ABC is a right angle triangle, right angled at A. A circle is inscribed in it. The length of two sides containing angle A is 12cm and 5cm find the radius.

Let ABC be the right angled triangle such that ∠A = 90° , AB = 5cm, AC = 12 cm. 

Let O be the centre and r  be the radius of the incircle.

AB, BC and CA are tangents to the circle at P, N and M.

 

∴ OP = ON = OM = r (radius of the circle)

 

By Pythagoras theorem,

 

 BC2 = AC2 + AB2

 ⇒ BC2 = 122 + 52

⇒ BC2 = 169

⇒ BC = 13 cm

 

Area of ∆ABC = Area ∆OAB + Area ∆OBC + Area ∆OCA

  • 95
Hi!
Here is the answer to your question.

Given: ABCD is a trapezium where AB||CD and AD = BC

To prove: ABCD is cyclic.
Construction: Draw DL⊥AB and CM⊥AB.
Proof: In ΔALD and ΔBMC,
AD = BC (given)
DL = CM (distance between parallel sides)
∠ALD = ∠BMC (90�)
ΔALD ≅ ΔBMC (RHS congruence criterion)
⇒ ∠DAL = ∠CBM (C.P.C.T) (1)
Since AB||CD,
∠DAL + ∠ADC = 180�(sum of adjacent interior angles is supplementary)
⇒ ∠CBM + ∠ADC = 180�(from (1))
⇒ ABCD is a cyclic trapezium (Sum of opposite angles is supplementary)

 

Cheers!
  • -13
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