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Answer is (C) 50°

Correct option is (D) Q, S

P & R and Q & S are opposite vertices in quadrilateral PQRS.

Correct option is  D) Q, S

ABCD is a rhombus in which ∠A = 60° then ZC = 60° 

[ ∵ ∠A, ∠C are the opposite angles, which are equal in a rhombus] 

∠A + ∠B = 180° [Sum of adjacent angles of a rhombus are supplementary] 

60° + ∠B = 180° 

∴ ∠B = 180° – 60° = 120°

(a) 8

Now let us assume number of sides of a regular polygon be n.

WKT, sum of all exterior angles of all polygons is equal to 360^o.

Form the question it is given that each exterior angle has a measure of 45^o.

Then,

n × 45^o = 360^o

n = 360^o/45^o

n = 8

(b) 120^o

We know that, sum of all interior angle of quadrilaterals is equal to 360^o.

Let us assume the fourth angle be x

Then,

80^o + 80^o + 80^o + x = 360^o

240^o + x = 360^o

x = 360^o – 240^o

x = 120^o

(b) 135^o

We know that, sum of interior angles of quadrilateral is equal to 360^o.

From the question it is given that, three angles of a quadrilateral are each equal to 75^o.

Let us assume the fourth angle be x.

Then, 75^o + 75^o + 75^o + x = 360^o

225 + x = 360^o

x = 360^o – 225

x = 135^o

(a) 180°

By property of the parallelogram, we know that, the sum of adjacent angles of a parallelogram is 180°.

The measure of each exterior angle of a regular polygon of 18 sides is 20^o.

We know that, the measure of each exterior angle of a regular polygon is 360^o/n.

Where ‘n’ is the number of sides in the polygon,

Then, polygon has 18 sides, i.e. n = 18

So, 360^o/18

= 20^o

Let x be one the three equal angles.

Sum of all the angles of a quadrilateral = 360^o

⇒ x + x + x + 120^o = 360^o

⇒ 3x = 360^o -120^o

⇒ 3x = 240^o

⇒ x = 80^o

Thus, the measure of each of the equal angle is 80^o.

Given: AO = 5, BO = 12 and AB = 13. 

To prove: ABCD is a rhombus.

Proof:

AO = 5, BO = 12, AB = 13 [Given] 

AO² + BO² = 5² + 12² = 25 + 144 

∴ AO² + BO² = 169 …..(i) 

AB² = 13² = 169 ….(ii) 

∴ AB² = AO² + BO² [From (i) and (ii)] 

∴ ∆AOB is a right-angled triangle. [Converse of Pythagoras theorem] 

∴ ∠AOB = 90° 

∴ seg AC ⊥ seg BD …..(iii) [A-O-C] 

∴ In parallelogram ABCD, 

∴ seg AC ⊥ seg BD [From (iii)] 

∴ ABCD is a rhombus. [A parallelogram is a rhombus perpendicular to each other]

Option : (D)

The two diagonals are equal in a rectangle

Option : (D)

Diagonals necessarily bisect opposite angles in a square.

Option : (B)

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a rhombus

Option : (B)

The figure formed by joining the mid-points of the adjacent sides of a rhombus is a rectangle.

Option : (B)

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a parallelogram.

Option : (B)

The figure formed by joining the mid-points of the adjacent sides of a square is a square.

Correct option is  C) 130°

Given, 

ABCD is a rectangle 

∠ABD = 40^∘ 

∠ABD + ∠DBC = 90^∘ 

(angles of rectangle) 

∠DBC = 90^∘ – 40^∘ = 50^∘

Each angle of a rectangle = 90°

So, ∠ABC = 90°

∠ABD = 40° (given) 

Now, ∠ABD + ∠DBC = 90°

40° + ∠DBC = 90°

or ∠DBC = 50°

ABCD is a rectangle and ∠BAC = 32° (given)

We know, diagonals of a rectangle bisects each other.

AO = BO

∠DBA = ∠BAC = 32° (Angles facing same side) 

Each angle of a rectangle = 90 degrees 

So, ∠DBC + ∠DBA = 90° 

or ∠DBC + 32° = 90° 

or ∠DBC = 58°

PQRS is a square. 

⇒ All side are equal, and each angle is 90° degrees and diagonals bisect the angles. 

So, ∠SRP = 1/2 (90°) = 45°

two included

We cap determine a quadrilateral uniquely, if three sides and two included angles are given.

5

To construct a unique quadrilateral, we require 5 measurements, i.e. four sides and one angle or three sides and two included angles or two adjacent sides and three angles are given.

True

A rhombus can be constructed uniquely, if both diagonals are given.

(b) rectangle

We know that, if in a parallelogram one angle is of 90°, then all angles will be of 90° and a parallelogram with all angles equal to 90° is called a rectangle.

True

A quadrilateral can be drawn, if all four sides and one angle is known.

False

Diagonals of a rectangle does not bisect each other.

True

A quadrilateral can be drawn, if three sides and two diagonals are given.

True

We can construct a unique quadrilateral with given three angles given and two included sides.

True

It is the property of a parallelogram.

Draw lines AD and DC such that AD||BC, AB||DC AD = BC, AB = DC

ABCD is a rectangle as opposite sides are equal and parallel to each other and all the interior angles are of 90º.

In a rectangle, diagonals are of equal length and also these bisect each other. Hence, AO = OC = BO = OD

Thus, O is equidistant from A, B, and C.

(i) Sum of all exterior angles of the given polygon = 360º Each exterior angle of a regular polygon has the same measure. Thus, measure of each exterior angle of a regular polygon of 9 sides 

= 360°/9 = 40°

(ii) Sum of all exterior angles of the given polygon = 360º Each exterior angle of a regular polygon has the same measure. Thus, measure of each exterior angle of a regular polygon of 15 sides

= 360°/15 = 24°

Sum of all exterior angles of the given polygon = 360º Measure of each exterior angle = 24º

Thus, number of sides of the regular polygon

= 360°/24° = 15

(i) Consider △ ABC and △ ADC

It is given that AB = AD and BC = DC

AC is common i.e. AC = AC

By SSS congruence criterion

△ ABC ≅ △ ADC ……… (1)

∠ BAC = ∠ DAC (c. p. c. t)

So we get

∠ BAE = ∠ DAE

We know that AC bisects the ∠ BAD i.e. ∠ A

So we get

∠ BCA = ∠ DCA (c. p. c. t)

It can be written as

∠ BCE = ∠ DCE

So we know that AC bisects ∠ BCD i.e. ∠ C

(ii) Consider △ ABE and △ ADE

It is given that AB = AD

AE is common i.e. AE = AE

By SAS congruence criterion

△ ABE ≅ ∠ ADE

BE = DE (c. p. c. t)

(iii) We know that △ ABC ≅ △ ADC

Therefore, by c. p. c. t ∠ ABC = ∠ ADC

(i) We know that the line segment QB can be written as

QB = BC – QC

Since ABCD is a square we know that BC = DC and QC = DR

So we get

QB = CD – DR

From the figure we get

QB = RC

(ii) Consider △ PBQ and △ QCR

It is given that PB = QC

Since ABCD is a square we get

∠ PBQ = ∠ QCR = 90^o

By SAS congruence criterion

△ PBQ ≅ △ QCR

PQ = QR (c. p. c. t)

(iii) We know that PQ = QR

Consider △ PQR

From the figure we know that ∠ QPR and ∠ QRP are base angles of isosceles triangle

∠ QPR = ∠ QRP

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

∠ QPR + ∠ QRP + ∠ PRQ = 180^o

By substituting the values in the above equation

∠ QPR + ∠ QRP + 90^o = 180^o

On further calculation

∠ QPR + ∠ QRP = 180^o – 90^o

By subtraction

∠ QPR + ∠ QRP = 90^o

We know that ∠ QPR = ∠ QRP

So we get

∠ QPR + ∠ QPR = 90^o

By addition

2 ∠ QPR = 90^o

By division

∠ QPR = 45^o

Therefore, it is proved that ∠ QPR = 45^o.