Geometry

Given:
The perimeter of the equilateral triangle ABC= 24 cm
Note:
Centroid divide each median in 2:1 ratio
∴ AG:GD= 2:1
Initially, AD should be found so that AG can be calculated.
An equilateral triangle is a triangle in which all three sides are equal.
From the above figure,
BC=24/3= 8 cm
AB=BC=CA=8cm
BD is half of BC
BD=8/2=4 cm
AD= √AB² - BD²
= √64-16
= 4√3
wkt,
AG:GD= 2:1
So, AG is the 2 parts of AD
AG= (4√3/3) × 2
AG= 8/√3

According to question,

PX || BD [mid point theorem]
∴ PX = $$\frac{1}{2}$$BD
Similarly, QY || BD
∴ QY = $$\frac{1}{2}$$BD
∴ PX : QY,           $$\frac{1}{2}$$BD : $$\frac{1}{2}$$BD
PX : QY = 1 : 1

According to question,
Circumradius of an equilateral triangle
I_R = 10 cm
AO = I_R = 10 cm
DO = I_r = ?

2 unit's → 10 cm
1 unit → 5 cm
∴ DO = I_r = 5 cm

According to question,
Given: ∠C = 90°

BC = AC = 5 cm (Isosceles triangle)
By Pythagoras theorem
AB² = AC² + BC²
AB² = 5² + 5²
AB² = 25 + 25
AB² = 50
AB = 5$$\sqrt 2 $$ cm

According to question,

Given: BAC is a right angle triangle
AD ⊥ BC
AD = 6 cm
BD = 4 cm
BC = ?
In ΔBAD
$$AB = \sqrt {B{D^2} + A{D^2}} $$
$$AB = \sqrt {{4^2} + {6^2}} $$
$$AB = \sqrt {52} \,cm$$
ΔBAC ∼ ΔBDA


$$\therefore \frac{{BC}}{{AB}} = \frac{{AB}}{{BD}}$$
$$\therefore \frac{{BC}}{{\sqrt {52} }} = \frac{{\sqrt {52} }}{4}$$
$$BC = \frac{{52}}{4}$$
$$BC = 13\,cm$$

According to question,

By internal bisector property
$$\frac{{AB}}{{AC}} = \frac{{BD}}{{DC}}$$
$$\frac{{AB}}{{AC}} = \frac{5}{{2.5}}$$
$$\frac{{AB}}{{AC}} = \frac{2}{1}$$
$$\therefore AB:AC = 2:1$$

According to question,

In right angle ΔBAC
∠A + ∠B + ∠C = 180°
∠B = 180° - 55° - 90°
∠B = 35°
In right angle ΔADB
∠ADB + ∠ABD + ∠BAD = 180°
∠BAD = 180° - 35° - 90°
∠BAD = 55°

According to question,

In ΔABC
AB² = AC² + BC² . . . . . . . (i)
ΔACD
AD² = AC² + CD²
AC² = AD² - CD² . . . . . . . (ii)
Put the value of AC² in equation (i)
AB² = AD² - CD² + BC²
AB² + CD² = AD² + BC²