Question

the areas of an isosceles triangle is 60 CM. The length of equal sides is 13 cm find the length of its base​

Answer 1

Answer:

The value of base cannot be negative the values of base BC are 10 and 24.

Step-by-step explanation:

Consider the isosceles triangle.

The equal sides are AB and AC.

The measure of sides AB and AC is 13 cm.

Let BC = b.

The line AD is a perpendicular to the base BC. AD divides BC divided into two equal parts.

The measure of AD using the Pythagoras theorem is:

$AB^{2}=AD^{2}+DB^{2}\\13^{2}=AD^{2}+(\frac{b}{2} )^{2}\\AD^{2}=169-\frac{b^{2}}{4}$

The area of a triangle is:

$Area=\frac{1}{2}\times base\times height$

The area of isosceles triangle ABC is 60 cm².

Compute the base as follows:

$Area=\frac{1}{2}\times BC\times AD\\60=\frac{1}{2}\times b\times\sqrt{169-\frac{b^{2}}{4} }\\120=b\times\sqrt{169-\frac{b^{2}}{4} }$

Square both sides and solve for b as follows:

$14400=b^{2}\times[169-\frac{b^{2}}{4} ]\\14400=169b^{2}-\frac{b^{4}}{4}\\b^{4}-676b^{2}+57600=0\\b^{4}-100b^{2}-576b^{2}+57600=0\\b^{2}(b^{2}-100)-576(b^{2}-100)=0\\(b^{2}-100)(b^{2}-576)=0$

• If (b² - 100) = 0,

        The value of b is ±10.

• If (b² - 576) = 0,

        The value of b is ±24.

Since the value of base cannot be negative the values of base BC are 10 and 24.