Geometrical Proofs

Here, we have given the length of two sticks, i.e.

First stick = 4cm and second stick = 6cm

Now, let length of the third stick be L

It is given that the figure made through these sticks will be like Triangle.

So, by Triangle Inequality Theorem

Now, we have a = 6cm and b = 4cm and we have to find c.

So,

a – b < c < a + b

⇒ 6 – 4 < c < 6 + 4

⇒ 2 < c < 10

Thus, c lies between 2 cm and 10 cm

Hence, the length of the third stick (L) lies between 2cm and 10 cm

Here, we have given the length of two sides, i.e.

First side = 6cm and second side = 3cm

Now, let length of the third side be L

So, by Triangle Inequality Theorem

Now, we have a = 6cm and b = 3cm and we have to find c.

So,

a – b < c < a + b

⇒ 6 – 3 < c < 6 + 3

⇒ 3 < c < 9

Thus, c lies between 3 cm and 9 cm

Hence, the length of the third side (L) lies between 3cm and 9cm

Let ΔABC in which AB, BC and CA are its sides.

To Prove: AB – BC < CA or BC – CA < AB or AC – AB < AB

Construction: Take a point D on AC such that AD = AB and Join B to D

Proof: In ΔABD,

∠3 is an exterior angle.

So, ∠3 > ∠1 …(i)

[Exterior angle of a triangle is greater than one of its opposite interior angle]

Since, AB = AD

⇒ ∠2 = ∠1 …(ii)

[Angle opposite to equal sides are equal]

⇒ ∠3 > ∠2 …(iii)

[from (i) and (ii)]

Now, In ΔBDC,

∠2 is an exterior angle.

So, ∠2 > ∠4 …(iv)

[Exterior angle of a triangle is greater than one of its opposite interior angle]

⇒ ∠3 > ∠4 [from (iii) and (iv)]

⇒ BC > CD

[greater angle has a longer side opposite to it]

or CD < BC

or AC – AD < BC [∵, AD + CD = AC ]

or AC – AB < BC [∵ by construction, AB = AD]

Hence, the difference of the lengths of any two sides of a triangle is less than the length of the third side.

Let

First side, = 3cm,

Second Side, = 6cm

Third Side, = 8cm

We know that,

The Triangle Inequality Theorem,

So, we have a = 3cm, b = 6cm and c = 8cm

Now, we have to check that these given sides satisfy the triangle inequality or not.

3 + 6 = 9 > 8

3 + 8 = 11 > 6

6 + 8 = 14 > 3

Yes, through the given lengths we can draw a triangle.

Let

First side, = 8cm,

Second Side, = 6cm

Third Side, = 15cm

We know that,

The Triangle Inequality Theorem,

So, we have a = 8cm, b = 6cm and c = 15cm

Now, we have to check that these given sides satisfy the triangle inequality or not.

8 + 6 = 14 ≯ 15

8 + 15 = 23 > 6

6 + 15 = 21 > 8

No, through the given lengths we can’t draw a triangle.

Let

First side, = 2.7cm,

Second Side, = 6.1cm

Third Side, = 8.8cm

We know that,

The Triangle Inequality Theorem,

So, we have a = 2.7cm, b = 6.1cm and c = 8.8cm

Now, we have to check that these given sides satisfy the triangle inequality or not.

2.7 + 6.1 = 8.8 ≯ 8.8

2.7 + 8.8 = 11.5 > 6.1

6.1 + 8.8 = 14.9 > 2.7

No, through the given lengths we can’t draw a triangle.

Let

First side, = 2.5cm,

Second Side, = 8cm

Third Side, = 6cm

We know that,

The Triangle Inequality Theorem,

So, we have a = 2.5cm, b = 8cm and c = 6cm

Now, we have to check that these given sides satisfy the triangle inequality or not.

2.5 + 8 = 10.5 > 6

2.5 + 6 = 8.5 > 8

6 + 8 = 14 > 2.5

Yes, through the given lengths we can draw a triangle.

Let us first draw a figure for above question.

Given:

D is any point on the segment BC

We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than the third side.

So let us first consider Δ ADB

AB + BD > AD …………………….. (i)

Similarly let us consider the Δ ADC

AC + DC >AD …………………….. (ii)

Let us now add the equations (i) and (ii)

AB + AC + BD + DC > AD + AD

From the figure we observe that total length BC = BD + DC

AB + AC + BC > 2 × AD

Hence Proved.

Let us first draw a diagram for the above proof.

We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than the third side.

So let us first consider Δ ABC

AB + AC > BC ……………………. (i)

Similarly let us consider the Δ OBC

OB + OC >BC ……………………. (ii)

Now let us subtract equation (ii) from equation (i)

(AB + AC) – (OB + OC) > BC – BC

(AB + AC) – (OB + OC) > 0

AB + AC > OB + OC

Hence Proved.

From the above proof, we have

AB + AC > OB + OC …….. (I)

Let us modify the above figure a bit by connecting OA

Now from Δ ABC

AC + BC > AB ……….. (i)

Similarly from Δ OAB

OA + OB > AB ………. (ii)

Now let us subtract equation (ii) from equation (i)

AC + BC – (OA + OB) > AB – AB

AC + BC – (OA + OB) > 0

AC + BC > OA + OB ………….. (II)

Now let us again consider Δ ABC,

AB + BC > AC ………….. (iii)

Similarly from Δ OAC

OA + OC > AC …………. (iv)

Now let us subtract equation (iv) from equation (iii)

AB + BC – (OA + OC) > AC – AC

AB + BC – (OA + OC) > 0

AB + BC > OA + OC ……. (III)

Now let us add equations (I), (II) and (III)

AB + BC + AC + BC + AB + AC > OB + OA + OA + OC + OC + OB

2AB + 2BC + 2AC > 2OB + 2OC + 2OA

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

(AB + BC + AC) > (OB + OC + OA)

Hence Proved.

Let us first draw a figure for above question.

ABCD is a quadrilateral. Quadrilateral is any figure with four sides.

AC is one of the diagonals.

The diagonal AC divides the quadrilateral into two triangles.

We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than the third side.

Let us first consider Δ ACD

AD + DC > AC ……….. (i)

Similarly let us consider the other Δ ABC

AB + BC > AC ………… (ii)

Now let us add the equations (i) and (ii)

AD + AC + BC + AB > AC + AC

AD + AC + BC + AB > 2 × AC

Hence it is proved that the perimeter of a quadrilateral is more than twice of the sum length of any diagonal.

Let us first draw a figure for the above question.

P is any point inside the Δ ABC.

We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than the third side.

Now let us consider the Δ APB

AP + BP > AB

Hence Proved.

From the above figure,

In Δ PAB,

PA + PB > AB ……… (i)

Similarly in Δ PBC

PB + PC > BC ……… (ii)

Also in Δ PAC,

PA + PC > AC …….. (iii)

Now let us add the equations (i), (ii) and (iii)

AP + BP + BP + CP + AP + CP > AB + BC + AC

2AP+ 2BP + 2CP > AB + BC + AC

2 (AP + BP + CP) > AB + BC + AC.

It can be written in other way as follows:

AB + BC + AC < 2 (AP + BP + CP)

Hence Proved

Median is any line in a triangle which divides the given triangle segment in two equal parts.

Let us first consider a median from point A dividing the line BC.

D is any point on the segment BC

We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than twice the median which is dividing the third side in two equal halves.

AB + AC > 2 × AM ……………. (I)

Now let us consider median from point B dividing the segment AC.

Similarly we can write here as follows:

AB + BC > 2 × BN …………… (II)

Now let us consider median from point C dividing the segment AB.

AC + BC > 2 × OC …………….. (III)

Now adding equations (I), (II) and (III)

AB + AC + BC + AB + AC + BC > (2AM + 2BN + 2OC)

2AB + 2BC + 2AC > 2 × (AM + BN + OC)

2 × (AB + AC + BC) > 2 × (AM + BN + OC)

AB + AC + BC > AM + BN + OC

Hence it is proved that the perimeter of a triangle is more than the sum of length of the medians.

Let us first construct the diagram for the question:

ABCD is a quadrilateral. Quadrilateral is any figure with four sides.

AC is one of the diagonals.

Diagonal BD also divides the quadrilateral into two equal triangles.

The diagonal AC divides the quadrilateral into two triangles.

Let the common point of intersection of two diagonals be O.

We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than the third side.

Let us first consider Δ OAB

OA + OB > AB ……….. (i)

Let us first consider Δ OBC

OB + OC > BC ………….. (ii)

Let us first consider Δ OCD

OD + OC > DC ………….. (iii)

Let us first consider Δ ODA

OD + OA > AD ………….. (iv)

Now let us add equations (i) and (iii)

OA + OB + OD + OC > AB + DC

We know that AC = OA + OC and BD = OB + OD

AC + BD > AB + DC

We can also get the same result for other two sides as well by adding equations (ii) and (iv).

OB + OC + OD + OA > AD + BC

AC + BD > AD + BC

Hence it is proved that the sum of length of two diagonals is greater than the sum of the length of any two opposite sides of a quadrilateral.

Semi perimeter of a quadrilateral means half of the total perimeter of a quadrilateral.

So semi-perimeter can be defined as the sum of any two sides of quadrilateral.

The solution of this question is similar to that of the above question only the question statement is different. The entire proof remains the same.

ABCD is a quadrilateral. Quadrilateral is any figure with four sides.

AC is one of the diagonals.

Diagonal BD also divides the quadrilateral into two equal triangles.

The diagonal AC divides the quadrilateral into two triangles.

Let the common point of intersection of two diagonals be O.

We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than the third side.

Let us first consider Δ OAB

OA + OB > AB ……….. (i)

Let us first consider Δ OBC

OB + OC > BC ………….. (ii)

Let us first consider Δ OCD

OD + OC > DC ………….. (iii)

Let us first consider Δ ODA

OD + OA > AD ………….. (iv)

Now let us add equations (i) and (iii)

OA + OB + OD + OC > AB + DC

We know that AC = OA + OC and BD = OB + OD

AC + BD > AB + DC

We can also get the same result for other two sides as well by adding equations (ii) and (iv).

OB + OC + OD + OA > AD + BC

AC + BD > AD + BC

Hence it is proved that the sum of lengths of two diagonals of a quadrilateral is more than the semi-perimeter of the quadrilateral.

In the figure shown below, ABCD is a quadrilateral and AC and BD are the two diagonals of quadrilateral.

Also,

E is a point inside the quadrilateral,

We need to prove that,

AE + BE + ED + EC > AC + BD

Now, we know that the straight line is the shortest distance between two points. Therefore, AC will be the shortest distance between A and C and thus AC > AE + EC….(1)

Also,

BD > BE + ED….(2)

Adding (1) and (2), we get,

AC + BD > AE + EC + BE + ED

Hence, Proved.

We can see that at the point where both the diagonals meet the lengths of the line segment obtained by the vertices of the quadrilateral with that point will be smallest.

Let ABCD is a quadrilateral.

Construction: Join A to C

To Prove: ∠A + ∠B + ∠C + ∠D = 360°

Proof: In ΔABC,

∠CAB + ∠ABC + ∠ACB = 180° [by Angle sum property]…(i)

Now, In ΔADC,

∠ADC + ∠ACD + ∠CAD = 180° [by Angle sum property]…(ii)

By adding (i) and (ii), we get

∠CAB + ∠ABC + ∠ACB + ∠ADC + ∠ACD + ∠CAD = 180° + 180°

⇒ (∠CAB + ∠CAD) + ∠ABC + (∠ACB + ∠ACD) + ∠ADC = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

Hence, the sum of four interior angles of a quadrilateral is 360°

We know that, if a polygon has ‘n’ sides, then it divides in (n – 2) triangles.

We also know that, by angle sum property that

Sum of the angles of a triangle = 180°

∴ the sum of angles of (n – 2) triangle = (n – 2) × 180°

Now, we have to find the sum of eight interior angles of octagon.

So, the number of sides in OCTAGON = 8

Sum of 8 interior angles of octagon = (n – 2) × 180°

= (8 – 2) × 180°

= 6 × 180°

= 1080°

Hence, the sum of measurement of eight interior angles of an octagon is 1080°

For Interior Angles

We know that, if a polygon has ‘n’ sides, then it divides in (n – 2) triangles.

We also know that, by angle sum property that

Sum of the angles of a triangle = 180°

∴, the sum of angles of (n – 2) triangle = (n – 2) × 180°

It is given that regular polygon has 10 sides

So,

= 8 × 18

= 144°

For Exterior Angles

It is given that regular polygon has 10 sides

Since, 10 sides of a polygon has 10 angles

and we know that, the sum of the exterior angles of a polygon is 360°

∴10 exterior angles of a regular polygon = 360°

= 36°

Given: Each angle = 120°

We know that, number of angles of a polygon = number of sides

And we know that, the sum of exterior angles of a polygon = 360°

∵, each angle = 120°

= 3

∴, Number of sides = 3

To find the sum of measurements of the interior angles of a polygon, we have formula which will make our computation very simpler and easy.

We know that if a polygon has n sides, then it can be easily divided into n – 2 triangles.

For example a quadrilateral can be divided into two triangles by a diagonal.

Sum of all angles in a triangle is 180°.

So sum of n – 2 angles will be = 180 × (n – 2)

= 2 right angles × (n – 2)

= (2n – 4) right angles

Measure of each angle in a polygon of n sides is given as = (2n – 4) / n

Now for a pentagon, we know it has five sides. So let us substitute n = 5 in the formula.

Sum of interior angles is given as = (2 × 5 – 4) × 90

= (10 – 4) × 90

= 6 × 90

= 540°

For a hexagon we know that it has six sides.

So let us substitute n = 6 in the above formula.

Sum of interior angles is given as = (2 × 6 – 4) × 90

= (12 – 4) × 90

= 8 × 90

= 720°

For a heptagon we know that it has seven sides.

So let us substitute n = 7 in the above formula.

Sum of interior angles is given as = (2 × 7 – 4) × 90

= (14 – 4) × 90

= 10 × 90

= 900°

For a octagon we know that it has eight sides.

So let us substitute n = 8 in the above formula.

Sum of interior angles is given as = (2 × 8 – 4) × 90

= (16 – 4) × 90

= 12 × 90

= 1080°

For a decagon we know that it has ten sides.

So let us substitute n = 10 in the above formula.

Sum of interior angles is given as = (2 × 10 – 4) × 90

= (20 – 4) × 90

= 16 × 90

= 1440°

So let us substitute n = 12 in the above formula.

Sum of interior angles is given as = (2 × 12 – 4) × 90

= (24 – 4) × 90

= 20 × 90

= 1800°

For quadrilateral, there are total four sides. So let us substitute n = 4 in the formula to find the sum of interior angles of quadrilateral.

Sum of interior angles = (2 × 4 – 4) × 90

= (8 – 4) × 90

= 4 × 90

= 360°

Let the fourth angle be x.

So let us add all the angles.

104.5 + 65 + 72.5 + x = 360

242 + x = 360

x = 360 – 242

x = 118°

Therefore the fourth angle is 118°.

Now for a pentagon, we know it has five sides. So let us substitute n = 5 in the formula.

Sum of interior angles is given as = (2 × 5 – 4) × 90

= (10 – 4) × 90

= 6 × 90

= 540°

Let the fifth angle be x.

Now let us add all the angles.

65 + 89 + 132 + 116 + x = 540

402 + x = 540

x = 540 – 402

x = 138°.

Now for a hexagon, we know it has six sides. So let us substitute n = 6 in the formula.

Sum of interior angles is given as = (2 × 6 – 4) × 90

= (12 – 4) × 90

= 8 × 90

= 720°

Let the sixth angle be x.

Now let us add all the angles.

120 + 70 + 95 + 78 + 160 + x = 720

523 + x = 720

x = 720 – 523

x = 197°.

Since the interior angle is greater than 180°, it is not possible to have a convex hexagon with these angles. The reason is that the sum of interior and exterior angles is 180°. So it is not possible to have an interior angle greater than 180°.

Now for a hexagon, we know it has six sides. So let us substitute n = 6 in the formula.

Sum of interior angles is given as = (2 × 6 – 4) × 90

= (12 – 4) × 90

= 8 × 90

= 720°

Let the sixth angle be x.

Now let us add all the angles.

120 + 70 + 95 + 78 + 160 + x = 720

523 + x = 720

x = 720 – 523

x = 197°.

Since the interior angle is greater than 180°, it is not possible to have a convex hexagon with these angles. The reason is that the sum of interior and exterior angles is 180°. So it is not possible to have an interior angle greater than 180°.

Tagging |||Maths||Geometrical Proofs||Geometrical Proofs

Difficulty ||| Medium

To find the sum of measurements of the interior angles of a polygon, we have formula which will make our computation very simpler and easy.

We know that if a polygon has n sides, then it can be easily divided into n – 2 triangles.

For example a quadrilateral can be divided into two triangles by a diagonal.

Sum of all angles in a triangle is 180°.

So sum of n – 2 angles will be = 180 × (n – 2)

= 2 right angles × (n – 2)

= (2n – 4) right angles

Measure of each angle in a polygon of n sides is given as = (2n – 4) / n

Now for a pentagon, we know it has five sides. So let us substitute n = 5 in the formula.

Sum of interior angles is given as = (2 × 5 – 4) × 90

= (10 – 4) × 90

= 6 × 90

= 540°

Measure of each interior angle in pentagon = Sum of all Interior Angles / n

= 540 / 5

= 108°

Sum of all exterior angles in a regular of n sides is always 360°.

Measure of each exterior angles for pentagon = 360 / 5
= 72° Tagging |||Maths||Geometrical Proofs||Geometrical Proofs

Difficulty ||| Medium

Now for a hexagon, we know it has six sides. So let us substitute n = 6 in the formula.

Sum of interior angles is given as = (2 × 6 – 4) × 90

= (12 – 4) × 90

= 8 × 90

= 720°

Measure of each interior angle in hexagon = Sum of all Interior Angles / n

= 720 / 6

= 120°

Sum of all exterior angles in a regular of n sides is always 360°.

Measure of each exterior angles for hexagon = 360 / 6
= 60°

Now for an octagon, we know it has eight sides. So let us substitute n = 8 in the formula.

Sum of interior angles is given as = (2 × 8 – 4) × 90

= (16 – 4) × 90

= 12 × 90

= 1080°

Measure of each interior angle in octagon = Sum of all Interior Angles / n

= 1080 / 8

= 135°

Sum of all exterior angles in a regular of n sides is always 360°.

Measure of each exterior angles for octagon = 360 / 8
= 45°

Now for a nonagon, we know it has nine sides. So let us substitute n = 9 in the formula.

Sum of interior angles is given as = (2 × 9 – 4) × 90

= (18 – 4) × 90

= 14 × 90

= 1260°

Measure of each interior angle in nonagon = Sum of all Interior Angles / n

= 1260 / 9

= 140°

Sum of all exterior angles in a regular of n sides is always 360°.

Measure of each exterior angles for nonagon = 360 / 9
= 40°

Now for a decagon, we know it has ten sides. So let us substitute n = 10 in the formula.

Sum of interior angles is given as = (2 × 10 – 4) × 90

= (20 – 4) × 90

= 16 × 90

= 1440°

Measure of each interior angle in decagon = Sum of all Interior Angles / n

= 1440 / 10

= 144°

Sum of all exterior angles in a regular of n sides is always 360°.

Measure of each exterior angles for decagon = 360 / 10
= 36°

Now for a polygon, with 18 sides let us substitute n = 18 in the formula.

Sum of interior angles is given as = (2 × 18 – 4) × 90

= (36 – 4) × 90

= 32 × 90

= 2880°

Measure of each interior angle = Sum of all Interior Angles / n

= 2880 / 18

= 160°

Sum of all exterior angles in a regular of n sides is always 360°.

Measure of each exterior angles = 360 / 18
= 20°

A regular polygon would be possible only if on dividing the sum of all exterior angles by the measure of each exterior angle gives a whole number.

Here measure of each exterior angle = 6°.

Sum of all exterior angles for regular polygon = 360°

∴ Number of sides for given polygon = 360 / 6

= 60

Since the number of sides of given polygon is a whole number, this exterior angle is a valid exterior angle.

A regular polygon would be possible only if on dividing the sum of all exterior angles by the measure of each exterior angle gives a whole number.

Here measure of each exterior angle = 10°.

Sum of all exterior angles for regular polygon = 360°

∴ Number of sides for given polygon = 360 / 10

= 36

Since the number of sides of given polygon is a whole number, this exterior angle is a valid exterior angle.

A regular polygon would be possible only if on dividing the sum of all exterior angles by the measure of each exterior angle gives a whole number.

Here measure of each exterior angle = 13°.

Sum of all exterior angles for regular polygon = 360°

∴ Number of sides for given polygon = 360 / 13

= 27.69

Since the number of sides of given polygon is not a whole number, this exterior angle is not a valid exterior angle.

A regular polygon would be possible only if on dividing the sum of all exterior angles by the measure of each exterior angle gives a whole number.

Here measure of each exterior angle = 18°.

Sum of all exterior angles for regular polygon = 360°

∴ Number of sides for given polygon = 360 / 18

= 20

Since the number of sides of given polygon is a whole number, this exterior angle is a valid exterior angle.

A regular polygon would be possible only if on dividing the sum of all exterior angles by the measure of each exterior angle gives a whole number.

Here measure of each exterior angle = 35°.

Sum of all exterior angles for regular polygon = 360°

∴ Number of sides for given polygon = 360 / 35

= 10.2857

Since the number of sides of given polygon is not a whole number, this exterior angle is not a valid exterior angle.

We know the basic formula for the sum of all interior angles of a polygon with n sides.

Here measure of each interior angle = 80°

n = 360 / 100

n = 3.6

Since the number of sides is not a whole number, the given interior angle is not possible for regular polygon.

We know the basic formula for the sum of all interior angles of a polygon with n sides.

Here measure of each interior angle = 100°

n = 360 / 80

n = 4.5

Since the number of sides is not a whole number, the given interior angle is not possible for regular polygon.

We know the basic formula for the sum of all interior angles of a polygon with n sides.

Here measure of each interior angle = 120°

n = 360 / 60

n = 6

Since the number of sides is a whole number, the given interior angle is possible for regular polygon.

The regular polygon is a hexagon as the total number of sides is 6.

We know the basic formula for the sum of all interior angles of a polygon with n sides.

Here measure of each interior angle = 144°

n = 360 / 36

n = 10

Since the number of sides is a whole number, the given interior angle is possible for regular polygon.

The regular polygon is a decagon as the total number of sides is 10.

We know the basic formula for the sum of all interior angles of a polygon with n sides.

Here measure of each interior angle = 155°

n = 360 / 25

n = 14.4

Since the number of sides is not a whole number, the given interior angle is not possible for regular polygon.

We know the basic formula for the sum of all interior angles of a polygon with n sides.

Here measure of each interior angle = 160°

n = 360 / 20

n = 18

Since the number of sides is a whole number, the given interior angle is possible for regular polygon.

The regular polygon has the total number of sides which is equal to 18.

Given:

Measurement of each exterior angle = 60°

Sum of all exterior angles of a regular polygon of n sides = 360°

Number of sides = 360 / 60

= 6

Hence this regular polygon with measure of each exterior angle of 60° is a hexagon.

Given:

Measurement of each exterior angle = 135°

Sum of all exterior angles of a regular polygon of n sides = 360°

Number of sides = 360 / 135

= 2.667

Since the number of sides is not a whole number, it is not possible to have regular polygon with a measure of each exterior angle of 135°.

We know the basic formula for the measure of each exterior and interior angles of a regular polygon.

Measure of each exterior angle = 360 / n

Ratio of interior angle to exterior angle = 3:2

2n – 4 = 3 × 2

2n – 4 = 6

2n = 6 + 4

2n = 10

n = 10 / 2

n = 5

Hence the regular polygon with the ratio of interior angle to exterior angle of 3:2 has total number of sides equal to 5 and it is a pentagon.

For a polygon with n sides we have a basic formula for the sum of interior angles of a polygon.

Sum of interior angles = 1800°

1800° = (2n – 4) × 90

1800/90 = (2n – 4)

20 = 2n – 4

2n = 20 + 4

2n = 24

n = 12

Therefore the number of sides of this polygon is 12.

Given:

Measure of each 5 angles = 172°

Measure of each remaining angles = 160°

Let the total number of sides for polygon be n.

Sum of all angles in given polygon = (5 × 172) + (n – 5) × 160

(5 × 172) + (n – 5) × 160 = (2n – 4) × 90

860 + 160n – 800 = (2n – 4) × 90

160n + 60 = 180n – 360

180n – 160n = 360 + 60

20n = 420

n = 420 / 20

n = 21

Therefore the given polygon has total number of sides equal to 21.

Let us first draw a diagram for the above question.

ABCD is the required quadrilateral.

OA is the angle bisector for ∠ DAB and OB is the angle bisector for ∠ ABC.

When we observe the above figure we know that,

∠ BAO = ∠ OAB = � ∠ BAD [as OA is the angle bisector of ∠ DAB].. (i)

∠ OBC = ∠ ABO = � ∠ ABC [as OB is the angle bisector of ∠ ABC] .. (ii)

To Prove: ∠ AOB = (∠ BCD + ∠ CDA) / 2

Proof:

For a quadrilateral we know that the sum of all interior angles is 360°.

Even we can verify it using the formula.

Sum of all interior angles = (2n – 4) × 90

= (2 × 4 – 4) × 90

= (8 – 4) × 90

= 4 × 90

= 360°

So ∠ DAB + ∠ ABC + ∠ BCD + ∠ CDA = 360° ………. (iii)

For a triangle the sum of all interior angles is 180°

So for Δ AOB

∠ ABO + ∠ OAB + ∠ AOB = 180°

From (i) we have ∠ OAB = � ∠ BAD and from (ii) we have ∠ ABO = � ∠ ABC.

So let us substitute these values in above equation.

� ∠ BAD + � ∠ ABC + ∠ AOB = 180°

∠ BAD + ∠ ABC + 2 × ∠ AOB = 2 × 180° [Removing the fraction part by multiplying throughout by 2]

∠ BAD + ∠ ABC + 2 × ∠ AOB = 360° …………. (iv)

Now for equation (iii) and (iv) the right hand side is same. Hence the left hand side has to be equal.

So let us equate the two left hand sides parts.

∠ BAD + ∠ ABC + 2 × ∠ AOB = ∠ DAB + ∠ ABC + ∠ BCD + ∠ CDA

∠ BAD and ∠ ABC gets eliminated from both the sides.

2 × ∠ AOB = ∠ BCD + ∠ CDA

∠ AOB = � (∠ BCD + ∠ CDA)

Hence proved that that the measurement of the angle made by the bisectors of two adjacent angles of quadrilateral is equal to the half of the sum of measurement of the other two angles.

For first part of the proof let us draw a figure.

ABCDE is a regular pentagon.

To Prove: Δ ABC is an isosceles triangle

Proof:

In the question it is mentioned that the given pentagon is a regular pentagon hence all the sides of this pentagon has equal lengths.

So in Δ ABC,

Length of AB = length of BC [as they both are sides of the regular pentagon]

Now for the given triangle, side AB = BC,

So ∠ BAC = ∠ ACB ………… (Angles opposite to the equal sides in a triangle are equal. One of the important property of isosceles triangle]

Hence it is proved that the given Δ ABC is an isosceles triangle.

Now for the second proof let us draw a diagram again.

To Prove: BE || CD

Proof:

In the question it is mentioned that the given pentagon is a regular pentagon hence all the sides of this pentagon has equal lengths.

So in Δ ABE,

Length of AB = length of AE [as they both are sides of the regular pentagon]

Now for the given triangle, side AB = AE,

So ∠ ABE = ∠ AEB ………… (Angles opposite to the equal sides in a triangle are equal. One of the important property of isosceles triangle]

Now we also know that BC = DE

Also ∠ ABC = ∠ AED ………. [For a regular pentagon all interior angles are equal]

Now we know that ∠ ABC = ∠ ABE + ∠ EBC …………. (i)

Similarly for ∠ AED = ∠ AEB + ∠ BED …………………… (ii)

Now subtracting equation (ii) from (i)

∠ ABC - ∠ AED = (∠ ABE + ∠ EBC) – (∠ AEB + ∠ BED) … [∠ ABC = ∠ AED]

0 = ∠ ABE + ∠ EBC - ∠ AEB - ∠ BED …… [∠ ABE = ∠ AEB]

∠ EBC - ∠ BED = 0

∠ EBC = ∠ BED

∠ BCD = ∠ CDE ………. [For a regular pentagon all interior angles are equal]

Now from the figure it is observed that BCDE is a quadrilateral.

For the given quadrilateral all its interior angles are equal and hence for the angles to be equal BE has to be parallel to CD.

Hence Proved.

Let us first draw a figure for the given question.

In the above figure ABCDEF is a regular hexagon.

AX is the angle bisector of ∠ BAF.

Now for a regular hexagon all its interior angles are equal and all its sides are also equal.

Now ∠ XAF = ∠ BAX = � ∠ BAF = � × 120

= 60°

AFEX is a quadrilateral.

So sum of all interior angles of quadrilateral is 360°.

∠ AFE = ∠ FED = 120° [interior angle of a regular hexagon]

∠ XAF + ∠ AFE + ∠ FEX + ∠ EXA = 360°

60 + 120 + 120 + ∠ EXA = 360°

300° + ∠ EXA = 360°

∠ EXA = 360° - 300°

∠ EXA = 60°

∠ AXD + ∠ EXA = 180° [As they form a linear pair]

∠ AXD = 180 - ∠ EXA

= 180 – 60

=120°

Hence ∠ AXD = 120°.

To solve this type of question we require a logical approach.

Before proceeding to the solution, we will have to see the diagram of cardinal directions of the route.

Now let us see the diagram of two persons starting with their journey in their respective directions.

Person labelled 1 is starting the journey from south direction and Person labelled 2 is starting the journey along the south-east direction.

All the direction are straight lines. The North-South line is perpendicular to West-East line.

Now to find out which person reaches the destination first we will have to draw a straight line joining the south direction with south east direction in the following manner.

Now we have got a triangle OAB and the ∠ OAB = 90°.

Person 1 will travel along the side OA of the triangle and the Person 2 will travel along the side OB of the triangle.

Now let us apply the Pythagoras Theorem to Δ OAB.

OB is the hypotenuse of the triangle [side opposite to 90° is the hypotenuse of the triangle]

So OB² = OA² + AB²

Now when we look at the above equation we observe that length of OB is far greater than length OA.

So it is observed that Person 1 travelling along OA side that is along south direction will have a shorter distance to reach the distance.

At the same time Person 2 travelling along OB side will have longer distance to travel to reach the distance as both Person 1 and Person 2 are starting from same place.

Hence Person 1 will reach the distance first and Person 2 will reach after some time.

Let us draw a figure for the question given.

ABCD is the required quadrilateral.

It is given that,

AB = AD and BC = CD.

AC is the diagonal of the quadrilateral.

DP is the shortest distance drawn from point D to AC.

Since it is given that AB = AD which means the point A is equidistant from point B and D that is point A is lying at equal distances from point B and D which means that A is a point on the right bisector of segment BD where BD is another diagonal of the quadrilateral.

Similar is the condition for C where C is equidistant from B and D and hence C is lying on the right bisector of segment BD.

This means that AC and BD intersect to give right angles [as both AC and BD are diagonal of a quadrilateral].

So it means that P which is the point on AC drawn from D is the intersection point of the two diagonals.

So it can observed that point P is lying on the diagonal BD.

Hence point B, P and D are collinear that is they are lying on the same line.

Let us first draw a diagram for the above given question.

In Δ BPD and Δ CQD

∠ BPD = ∠ CQD ………….. [As both the angles are 90°]

∠ BDP = ∠ CDQ ………….. [Vertically opposite angles]

BP = CD ………………………. [As AD is median to BC so BC is divided into two equal halves]

Hence Δ BPD is congruent to Δ CQD by Angle Angle Side test.

Since both the triangles are congruent, side BP = side CQ [By congruent parts of congruent triangle are equal]

Hence BP = CQ.

Hence Proved.