Answer:
Area of Δ ABC = 21.86 units square
Perimeter of Δ ABC = 24.59 units
Step-by-step explanation:
Given:
In Δ ABC
∠A=45°
∠C=30°
Height of triangle = 4 units.
To find area and perimeter of triangle we need to find the sides of the triangle.
Naming the end point of altitude as 'O'
Given [tex]BO\perp AC[/tex]
For Δ ABO
Since its a right triangle with one angle 45°, it means it is a special 45-45-90 triangle.
The sides of 45-45-90 triangle is given as:
We are given BO (Leg 1) [tex]x=4[/tex]
∴ AO (Leg2) [tex]=x=4[/tex]
∴ AB (hypotenuse) [tex]=x\sqrt2=4\sqrt2=5.66 [/tex]
For Δ CBO
Since its a right triangle with one angle 30°, it means it is a special 30-60-90 triangle.
The sides of 30-60-90 triangle is given as:
We are given BO (side opposite 30° angle) [tex]=x=4[/tex]
CO (side opposite 60° angle) [tex]=x\sqrt3=4\sqrt3=6.93[/tex]
BC (Hypotenuse) [tex]=2x=2\times 4 =8[/tex]
Length of side AC is given as sum of AO and CO
[tex]AC=AO+CO=4+6.93=10.93[/tex]
Perimeter of Δ ABC= Sum of sides of triangle
⇒ AB+BC+AC
⇒ [tex]5.66+8+10.93[/tex]
⇒ [tex]24.59[/tex] units
Area of Δ ABC = [tex]\frac{1}{2}\times base\times height[/tex]
⇒ [tex]\frac{1}{2}\times 10.93\times 4[/tex]
⇒ [tex]21.86[/tex] units square