New Numbers

Applying Pythagoras Theorem for right – angled triangle,

Base² + Perpendicular² = Hypoteneous²

Which gives us in following right – angled triangles: –

In Δ BAD,

BA⊥ AD

⇒ BA² + AD² = BD²

⇒ BD

⇒ = √2 metre

In Δ DBC, DB⊥ BC

⇒ DB² + BC² = DC²

⇒ DC = = √3 metre

In Δ CDE, DC⊥ CE

⇒ DC² + CE² = DE²

⇒ = 2 metre

In Δ DEF, DE⊥ EF

⇒ DE² + EF² = DF²

⇒ DF = = √5 metre….(Ans.)

The area of the square = length of any of its side

= DF²

= (√5)² = 5 metre²

(Avoid the very naming given in the following figures. As it was not there in the main problem,it is solver’s own choice)

(i) The figure is given below:

In Δ ABC,AB = BC = CA = 2 metre.

CD is the altitude. As it is a equilateral triangle, CD is also a median.

(proof: In ΔACD and ΔBCD, –

∠ADC = ∠BDC = 90⁰

AC = BC

AD is the common side

So, ΔACD≅ Δ BCD(R – H – S)

⇒ AD = BD…hence, CD is a median ,too. )

Thus, AD = BD = AB/2 = 2/2 = 1 metre

In right – angled Δ BCD,

Using Pythagoras theorem , –

….(1)

So, area of the square

(ii) altitude of the Δ ABC = CD = √3 metre [from (1)]

(iii)

Let’s name it Δ ABC, where the angles and sides are shown in the figure.

As, we know, opposite side ofistwice than that of (can be proved by trigonometry,to be learnt in future)

⇒AC = 2BC

As ∠ACB = , using Pythagoras Theorem, –

AC² + BC² = 4BC² + BC²

= 5BC²

= BA²

= 2²

= 4

⇒ BC = √(0.8) metre

AC = 2 AC = 2 × √(0.8) metre…

→ Any odd number can be written in the form 2k – 1,k being a natural number.

Now,2k – 1 = k² – (k – 1)²

● For,

2k – 1 = 7

⇒2k = 8

⇒ k = 4

So, √7 = 4² – 3²

So, if we draw a line of 4 units, say CD, then,

Then,

Draw a random line and cut a length of 3c.m. s off it, to have AB.

→ Then draw a normal BA’ on AB.

→ Draw a circular arc with A as the centre and CD as centre, which eventually intersects BA’ at E.

→ Join B and E; and, A and E.

See, In Δ ABE, –

AB⊥ BE

BE² = AE² – AB² = 4² – 3² = 7

⇔ E = √7 c.m.

● And,

2k – 1 = 11

⇒ 2k = 12

⇒ k = 6

So, √11 = √( 6² – 5²)

Draw a random line and cut a length of 5 c.m. s off it, to have AB.

→ Then draw a normal BA’ on AB.

→ Draw a circular arc with A as the centre and CD as centre, which eventually intersects BA’ at E.

→ Join B and E; and,A and E.

See, In Δ ABE, –

AB⊥ BE

BE² = AE² – AB² = 6² – 5² = 11

óBE = √11 c.m.

(i) As, any odd number, 2k – 1 = k² – (k – 1)²

For,13 = 2k – 1 ⇔ 2k = 14 ⇔ k = 7

So,

√13 = √(7² – 6²)

(units are in c.m.)

Draw a random line and cut a length of 6 c.m. s off it, to have AB.

→ Then draw a normal BA’ on AB.

→ Draw a circular arc with A as the centre and CD as centre, which eventually intersects BA’ at E.

→ Join B and E; and, A and E.

See, In Δ ABE, –

AB⊥ BE

BE² = AE² – AB² = 7² – 6² = 13

óBE = √13 c.m.

(ii) Also observe,

ó

(units are in c.m.)

Draw a random line and cut a length of 3 c.m. off it, to have AB.

→ Then draw a normal BA’ on AB.

→ Draw a circular arc with B as the centre and radius of CD (2 c.m.), which eventually intersects BA’ at E.

→ Join B and E ; and , A and E.

See, In Δ ABE, –

AB⊥ BE

BE² + AB² = AE² = 2² + 3² = 13

óAE = √13

→ Say, in general a is a fraction such that,

√2<a< √3[note, a>0]

⇒

⇒,for k being a natural number.

So, for a particular k, our job is to find a perfect square(of a fraction) between and .

Put, k = 2,3,5 (as three fractions are asked,three primes are chosen, else, like the case of k = 2 and k = 4 same fraction a may occur) – – –

●

[ 0]

●

●

⇒

]

●

So, the three fractions are:

In the given figure, hypotenuse, BC = 1.5 m and AB = 0.5 m.

We know that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

⇒ In ΔABC, BC² = AB² + AC²

⇒ 1.5² = 0.5² + AC²

⇒ 2.25 = 0.25 + AC²

⇒ AC² = 2.25 – 0.25 = 2

∴ AC = √2 m

⇒ √2 ≈ 1.41 m (correct to a centimetre)

We know that perimeter of a polygon is the sum of all its sides.

⇒ Perimeter of given triangle = 1.5 + 0.5 + √2

= 2 + √2

⇒ 2 + √2 ≈ 2 + 1.41 = 3.41 m

∴ Perimeter = 3.41 metres (correct to a centimetre)

Consider ΔACD,

We know that in a right angled triangle, the square of tshe hypotenuse is equal to the sum of the squares of the other two sides.

⇒ AC² = AD² + CD²

⇒ 2² = 1² + CD²

⇒ 4 = 1 + CD²

⇒ CD² = 4 – 1 = 3

∴ AC = √3 m

⇒ √3 ≈ 1.73 m

We know that perimeter of a polygon is the sum of all its sides.

(i) Here, ΔACD is a part.

⇒ Perimeter = 2 + 1 + √3

= 3 + √3

⇒ 3 + √3 = 3 + 1.73 = 4.73 m

∴ Perimeter of a part of the given equilateral triangle = 4.73 metres

(ii) Perimeter of the whole triangle = 2 + 2 + 2 = 6 metres

The perimeter of a part is less than the whole by,

⇒ 6 – 4.73 = 1.27 metres

We know that sum of angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 30° + 105° + ∠C = 180°

⇒ ∠C = 180° - 135°

∴ ∠C = 45°

We know that the sine rule is = =.

In the triangle above, we have

⇒ c = 2 m;

a, b =?

Consider

∴ a = √2 ≈ 1.41 m

Now consider = ,

⇒ b

= √3 + 1

∴ b = √3 + 1 ≈ 1.73 + 1 = 2.73 m

We know that perimeter of a polygon is the sum of all its sides.

∴ Perimeter = a + b + c

= 1.41 + 2.73 + 2

= 6.14 m

We know that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Consider the 1^st triangle.

⇒ Hypotenuse of 1^st triangle, x² = 1² + 1²

= 2

∴ Hypotenuse, x = √2 m

Consider 2^nd triangle.

⇒ Hypotenuse of 2^nd triangle, y² = (√2)² + 1²

= 2 + 1

= 3

∴ Hypotenuse, y = √3 m

Consider 3^rd triangle.

⇒ Hypotenuse of 3^rd triangle, z² = (√3)² + 1²

= 3 + 1

= 4

∴ Hypotenuse, y = √4 m

And so on.

(i) The lengths of the 10^th triangle will be √10 m, 1 m and hypotenuse.

Let hypotenuse be a.

⇒ a² = (√10)² + 1²

= 10 + 1

= 11

⇒ a = √11 m

∴ The lengths of the sides of the 10^th triangle are √10 m, 1 m and √11 m.

(ii) We know that perimeter of a polygon is the sum of all its sides.

The perimeter of 10^th triangle = √10 + 1 + √11

[√10 ≈ 3.16; √11 ≈ 3.31]

∴ Perimeter = 3.16 + 1 + 3.31

= 7.47 m

The perimeter of 9^th triangle = √9 + 1 + √10

= 3 + 1 + 3.16

= 7.16 m

Perimeter of 10^th triangle – Perimeter of 9^th triangle = 7.47 – 7.16 = 0.31

∴ The perimeter of the 10^th triangle is 0.31 m more than the perimeter of 9^th triangle.

(iii) Perimeter of nth triangle = √n + 1 + √(n+1)

Perimeter of (n-1)th triangle = √(n – 1) + 1 + √(n – 1 + 1)

= √(n - 1) + 1 + √n

Difference between perimeters of nth triangle and (n – 1)th triangle = √n + 1 + √(n+1) – [√(n - 1) + 1 + √n]

= √n + 1 + √(n+1) – √(n - 1) - 1 - √n

= √(n + 1) - √(n – 1)

We know that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

⇒ In ΔABC, BC² = AB² + AC²

⇒ BC² = (√3)² + (√2)²

⇒ BC² = 3 + 2

⇒ BC² = 5

∴ BC = √5 cm (Hypotenuse)

⇒ √5 ≈ 2.23 cm,

√2 ≈ 1.41 ,

and √3 ≈ 1.73

Sum of perpendicular sides = 1.41 + 1.73 = 3.14 cm

⇒ 3.14 – 2.23 = 0.91 cm

∴ The sum of perpendicular sides is 0.91 cm more than the hypotenuse.

First we find the sides of the equally cut triangles formed from the original equilateral triangle.

Breadth of the rectangle = Side of equilateral triangle = 1m

Length of rectangle = () = √3

Perimeter of rectangle = 1 + √3 + 1 + √3 = (2 + 2 √3) m

Area of rectangle = Length × Breadth

⇒ Area of rectangle = √3 × 1 = √3 m²

First we find the sides of the equally cut triangles formed from the original equilateral triangle.

Slant height of trapezium = Diagonal of square = 2 √2 cm

Smaller side of trapezium = 2 √3 cm

Perimeter of trapezium = 2√2 + 2√3 + 2√2 + (2 + 2√3 + 2)

⇒ Perimeter of trapezium = (4√2 + 4√3 + 4) cm

Area of trapezium =

⇒ Area of trapezium =

⇒ Area of trapezium = (4√3 + 4) cm²

We name the vertices of the triangle as A, B and C. We draw a perpendicular from B on AC as BD.

In ΔABD,

⇒ ∠A + ∠ABD + ∠ADB = 180° (Sum of all angles of a triangle)

⇒ 60 + ∠ABD + 90 = 180°

⇒ ∠ABD + 150 = 180°

⇒ ∠ABD = 30°

⇒ AD = AB cos60°

⇒ AD = 4×(1/2) = 2cm …………………….(1)

⇒ BD = AB sin60°

⇒ BD = 4×(√3/2) = 2√3 cm ……………………(2)

∠B = ∠ABD + ∠CBD

⇒ 75 = 30 + ∠CBD

⇒ 45 = ∠CBD

In Δ BCD,

∠BCD = ∠CBD = 45°

CD = BD ( Sides opposite to equal angles are equal)

⇒ CD = 2√3 cm (From eq(2) ) ……………………(3)

BC = √2 BD ( ΔBCD is a right isosceles triangle)

⇒ BC = √2(2√3) = 2√6 …………………….(4)

AC = AD + DC

⇒ AC = (2 + 2√3) cm

Perimeter = AB + BC + CA

⇒ Perimeter = 4 + 2√6 + (2 + 2√3)

⇒ Perimeter = 6 + 2√6 + 2√3

Area =

⇒ Area =

⇒ Area =

⇒ Area = (2 + 2√3)(√3)

⇒ Area = (6 + 2√3) cm²

Let the side of red equilateral triangles be 1m.

In such rectangles, breadth = 1m

And length = (√3/2) + (√3/2) = √3 m (Sum of height of red triangle)

Side of outer square = (√3 + 1) m

Side of inner square = (√3 – 1) m

∴ Ratio of side of outer to inner squares =

We know that √x × √y = √xy

(i ) √3 × √12 = √3×12 = √36 = 6 Natural number

(ii) √3 × √1.2 = √3×1.2 = √3.6 = 6√0.1 Not a natural number or fraction

(iii ) √5 × √8 = √5×8 = √40 = 4√10 Not a natural number or fraction

(iv) √0.5 × √8 = √0.5×8 = √4 = 2 Natural number

(v ) √ × √ = Natural number

The figure is attached below:

An equilateral triangle has all sides of equal length,

i.e.,

AB = BC = AC

AD is perpendicular to BC,

∴ D is the midpoint of BC

⇒ BD = DC = 1/2 BC = 1/2 AB =1/2 AC

Now using Pythagoras theorem,

AD² + BD² = AB²

⇒ 4² + (1/2 AB)² = AB²

⇒ AB² + 1/4 AB² = 4²

Hence, sides of triangle are 3.6 cm

i) A regular hexagon is made up of 6 equilateral triangles.

Therefore, the green coloured triangles inside the yellow hexagon are equilateral triangles, that is all their sides are equal.

⇒ All the sides that comprises the hexagon are equal.

Hence, the inner red hexagon is also regular.

ii) Let the side of the inner red hexagon be x

⇒ the sides of the triangle will also be x

Let the side of outer yellow hexagon be y

Using Pythagoras theorem,

x² + y² = (2x)²

⇒ y² = (2x)² – x²

⇒ y² = 4x² – x²

⇒ y² = 3x²

⇒ y= x√3

Hence, the outer hexagon’s side is √3 times the inner hexagon’s side.

iii) Now

Area of larger hexagon

⇒ Area of larger hexagon

And,

Area of smaller hexagon

Hence, the outer hexagon’s area is 3 times the inner hexagon’s area.

(√2 + 1) (√2 – 1) = (√2)² – (1)² [∵ (a+b)(a-b) = a²-b²]

⇒ (√2 + 1) (√2 – 1) = 2 – 1

⇒ (√2 + 1) (√2 – 1) = 1

Hence, (√2 + 1) (√2 – 1) = 1

Now,

Now,

[∵ (√2 + 1) (√2 – 1) = 1]

Similarly,

AC = CE = EF = AF = 4 cm

∵ B is midpoint of AC

AB = BC = 1/2 AC

⇒ AB = BC = 1/2 × 4 cm

⇒ AB = BC = 2 cm

Similarly,

∵ D is midpoint of AC

CD = DE = 1/2 CE

⇒ CD = DE = 1/2 × 4 cm

⇒ CD = DE = 2 cm

AE is diagonal of ACEF,

∴ By Pythagoras theorem,

AF² + EF² = AE²

⇒ 4² + 4² = AE²

⇒ AE² = 16 + 16

⇒ AE² = 32

⇒ AE= √32

⇒ AE= 4√2 cm

∵ J is midpoint of AE,

∴ AJ = JE = 1/2 AE

⇒ AJ = JE = 1/2 × 4√2 cm

⇒ AJ = JE = 2√2 cm

∵ G is midpoint of AJ,

∴ AG = JG = 1/2 AJ

⇒ AG = JG = 1/2 × 2√2 cm

⇒ AG = JG = √2 cm

∵ BGJH is a square

BG = JH = HB = JG = √2 cm

Also,

HD = HB = √2 cm

And,

IE = IJ = GJ =√2 cm

∵ HD and IE are equal and parallel

∴ HDEI is a parallelogram

⇒ HI = DE = 2 cm