Question

2
2.6 पाइथागोरस प्रमेय के अनुसार,
(आधार)2 + (लम्ब)2 3
a) (केंद्र)
b) (कोण)2
c) (कर्ण)
According to Pythagoras theorem, (Base)2 + (Perpend
a) (Centre)
b) (Angle)
c) (
7 A(2,4) तथा B(6,8) बिंदओं का मध्य बिंदु है-
The mid point of the points A(2,4) and B(6,8) is -
a) (4,6)
c) (3,5)
b) (6,4)
2.8 त्रिकोणमितीय सर्वसमिका :-
Trigonometric identity: -
sin? A + cos? A =
sin2 A + cos2 A =​

Answer 1

Question 2.6) :- According to Pythagoras theorem, (Base)² + (Perpendicular)² :-

a) (Centre)²

b) (Angle)²

c) (Hypotenuse)²

Solution :-

A right angle ∆ , has three sides as Perpendicular, Base and hypotenuse.

Pythagoras theorem :- The sum of the squares of the lengths of the legs of a right triangle is equal to the square of the length of the hypotenuse .

Or,

→ (Base)² + (Perpendicular)² = (Hypotenuse)²

∴ option (C) (Hypotenuse)² is right Answer.

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Question 2.7) The mid point of the points A(2,4) and B(6,8) is :-

a) (4,6)

c) (3,5)

b) (6,4)

Solution :-

The mid - Point of (x1 , y1) and (x2, y2) is given by :-

→ x = (x1 + x2) / 2

→ y = (y1 + y2) / 2

we have given that,

• x1 = 2
• y1 = 4
• x2 = 6
• y2 = 8

Putting values we get :-

→ x = (2 + 6)/2 = 8/2 = 4 .

→ y = (4 + 8) / 2 = 12/2 = 6.

Therefore,

Mid - Points of the given Points is (4,6) . (Option A)

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Question 2.8) Trigonometric identity : -

sin²A + cos²A = ?

Solution :-

we know that,

• sinθ = (Perpendicular) / (Hypotenuse) = P/H
• cosθ = (Base) / (Hypotenuse) = B/H

So,

→ sin²θ + cos²θ

→ (P/H)² + (B/H)²

→ (P²/H²) + (B²/H²)

→ (P² + B²) / H²

Now, According to Pythagoras theorem, (Base)² + (Perpendicular)² = (Hypotenuse)²

→ (P² + B²) = H²

So,

→ (P² + B²) / H²

→ H² / H²

→ 1 .

Therefore,

→ sin²A + cos²A = 1. (Ans.)

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