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Lesson Posted on 06 Feb CBSE/Class 8 CBSE/Class 9

ABCD Square, AE = EB. Find Angle EDC.

Sujoy D.

1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....

Question given below: Solution (Method 1): solution (Method 2): read more

Question given below:

Solution (Method 1):

solution (Method 2):

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Comments Lesson Posted on 06 Feb CBSE/Class 8

Sujoy D.

1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....

Question: Solution: CDB = 30, so DAB = 30 (alternate segment theorem)ADB = 90So DBE = 60 degrees.AD = 6. sin EAD = sin 30 = DE/ADSo DE = AD/2 = 3. So cot 60 = BE/DE..BE = 3/ root 3 = root3So are DEB = 1/2* DE*EB = 1/2*3*root 3 = 3 root 3/2 read more

Question:

Solution:

CDB = 30, so DAB = 30 (alternate segment theorem)

ADB = 90

So DBE = 60 degrees.

AD = 6. sin EAD = sin 30 = DE/AD

So DE = AD/2 = 3.

So cot 60 = BE/DE..BE = 3/ root 3 = root3

So are DEB = 1/2* DE*EB = 1/2*3*root 3 = 3 root 3/2

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Comments Lesson Posted on 06 Feb CBSE/Class 8 CBSE/Class 9

Sujoy D.

1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....

Question: Solution: read more

Question:

Solution:

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Comments Lesson Posted on 06 Feb CBSE/Class 8 CBSE/Class 9

Find CD (Circle Problem) Oympiad Level Question

Sujoy D.

Question: Solution : Join center of third circle (say X) and B Join center of middle circle (say Y) and mid point of CD (say Q)Then, AYQ and AXB are similar, => YQ/AY = BX/AX => YQ = 15×5/25 = 3=> CQ = √(CY² - YQ²) = √(5² - 3²)... read more

Question:

Solution :

Join center of third circle (say X) and B

Join center of middle circle (say Y) and mid point of CD (say Q)

Then, AYQ and AXB are similar,

=> YQ/AY = BX/AX

=> YQ = 15×5/25 = 3

=> CQ = √(CY² - YQ²) = √(5² - 3²) = 4

=> CD = 8

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Comments Lesson Posted on 06 Feb CBSE/Class 7 CBSE/Class 8

Basics Property Of Line And Plane

Sujoy D.

Basic properties of lines and planes: 1. One and only one line passes through two distinct points. 2. Infinite number of lines pass through a given point. These lines are called concurrent lines. 3. The intersection of two distinct lines is a point. 4. Three or more points are said to be collinear... read more

Basic properties of lines and planes:

1. One and only one line passes through two distinct points.

2. Infinite number of lines pass through a given point. These lines are called concurrent lines.

3. The intersection of two distinct lines is a point.

4. Three or more points are said to be collinear if they lie on a line, otherwise they are said to be non collinear.

5. There are infinite number of planes passing through any given lines.

6. There is exactly one plane passing through three non - collinear points.

7. The intersection of two planes is a line.

8. If two lines are perpendicular to the same line they are parallel to each other.

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Comments Lesson Posted on 06 Feb CBSE/Class 9 CBSE/Class 8

Calculate AB In Terms Of A And B (Conceptual)

Sujoy D.

Question: Solution: DC =√ (AD² + AC² )=> DC = √(a²+b²)AC × DB = AB × DC + AD×BC =>a² = AB × √(a²+b²) + b² => AB = (a² - b²)/√(a²+b²) read more

Question:

Solution:

DC =√ (AD² + AC² )

=> DC = √(a²+b²)

AC × DB = AB × DC + AD×BC

=>a² = AB × √(a²+b²) + b²

=> AB = (a² - b²)/√(a²+b²)

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Comments Lesson Posted on 06 Feb CBSE/Class 8

Find The Height Of The Trapezium (Important Concept)

Sujoy D.

Question: Solution: Concept = if a cicrcle is inscribed, sum of parallel sides= sum of non parallel sides. x, 4x. So each of non parallel sides = 5x/2 So 10x = 60, x = 6. parallel sides = 6, 24. non parallel = 15,15. So height = root (225-81) = 12 read more

Question:

Solution:

Concept = if a cicrcle is inscribed, sum of parallel sides= sum of non parallel sides.

x, 4x.

So each of non parallel sides = 5x/2

So 10x = 60, x = 6. parallel sides = 6, 24. non parallel = 15,15.

So height = root (225-81) = 12

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Comments Lesson Posted on 06 Feb CBSE/Class 8

Sujoy D.

1. Tangents: For any straight line we have only three possibilities: - it doesn’t intersect the circle, it intersects the circle at two different points and it touches the circle at just one point. 2. Secant: When a straight line intersects the circle as two points it is known as secant. 3.... read more

1. Tangents:

For any straight line we have only three possibilities: -

it doesn’t intersect the circle,

it intersects the circle at two different points and it touches the circle at just one point.

2. Secant:

When a straight line intersects the circle as two points it is known as secant.

3. Tangent:

When the straight line touches a circle at exactly one point, it is known as tangent to the circle at that point.

A circle can have infinite number of tangents.

There can be at most two tangents parallel to a secant.

The common point of the circle and the tangent is known as point of contact.

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Number of tangents to a circle:

i) There is no tangent to a circle passing through a point lying inside the circle.

ii) There is one and only one tangent to a circle passing through a point lying on the circle.

iii) There are exactly two tangents to a circle through a point lying outside the circle

Length of the segment of the tangent from an external point to the point of contact is length of the tangent. The lengths of tangents drawn from an external point to a circle are equal.

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Comments Lesson Posted on 06 Feb CBSE/Class 7 CBSE/Class 8

Sujoy D.

Question: In a triangle ABC, AB = 8 units, BC = 10 units and CA = 12 units. AD is angle bisector of angle A and I is the incenter, find AI:ID. Solution: Solution : AB/AC = BD/CD (angle bisector theorem, as AD is angle bisector of angle A)=> BD = 4 and CD = 6Now, in triangle BAD, BI is angle bisector =>... read more

Question:

In a triangle ABC, AB = 8 units, BC = 10 units and CA = 12 units. AD is angle bisector of angle A and I is the incenter, find AI:ID.

Solution:

Solution : AB/AC = BD/CD (angle bisector theorem, as AD is angle bisector of angle A)

=> BD = 4 and CD = 6

Now, in triangle BAD, BI is angle bisector

=> BD/AB = DI/AI (angle bisector theorem)

=> AI/DI = 8/4 = 2/1

So, 2:1

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Comments Lesson Posted on 06 Feb CBSE/Class 8

Sujoy D.

Question: Consider a right angled triangle with in radius 2cm and circumradius of 7cm . What is the area of the triangle ? Solution: Method 1: For right tringle, area = r (r+2R) = 2*(2+14) = 32 Method 2: Use delta = r^2+2R×r = 4+2×2×7 = 32 Method 3: C = 7 H = 14 1/2(a+b-c) = 2 A+b-14... read more

Question:

Consider a right angled triangle with in radius 2cm and circumradius of 7cm . What is the area of the triangle ?

Solution:

Method 1:

For right tringle, area = r (r+2R) = 2*(2+14) = 32

Method 2:

Use delta = r^2+2R×r = 4+2×2×7 = 32

Method 3:

C = 7

H = 14

1/2(a+b-c) = 2

A+b-14 = 4

A+b = 18

2ab + 196 = 324

2ab = 128

Ab/2 = 128/4 = 32

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