Advertisements
Advertisements
Question
ABC is a right triangle in which ∠B = 90°. If AB = 8 cm and BC = 6 cm, find the diameter of the circle inscribed in the triangle.
Solution
We have given that a circle is inscribed in a triangle
Using pythagoras theorem
(AC)2 = (AB)2 + (BC)2
(AC)2 = (8)2 + (6)2
(AC)2 = 64 + 36
(AC)2 = 100
⇒ AC = 10
Area of △ABC = area of △APB + area of △BPC + area of △APC
`1/2 xx b xx h = 1/2 xx b_1 xx h_1 + 1/2 xx b_2 xx h_2 + 1/2 xx b_3 xx h_3`
`1/2 xx 6 xx 8 = 1/2 xx 8 xx r + 12 xx 6 xx r + 12 xx 10 xx r`
24 = 4r + 3r + 5r
24 = 12r
⇒ r = 2
∵ d = 2r
⇒ d = 2 x 2
⇒ d = 4 cm.
APPEARS IN
RELATED QUESTIONS
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.
In Figure, DE || BC If DE = 4 cm, BC = 6 cm and Area (ΔADE) = 16 cm2, find the area of ΔABC.
In ΔABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ΔADE and ΔABC
In ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ΔAPQ and trapezium BPQC.
In ΔABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ΔABC into two parts equal in area. Find `(BP)/(AB)`
If ∆ABC ~ ∆PQR, A (∆ABC) = 80, A (∆PQR) = 125, then fill in the blanks. \[\frac{A\left( ∆ ABC \right)}{A\left( ∆ . . . . \right)} = \frac{80}{125} \therefore \frac{AB}{PQ} = \frac{......}{......}\]
∆LMN ~ ∆PQR, 9 × A (∆PQR ) = 16 × A (∆LMN). If QR = 20 then Find MN.
In the given figure DE || AC which of the following is true?
In the given figure, ΔACB ~ ΔAPQ. If AB = 6 cm, BC = 8 cm, and PQ = 4 cm then AQ is equal to ______.
