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| [[File:fig3.png|200px]] | | [[File:fig3.png|200px]] |
| ==Interpretation of problem== | | ==Interpretation of problem== |
− | # In the quadrilateral ABCD sides BC , DC & QB are given .
| + | # In the quadrilateral ABCD sides BC , DC & QB are given . |
− | #AD⊥DC.
| + | #AD⊥DC. |
− | # Asked to find the radius OS or OP
| + | # Asked to find the radius OS or OP |
− | '''==Concepts used=='''
| + | ==Concepts used== |
| #Tangents drawn from an external point to a circle are equal. | | #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 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 | | #In a circle, the radius drawn at the point of contact is perpendicular to the tangent |
− | '''==Algorithm=='''
| + | ==Algorithm== |
− | In the fig BC=38 cm and BQ=27 cm <br>
| + | In the fig BC=38 cm and BQ=27 cm <br> |
− | BQ=BR=27 cm (because by concept 1) <br>
| + | BQ=BR=27 cm (because by concept 1) <br> |
− | Therefore, CR=BC-BR=38-27=11 cm <br>
| + | ∴CR=BC-BR=38-27=11 cm <br> |
− | CR=SC=11 cm (because by concept 1) <br>
| + | CR=SC=11 cm (because by concept 1) <br> |
− | DC=25 cm <br>
| + | DC=25 cm <br> |
− | therefore , DS=DC-SC=25-11=14 cm <br>
| + | ∴ DS=DC-SC=25-11=14 cm <br> |
− | DS=DP=14 cm (because by concept 1) <br>
| + | DS=DP=14 cm (because by concept 1) <br> |
− | Also {AD ortho DC} , {OP ortho AD} and {OS ortho DC} <br>
| + | Also {AD ortho DC} , {OP ortho AD} and {OS ortho DC} <br> |
| ∠D=∠S=∠P=90˚ <br> | | ∠D=∠S=∠P=90˚ <br> |
− | ⇒ ∠O=90˚ <br>
| + | ⇒ ∠O=90˚ <br> |
| therefore DSOP is a Square <br> | | therefore DSOP is a Square <br> |
− | SO=OP=14 cm <br> hence Radius of given circle is 14 cm
| + | SO=OP=14 cm <br> hence Radius of given circle is 14 cm |
| | | |
| =Problem 5= | | =Problem 5= |