Madhyamik 2024 Math Paper English Version Solution

Get the detailed solution of Madhyamik Mathematics Paper 2024 in English Version. 

WB Board Class 10 Maths Question Paper PDF: Get here free PDF download of the West Bengal Board Class 10 Maths question paper for board exam 2024. Also, check the WBBSE Class 10 Maths answer key by experts here.

Surds Practice Problem Set

In Mathematics, surds are the values in square root that cannot be further simplified into whole numbers or integers. Surds are irrational numbers. The examples of surds are √2, √3, √5, etc., as these values cannot be further simplified. If we further simply them, we get decimal values, such as:

√2  = 1.4142135…

√3 = 1.7320508…

√5 = 2.2360679…

Surds Definition

Surds are the square roots  (√) of numbers that cannot be simplified into a whole or rational number. It cannot be accurately represented in a fraction. In other words, a surd is a root of the whole number that has an irrational value. Consider an example, √2 ≈ 1.414213. It is more accurate if we leave it as a surd √2.

Surds Worksheet

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Surd-Practice-III

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Geometry Solved Problems : Circles

Solved problems on circles for class X. Get solutions of problems shown in the post and many more. Visit the link Solved Problems: Circles

TRIGONOMETRY : TRIGONOMETRICAL RATIOS SOLVED PROBLEMS FOR 10TH GRADE CBSE ICSE OTHER STATE BOARDS

 

TRIGONOMETRY : TRIGONOMETRICAL RATIOS SOLVED PROBLEMS FOR 10TH GRADE ( CBSE, ICSE AND OTHER STATE BOARDS). SUBSCRIBE FOR MORE SOLVED PROBLEMS WHICH WILL BE UPLOADED SOON.

Trigonometric ratios are the ratios between edges of a right triangle. These ratios are given by the following trigonometric functions of the known angle A, where a, b and c refer to the lengths of the sides in the accompanying figure:

• Sine function (sin), defined as the ratio of the side opposite the angle to the hypotenuse.

${\displaystyle \sin A={\frac {\textrm {opposite}}{\textrm {hypotenuse}}}={\frac {a}{c}}.}$

• Cosine function (cos), defined as the ratio of the adjacent leg (the side of the triangle joining the angle to the right angle) to the hypotenuse.

${\displaystyle \cos A={\frac {\textrm {adjacent}}{\textrm {hypotenuse}}}={\frac {b}{c}}.}$

• Tangent function (tan), defined as the ratio of the opposite leg to the adjacent leg.

${\displaystyle \tan A={\frac {\textrm {opposite}}{\textrm {adjacent}}}={\frac {a}{b}}={\frac {a/c}{b/c}}={\frac {\sin A}{\cos A}}.}$

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Geometry Inequality

Post by Maths.

Prove that in any quadrilateral, the sum of the diagonals is greater than the half of its perimeter.

Consider the quad. In the above diagram. Let E be the point of the intersection.

Now, AE+EB > AB

EB+EC > BC

AE+ED > AD

EC+ED > DC (Using Triangle Inequality)

Adding the above four inequalities we get

2(AE+EC+EB+ED) > AB+BC+AD+DC

=> AC + BD > ½(AB+BC+AD+DC)

Thus sum of the diagonals is greater than the half of its perimeter Q.E.D

In any triangle four times the sum of its medians is greater than 3 times its perimeter.

We know that difference of any two sides of a triangle is less than the third side (prove it)

In triangle ABE,

AE > AB-BE

In triangle ACE,

AE > AC-CE

Adding above two inequalities we get,

2AE > AB + AC -(BE+CE)

=> AE > ½(AB+AC-BC)

=> 4AE > 2(AB+AC-BC).........(1)

Similarly,

4BD > 2(AB+BC-AC).............(2) and 4CF > 2(AC+BC-AB)........(3)

Adding (1),(2) and (3) we have,

4(AE+BD+CF) > 2(AB+AC-BC+AB+BC-AC+AC+BC-AB)

=> 4(AE+BD+CF) > 2(AC+AB+BC)

=> sum of the lengths of the medians is greater than half the perimeter

We can strengthen the inequality by using the fact that the point 'O' divides the medians AE,BD,CF internally in the ration 2:1

Therefore, OD:OB = 1:2

=> (OB+OD):OB = (1+2):2

=> BD:OB=3:2

=> OB = 2/3 BD........(a)

Similarly, OC = 2/3 CF.........(b) and OA = 2/3 AE.......(c)

Now in triangle OBC, OB+OC> BC

=>2/3(BD+CF)>BC [using (a) and (b)]

=> 2(BD+CF)>3BC

Similarly, 2(CF+AE)>3AC and 2(BD+AE)>3AB

Adding the last three inequalities we get 4(AE+BD+CF) > 3(AB+BC+CA)

In the triangle ABC, AE,BD and CF are the medians where O is the point of there intersection