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Given angles are 38°, 102°, 40° 

Sum of the angles = 38° + 102° + 40° = 180° 

So, 38°, 102°, 40° can form a triangle.

Given angles are 60°, 70°, 80°. 

Sum of the angles = 60° + 70° + 80° 

= 210° >180° 

So, 60°, 70°, 80° cannot form a triangle.

Correct answer is (c) 120°

(c) 120^o

We know that, exterior angle is equal to sum of opposite interior angle.

From the figure,

∠ACD = ∠A + ∠B

∠ACD = 25^o + 35^o

= 60°

Then, in another triangle

x is exterior angle

∴ x = 60° + ∠ACD

x = 60° + 60°

x = 120°

(c) 3 cm and 16 cm

From the question it is given that, the length of two sides of a triangle are 7 cm and 9 cm.

Let us assume the length of the third side of the triangle be ‘P’.

We Know that, the sum of the two sides of the triangle is greater than the third side.

So, 7 + 9 > P

16 > P

Now, difference between two sides = 9 – 7 = 2

Therefore, the third side is greater than 2 and smaller than 16.

i.e. 3 cm and 16 cm

(b) 80^o

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

So, ∠ACD = ∠A + ∠B

As ΔACB is isosceles triangle with AC = BC

Therefore, ∠A must be equal to ∠B

 ∠ACD = 40^o + 40^o

= 80^o

∠PRS is an exterior angle of ∆PQR. 

So from the theorem of remote interior angles, 

∠PRS = ∠PQR + ∠QPR 

= 40° + 30° 

∴ ∠PRS = 70° 

∴ ∠TRS=70° …[P – T – R]

 In ∆RTS, 

∠TRS + ∠RTS + ∠TSR = 180° …[Sum of the measures of the angles of a triangle is 180°] 

∴ 70° + ∠RTS + 20° = 180° 

∴ ∠RTS + 90° = 180°

∴ ∠RTS = 180° 

∴ ∠RTS = 90°

Correct answer is (b) 80°

Equilateral

A triangle is said to be equilateral, if each one of its sides has the same length.

Correct answer is 50°

PQO: 

In ∆ARO and ∆PQO,

∠ARO = ∠PQO = 55°

∠AOR = ∠POQ [Vertically opposite angles]

∴ ∠RAQ = ∠QPO

[∵ If two angles of a triangle are equal to two angles of another triangle then third angle is also equal]

AO = PO = 2.5 cm

∴ ∆ARO ≅ ∆PQO [ASA criterion]

Correct answer is Triangle, Obtuse angled triangle

It is given that AD divides ∠BAC in the ratio 1: 3

So let us consider ∠BAD and ∠DAC as y and 3y

According to the figure we know that BAE is a straight line

From the figure we know that ∠BAC and ∠CAE form a linear pair of angles

So we get

∠BAC + ∠CAE = 180^o

We know that

∠BAC = ∠BAD + ∠DAC

So it can be written as

∠BAD + ∠DAC + ∠CAE = 180^o

By substituting the values we get

y + 3y + 108^o = 180^o

On further calculation

4y = 180^o – 108^o

By subtraction

4y = 72^o

By division

y = 72/4

y = 18^o

We know that the sum of all the angles in triangle ABC is 180^o.

So we can write it as

∠ABC + ∠BCA + ∠BAC = 180^o

It is given that AD = DB so we can write it as ∠ABC = ∠BAD

From the figure we know that ∠BAC = y + 3y = 4y

By substituting the values

y + x + 4y = 180^o

On further calculation

5y + x = 180^o

By substituting the value of y

5 (18^o) + x = 180^o

By multiplication

90^o + x = 180^o

x = 180^o – 90^o

By subtraction we get

x = 90^o

Therefore, the value of x is 90.

Correct answer is y = 51°, x = 129°

Correct option is (B) 135°

\(\because\) An exterior angle of a triangle is equal to the sum of its opposite interior angle.

\(\therefore\) \(x=55^\circ+80^\circ\)

\(=135^\circ\)

Correct option is: (B) 135°

Correct answer is (A) 10 cm

Correct answer is (C) PR² + QR² = PQ²

Here AB² = 576, BC² = 100 and AC² = 676. So, AC² = AB² + BC²

Hence, the given triangle is a right triangle.

(c) BC. DE = AB. EF 

∆ABC ~ ∆EDF

Therefore, 

AB/DE = AC/EF = BC/DF 

⇒ BC. DE ≠ AB. EF

False

Since, two acute angles may vary from 0° to 89°. So, two acute angles of different measure are not congruent.

True

Two right angles are congruent.

False

Since, there is no congruency criterion of three angles.

Two squares are congruent, if they have same side.

Correct option is (B) Circles

Two circles are always similar but two triangles, trapezium and rectangles need not be always similar.

Correct option is: (B) Circles

True

If the areas of two squares is same, they are congruent.

False

Since, two figures are congruent, if they have the same shape and same size.

True

Two circles having the same circumference are congruent.

True

Two squares having the same perimeter are congruent.

Correct option is (A) congruent

Figures having same shape and size are called congruent figures.

A) congruent

Correct option is (A) shape

Two figures are said to be similar if they have same shape.

Correct option is: (A) shape

Correct option is (C) ∠C

If \(\triangle PQR\) \(\cong\) \(\triangle ABC\)

Then \(\angle P=\angle A,\angle Q=\angle B\;and\;\angle R=\angle C\)

Correct option is  C) ∠C

Let the first angle be x, second angle is x +10° and third angle is x + 20° x + x+ 10° + x + 20° = 180° 

[Sum of the angles of a triangle 180° ] 

3x + 30° = 180° 

3x = 180° – 30° 

3x = 150° 

x = \(\frac{150^o}{3}\)= 50° 

x = 50° 

∴ First angle x = 50° 

Second angle = x + 10° = 50° + 10° = 60° 

Third angle = x + 20° = 50° + 20° = 70° 

Three angles are 50°, 60° & 70°

∠A + ∠B + ∠C = 180° 

[Sum of the angles of a triangle 180° ] 

x + 15° + x – 15° + x + 30° = 180° 

3x + 30° = 180° 

3x = 180° – 30° 

3x = 150° x = \(\frac{150^o}{3}\)= 50° 

∠A = x +15° = 50° + 15° = 65° 

∠B = x – 15° = 50° – 15° = 35° 

∠C = x + 30° = 50° + 30° = 80° 

∴ The angles are 65°, 35° and 80°

Let the common ratio be ‘x’ 

The three angles are x, 2x and 3x 

x + 2x + 3x = 180° 

[Sum of the angles of a triangle 180°] 

6x = 180° 

x = \(\frac{180^o}{6}\) = 30° 

x = 30° 

2x = 2 × 30° = 60° 

3x = 3 × 30° = 90° 

∴ The angles are 30°, 60° and 90°

In an isosceles triangle, the base angles are equal, Let the each base angle be ‘x’ 

∠A +∠B +∠C = 180° 

50° + x + x = 180° 

[Sum of the angles of a traingle 180°] 

50° + 2x = 180° 

2x = 180° – 50° 

2x = 130° 

x = \(\frac{130^o}{2}\)

 x = 65° 

∴The other two angles are equal to 65° & 65°

Let the angles be ∠A, ∠B and ∠C 

Then ∠A = 90°,∠B = 35° and∠C = ? 

∠A + ∠B + ∠C = 180° 

90° + 35° + ∠C = 180° 

125° + ∠C = 180° 

∠C = 180° – 125° = 55° 

∠C = 55°

1. – c 

2. – d 

3. – a 

4. – b

∠A +∠B + ∠C = 180° 

[Sum of the angles of a triangle 180°] 

55 + 40 + ∠C = 180° 

95 + ∠C = 180“ 

∠C = 180 – 95 

∠C = 85°

 From the above table, we can conclude that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

False

Let two angle of a triangle be 90° and third angle be x. Then

90° + 90° + x = 180° (Angle sum property)

⇒ x = 180° – 180° = 0°

which is not possible.

False

Let two angles x and y of a triangle are obtuse.

Then x > 90° ………..(i)

and y > 90° …………. (ii)

From (i) and (ii), we have x + y > 180°

But in a triangle sum of all angles can not be greater than 180° .

∴ Triangle is not possible.

In the given figure, 

∠HAC = ∠TAE = a (∵ Vertically opposite angles) 

In ∆AHC, 

we know, ∠H + ∠C + ∠HAC 180° 

⇒ 60° + 80° + a = 180°

⇒ 140° + a – 140° = 180°- 140° … a = 40°. 

∴ ∠HAC = ∠TAE = a = 40° 

In ∆ATE, we know, 

∠T + ∠TAE + ∠E = 180° 

⇒ x + a + 70° = 180° 

⇒ x + 40° + 70° = 180° 

⇒ x + 110°- 110° = 180°- 110° 

∴ x = 70°