Question

Q1 Two times a number is equal to six less than three times the number what the number is.
Q2 ABC is an isosceles triangle right angled at C. Prove that AB²=2AC²​

Answer 1

$\underline{\bf{\dag} \:\mathfrak{Question-1 }}$⠀⠀⠀⠀

• Two times a number is equal to six less than three times the number what the number is.

❍ Let's say, the number be n respectively.

A/Q,

• Two times a number is equal to six less than three times the number '(n)'.

Therefore,

↠ 2n = 3n – 6

↠ 2n - 3n = - 6

↠- n = - 6

• (Cancelling –ve sign from both sides).

↠ n = 6

$\therefore$ Hence, the needed number is 6.

$\rule{250px}{.3ex}$

$\underline{\bf{\dag} \:\mathfrak{Question-2}}$⠀⠀⠀⠀

• ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

AnswEr :

Here, In ∆ ACB, we'll use Pythagoras theorem.

• AC = BC Because, in isosceles triangle two sides are always equal and ∆ACB is an isosceles triangle.
• And, AB is Hypotenuse (longest side).

↠(Hypotenuse)² = (Height)² + (Base)²

↠(AB)² = (AC)² + (BC)²

• Here, in triangle (AC = BC). Hence, we can write AC instead of BC.

↠(AB)² = 2AC² ⠀⠀⠀⠀⠀⠀ ∴ (Hence, Proved!)

Answer 2

1

Given :-

Two  times a number is equal to six less than three times the number what the number is.

Solution :-

Let the number be 'l'

$\sf 2\ell = 3\ell - 6$

$\sf 3\ell - 2\ell = 6$

$\sf \ell = 6$

Verification

$\sf 2(6) = 3(6) - 6$

$\sf 12 = 18 - 6$

$\sf 12 =12$

2

Given :

ABC  is an isosceles triangle right angled at C. Prove that AB²=2AC²​

Solution :

At first In △ABC,

We need to use Pythagorean theorem

$\sf H^{2} = B^{2} + P^{2}$

H = AB

B = AC

P = BC

$\sf AB^{2} = AC^{2} + BC^{2}(1)$

Then,

ABC is an isosceles triangle so,

AC = BC(2)

From 1 and 2 we get

AB²=2AC²​