm.** **

**Question 11. The two legs of a right triangle are equal and the square of the hypotenuse is 50. Find the length of each leg. **

**Solution 11:**

According to the Pythagoras theorem,

(Hypotenuse)^{2} = (Base)^{2} + (Height)^{2}

Given that the square of the hypotenuse is 50 and the two legs of a right triangle are equal,

Each leg of the given triangle should be x units long.

Using Pythagoras' theorem

x^{2 }+ x^{2 }= (Hypotenuse)^{2}

x^{2 }+ x^{2 }= 50

2x^{2 }= 50

x^{2} = 25

x = 5

Hence, the length of each leg is 5 units.** **

**Question 12. Verity that the following numbers represent Pythagorean triplet:**** **

**(i) 12, 35, 37**** **

**(ii) 7, 24, 25**** **

**(iii) 27, 36, 45**** **

**(iv) 15, 36, 39**

**Solution 12:**

(i) The square of the largest side must equal the number of the squares of the other two sides in a Pythagorean triplet.

37^{2} =1369

12^{2} + 35^{2} = 144 + 1225 = 1369

12^{2} + 35^{2} = 37^{2}

Yes, they represent a Pythagorean triplet.

(ii) The square of the largest side must equal the number of the squares of the other two sides in a Pythagorean triplet.

25^{2 }= 625

7^{2} + 24^{2 }= 49 + 576 = 625

7^{2} + 24^{2} = 25^{2}

Yes, they represent a Pythagorean triplet.

(iii) The square of the largest side must equal the number of the squares of the other two sides in a Pythagorean triplet.

45^{2 }= 2025

27^{2} + 36^{2} = 729 + 1296 = 2025

27^{2} + 36^{2} = 45^{2}

Yes, they represent a Pythagorean triplet.

(iv) The square of the largest side must equal the number of the squares of the other two sides in a Pythagorean triplet.

39^{2} = 1521

15^{2 }+ 36^{2} = 225 + 1296 = 1521

15^{2} + 36^{2} = 39^{2}

Yes, they represent a Pythagorean triplet.** **

**Question 13. In △ABC, ∠ABC = 100°, ∠BAC = 35° and BD ⊥ AC meets side AC in D. If BD = 2 cm, find ∠C, and length DC.**

**Solution 13:**

As we know the sum of all angles of a triangle is 180^{o}

Thus, for the given △ABC,

∠ABC + ∠BAC + ∠ACB = 180^{o}

100^{o} + 35^{o} + ∠ACB = 180^{o}

∠ACB = 180^{o} –135^{o}

∠ACB = 45^{o}

∠C = 45^{o}

By applying same way for the △BCD,

∠BCD + ∠BDC + ∠CBD = 180^{o}

45^{o} + 90^{o} + ∠CBD = 180^{o} (∠ACB = ∠BCD and BD is perpendicular to AC)

∠CBD = 180^{o }– 135^{o}

∠CBD = 45^{o}

The sides opposite to equal angles have equal length.

Thus, BD = DC

DC = 2 cm

** **

**Question 14. 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.**** **

**Solution 14:**

For △ADC,

∠ADC = 90^{o} [AD is an altitude on BC]

Using the Pythagoras theorem,

12^{2 }+ 16^{2 }= AC^{2}

AC^{2} = 144 + 256

= 400

AC = 20 cm

Again, for △ADB,

∠ADB = 90^{o} [AD is an altitude on BC]

Using the Pythagoras theorem,

12^{2 }+ 9^{2 }= AB^{2}

AB^{2 }= 144 + 81 = 225

AB = 15 cm

For △ABC,

BC^{2} = 25^{2} = 625

AB^{2} + AC^{2} = 15^{2} + 20^{2 }= 625

AB^{2} + AC^{2} = BC^{2}

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

**Question 15. Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 105 ^{o}. Measure AB. Is (AB)^{2 }= (AC)^{2 } + (BC)^{2}? If not which one of the following is true:**

**(AB) ^{2} > (AC)^{2} + (BC)^{2} or (AB)^{2 }< (AC)^{2 }+ (BC)^{2}?**

**Solution 15:**

Draw △ABC as shown in the figure according to following steps.

Draw a line BC = 3 cm.

At point C, draw a line at 105^{o} angle with BC.

Take an arc of 4 cm from point C, which will cut the line at point A.

Now, join AB, which will be 5.5 cm.

AC^{2 }+ BC^{2} = 4^{2} + 3^{2 }

= 9 + 16

= 25

AB^{2} = 5.5^{2} = 30.25

AB^{2} is not equal to AC^{2}+ BC^{2}

Thus, AB^{2} > AC^{2 }+ BC^{2}** **

**Question 16. Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 80°. Measure AB. Is (AB) ^{2 }= (AC)^{2 } + (BC)^{2}? If not which one of the following is true:**

**(AB) ^{2 }> (AC)^{2} + (BC)^{2} or (AB)^{2 }< (AC)^{2 }+ (BC)^{2}?**

**Solution 16:**

Draw △ABC as shown in the figure according to following steps.

Draw a line BC = 3 cm.

At point C, draw a line at 80° angle with BC.

Take an arc of 4 cm from point C, which will cut the line at point A.

Now, join AB, it will be approximately 4.5 cm.

AC^{2} + BC^{2} = 4^{2} + 3^{2}

= 9 +16

= 25

AB^{2} = (4.5)^{2}

= 20.25

AB^{2} not equal to AC^{2 }+ BC^{2}

Thus, AB^{2 }< AC^{2 }+ BC^{2}