Question

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

Answer 1

Answer:

Perimeter of $\triangle ABC$ is $48+16\sqrt{3}$.

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,

$\tan = \frac{perpendicular}{Adjacent}$

then;

$\tan 30 = \frac{8}{CD}$ or

$\frac{1}{\sqrt{3}} =\frac{8}{CD}$

Simplify:

$CD=8\sqrt{3}$ cm.

Also, find the length of AC, we use sine ratio i.e, $\sin=\frac{perpendicular}{Hypotenuse}$.

Then,

$\sin 30 =\frac{AD}{AC}$ or

$\frac{1}{2} =\frac{8}{AC}$

On simplify:

$AC=2\times 8 =16$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;

$\tan A = \frac{BC}{AC}$     [∴$\ tan= \frac{perpendicular}{adjacent}$ ]

therefore,

$\tan 60 = \frac{BC}{16}$

or

$\sqrt{3}= \frac{BC}{16}$

Simplify:

$BC=16\sqrt{3}$cm.

Using Pythagoras theorem in triangle ACB;

Let BD be x cm

AB = 8+x cm ,  AC = 16 cm and $BC=16\sqrt{3}$ cm

$AB^2=AC^2+BC^2$

$(8+x)^2=(16)^2+(16\sqrt{3} )^2$ or

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

$8+x=\sqrt{1024} =32$

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 = $32+16\sqrt{3} +16=48+16\sqrt{3}$ cm