The document shows the steps to solve a geometry problem about finding the area of a right-angled triangle given its inradius and circumradius. It uses the properties that the inradius r = 2 cm and circumradius R = 7 cm to find that the hypotenuse is 14 cm. It then sets up and solves equations relating the leg lengths a and b to find that a = 12 and b = 6. The area is then calculated as 1/2 × the product of the roots of the equation, which is 32 sq cm.

1. Geometry

2. Geometry
Consider a right–angled triangle with inradius 2 cm and circumradius
of 7 cm. What is the area of the triangle?
(a) 32 sq cms (b) 31.5 sq cms
(c) 32.5 sq cms (d) 33 sq cms

3. Geometry
r = 2
R = 7 =
1
2
of hypotenuse
Hypotenuse = 14
r =
a + b −h
2
2 =
a + b −14
2
a + b – 14 = 4
Consider a right–angled triangle with inradius 2 cm and circumradius
of 7 cm. What is the area of the triangle?

4. Geometry
a + b = 18
a2 + b2 = 142
a2 + (18 – a)2 = 196
a2 + 324 + a2 – 36a = 196
2a2 – 36a + 128 = 0
a2 – 18a + 64 = 0
Consider a right–angled triangle with inradius 2 cm and circumradius
of 7 cm. What is the area of the triangle?

5. Geometry
Now the 2 roots to this equation will effectively be a, 18 – a. Product
of the roots = 64.
Area =
1
2
× product of roots (How? Come on, you can figure this out.)
= 32 sq. cms
Answer choice (a)
Consider a right–angled triangle with inradius 2 cm and circumradius
of 7 cm. What is the area of the triangle?

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