Question

In an isosceles $△ABC,AB=BC=4.5\text{cm, and}\angle BAC={30}^{\circ }$. What will be the perimeter of the triangle?

• A. $16\text{cm}$
• B. $23.4\text{cm}$
• C. $15.2\text{cm}$
• D. $16.8\text{cm}$

Answer 1

The correct option is D $16.8\text{cm}$

Given:

• $AB=BC=4.5\text{cm}$
• $\angle BAC={30}^{\circ }$

We can construct the triangle using SAS criterion, but there is a problem, $\angle BAC$ is not the included angle between sides AB and BC.

The actual included angle is $\angle ABC$.
We can find $\angle ABC$ using the angle sum property of a triangle.

$\angle BAC=\angle BCA={30}^{\circ }$….(Opposite angles of equal sides are equal)

Now, $\angle BAC+\angle BCA+\angle ABC={180}^{\circ }$(angle sum property)
$\Rightarrow \angle ABC+{60}^{\circ }={180}^{\circ }\Rightarrow \angle ABC={120}^{\circ }$

Now that we know the included angle, we can construct the triangle using the SAS criterion.

1. First, draw a line segment $AB$ of length $4.5\text{cm}$.


2. Next, take any radius and draw an arc with $B$ as the center.


3. Next, take the point where the arc intersects with $AB$ as the center and draw another arc with the same radius.


4. Next, take the new intersection point as the center and draw an arc with the same radius.

5. Join the last intersection point with $B$ to get ${120}^{\circ }$ angle at $B$.


6. Cut an arc of $4.5\text{cm}$ with $B$ as the center to get $C$.


7.Join $AC$.


Measuring the length of $AC$ using a ruler, we get: $AC=7.8\text{cm}$

Perimeter of $△ABC=AB+BC+AC=4.5+4.5+7.8=16.8\text{cm}$