wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In the figure below, the triangle is a 3−4−5 sides triangle. Two equal circle are placed as in the figure. The radius of each of circle is
1259378_8f5c0297c5c241c28d4aad1492ea6daf.png

A
45
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
58
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
611
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
57
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 57
figure shows right angled triangle ABC.
Let AB=4, BC=3 and AC=5

Points P and Q are centers of identical circles as shown in figure. Join them and construct right angled triangle PQR as shown in figure.
Let r= radius of each identical circle.
As circles are touching each other, we can say from the figure that,
PR=r+r
PR=2r

As ABC and PQR both are right angled triangles, both are similar triangles.
Thus, corresponding angles are congruent.

Now, Let ACB=θ
sinθ=ABAC
sinθ=45

By property of similar triangles, PRQ=θ
sinθ=PQPR
45=PQ2r

PQ=45×2r
PQ=8r5 Equation (1)

Now, In ABC, cosθ=BCAC
cosθ=35

By property of similar triangles, cosθ=QRPR
35=QR2r

QR=35×2r
QR=6r5 Equation (2)

Now, As shown in figure,
R´T=r
`SB=PQ+R´T
`SB=8r5+r
`SB=13r5

Now, A`S=AB`SB
A`S=413r5

AS=A`S=413r5 (Tangents from same point to same circle) Equation (3)

Similarly, B`T=`SP+QR
B`T=r+6r5
B`T=11r5

Now, C`T=BCB`T
C`T=311r5

CT=C`T=311r5 (Tangents from same point to same circle) Equation (4)

Similarly, ST=PR=2r

Now, we can write from the figure that,
AC=AS+ST+CT

From equations (3) and (4), we get,

AC=413r5+2r+311r5

5=413r5+2r+311r5

2=13r5+2r11r5

2=13r11r+10r5

2=14r5

14r5=2

r=1014

r=57

Thus, answer is option (D)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Matter Waves
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon