The Internet is a worldwide system of computer. networks, i.e. network of networks. Through Internet, computers become able to exchange information with each other and find diverse perspective on issues from a global audience.

1. the ship wavewalker was almost sank because of the water wave
2. her daughter has injured
3. their ship almost sank and Gordon cook and his wife had lose their hope to survive

Construction Procedure:

1. Draw a line segment BC with the measure of 8 cm.

2. Now draw the perpendicular bisector of the line segment BC and intersect at the point D

3. Take the point D as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point A

4. Now join the lines AB and AC and the triangle is the required triangle.

5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.

6. Locate the 3 points B₁, B₂ and B₃ on the ray BX such that BB₁ = B₁B₂ = B₂B₃

7. Join the points B₂C and draw a line from B₃ which is parallel to the line B₂C where it intersects the extended line segment BC at point C’.

8. Now, draw a line from C’ the extended line segment AC at A’ which is parallel to the line AC and it intersects to make a triangle.

9. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

A’B = (3/2)AB

BC’ = (3/2)BC

A’C’= (3/2)AC

From the construction, we get A’C’ || AC

∴ ∠ A’C’B = ∠ACB (Corresponding angles)

In ΔA’BC’ and ΔABC,

∠B = ∠B (common)

∠A’BC’ = ∠ACB

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Therefore, A’B/AB = BC’/BC= A’C’/AC

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

A’B/AB = BC’/BC= A’C’/AC = 3/2

Hence, justified.

5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Construction Procedure:

1. Draw a ΔABC with base side BC = 6 cm, and AB = 5 cm and ∠ABC = 60°.

2. Draw a ray BX which makes an acute angle with BC on the opposite side of vertex A.

3. Locate 4 points (as 4 is greater in 3 and 4), such as B1, B2, B3, B4, on line segment BX.

4. Join the points B4C and also draw a line through B3, parallel to B4C intersecting the line segment BC at C’.

5. Draw a line through C’ parallel to the line AC which intersects the line AB at A’.

6. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

Since the scale factor is 3/4 , we need to prove

A’B = (3/4)AB

BC’ = (3/4)BC

A’C’= (3/4)AC

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

∴ ∠ A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, A’B/AB = BC’/BC= A’C’/AC

So, it becomes A’B/AB = BC’/BC= A’C’/AC = 3/4

Hence, justified.

6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.

To find ∠C:

Given:

∠B = 45°, ∠A = 105°

We know that,

Sum of all interior angles in a triangle is 180°.

∠A+∠B +∠C = 180°

105°+45°+∠C = 180°

∠C = 180° − 150°

∠C = 30°

So, from the property of triangle, we get ∠C = 30°

Construction Procedure:

The required triangle can be drawn as follows.

1. Draw a ΔABC with side measures of base BC = 7 cm, ∠B = 45°, and ∠C = 30°.

2. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.

3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, B4, on the ray BX.

4. Join the points B3C.

5. Draw a line through B4 parallel to B3C which intersects the extended line BC at C’.

6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’.

7. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

Since the scale factor is 4/3, we need to prove

A’B = (4/3)AB

BC’ = (4/3)BC

A’C’= (4/3)AC

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

∴ ∠A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, A’B/AB = BC’/BC= A’C’/AC

So, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3

Hence, justified.

7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Given:

The sides other than hypotenuse are of lengths 4cm and 3cm. It defines that the sides are perpendicular to each other

Construction Procedure:

The required triangle can be drawn as follows.

1. Draw a line segment BC =3 cm.

2. Now measure and draw ∠= 90°

3. Take B as centre and draw an arc with the radius of 4 cm and intersects the ray at the point B.

4. Now, join the lines AC and the triangle ABC is the required triangle.

5. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.

6. Locate 5 such as B1, B2, B3, B4, on the ray BX such that such that BB₁ = B₁B₂ = B₂B₃= B₃B₄ = B₄B₅

7. Join the points B3C.

8. Draw a line through B5 parallel to B3C which intersects the extended line BC at C’.

9. Through C’, draw a line parallel to the line AC that intersects the extended line AB at A’.

10. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

Since the scale factor is 5/3, we need to prove

A’B = (5/3)AB

BC’ = (5/3)BC

A’C’= (5/3)AC

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

∴ ∠ A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, A’B/AB = BC’/BC= A’C’/AC

So, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3

Hence, justified.

Boyle’s Law

• Boyle’s law was stated by Robert Boyle.
• It states that at constanttemperature, the pressure of a fixednumber of moles nof gasvaries inversely with its volume.

k₁ = Proportionality constant.

• It depends upon the amount and temperature of gas. It also depends upon the units in which p as well v are expressed.
• Let volume V₁ is occupied at pressure P₁ and temperature T₁.
• Again volume V₂ is occupied at pressure P₂ and temperature T₂. Mathematically, as per Boyle’s law:

P₁ V₁= P₂ V₂ = Constant

P_1/P₂  = V_2/V₁ 

There were many factors that support Britain to become the epicentre of the Industrial Revolution. Important factors are given below:

1. Continental European countries could not adopt Industrial Revolution as early as Britain because Europe faced political instability.

2. French Revolution and Napoleonic wars kept the Europe engaged. Belgium was first to follow British footsteps. Germany also has its ‘Ruhr’ are Industrialized. Soon other European nations followed suit.

3. Britain had adequate capital which was accumulated through colonialism.

4. Disappearance of serfdom (a person who is bound to the land and owned by the feudal lord) and ‘enclosure movement’ provided huge surplus agricultural labour which looked for employment and became source of cheap labour.

संसाधन

• हमारे पर्यावरण में उपलब्ध प्रत्येक वस्तु जो हमारी आवश्यकताओं को पूरा करने में उपयोग की जाती है और जिसको बनाने के लिए तकनीक उपलब्ध है, जो आर्थिक रूप से संभाव्य तथा सांस्कृतिक रूप से मान्य है, एक संसाधन कहलाता है।

संसाधनों का वर्गीकरण

• संसाधनों का वर्गीकरण इस प्रकार किया जा सकता है-
→ उत्पत्ति के आधार पर- जैव संसाधन, अजैव संसाधन

→ समाप्यता के आधार पर - नवीकरणीय संसाधन, अनवीकरणीय संसाधन

→ स्वामित्व के आधार पर - व्यक्तिगत संसाधन, सामुदायिक स्वामित्व वाले संसाधन, राष्ट्रीय संसाधन

→ विकास के स्तर के आधार पर- संभावी संसाधन, विकसित संसाधन, भंडार, संचित कोष

अधिक के लिए दिए गए लिंक पर क्लिक करें:

Lemon juice is a good conductor of electricity because it contains citrus acid having H+ ions which are responsible for the conduction.