Properties of Triangles

(i) 8 cm, 6 cm, 15 cm

Sum of smaller line segments: 8 cm + 6 cm = 14 cm

The length of longest line segment = 15 cm

14 cm < 15 cm

The sum of two smaller line segments is not greater than the longest line segment.

∴ Triangle cannot be formed.

(ii) 6 cm, 8 cm, 14 cm

Sum of smaller line segments: 6 cm + 8 cm = 14 cm

The length of longest line segment = 14 cm

14 cm = 14 cm

The sum of two smaller line segments is not greater than the longest line segment.

∴ Triangle cannot be formed.

(iii) 5 cm, 6 cm, 10 cm

Sum of smaller line segments: 5 cm + 6 cm = 11 cm

The length of longest line segment = 10 cm

11 cm > 10 cm

The sum of two smaller line segments is greater than the longest line segment.

∴ Triangle can be formed.

(iv) 5.6 cm, 4.4 cm, 9 cm

Sum of smaller line segments: 5.6 cm + 4.4 cm = 10 cm

The length of longest line segment = 9 cm

10 cm > 9 cm

The sum of two smaller line segments is greater than the longest line segment.

∴ Triangle can be formed.

(v) 6 cm, 6 cm, 6 cm

Sum of smaller line segments: 6 cm + 6 cm = 12 cm

The length of longest line segment = 6 cm

12 cm > 6 cm

The sum of two smaller line segments is greater than the longest line segment.

∴ Triangle can be formed.

(vi) 10 cm, 7 cm, 4 cm

Sum of smaller line segments: 7 cm + 4 cm = 11 cm

The length of longest line segment = 10 cm

11 cm > 10 cm

The sum of the two smaller line segments is greater than the longest line segment.

∴ Triangle can be formed.

(i) 25°, 65°, 90°

Let's calculate the sum of the angles:

25° + 65° + 90° = 180°

The sum of the angles of a triangle is 180°.

∴ Triangle can be formed.

(ii) 50°, 120°, 30°

Let's calculate the sum of the angles:

50° + 120° + 30° = 200°

But the sum of the angles of a triangle is 180°.

Since 200° > 180°,

∴ Triangle cannot be formed.

(iii) 130°, 40°, 10°

Let's calculate the sum of the angles:

130° + 40° + 10° = 180°

The sum of the angles of a triangle is 180°.

∴ Triangle can be formed.

(iv) 70°, 70°, 50°

Let's calculate the sum of the angles:

70° + 70° + 50° = 190°

But the sum of the angles of a triangle is 180°.

Since 190° > 180°,

∴ Triangle cannot be formed.

(v) 30°, 90°, 90°

Let's calculate the sum of the angles:

30° + 90° + 90° = 210°

But the sum of the angles of a triangle is 180°.

Since 210° > 180°,

∴ Triangle cannot be formed.

(vi) 60°, 50°, 60°

Let's calculate the sum of the angles:

60° + 50° + 60° = 170°

But the sum of the angles of a triangle is 180°.

Since 170° < 180°,

∴ Triangle cannot be formed.

Given:

∠A = 72°, ∠B = 65°, ∠C = ?

The sum of all three angles of a triangle is 180°.

∴ ∠A + ∠B + ∠C = 180°

⇒ 72° + 65° + ∠C = 180°

⇒ 137° + ∠C = 180°

⇒ ∠C = 180° - 137°

⇒ ∠C = 43°

∴ The measure of ∠C is 43°.

Given:

∠P = 90° and ∠Q = ∠R

Let the measure of each of these equal angles be x.

The sum of all three angles of a triangle is 180°.

∴ ∠P + ∠Q + ∠R = 180°

Substituting the values in above, we get:

⇒ 90° + x + x = 180°

⇒ 90° + 2x = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\frac{90^\circ}{2}$

⇒ x = 45°

∴ The measure of ∠Q = 45° and ∠R = 45°

Let the measure of given angles be 3x, 5x and 4x.

Since the sum of the angles of a triangle is 180°, we have:

3x + 5x + 4x = 180°

⇒ 12x = 180°

⇒ x = $\frac{180^\circ}{12}$

⇒ x = 15°

Let's find the measure of each angle by substituting the value of x:

3x = 3 x 15° = 45°

5x = 5 x 15° = 75°

4x = 4 x 15° = 60°

∴ The three angles of a triangle are 45°, 75° and 60°.

Given:

First angle = 30°

The other two angles are in the ratio 3 : 2.

∴ Second angle = 3x

And

Third angle = 2x

Since the sum of the angles of a triangle is 180°, we have:

30° + 3x + 2x = 180°

⇒ 30° + 5x = 180°

⇒ 5x = 180° - 30°

⇒ 5x = 150°

⇒ x = $\frac{150^\circ}{5}$

⇒ x = 30°

Let's find the measure of other two angles by substituting the value of x:

Second angle = 3x = 3 x 30° = 90°

Third angle = 2x = 2 x 30° = 60°

∴ The other two angles are 90° and 60°.

Given:

The ratio of two acute angles = 4 : 5

∴ First acute angle = 4x

And

Second acute angle = 5x

As it is a right-angled triangle, one angle = 90°.

Since the sum of the angles of a triangle is 180°, we have:

90° + 4x + 5x = 180°

⇒ 90° + 9x = 180°

⇒ 9x = 180° - 90°

⇒ 9x = 90°

⇒ x = $\frac{90^\circ}{9}$

⇒ x = 10°

Let's find the measure of two acute angles by substituting the value of x:

First acute angle = 4x = 4 x 10° = 40°

Second acute angle = 5x = 5 x 10° = 50°

∴ The measure of two acute angles is 40° and 50°.

Given:

One of the two equal angles of an isosceles triangle = 65°

By the property of an isosceles triangle, two angles are equal.

∴ First equal angle = 65° = Second equal angle

Let the measure of the third angle be x.

Since the sum of the angles of a triangle is 180°, we have:

65° + 65° + x = 180°

⇒ 130° + x = 180°

⇒ x = 180° - 130°

⇒ x = 50°

∴ Third angle = 50°

∴ The measure of each angle of the triangle is 65°, 65° and 50°.

One angle in isosceles right-angled triangle = 90°

The other two angles are equal.

Let each equal angle be x.

Since the sum of the angles of a triangle is 180°, we have:

90° + x + x = 180°

⇒ 90° + 2x = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\frac{90^\circ}{2}$

⇒ x = 45°

∴ First equal angle = 45° and Second equal angle = 45°

∴ The measure of each of the two equal angles of an isosceles right-angled triangle is 45°.

Given:

All three angles of a triangle are of the same measure.

Let the measure of each angle be x.

Since the sum of the angles of a triangle is 180°, we have:

x + x + x = 180°

⇒ 3x = 180°

⇒ x = $\frac{180^\circ}{3}$

⇒ x = 60°

∴ The measure of each of the angles is 60°.

Given:

One acute angle = 40°

Since it is a right-angled triangle, one angle = 90°

Let the measure of the other acute angle be x.

Since the sum of the angles of a triangle is 180°, we have:

90° + 40° + x = 180°

⇒ 130° + x = 180°

⇒ x = 180° - 130°

⇒ x = 50°

∴ The measure of the other acute angle is 50°.

Given:

∠B = 80°, ∠C = 25°, ∠A = x°

∠B + ∠C + ∠A = 180° $\quad$[Sum of the angles of a triangle]

∴ 80° + 25° + x° = 180°

⇒ 105° + x° = 180°

⇒ x° = 180° - 105°

⇒ x° = 75°

∴ ∠A = 75°

Given:

∠P = 2x°, ∠Q = 6x°, ∠R = x°

∠P + ∠Q + ∠R = 180° $\quad$[Sum of the angles of a triangle]

∴ 2x° + 6x° + x° = 180°

⇒ 9x° = 180°

⇒ x° = $\frac{180^\circ}{9}$

⇒ x° = 20°

∠R = x° = 20°

∠P = 2x° = 2 x 20° = 40°

∠Q = 6x° = 6 x 20° = 120°

∴ ∠P = 40°, ∠Q = 120° and ∠R = 20°

Given:

∠X = x°, ∠Y = x°, ∠Z = 40°

∠X + ∠Y + ∠Z = 180° $\quad$[Sum of the angles of a triangle]

∴ x° + x° + 40° = 180°

⇒ 2x° + 40° = 180°

⇒ 2x° = 180° - 40°

⇒ 2x° = 140°

⇒ x° = $\frac{140^\circ}{2}$

⇒ x° = 70°

∴ ∠X = 70° and ∠Y = 70°

Given:

∠L = 3x°, ∠M = 90°, ∠N = 2x°

∠L + ∠M + ∠N = 180° $\quad$[Sum of the angles of a triangle]

∴ 3x° + 90° + 2x° = 180°

⇒ 5x° + 90° = 180°

⇒ 5x° = 180° - 90°

⇒ 5x° = 90°

⇒ x° = $\frac{90^\circ}{5}$

⇒ x° = 18°

∠L = 3x° = 3 x 18° = 54°

∠N = 2x° = 2 x 18° = 36°

∴ ∠L = 54° and ∠N = 36°

Given:

∠BAE = 110°, ∠ACD = 120°, ∠ABC = x°

Since, CE is a straight line:

∠BAE + ∠BAC = 180° $\quad$[Linear pair]

⇒ 110° + ∠BAC = 180° $\quad$[Substituting the value of ∠BAE]

⇒ ∠BAC = 180° - 110°

⇒ ∠BAC = 70°

Since BD is a straight line:

∠ACD + ∠ACB = 180° $\quad$[Linear pair]

⇒ 120° + ∠ACB = 180° $\quad$[Substituting the value of ∠ACD]

⇒ ∠ACB = 180° - 120°

⇒ ∠ACB = 60°

In △ABC, we have:

∠ABC + ∠BAC + ∠ACB = 180° $\quad$[Sum of the angles of a triangle]

∴ x° + 70° + 60° = 180°

⇒ x° + 130° = 180°

⇒ x° = 180° - 130°

⇒ x° = 50°

∴ ∠B = 50°

Given:

∠A = 75°, ∠B = 65°, ∠D = x°, ∠E = 110°

In △ABC, we have:

∠A + ∠B + ∠ACB = 180° $\quad$[Sum of the angles of a triangle]

∴ 75° + 65° + ∠ACB = 180°

⇒ 140° + ∠ACB = 180°

⇒ ∠ACB = 180° - 140°

⇒ ∠ACB = 40°

Now,

∠ACB = ∠DCE $\quad$[Vertically opposite angles are equal]

Now, in △CDE, we have:

∠DCE + ∠D + ∠E = 180° $\quad$[Sum of the angles of a triangle]

∴ 40° + x° + 110° = 180°

⇒ 150° + x° = 180°

⇒ x° = 180° - 150°

⇒ x° = 30°

∴ ∠D = 30°

Given interior angles:

∠A = 50°, ∠B = x°

Exterior angle: ∠BCD = 125°

According to exterior angle property, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ ∠A + ∠B = ∠BCD

⇒ 50° + x° = 125°

⇒ x° = 125° - 50°

⇒ x° = 75°

⇒ ∠B = 75°

In △ABC, we have:

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ 50° + 75° + ∠C = 180°

⇒ 125° + ∠C = 180°

⇒ ∠C = 180° - 125°

⇒ ∠C = 55°

∴ The angles of the triangle are ∠A = 50°, ∠B = 75°, ∠C = 55°

Given interior angles:

∠A = 2x°, ∠C = x°

Exterior angle: ∠CBD = 120°

According to exterior angle property, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ ∠A + ∠C = ∠CBD

2x° + x° = 120°

3x° = 120°

x° = $\frac{120^\circ}{3}$

x° = 40°

⇒ ∠A = 2x° = 2 x 40° = 80°

⇒ ∠C = x° = 40°

In △ABC, we have:

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ 80° + ∠B + 40° = 180°

⇒ 120° + ∠B = 180°

⇒ ∠B = 180° - 120°

⇒ ∠B = 60°

∴ The angles of the triangle are ∠A = 80°, ∠B = 60°, ∠C = 40°

Given interior angles:

∠B = 2x°, ∠C = 3x°

Exterior angle: ∠BAD = 80°

According to exterior angle property, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ ∠B + ∠C = ∠BAD

2x° + 3x° = 80°

5x° = 80°

x° = $\frac{80^\circ}{5}$

x° = 16°

⇒ ∠B = 2x° = (2 x 16°) = 32°

⇒ ∠C = 3x° = (3 x 16°) = 48°

In △ABC, we have:

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ ∠A + 32° + 48° = 180°

⇒ ∠A + 80° = 180°

⇒ ∠A = 180° - 80°

⇒ ∠A = 100°

∴ The angles of the triangle are ∠A = 100°, ∠B = 32°, ∠C = 48°

Given interior angles:

∠A = x°, ∠B = y°, ∠C = z°

Exterior angle: ∠CBE = 140°, ∠ACD = 130°

Since AE is a straight line:

∠CBE + ∠CBA = 180° $\quad$[Linear pair]

⇒ 140° + y° = 180°

⇒ y° = 180° - 140°

⇒ y° = 40°

⇒ ∠B = 40°

Since BD is a straight line:

∠ACD + ∠ACB = 180°

⇒ 130° + z° = 180°

⇒ z° = 180° - 130°

⇒ z° = 50°

⇒ ∠C = 50°

In △ABC, we have:

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ x° + 40° + 50° = 180°

⇒ x° + 90° = 180°

⇒ x° = 180° - 90°

⇒ x° = 90°

⇒ ∠A = 90°

∴ The angles of the triangle are ∠A = 90°, ∠B = 40°, ∠C = 50°

Given interior angles:

∠A = x°, ∠B = 55°, ∠C = y°

Exterior angle: ∠ACD = 125°

According to exterior angle property, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ ∠A + ∠B = ∠ACD $\quad$[∠A and ∠B are opposite interior angles to ∠ACD]

⇒ x° + 55° = 125°

⇒ x° = 125° - 55°

⇒ x° = 70°

⇒ ∠A = x° = 70°

In △ABC, we have:

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ 70° + 55° + y° = 180°

⇒ 125° + y° = 180°

⇒ y° = 180° - 125°

⇒ y° = 55°

⇒ ∠C = y° = 55°

∴ The angles of the triangle are ∠A = 70°, ∠B = 55°, ∠C = 55°

Given interior angles:

∠A = x°, ∠C = x°

Exterior angle: ∠CBD = 130°

According to exterior angle property, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ ∠A + ∠C = ∠CBD

x° + x° = 130°

2x° = 130°

x° = $\frac{130^\circ}{2}$

x° = 65°

⇒ ∠A = x° = 65°

⇒ ∠C = x° = 65°

In △ABC, we have:

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ 65° + ∠B + 65° = 180°

⇒ ∠B + 130° = 180°

⇒ ∠B = 180° - 130°

⇒ ∠B = 50°

∴ The angles of the triangle are ∠A = 65°, ∠B = 50°, ∠C = 65°

From the figure, we have:

∠BAC = x°, ∠ABD = 108° and ∠ACE = 116°

Since, DE is a straight line and AB is a transversal:

∠ABD + ∠ABC = 180° $\quad$[Linear pair]

⇒ 108° + ∠ABC = 180°

⇒ ∠ABC = 180° - 108°

⇒ ∠ABC = 72°

Since, DE is a straight line and AC is a transversal:

∠ACE + ∠ACB = 180° $\quad$[Linear pair]

⇒ 116° + ∠ACB = 180°

⇒ ∠ACB = 180° - 116°

⇒ ∠ACB = 64°

Now, consider △ABC:

∠ABC + ∠BAC + ∠ACB = 180° $\quad$[Sum of the angles of a triangle]

∴ 72° + x° + 64° = 180°

⇒ x° + 136° = 180°

⇒ x° = 180° - 136°

∴ x = 44°

From figure, we have:

∠ABC = x°, ∠ACD = 115° and ∠FAE = 60°

∠BAC = ∠FAE $\quad$[Vertically opposite angles]

⇒ ∠BAC = 60°

x° + ∠BAC = ∠ACD $\quad$[Exterior angle = sum of interior opposite angles]

⇒ x° + 60° = 115°

⇒ x° = 115° - 60°

∴ x = 55°

From the figure, we have:

∠ADE = 65°, ∠ACB = 35° and ∠BAC = x°

Since ED || BC and BD is a transversal:

∠ABC = ∠ADE $\quad$[Interior alternate angle]

⇒ ∠ABC = 65°

Now, consider △ABC:

∠ABC + ∠BAC + ∠ACB = 180° $\quad$[Sum of the angles of a triangle]

∴ 65° + x° + 35° = 180°

⇒ x + 100° = 180°

⇒ x = 180° - 100°

∴ x = 80°

From the figure, we have:

∠ACB = x°
∠ACD = 87°
∠ADC = 75°
∠AED = 42°
∠ABF = 126°
∠BAC = y°
∠EAD = z°

At vertex D:

75° = z° + ∠AED $\quad$[Exterior angle = sum of interior opposite angles]

⇒ 75° = z° + 42°

⇒ z° = 75° - 42°

⇒ z° = 33°

Since EF is a straight line:

∠ACB + ∠ACD = 180° $\quad$[Linear pair]

⇒ x° + 87° = 180°

⇒ x° = 180° - 87°

⇒ x° = 93°

At vertex B:

126° = x° + y° $\quad$[Exterior angle = sum of interior opposite angles]

⇒ 126° = 93° + y°

⇒ y° = 126° - 93°

⇒ y° = 33°

∴ x° = 93°, y° = 33°, z° = 33°

From the figure, we have:

∠ABC = 70°
∠ACB = 55°
∠BAC = x°
∠ADE = y°
∠AED = z°

Consider △ABC,

∠ABC + ∠BAC + ∠ACB = 180° $\quad$[Sum of the angles of a triangle]

∴ 70° + x° + 55° = 180°

x° + 125° = 180°

x° = 180° - 125°

x° = 55°

Now Consider △ADE,

x° + y° = 136° $\quad$[Exterior angle = sum of interior opposite angles]

⇒ 55° + y° = 136°

⇒ y° = 136° - 55°

⇒ y° = 81°

Since AC is a straight line:

z° + 136° = 180°

z° = 180° - 136°

z° = 44°

∴ x° = 55°, y° = 81°, z° = 44°

(i)

Consider △ABC,

The angle f is an exterior angle at vertex C

Interior opposite angles are a and b

According to exterior angle property, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ a + b = f

a = f - b

(ii)

Consider △GCD,

The angle g is an exterior angle at vertex D

Interior opposite angles are e and f

According to exterior angle property, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ e + f = g

e = g - f

(iii)

Consider △AFG,

The angle d is an exterior angle at vertex F

Interior opposite angles are a and ∠AGF

∠AGF and e are vertically opposite angles, so ∠AGF = e.

According to exterior angle property, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ d = a + e

(iv)

Consider △GCD,

The angle c is an exterior angle at vertex C

The interior opposite angles are e and ∠GDC.

Since CE is a straight line:

∠GDC + ∠GDE = 180° $\quad$[Linear pair]

∠GDC + g = 180°

∠GDC = 180° - g

According to exterior angle property, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ c = e + ∠GDC

c = e + 180° - g

Given:

AB = AC

∠A = 50°

∠B = ∠C = x [∵ angles opposite to equal sides of a triangle are equal]

Consider △ABC,

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ 50° + x + x = 180°

⇒ 2x + 50° = 180°

⇒ 2x = 180° - 50°

⇒ 2x = 130°

⇒ x = $\frac{130^\circ}{2}$

⇒ x = 65°

∴ ∠B = ∠C = 65°

Given:

PQ = PR

∠Q = 70°

∠R = ∠Q [∵ angles opposite to equal sides of a triangle are equal]

⇒ ∠R = 70°

Consider △PQR,

∠P + ∠Q + ∠R = 180° $\quad$[Sum of the angles of a triangle]

∴ ∠P + 70° + 70° = 180°

⇒ ∠P + 140° = 180°

⇒ ∠P = 180° - 140°

∴ ∠P = 40°

Given:

Since, AB || CD and BC is a transversal:

∠DCO = ∠ABO $\quad$[Interior alternate angles]

∴ ∠DCO = y°

Now, consider △OCD,

z° is an exterior angle to this triangle

∴ z° = ∠ODC + ∠DCO $\quad$[Exterior angle = sum of interior opposite angles]

z° = x° + y°

⇒ Hence proved

Let the measure of the vertical angle be x°.

According to the problem, each base angle is 15° more than the vertical angle.

Vertical angle = x°

First base angle = (x + 15)°

Second base angle = (x + 15)° (since base angles of an isosceles triangle are equal)

Consider △ABC,

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ x° + (x + 15)° + (x + 15)° = 180°

⇒ 3x° + 30° = 180°

⇒ 3x° = 180° - 30°

⇒ 3x° = 150°

⇒ x° = $\frac{150^\circ}{3}$

⇒ x° = 50°

∠A = x° = 50°

∠B = (x + 15)° = (50 + 15)° = 65°

∠C = (x + 15)° = (50 + 15)° = 65°

∴ The angles of triangle are 50°, 65° and 65°.

In an isosceles triangle, the two base angles are equal.

Let the measure of each base angle be x.

First base angle = x

Second base angle = x

According to the problem, the vertical angle is twice the sum of its base angles:

Vertical angle = 2(x + x) = 2(2x) = 4x

Consider △ABC,

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ 4x + x + x = 180°

⇒ 6x = 180°

⇒ x = $\frac{180^\circ}{6}$

⇒ x = 30°

∠A = 4x = 4 x 30° = 120°

∠B = x = 30°

∠C = x = 30°

∴ The angles of triangle are 120°, 30° and 30°.

Let the measure of the vertical angle be x°.

According to the problem, each base angle is four times the vertical angle.

Since base angles of an isosceles triangle are equal:

Vertical angle = x°

First base angle = 4x°

Second base angle = 4x°

Consider △ABC,

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ x° + 4x° + 4x° = 180°

⇒ 9x° = 180°

⇒ x° = $\frac{180^\circ}{9}$

⇒ x° = 20°

∠A = x° = 20°

∠B = 4x° = 4 x 20° = 80°

∠C = 4x° = 4 x 20° = 80°

∴ The angles of the triangle are 20°, 80°, and 80°.

Given:

The ratio between a base angle and the vertical angle is 2 : 5.

In an isosceles triangle, the two base angles are equal.

First base angle: 2x

Second base angle: 2x

Vertical angle: 5x

Consider △ABC,

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ 5x + 2x + 2x = 180°

⇒ 9x = 180°

⇒ x = $\frac{180^\circ}{9}$

⇒ x = 20°

∠A = 5x = 5 x 20° = 100°

∠B = 2x = 2 x 20° = 40°

∠C = 2x = 2 x 20° = 40°

∴ The three angles of the triangle are 40°, 40°, and 100°.

Since △ABC is an equilateral triangle, all its interior angles are equal and each measures 60°.

∠ABC = 60°.

∠ACB = 60°.

At vertex B:

∠ABE + ∠ABC = 180° $\quad$[Linear pair]

⇒ ∠ABE + 60° = 180°

⇒ ∠ABE = 180° - 60°

⇒ ∠ABE = 120° .....(Equation 1)

At vertex C:

∠ACD + ∠ACB = 180° $\quad$[Linear pair]

⇒ ∠ACD + 60° = 180°

⇒ ∠ACD = 180° - 60°

⇒ ∠ACD = 120° .....(Equation 2)

Comparing Equation 1 and Equation 2:

∠ABE = 120°

∠ACD = 120°

∴ ∠ACD = ∠ABE

⇒ Hence proved

From figure, we have:

AB = AC

∠A = 70°

∠B = ∠C = x° [∵ angles opposite to equal sides of a triangle are equal]

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ 70° + x° + x° = 180°

⇒ 70° + 2x° = 180°

⇒ 2x° = 180° - 70°

⇒ 2x° = 110°

⇒ x° = $\frac{110^\circ}{2}$

x° = 55°.

From figure, we have:

AB = BC

∠B = 90°

∠A = ∠C = x° [∵ angles opposite to equal sides of a triangle are equal]

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

∴ x° + 90° + x° = 180°

⇒ 90° + 2x° = 180°

⇒ 2x° = 180° - 90°

⇒ 2x° = 90°

⇒ x° = $\frac{90^\circ}{2}$

x° = 45°.

From figure, we have:

AB = BC

∠B = 90°

∠CAD = x°

∠CAB = ∠ACB [∵ angles opposite to equal sides of a triangle are equal]

∠ABC + ∠CAB + ∠ACB = 180° $\quad$[Sum of the angles of a triangle]

∴ 90° + ∠ACB + ∠ACB = 180° $\quad$[∵ ∠CAB = ∠ACB]

90° + 2∠ACB = 180°

2∠ACB = 180° - 90°

2∠ACB = 90°

∠ACB = $\frac{90^\circ}{2}$

∠ACB = 45°

⇒ ∠CAB = 45°

Since BD is a straight line and AC is a transversal:

∠CAD + ∠CAB = 180° $\quad$[Linear pair]

x° + 45° = 180°

x° = 180° - 45°

x° = 135°.

From figure, we have:

AB = BC

∠CBE = 110°

Since AE is a straight line and BD is a transversal:

∠ABC + ∠CBE = 180° $\quad$[Linear pair]

∠ABC + 110° = 180°

∠ABC = 180° - 110°

∠ABC = 70°

Since, AB = BC we have:

∠ACB = ∠ABC [∵ angles opposite to equal sides of a triangle are equal]

⇒ ∠ACB = 70°

In △ABC,

∠ABC + ∠CAB + ∠ACB = 180° $\quad$[Sum of the angles of a triangle]

∴ 70° + y° + 70° = 180°

⇒ 140° + y° = 180°

⇒ y° = 180° - 140°

⇒ y° = 40°

⇒ x° = y° + ∠ABC $\quad$[Exterior angle = sum of interior opposite angles]

⇒ x° = 40° + 70°

⇒ x° = 110°

x° = 110°, y° = 40°

From figure,

∠CAB = 70°, ∠ACB = y° and ∠BDA = x°

Consider △ABC, we have:

AC = BC

∠ABC = ∠CAB [∵ angles opposite to equal sides of a triangle are equal]

⇒ ∠ABC = 70°

∠ABC + ∠CAB + ∠ACB = 180° $\quad$[Sum of the angles of a triangle]

∴ 70° + 70° + y° = 180°

⇒ 140° + y° = 180°

⇒ y° = 180° - 140°

⇒ y° = 40°

Consider △ABD, we have:

AB = BD

∠BDA = ∠BAD = x° [∵ angles opposite to equal sides of a triangle are equal]

∠BDA + ∠BAD = ∠ABC $\quad$[Exterior angle = sum of interior opposite angles]

⇒ x° + x° = 70°

⇒ 2x° = 70°

⇒ x° = $\frac{70^\circ}{2}$

⇒ x° = 35°

x° = 35°, y° = 40°

From figure, we have:

∠BAC = 40°, ∠ABE = x° and ∠ACD = y°

Consider △ABC, we have:

AB = AC

∠ABC = ∠ACB [∵ angles opposite to equal sides of a triangle are equal]

∠BAC + ∠ABC + ∠ACB = 180° $\quad$[Sum of the angles of a triangle]

∴ 40° + ∠ACB + ∠ACB = 180° $\quad$[∵ ∠ABC = ∠ACB]

40° + 2∠ACB = 180°

2∠ACB = 180° - 40°

2∠ACB = 140°

∠ACB = $\frac{140^\circ}{2}$

∠ACB = 70°

⇒ ∠ABC = 70°

At vertex B,

x° = ∠ACB + ∠BAC $\quad$[Exterior angle = sum of interior opposite angles]

⇒ x° = 70° + 40°

⇒ x° = 110°

At vertex C:

y° = ∠BAC + ∠ABC $\quad$[Exterior angle = sum of interior opposite angles]

⇒ y° = 40° + 70°

⇒ y° = 110°

x° = 110°, y° = 110°

Given:

The other two sides of the right-angled triangle are 12 cm and 35 cm.

Let a = 12 and b = 35.

Let the hypotenuse be c.

c² = a² + b² $\quad$[Pythagoras' theorem]

c² = (12² + 35²) cm²

c² = (144 + 1225) cm²

c² = 1369 cm²

c = $\sqrt{1369}$ cm

c = 37 cm

∴ The length of the hypotenuse is 37 cm.

Given:

The length of one side of a right triangle: a = 24 cm

The length of its hypotenuse: c = 40 cm

Let the length of its third side = b.

c² = a² + b² $\quad$[Pythagoras' theorem]

⇒ b² = c² - a²

b² = (40² - 24²) cm²

b² = (1600 - 576) cm²

b² = 1024 cm²

b = $\sqrt{1024}$ cm

b = 32 cm

∴ The length of the third side is 32 cm.

Given:

The two legs of a right triangle are equal: a = b

The square of its hypotenuse: c² = 50 cm²

c² = a² + b² $\quad$[Pythagoras' theorem]

50 cm² = a² + a² $\quad$[∵ a = b]

50 cm² = 2a²

⇒ a² = $\frac{50}{2}$ cm²

a² = 25 cm²

a = $\sqrt{25}$ cm

a = 5 cm

∴ The length of each leg is 5 cm.

(i) a = 32 cm, b = 60 cm

Let the hypotenuse be c.

c² = a² + b² $\quad$[Pythagoras' theorem]

c² = (32² + 60²) cm²

c² = (1024 + 3600) cm²

c² = 4624 cm²

c = $\sqrt{4624}$ cm

c = 68 cm

∴ The length of the hypotenuse is 68 cm.

(ii) a = 28 m, b = 45 m

Let the hypotenuse be c.

c² = a² + b² $\quad$[Pythagoras' theorem]

c² = (28² + 45²) m²

c² = (784 + 2025) m²

c² = 2809 m²

c = $\sqrt{2809}$ m

c = 53 m

∴ The length of the hypotenuse is 53 m.

(iii) a = 32 cm, b = 24 cm

Let the hypotenuse be c.

c² = a² + b² $\quad$[Pythagoras' theorem]

c² = (32² + 24²) cm²

c² = (1024 + 576) cm²

c² = 1600 cm²

c = $\sqrt{1600}$ cm

c = 40 cm

∴ The length of the hypotenuse is 40 cm.

(i) AB = 40 cm, BC = 58 cm, CA = 44 cm

Hypotenuse = The longest side = BC

BC² = AB² + CA²

BC² = (40² + 44²) cm²

BC² = (1600 + 1936) cm²

BC² = 3536 cm²

But, BC² = (58 cm)² = 3364 cm²

Clearly, AB² + CA² ≠ BC²

∴ △ABC is not right-angled.

(ii) AB = 43 cm, BC = 35 cm, CA = 12 cm

Hypotenuse = The longest side = AB

AB² = BC² + CA²

AB² = (35² + 12²) cm²

AB² = (1225 + 144) cm²

AB² = 1369 cm²

But, AB² = (43 cm)² = 1849 cm²

Clearly, BC² + CA² ≠ AB²

∴ △ABC is not right-angled.

(iii) AB = 55 cm, BC = 73 cm, CA = 48 cm

Hypotenuse = The longest side = BC

BC² = AB² + CA²

BC² = (55² + 48²) cm²

BC² = (3025 + 2304) cm²

BC² = 5329 cm²

And, BC² = (73 cm)² = 5329 cm²

Clearly, BC² = AB² + CA²

∴ △ABC is a right-angled triangle.

Given:

A man travels 90 km due East and then 56 km due South.

Let O be the starting point of the man.

He moves from O to A due East such that OA = 90 km.

Then, he moves from A to B due South such that AB = 56 km.

Then, B is his final position. Join OB.

Now, in right △OAB, by Pythagoras Theorem, we have:

OB² = OA² + AB²

OB² = (90² + 56²) km²

OB² = (8100 + 3136) km²

OB² = 11236 km²

OB = $\sqrt{11236}$ km

OB = 106 km

∴ Distance of the man from the starting point = 106 km.

Given:

Distance of the foot of the pole from the wall (base): BC = 16 m

Height of the window from the ground (height): AB = 12 m

BC is the distance of the foot of the pole from the wall.

AB is the height of the window from the ground.

Let AC be the length of the pole.

Since the wall stands vertically to the ground, △ABC is a right-angled triangle with the right angle at B.

By Pythagoras Theorem, we have:

AC² = AB² + BC²

AC² = (12² + 16²) m²

AC² = (144 + 256) m²

AC² = 400 m²

AC = $\sqrt{400}$ m

AC = 20 m

∴ The length of the pole is 20 m.

Given:

The sides of the rectangle are 21 cm and 20 cm.

Let ABCD be the rectangle where AB = 21 cm and BC = 20 cm.

Let AC be the diagonal.

Since every interior angle of a rectangle is a right angle (90°), △ABC is a right-angled triangle with the right angle at B. The diagonal AC acts as the hypotenuse.

By Pythagoras Theorem, we have:

AC² = AB² + BC²

AC² = (21² + 20²) cm²

AC² = (441 + 400) cm²

AC² = 841 cm²

AC = $\sqrt{841}$ cm

AC = 29 cm

∴ The length of the diagonal of the rectangle is 29 cm.

(i) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

(ii) If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is right-angled.

(iii) Of all the line segments that can be drawn to a given line from a given point outside it, the perpendicular is the shortest.

Pythagoras' Theorem states that:

In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Given:

∠A = 15° and ∠B = 55°

∠A + ∠B + ∠C = 180° $\quad$[Sum of the angles of a triangle]

15° + 55° + ∠C = 180°

70° + ∠C = 180°

∠C = 180° - 70°

∠C = 110°

Hence, option 4 is the correct option.

Given:

Two angles of a triangle are equal.

Third angle is three times each equal angle.

Let each equal angle be x.

The third (largest) angle = 3x.

x + x + 3x = 180° $\quad$[Sum of the angles of a triangle]

5x = 180°

x = $\frac{180^\circ}{5}$

x = 36°

Largest angle = 3x = 3 x 36° = 108°

Hence, option 2 is the correct option.

For three angles to form a triangle, their sum must be exactly 180°.

Let's check the options:

1. 56° + 64° + 60° = 180°

2. 42° + 67° + 81° = 190°

3. 45° + 70° + 65° = 180°

4. 63° + 35° + 82° = 180°

The angles in option 2 cannot be the angles of a triangle, because it sums up to 190°.

Hence, option 2 is the correct option.

Given:

Exterior angle = 136°

One interior angle = 69°

Let the other interior angle be x.

x + 69° = 136° $\quad$[Exterior angle = sum of interior opposite angles]

x = 136° - 69°

x = 67°

The other interior angle of a triangle is 67°.

Hence, option 1 is the correct option.

A triangle cannot have two straight angles (180° each), as the total would exceed 180°.

A triangle cannot have two obtuse angles (> 90° each) or two right angles (90° each) because the sum of just those two would be ≥ 180°, leaving no room for the third angle.

Every triangle must have at least two acute angles (< 90°).

Hence, option 4 is the correct option.

From figure, we have:

Exterior angle: ∠ACD = 105°

Interior angles: ∠BAC = 2x and ∠ABC = x

∠ABC + ∠BAC = ∠ACD $\quad$[Exterior angle = sum of interior opposite angles]

x + 2x = 105°

3x = 105°

x = $\frac{105^\circ}{3}$

x = 35°

Hence, option 3 is the correct option.

From figure, we have:

Exterior angle: ∠QRS = 111°

Interior angles: ∠PQR = (7x - 3°) and ∠QPR = (4x - 7°)

∠PQR + ∠QPR = ∠QRS $\quad$[Exterior angle = sum of interior opposite angles]

(7x - 3°) + (4x - 7°) = 111°

11x - 10° = 111°

11x = 111° + 10°

11x = 121°

x = $\frac{121^\circ}{11}$

x = 11°

Hence, option 3 is the correct option.

To be a right-angled triangle, the sides must satisfy the Pythagoras Theorem (a² + b² = c²), where c is the longest side.

Let's check the options:

1. 4² + 3² = (16 + 9) = 25 and 5² = 25

2. 12² + 5² = (144 + 25) = 169 and 13² = 169

3. 15² + 21² = (225 + 441) = 666 and 27² = 729

4. 35² + 12² = (1225 + 144) = 1369 and 37² = 1369

The sides in option 3 cannot be the sides of a right-angled triangle. Because, it does not satisfy a² + b² = c².

Hence, option 3 is the correct option.

Hypotenuse (Ladder) = 15 m

Base (Distance from wall) = 12 m

Height (Wall) = h

Using Pythagoras Theorem:

h² + 12² = 15²

h² + 144 = 225

h² = 225 - 144

h² = 81

h = $\sqrt{81}$

h = 9 m

Hence, option 3 is the correct option.

In an isosceles right-angled triangle:

One angle is 90°.

The other two angles are equal. Let each be x.

x + x + 90° = 180° $\quad$[Sum of the angles of a triangle]

2x + 90° = 180°

2x = 180° - 90°

2x = 90°

x = $\frac{90^\circ}{2}$

x = 45°

Hence, option 2 is the correct option.

(i) In a scalene triangle the measures of all the three angles are different.

(ii) The longest side of a right-angled triangle is hypotenuse.

(iii) If two angles of a triangle are 36° and 63°, then the third angle is 81°.

(iv) The sum of the two acute angles in a right-angled triangle is 90°.

(v) If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is right-angled.

Explanation

(iii)

Given:

Two angles of a triangle are 36° and 63°.

Let the third angle be x.

36° + 63° + x = 180° $\quad$[Angle sum property]

99° + x = 180°

x = 180° - 99°

x = 81°

(iv)

Since a right-angled triangle already has one 90° angle, the remaining two angles must add up to 90° to reach the 180° total.

(i) False
Reason — A triangle can have at most one angle greater than 90°. If it had two obtuse angles (e.g., 91° and 91°), their sum alone would be 182°, which exceeds the Angle Sum Property limit of 180° for a triangle.

(ii) False
Reason — If all three angles were less than 60° (e.g., 59° each), their sum would be less than 177°. For a triangle to exist, the sum must be exactly 180°. In an acute triangle, at least one angle must be 60° or greater.

(iii) False
Reason — According to Pythagoras' Theorem, the side that is by itself in the equation ($BC^2$) is the hypotenuse. The right angle is always opposite the hypotenuse. Since the hypotenuse is BC, the vertex opposite to it is A. Therefore, the right angle is ∠A.

(iv) True
Reason —

Given:

The two angles of triangle are 30° and 70°.

Let the third angle be x.

30° + 70° + x = 180° $\quad$[Angle sum property]

100° + x = 180°

x = 180° - 100°

x = 80°

Since all three angles (30°, 70° and 80°) are different, all three sides must also be different lengths. A triangle with no equal sides is a scalene triangle.

(v) False
Reason — In a right-angled triangle, one angle is 90°.
Because the total sum is 180°, the other two acute angles must add up to 90° (180° - 90° = 90°). These two angles are complementary, not 45° unless the triangle is also isosceles.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Explanation

Given:

Let the angles be x, y and z.

According to the given statement, One angle is equal to the sum of other two.

∴ x = y + z

We know that,

x + y + z = 180° $\quad$[Angle sum property]

x + x = 180° $\quad$[∵ x = y + z]

2x = 180°

x = $\frac{180^\circ}{2}$

x = 90°

So, it is a right-angled triangle.

The statement in reason is correct according to angle sum property.

Both statements are true, and the Reason explains why the Assertion is true.

Hence, option 1 is the correct option.

Assertion (A) is false but Reason (R) is true.

Explanation

The equation AB² + AC² = BC² is the Pythagoras Theorem. According to this theorem, the angle opposite to the longest side (BC) must be exactly 90° (a right angle), not an obtuse angle. Therefore, the Assertion is false.

The Reason is a correct statement of the Pythagoras theorem.

Hence, option 4 is the correct option.

Assertion (A) is true but Reason (R) is false.

Explanation

In △ABC, x + 30° + y = 180° $\quad$[Angle sum property]

Since AB || DC,

z = ∠ABC $\quad$[Corresponding angles]

z = 30°

Again, x + y + z = 180°. So, the Assertion is true.

The exterior angle property states an exterior angle is equal to the sum of its two interior opposite angles. The phrase "any two interior angles" is mathematically incorrect.

Thus, the Reason is false.

Hence, option 3 is the correct option.

For three lengths to form a triangle:

(a − 3) + 2a > a + 5

3a − 3 > a + 5

2a > 8

a > 4

Also,

a − 3 > 0 ⇒ a > 3

So a must be greater than 4.

Among the options, only 5 satisfies this.

Hence, option 4 is the correct option.

For any triangle:

Difference of two sides < third side < sum of two sides

Given sides:

20 cm and 21 cm

So the third side lies between:

21 − 20 and 21 + 20

Hence, option 3 is the correct option.

From the figure:

Angles 108° and (5x+10) form a linear pair.

108° + (5x + 10)° = 180°

5x + 118° = 180°

5x = 180° - 118°

5x = 62°

x = $\frac{62°}{5}$

x = 12.4

Angles 40° and (4y+12) form a linear pair.

40° + (4y + 12)° = 180°

4y + 52° = 180°

4y = 180° - 52°

4y = 128°

y = $\frac{128°}{4}$

y = 32

Angles 32° and (2z+8) form a linear pair.

32° + (2z + 8)° = 180°

2z + 40° = 180°

2z = 180° - 40°

2z = 140°

z = $\frac{140°}{2}$

z = 70

Finding x + y + z:

x + y + z = 12.4 + 32 + 70

= 114.4

Among these options no option matches the answer. So, none of these options is correct.

From the figure,

In △BSD,

∠SBD + ∠BSD + ∠BDS = 180° $\quad$[Sum of angles in a triangle]

30° + ∠BSD + 36° = 180°

66° + ∠BSD = 180°

∠BSD = 180° - 66°

∠BSD = 114°

∠BSD + ∠AST = 180° $\quad$[Linear pair]

114° + ∠AST = 180°

∠AST = 180° - 114°

∠AST = 66°

In △CTE,

∠ECT + ∠CET + ∠CTE = 180° $\quad$[Sum of angles in a triangle]

35° + 31° + ∠CTE = 180°

66° + ∠CTE = 180°

∠CTE = 180° - 66°

∠CTE = 114°

∠CTE + ∠ATS = 180° $\quad$[Linear pair]

114° + ∠ATS = 180°

∠ATS = 180° - 114°

∠ATS = 66°

∠ATS = x

∴ x = 66°

In △ATS,

∠ATS + ∠SAT + ∠AST = 180°

x + y + ∠AST = 180°

66° + y + 66° = 180°

132° + y = 180°

y = 180° - 132°

y = 48°

⇒ x = 66° and y = 48°

Hence, option 2 is the correct option.

In the figure, QS is extended to T and RQ is extended to P, so ∠RST and ∠PQS are exterior angles of triangle QRS at vertices S and Q respectively.

Exterior angle at S:

∠RST = ∠RQS + ∠QRS

Given ∠RST = 2∠QRS:

∠RQS + ∠QRS = 2∠QRS

⇒ ∠RQS = ∠QRS

Exterior angle at Q:

∠PQS = ∠QRS + ∠QSR

Given ∠PQS = 2∠QSR:

∠QRS + ∠QSR = 2∠QSR

⇒ ∠QRS = ∠QSR

Combining the two results:

∠RQS = ∠QRS = ∠QSR

Since all three angles are equal (each = 60°), triangle QRS is equilateral.

Hence, option 3 is the correct option.

In a right-angled isosceles triangle, the two legs are equal.

Let each leg be a.

By the Pythagoras theorem:

(hypotenuse)² = a² + a²

= 2a²

So,

square on hypotenuse = 2 × (square on one leg)

Hence, option 1 is the correct option.

In an isosceles triangle, two angles are equal.

If each equal angle is 52°, then the third angle is:

180° - (52° + 52°)

= 180° - 104°

= 76°

The greatest angle is 76°

So, Statement–1 is true.

An exterior angle of a triangle equals the sum of the two interior opposite angles.

If the exterior angle is 90°, then:

sum of interior opposite angles = 90°

Hence, each of them must be less than 90°, i.e., acute.

So, Statement–2 is also true.

Hence, option 3 is the correct option.

In △ADB,

a + c + f = 180° ...(i) $\quad$[Sum of angles in a triangle]

In △ADC,

b + e + d = 180° ...(ii) $\quad$[Sum of angles in a triangle]

Adding (i) and (ii):

a + c + f + b + e + d = 180° + 180°

a + b + c + d + e + f = 360°

Hence, option 3 is the correct option.