Question

ABC is an isosceles triangle with an inscribed circle with center O. Let P be the midpoint of BC. If AB=AC=15 and BC=10,then OP equals

• A. \(\frac{\sqrt5}{\sqrt2}\) unit
• B. \(\frac{5}{\sqrt2}\) unit
• C. \(\frac{2}{\sqrt5}\) unit
• D. \(5\sqrt2\) unit

Answer 1

The Correct Option is B

Given:

Isosceles triangle ABC with AB = AC = 15 and BC = 10. 

O is the center of the inscribed circle.

P is the midpoint of BC.

We want to find the length of OP.

Step 1: Draw a diagram.

Draw isosceles triangle ABC with AB = AC = 15 and BC = 10. Draw the inscribed circle with center O. Since ABC is isosceles, the altitude from A to BC will bisect BC at point P. Thus, AP is the altitude and median to BC. Also, O will lie on AP. OP is the radius (r) of the inscribed circle.

Step 2: Find the length of AP.

Triangle ABP is a right-angled triangle with AB = 15 and BP = BC/2 = 10/2 = 5.

Using the Pythagorean theorem:

\(AP^2 + BP^2 = AB^2\)

\(AP^2 + 5^2 = 15^2\)

\(AP^2 + 25 = 225\)

\(AP^2 = 200\)

\(AP = \sqrt{200} = 10\sqrt{2}\)

Step 3: Find the semi-perimeter (s) of triangle ABC.

\(s = \frac{AB + AC + BC}{2} = \frac{15 + 15 + 10}{2} = \frac{40}{2} = 20\)

Step 4: Find the area of triangle ABC.

\(Area = \frac{1}{2} \cdot BC \cdot AP = \frac{1}{2} \cdot 10 \cdot 10\sqrt{2} = 50\sqrt{2}\)

Step 5: Find the inradius (r) of the inscribed circle.

The area of a triangle can also be expressed as \(Area = r \cdot s\), where r is the inradius and s is the semi-perimeter.

\(r = \frac{Area}{s} = \frac{50\sqrt{2}}{20} = \frac{5\sqrt{2}}{2}\)

Since OP is the radius of the inscribed circle, OP = r.

\(OP = \frac{5\sqrt{2}}{2} = \frac{5}{\sqrt{2}}\)

Therefore, OP = \(\frac{5}{\sqrt{2}}\) units.