Q 26: Construct a pair of tangents to a circle of 5 cm radius from a point 13 cm away from
the centre. Measure the length of each tangent.
Answers
Step-by-step explanation:
Given:-
Radius of the circle = 5 cm
To find:-
Construct a pair of tangents to a circle of 5 cm radius from a point 13 cm away from the centre. Measure the length of each tangent ?
Solution:-
1) Rough diagram:-
See the first image in the attachment
2) Construction:-
See the second image in the above attachment
3) Steps of Construction:-
- Draw a circle or radius 5 cm and mark its centre O
- Take a point P of the distance 13 cm away from the centre of the circle.
- Join O and P
- Draw perpendicular bisectors X and Y of the linesegment OP.
- Join X and Y
- Mark the Intersecting point M of X nd Y.
- Draw another circle with the radius OM= MP
- It touches the first circle at the points A and B.
- Join P and A
- Join P and B
- PA and PB are the pair of tangents to the given circle.
- Join O and A, O and B.
Finding the measures of Tangents :-
In ∆OAP and ∆OBP,
OA = OB(radius)
angle OAP = angle OBP = 90°
OP = OP (common side)
By RHS Property , ∆OAP and ∆OBP are congruent triangles
PA = PB ( Congruent parts in the Congruent
triangles)
Now , In ∆ OAP , a right angled triangle
By Pythagoras theorem,
In a right angled triangle ,The square of the hypotenuse is equal to the sum of the squares of the other two sides
=>OA^2+AP^2=OP^2
=>5^2+PA^2=13^2
=>25+PA^2=169
=>PA^2=169-25
=>PA^2=144
=>PA=√144
=>PA=12 cm
Since PA = PB = 12 cm
The lengths of the tangents each 12 cm
Given :-
A circle with radius as 5 cm .
To Construct :-
A pair of tangents to the circle from a point 13cm away from the radius of the given Circle and measure the length of each tangent
Used Concepts :-
- The pair of tangents to a circle touches the circle at two different points known as point of contact.
- The length of each tangent to a circle is equal .
- The tangent makes an angle of 90° with the point of contact and the radius of the circle.
- There are two methods to measure the length of tangent to a circle i.e i ) By actual measurement. ii ) As the point of contact makes an angle of 90° with the radius of the circle we will apply Pythagoras Theorem and get the length of the tangents .
- A identity " a² - b² = ( a + b ) ( a - b )".
Solution :-
Diagram :-
For diagram Refer the above attachment .
Steps of Construction :-
- Draw a circle with radius ( OX ) 5 cm and Centre "O".
- Exceed the radius "OX" to a point "M" which is 13 cm away from the centre of the circle "O" .
- As "OM" is of length 13 cm open the compass more than half of length of "OM" and draw the perpendicular bisectors " K and L " of "OM" .
- Join " K to L " .
- The points " K and L " Intersect the line segment " OX " at a point "N".
- Taking "N" as centre Draw another circle with radius " ON = NM " .
- Let , The points at which The two circles cuts each other be " A and B " .
- Join " M to A and B " respectively .
- "A" and "B" are the required tangents .
Measurement of the tangents :-
i ) By actual measurement :-
By actual measurement the length of the tangents is 12 cm .
ii ) By Pythagoras Theorem :-
As point of contact makes an angle of 90° with the radius of the circle . So we conclude triangle
" AOB " as right angled triangle .
Now ,
In right angled triangle" AOM " :-
As "OM" is opposite to the 90° angle we take "OM" as Hypotenuse of the triangle .
Hypotenuse of the triangle = OM = 13 cm .
Let , Base of the triangle = OA = 5 cm
Perpendicular of the triangle = AM .
Now , By Pythagoras Theorem ,
OM² = OA² + AM²
AM² = OM² - OA²
AM² = ( 13 )² - ( 5 )²
AM² = 169 - 25
AM² = 144
AM² - 144 = 0
AM² - ( 12 )² = 0
Now , we will use the identity discussed above.
( AM + 12 ) ( AM - 12 ) = 0
Either , AM + 12 = 0 or AM - 12 = 0
AM = 12 , - 12 cm
But , we can't measure lengths " <0 " . So , we neglect the length of the tangent as " - 12 cm " .
As The length of the tangents to a circle are equal .
So , The Length of the pair of tangents ( AM and BM ) is 12 cm .
