Question

Given: ΔАВС, m∠ACB = 90° CD ⊥ AB , m∠ACD = 30°,AD = 8 cm Find: Perimeter of ΔABC

Answer 1

To find the perimeter of triangle ABC, calculate the lengths of the sides AB, AC, and BC and add them together. Use trigonometric functions and the Pythagorean theorem to calculate the lengths of AC and BC. Finally, calculate the perimeter by adding the three side lengths.

Step-by-step explanation:

To find the perimeter of triangle ABC, we need to find the lengths of all three sides and then add them together. Since AC is the hypotenuse of a right triangle, we can use trigonometric functions to find its length. Given that angle ACD is 30 degrees, we can use the sine function to find the length of AC:

sin(30) = AD/AC

sin(30) = 8/AC

AC = 8/sin(30) = 16 cm

Now, we can use the Pythagorean theorem to find the length of BC:

BC^2 = AC^2 - AB^2

BC^2 = 16^2 - 8^2

BC = sqrt(256 - 64) = 12 cm

Finally, we can find the length of AB using the fact that AB = 3x:

AB = 3(8) = 24 cm

Now, we can calculate the perimeter:

Perimeter = AB + AC + BC = 24 + 16 + 12 = 52 cm

Answer 2

Perimeter of
`\triangle ABC` is
`<strong>48+16√(3)</strong>`.

Step-by-step explanation:

Given: In
`\triangle ABC` ,
`\angle ACB = 90^(\circ)` ,
`\angle ACD= 30^(\circ)` and AD= 8cm.

In
`\triangle ADC`

Sum of the measure of the angles of triangles is 180 degree.

`\angle CAD+\angle CDA+\angle ACD=180^(\circ)`

`\angle CAD+90^(\circ)+30^(\circ)=180^(\circ)` or

`\angle CAD+120^(\circ)=180^(\circ)`

Simplify:

`\angle CAD=60^(\circ)`

∵ AD= 8cm and
`\angleACD= 30^(\circ)` to calculate the length of CD we use tangent ratio i.e,

`\tan = (perpendicular)/(Adjacent)`

then;

`\tan 30 = (8)/(CD)` or

`(1)/(√(3)) =(8)/(CD)`

Simplify:

`CD=8√(3)` cm.

Also, find the length of AC, we use sine ratio i.e,
`\sin=(perpendicular)/(Hypotenuse)`.

Then,

`\sin 30 =(AD)/(AC)` or

`(1)/(2) =(8)/(AC)`

On simplify:

`AC=2* 8 =16`cm.

Now, in triangle ABC;

`\angle BAC= 60^(\circ)` ,
`\angle ACB= 90^(\circ)` then

`\angle ABC= 30^(\circ)` [Sum of the measure of the angles in the triangle is 180 degree]

To calculate the length of BC;

`\tan A = (BC)/(AC)` [∴
`\ tan= (perpendicular)/(adjacent)` ]

therefore,

`\tan 60 = (BC)/(16)`

or

`√(3)= (BC)/(16)`

Simplify:

`BC=16√(3)`cm.

Using Pythagoras theorem in triangle ACB;

Let BD be x cm

AB = 8+x cm , AC = 16 cm and
`BC=16√(3)` cm

`AB^2=AC^2+BC^2`

`(8+x)^2=(16)^2+(16√(3) )^2` or

`(8+x)^2=256+768=1024` or

`8+x=√(1024) =32`

Simplify:

x=24 cm

Therefore, the length of AB = 8+x= 8+24=32 cm

Perimeter(P) of triangle ABC is equal to the sum of the sides of the triangle.

⇒ P= AB+BC+AC =
`32+16√(3) +16=48+16√(3)` cm