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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
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
Updated On: Jun 25, 2024
- \(\frac{\sqrt5}{\sqrt2}\) unit
\(\frac{5}{\sqrt2}\) unit
- \(\frac{2}{\sqrt5}\) unit
- \(5\sqrt2\) unit
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The Correct Option is B
Approach Solution - 1
The correct answer is option (B) : \(\frac{5}{\sqrt2}\) unit.
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Approach Solution -2
Geometric Solution Simplified
1. Draw a perpendicular line AD to BC.
2. D is the midpoint of BC, so BD = DC.
3. Given BC = 14 cm, then BD = DC = 7 cm.
4. Using the Pythagorean theorem in triangle ADB:
AB2 = AD2 + BD2
252 = AD2 + 72
625 = AD2 + 49
AD2 = 625 - 49 = 576
AD = 24 cm
5. Using the Pythagorean theorem in triangle ODB:
OB2 = OD2 + BD2
Here OD = AD - AO = 24 - r, where r is the radius of the circle:
(24 - r)2 + 72 = r2
(24 - r)2 + 49 = r2
Expanding and simplifying:
576 - 48r + r2 + 49 = r2
576 + 49 - 48r = 0
625 - 48r = 0
48r = 625
r = 625 / 48 ≈ 13.02 cm
1. Draw a perpendicular line AD to BC.
2. D is the midpoint of BC, so BD = DC.
3. Given BC = 14 cm, then BD = DC = 7 cm.
4. Using the Pythagorean theorem in triangle ADB:
AB2 = AD2 + BD2
252 = AD2 + 72
625 = AD2 + 49
AD2 = 625 - 49 = 576
AD = 24 cm
5. Using the Pythagorean theorem in triangle ODB:
OB2 = OD2 + BD2
Here OD = AD - AO = 24 - r, where r is the radius of the circle:
(24 - r)2 + 72 = r2
(24 - r)2 + 49 = r2
Expanding and simplifying:
576 - 48r + r2 + 49 = r2
576 + 49 - 48r = 0
625 - 48r = 0
48r = 625
r = 625 / 48 ≈ 13.02 cm
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Consider a line L that is passing through the three-dimensional plane. Now, x,y and z are the axes of the plane and α,β, and γ are the three angles the line makes with these axes. These are commonly known as the direction angles of the plane. So, appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the given line L.
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