Question

Plz solve this Q.4 .1

Answer 1

Answer:

Step-by-step explanation:

The question that you have asked is very easy so try toso please try to do it yourself or we use side book like rs aggarwal in rs aggarwal in triangle chapter it is clearly or IT solution is kill written in the hint section so please check it out

Answer 2

Step-by-step explanation:

Given:

AD is an altitude of an equilateral triangle ABC. On AD as a base,another equilateral triangle ADE is constructed.

Prove:

area(ADE):area(ABC)=3:4.

Solution:

ABC is an equilateral triangle.

Let one side AB be 2x.

In equilateral triangle all the sides are of equal length.

==> AB=BC=AC= 2X

∵ AD⊥BC Since perpendicular bisects the given side into two equal parts,then BD = DC = x.

In ΔADB,

By Pythagoras theorem,AB² = AD² + BD²

==> AD² = AB² - BD²

==> AD² = (2x)² - x²

==> AD² =3x²

==> AD = √3x cm

By the theorem which states that the areas of two similar triangles are in the ratio of the squares of the any two corresponding sides.

A(ΔADE)/A(ΔABC)

==> AD²/AB²

==> (√3x)²/2x²

==> 3/4

Result:

A(ΔADE) : A(ΔABC) = 3 : 4