Question

let ∆ be the Area of a triangle. find the area of a triangle whose each side is twice the side of the given triangle.​

Answer 1

Answer:it is solved below

Step-by-step explanation:

In paper.

Answer 2

Step-by-step explanation:

Let thsideof these Original triangle are a, b, c respectively...

Now area

Del(1)=

$\sqrt{ \frac{a + b + c}{2}( \frac{a + b + c}{2} - a)( \frac{a + b + c}{2} - b)( \frac{a + b + c}{2} - c) } \\ \sqrt{ \frac{a + b + c}{2}( \frac{b + c - a}{2})( \frac{a + c - b}{2} )( \frac{a + b - c}{2} ) } \\ \frac{ \sqrt{(a + b + c)(b + c - a)(a + c - b)(a + b - c)} }{4}$

Now, the sides of the new triangle are 2a,2b and 2c

So the area Del(2)=

$\sqrt{ \frac{2(a + b + c)}{2}( a + b + c - 2a)(a + b + c - 2b)(a + b + c - 2c) } \\ \sqrt{(a + b + c)(b + c - a)(a + c - b)(a + b - c) }$

Del(2)=(4)Del(1)

Therefore the area of the new one will be 4 times the area of the previous one........