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The number of right angles in a straight angle is Two  and that in a complete angle is Four.

The number of common points in the two angles marked in Fig. 2.21 is Four. 

(D) 14

From the formula, diagonal = n(n – 3)/2

Where n = number of sides in a polygon.

There are 7 sides in a heptagon.

So, d = 7(7 – 3)/2

= 7(4)/2

= 28/2

= 14

The number of diagonals in a hexagon is 9.

We know that, hexagon has 6 sides.

By using the formula, number of diagonals = n(n – 3)/2

= 6(6 – 3)/2

= 6(3)/2

= 18/2

= 9

An angle greater than 180° and less than a complete angle is called Reflex angle

A pair of opposite sides of a trapezium are  Parallel

Correct answer is (D)-14

The number of diagonals in a hexagon is 9

From the figure, by Pythagoras theorem,

x² = 20² + 15² = 400 + 225 = 625

x² = 25² ⇒ x = 25ft.

∴ The length of the support cable required to support the tower with the floor is 25 ft.

1. It is a parallelogram.
2. Yes
3. Yes
4. no
5. yes

In this right angled triangle ΔABC, length of the altitude drawn from vertex A is the leg AB itself. By Pythagoras theorem.

AC² = AB² + BC²

13² = AB² + 12²

169 = AB² + 144

AB² = 169 – 144 = 25

AB² = 52

AB = 5cm

(i) AD = Altitude

(ii) l₁ = Perpendicular bisector

(iii) BD = Median

(iv) CD = Angular bisector

Exterior of the triangle.

(i) Sum of two smaller sides of the triangle
= 6 + 4 = 10 cm > 8 cm

It is greater than the third side. So, a triangle can be formed scalene triangle.

(ii) Sum of two smaller sides of the triangle
= 8 + 5 = 13 cm > 10 cm

It is greater than the third side. So, a triangle can be formed scalene triangle.

(iii) Sum of two smaller sides of the triangle
= 1.3 + 3.5 = 4.8 cm < 6.2 cm

It is not greater than the third side. So, a triangle cannot be formed.

(iv) Two sides are equal.

So, a triangle can be formed. Isosceles triangle.

(v) Three sides are equal.

So, a triangle can be formed equilateral triangle.

(vi) Sum of two smaller sides of the triangle
= 4 + 5 = 9 cm = 9 cm

It is equal to the third side. No, a triangle cannot be formed.

Vertex containing 90°

(a) AB, BC, CA

(b) ∠ABC, ∠BCA, ∠CAB or ∠A, ∠B, ∠C

(c) A, B, C

(i) Equilateral Triangle

(ii) Scalene Triangle

(iii) Isosceles Triangle

(iv) Scalene Triangle

(a) Isosceles triangle

(b) Equilateral triangle

(c) Right angled triangle

(d) Isosceles triangle

(i) Acute angled triangle

(ii) Right angled triangle

(iii) Obtuse angled triangle

(iv) Acute angled triangle

New adjacent angles are formed. 

The new angles become smaller in measure. But their sum is 180° as it is a linear angle.

(i) Given one angle 84° 

∴ Other angle of the linear pair is 180° – 84° = 96° 

(ii) One angle is given as 86° 

Other angle of linear pair is 180° – 86° = 94° 

(iii) Given one angle = 159° 

Other angle of the linear pair = 180° – 159° = 21°

Sum of angles at a point = 360° 

∴ x° + 2x° + 3x° + 4x° + 5x° = 360° 

15x° = 360° 

x° = \(\frac{360°}{15}\)

x° = 24°. 

∴ The largest angle = 5x° 

= 5 x 24° = 120° 

The largest angle is 120°

(iii) SSA rule

(iv) Interior angles on the same side of the transversal are supplementary.

Given ∠PQR = 150° 

∠QPS = 30°

They are supplementary angles, 

But they are not adjacent angles as they don’t have common vertex or common arm. 

∴ They are not a linear pair.

(i) Reflection

(ii) Rotation

(iii) Translation

From the figure x° and 125° are vertically opposite angles. 

So they are equal i.e. x° = 125° 

Also y° and 125° are linear pair of angles. 

∴ y° + 125° = 180° 

y° + 125° – 125° = 180°- 125° 

y° = 55° 

x° = 125°

(iv) equal in measure

(iii) one common arm, one common vertex, no common interior

(i) It is translation 

(ii) Reflection about horizontal line. 

(iii) Reflection about vertical line. 

(iv) Rotation about the heel.

In ∆ABC and ∆EBD,

AB = EB 

BC = BD 

∠ABC = ∠EBD [∵ Vertically opposite angles] 

By SAS congruency criteria. ∆ABC ≅ ∆EBD. 

We know that corresponding parts of congruent triangles are congruent. 

∴ ∠BCA ≅ ∠BDE and 

∠BAC ≅ ∠BED 

∠BCA ≅ ∠BDE means that alternate interior angles are equal if CD is the transversal to lines AC and DE. 

Similarly, if AE is the transversal to AC and DE, we have ∠BAC ≅ ∠BED 

Again interior opposite angles are equal. 

We can conclude that AC is parallel to DE.

Given AB || CD 

∴ Z = 42 (∵ Alternate interior angles) 

Also y = 42° [vertically opposite angles] 

Also x° + 63° + z° = 180° 

x° + 63° + 42° = 180° 

x° + 105° = 180° 

x° + 105° – 105° = 180° – 105° 

x° = 75° 

∴ x = 75°; y = 42°; z = 42°

Given GH || IZ 

∠1 = 108° 

∠2 = 123° 

∠1 + ∠KGH = 180 [linear pair] 

108° + ∠KGH = 180°

108° + ∠KGH – 108° = 180° – 108° 

∠KGH = 72° 

∠KGH = x° (corresponding angles if KG is a transversal)

∴ x° = 72° 

Similarly 

∠2 + ∠GHK = 180° (∵ linear pair) 

123° + ∠GHK = 180° 

123° + ∠GHK – 123° = 180° – 123° 

∠GHK = 57° 

Again ∠GHK = y° (corresponding angles if KH is a transversal)

y = 57° 

x° + y° + z° = 180° (sum of three angles of a triangle is 180°)

72° + 57° + z° = 180° 

129° + z° = 180° 

129° + z° – 129° = 180° – 129° 

z = 51° 

x = 72°, y = 57°, z = 51°

Answer is (ii) 68°

(iii) Linear pair

From the picture 

∠2 + 65° = 180° [Sum of interior angles on the same side of a transversal] 

x + 39° + 65° = 180° 

x + 104° = 180° 

x + 104° – 104° = 180° – 104° 

x = 76° 

Also from the picture 

∠1 = 65° [alternate exterior angles]

2x – 3y = 65° 

2 (76) – 3y = 65°

152° – 3y = 65° 

152° – 3y – 152° = 65 – 152° 

-3y = -87

y = \(\frac{-87}{-3}\)

y = 29°

x = 76°; y = 29° 

(c) 180°

No, they are not adjacent pairs.

No, two lines cannot intersect in more than one point.

There is no right, left, up or down movement took place.

The rule bind here in 3→ , 1↓

(i) Image of ∠L is ∠L’ , Image of ∠M is ∠M’ , 

Image of ∠N is ∠N’ , Image of ∠O is ∠O’ 

Image of vertex L is L’ , Image of vertex M is ∠M’ 

Image of vertex N is ∠N’ , Image of vertex O is O’ 

(ii) Corresponding sides are LM and L’M’ , MN and M’N’ , NO and N’O’ and OL and O’L

(iv) same shape and same size

From the given figures 

∠ABC = 50° 

∠EFG = 120° 

∠HIJ = 120° 

∠KLH = 90° 

∠PON = 50° 

∠RST = 90°

From the above measures, we can conclude that

(i) ∠ABC = ∠PON 

(ii) ∠EFG = ∠HIJ 

(iii) ∠KLH ≅ ∠RST