Question

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Answer 1

Answer:

Step-by-step explanation:

→ The area is${\displaystyle A={\frac {\sqrt {3}}{4}}a^{2}}A=\frac{\sqrt{3}}{4} a^2,$

→The perimeter is ${\displaystyle p=3a\,\!}p=3a\,\!$

→The radius of the circumscribed circle i$s {\displaystyle R={\frac {a}{\sqrt {3}}}}R = \frac{a}{\sqrt{3}}$

→The radius of the inscribed circle is${\displaystyle r={\frac {\sqrt {3}}{6}}a}r=\frac{\sqrt{3}}{6} a or {\displaystyle r={\frac {R}{2}}} r=\frac{R}{2}$

→The geometric center of the triangle is the center of the circumscribed and inscribed circles

→The altitude (height) from any side is${\displaystyle h={\frac {\sqrt {3}}{2}}a}h=\frac{\sqrt{3}}{2} a$

→Denoting the radius of the circumscribed circle as R, we can determine using trigonometry that:

→The area of the triangle is${\displaystyle \mathrm {A} ={\frac {3{\sqrt {3}}}{4}}R^{2}}{\displaystyle \mathrm {A} ={\frac {3{\sqrt {3}}}{4}}R^{2}}$

→Many of these quantities have simple relationships to the altitude ("h") of each vertex from the opposite side:

→The area is ${\displaystyle A={\frac {h^{2}}{\sqrt {3}}}}A=\frac{h^2}{\sqrt{3}}$

→The height of the center from each side, or apothem, is  ${\displaystyle {\frac {h}{3}}}\frac{h}{3}$

→The radius of the circle circumscribing the three vertices is ${\displaystyle R={\frac {2h}{3}}}R=\frac{2h}{3}$

→The radius of the inscribed circle is ${\displaystyle r={\frac {h}{3}}}r=\frac{h}{3}$

→In an equilateral triangle, the altitudes, the angle bisectors, the perpendicular bisectors, and the medians to each side coincide.

HOPE IT HELPS

Answer 2

Answer:

Step-by-step explanation:

when an altitude is dropped on base in an equilateral triangle it bisects the base as well as make right angle on both side hence

bd=cd

also bd+dc=bc

2bd=bc

bd=1/2bc

now in triangle abd

ab^2=bd^2+ad^2 (pythagores theorem)

ab^2=1/4ab^2+ad^2                (ab=bc and also ab^2=bc^2 and 1/4bc^2=bd^2)

ab^2=ab^2+4ad^2/4

4ab^2-ab^2=4ad^2

3ab^2=4ad^2

hence proved