Geometry

According to question,
Length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, this is right angle triangle.

Note: In right angle triangle median divides the hypotenuse in two equal parts
∴ BD = $$\frac{{\text{H}}}{2}$$
BD = $$\frac{{10}}{2}$$
BD = 5 cm

According to question,
Given: AC = 9 cm

area of ΔABC = 40 cm²
area of ΔADC = 10 cm²
ΔABC ∼ ΔADC


$$\frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ADC}} = \frac{{A{B^2}}}{{A{D^2}}} = \frac{{B{C^2}}}{{A{C^2}}}$$
(In similar Δratio of their area is square of ratio of corresponding sides)
$$\frac{{40}}{{10}} = \frac{{B{C^2}}}{{{{\left( 9 \right)}^2}}}$$
$$\frac{{40}}{{10}} \times 81 = B{C^2}$$
$$BC = 18\,{\text{cm}}$$

Details

Given, three points are
(x1,y1) = (2,3)
(x2,y2) = (4,5)
(x3,y3) = (6,3)
Then, Area of triangle = ($$\frac{1}{2}$$) × | (x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)|
= ($$\frac{1}{2}$$) × | (2x5 + 4x3 + 6x3) – (4x3 + 6x5 + 2x3)|
= ($$\frac{1}{2}$$) × |(10 + 12 + 18) – (12 + 30 + 6)|
= ($$\frac{1}{2}$$) × |40 – 48|
= ($$\frac{1}{2}$$) × |(-8)|
= ($$\frac{1}{2}$$) × 8
= 4
∴ Area of triangle = 4 sq.unit's

Details

According to question,

AO = I_R = Circumradius
DO = I_r = Inradius = 3 cm
Median AD = 3 unit's
1 unit = 3 cm
3 unit's = 3 × 3 = 9 cm
∴ AD = 9 cm

According to question,
Let sides of the triangle be 3x, 4x, 6x
Now check the square of biggest side and sum of square of two smallest side and check which is greater.
∴ (3x)² + (4x)² < (6x)²
⇒ 25x² > 36x²
∴ The triangle will be obtuse angled triangle.

If line DE is parallel to base BC of triangle ABC to internally divide the remaining two sides, then
AD/DB = AE/EC
As ∆ADE ~ ∆ABC

According to question,
We know ∠BAC = 85°
∠BCA = 75°
∠OAC = ?

∠ABC + ∠BCA + ∠CAB = 180°
∠ABC = 20°
∴ ∠COA = 2 × ∠ABC
    ∠COA = 2 × 20 = 40°
In ΔAOC
We know OC = OA
∴ ∠OAC = ∠OCA
∴ ∠OAC + ∠OCA + ∠COA = 180°
2∠OAC = 180° - 40°
2∠OAC = 140°
∠OAC = 70°