Changes
From Karnataka Open Educational Resources
1,044 bytes added
, 05:33, 14 August 2014
Line 96: |
Line 96: |
| [[File:fig3.png|200px]] | | [[File:fig3.png|200px]] |
| ==Interpretation of problem== | | ==Interpretation of problem== |
| + | # In the quadrilateral ABCD sides BC , DC & QB are given . |
| + | #AD⊥DC. |
| + | # Asked to find the radius OS or OP |
| + | '''==Concepts used==''' |
| + | #Tangents drawn from an external point to a circle are equal. |
| + | #In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square |
| + | #In a circle, the radius drawn at the point of contact is perpendicular to the tangent |
| + | '''==Algorithm==''' |
| + | In the fig BC=38 cm and BQ=27 cm <br> |
| + | BQ=BR=27 cm (because by concept 1) <br> |
| + | Therefore, CR=BC-BR=38-27=11 cm <br> |
| + | CR=SC=11 cm (because by concept 1) <br> |
| + | DC=25 cm <br> |
| + | therefore , DS=DC-SC=25-11=14 cm <br> |
| + | DS=DP=14 cm (because by concept 1) <br> |
| + | Also {AD ortho DC} , {OP ortho AD} and {OS ortho DC} <br> |
| + | ∠D=∠S=∠P=90˚ <br> |
| + | ⇒ ∠O=90˚ <br> |
| + | therefore DSOP is a Square <br> |
| + | SO=OP=14 cm <br> hence Radius of given circle is 14 cm |
| + | |
| =Problem 5= | | =Problem 5= |
| A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC]. | | A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC]. |