In figure, a circle touches all the four sides of a quadrilateral
ABCD whose sides are AB = 6 cm, BC = 7 cm and CD = 4
cm. Find the length of AD .
Answers
Answer:
AD=3cm
Step-by-step explanation:
First draw the diagram properly
The length of two tangents drawn from an external point of the circle are equal
so, AP=AS
BP=BQ
DR=DS
RC=CQ
Then add all
(AP+BP)+(DR+RC)=AS+BQ+DS+CQ
(AP+BP)+(DR+RC)
(AS+DS)+(BQ+CQ)
AB+DC=AD+BC
6+4=AD+7
10=AD+7
AD=10-7
AD=3
Hope this helps you
Answer:
AD=3
Step-by-step explanation:
Given : A circle inscribed in a quadrilateral ABCD,AB=6cm,BC=7cm,CD=4cm.
To find : AD
Proof : Let the circle touch the sides AB,BC,CD,DA, at P,Q,R and S, respectively.
AP=AS
BP=BQ
DR=DS
CR=CQ {Lengths of two tangents drawn from an external point of circle, are equal}
Adding all these, we get
(AP+BP)+(CR+RD)=(BQ+QC)+(DS+SA)
AB+CD=BC+DA
⇒6+4=7+AD
⇒AD=10-7=3
I HOPE THIS ANSWER HELPFUL TO YOU