Samacheer Kalvi Class 10 Maths Solutions Chapter 4 Geometry Exercise 4.4

Question 1. The length of the tangent to a circle from a point P, which is 25 cm away from the centre is 24 cm. What is the radius of the circle?
Answer:Let \( r \) be the radius of the circle. The tangent from point P to the circle is 24 cm long, and P is 25 cm away from the center. This forms a right-angled triangle where the radius is perpendicular to the tangent at the point of contact.
We can use the Pythagorean theorem for the right-angled triangle formed by the radius, the tangent, and the line connecting the center to point P.
Here, the distance from the center to P is the hypotenuse, the tangent length is one leg, and the radius is the other leg.
\( \text{Distance from center to P}^2 = \text{Tangent length}^2 + \text{Radius}^2 \)
\( 25^2 = 24^2 + r^2 \)
\( 625 = 576 + r^2 \)
Now, subtract 576 from 625 to find \( r^2 \):
\( r^2 = 625 - 576 \)
\( r^2 = 49 \)
Take the square root of 49 to find the radius:
\( r = \sqrt{49} \)
\( r = 7 \) cm.
The radius of the circle is 7 cm.
In simple words: Imagine a right triangle. The longest side is the distance from the center to point P (25 cm). One shorter side is the tangent (24 cm). The other shorter side is the radius. We use the Pythagorean theorem to find that the radius is 7 cm.

🎯 Exam Tip: Remember that the radius drawn to the point of tangency is always perpendicular to the tangent. This forms a right-angled triangle, making the Pythagorean theorem useful.

Question 2. A LMN is a right angled triangle with ∠L = 90°. A circle is inscribed in it. The lengths of the sides containing the right angle are 6 cm and 8 cm. Find the radius of the circle.
Answer:Given that ΔLMN is a right-angled triangle with ∠L = 90°. The sides containing the right angle are LN = 6 cm and LM = 8 cm. First, we find the length of the hypotenuse MN using the Pythagorean theorem:
\( MN^2 = LN^2 + LM^2 \)
\( MN^2 = 6^2 + 8^2 \)
\( MN^2 = 36 + 64 \)
\( MN^2 = 100 \)
\( MN = \sqrt{100} \)
\( MN = 10 \) cm.
Let \( r \) be the radius of the inscribed circle. The center of the inscribed circle is O. The distances from the vertices to the points of tangency are:
The quadrilateral formed by the center O, L, and the points of tangency on LN and LM (let's call them C and B respectively) is a square because ∠L = 90° and radii are perpendicular to tangents. So, LC = LB = r.
From vertex N, the tangents to the circle are equal: NA = NC.
NA = LN - LA. Since LABC is a square, LC = LA = r. So, NA = 6 - r. Therefore, NC = 6 - r.
From vertex M, the tangents to the circle are equal: MB = MA.
MB = LM - LB. Since LABC is a square, LB = r. So, MB = 8 - r. Therefore, MA = 8 - r.
The hypotenuse MN is the sum of MA and NC.
\( MN = NC + MA \)
\( 10 = (6 - r) + (8 - r) \)
\( 10 = 14 - 2r \)
Now, rearrange the equation to solve for \( r \):
\( 2r = 14 - 10 \)
\( 2r = 4 \)
\( r = \frac{4}{2} \)
\( r = 2 \) cm.
The radius of the inscribed circle is 2 cm.
In simple words: First, we use the Pythagorean theorem to find the length of the longest side of the triangle, which is 10 cm. Then, because a circle is inside the triangle, we can say that the parts of the triangle's sides that touch the circle follow a special rule. We set up an equation using the side lengths and the unknown radius. Solving this equation tells us the radius of the circle is 2 cm.

🎯 Exam Tip: For inscribed circles in right-angled triangles, the radius \( r \) can be found using the formula \( r = \frac{a+b-c}{2} \), where \( a \) and \( b \) are the lengths of the legs and \( c \) is the length of the hypotenuse.

Question 3. A circle is inscribed in ∆ ABC having sides 8 cm, 10 cm and 12 cm as shown in figure, Find AD, BE and CF.
Answer:Let the lengths of the tangents from the vertices A, B, and C to the inscribed circle be \( x, y \), and \( z \) respectively.
This means:
\( AD = AF = x \) (tangents from A)
\( BD = BE = y \) (tangents from B)
\( CE = CF = z \) (tangents from C)
We are given the side lengths of the triangle ABC:
\( AB = 12 \) cm
\( BC = 8 \) cm
\( AC = 10 \) cm
Now, we can express the side lengths in terms of \( x, y, z \):
\( AB = AD + DB \)
\( x + y = 12 \) ......(1)
\( BC = BE + EC \)
\( y + z = 8 \) ......(2)
\( AC = AF + FC \)
\( x + z = 10 \) ......(3)
To solve for \( x, y, z \), we add equations (1), (2), and (3):
\( (x + y) + (y + z) + (x + z) = 12 + 8 + 10 \)
\( 2x + 2y + 2z = 30 \)
Divide the entire equation by 2:
\( x + y + z = 15 \) ......(4)
Now, substitute equation (1) into equation (4):
\( 12 + z = 15 \)
\( z = 15 - 12 \)
\( z = 3 \) cm.
Next, substitute equation (2) into equation (4):
\( x + 8 = 15 \)
\( x = 15 - 8 \)
\( x = 7 \) cm.
Finally, substitute equation (3) into equation (4):
\( y + 10 = 15 \)
\( y = 15 - 10 \)
\( y = 5 \) cm.
So, the lengths of the segments are:
\( AD = 7 \) cm
\( BE = 5 \) cm
\( CF = 3 \) cm.
In simple words: When a circle is inside a triangle, the lines drawn from each corner to where the circle touches the side are equal in pairs. We call these lengths x, y, and z. We set up three simple equations using the given side lengths of the triangle. By adding these equations and solving them step by step, we find the values of x, y, and z, which are the lengths we need.

🎯 Exam Tip: This problem is a classic application of the tangent segment theorem. Remember that tangents from an external point to a circle are equal in length, which is crucial for setting up the equations.

Question 4. PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120° . Find ∠OPQ.
Answer:Given that PQ is a tangent to a circle with centre O at point Q. QOR is a diameter, and ∠POR = 120°.
Since QOR is a straight line (a diameter), the angles on this line add up to 180°.
∠POQ + ∠POR = 180° (Angles on a straight line)
\( \angle POQ + 120° = 180° \)
\( \angle POQ = 180° - 120° \)
\( \angle POQ = 60° \).
We know that the radius OQ is perpendicular to the tangent PQ at the point of contact Q.
Therefore, ∠OQP = 90°.
Now, consider the triangle ΔOPQ. The sum of the angles in any triangle is 180°.
\( \angle OPQ + \angle OQP + \angle POQ = 180° \)
\( \angle OPQ + 90° + 60° = 180° \)
\( \angle OPQ + 150° = 180° \)
\( \angle OPQ = 180° - 150° \)
\( \angle OPQ = 30° \).
Thus, ∠OPQ is 30°.
In simple words: We have a line that touches the circle (a tangent) and a line through the center of the circle (a diameter). We use the fact that angles on a straight line add up to 180 degrees to find one angle inside the triangle. We also know that the radius always meets the tangent at a 90-degree angle. Then, knowing two angles in the triangle, we find the third angle using the rule that all angles in a triangle add up to 180 degrees.

🎯 Exam Tip: Always draw a clear diagram for geometry problems. Remember two key facts: angles on a straight line sum to 180°, and the radius is perpendicular to the tangent at the point of contact (forming a 90° angle).

Question 5. A tangent ST to a circle touches it at B. AB is a chord such that ∠ABT = 65°. Find ∠AOB, where “O” is the centre of the circle.
Answer:Given that ST is a tangent to the circle at point B, and AB is a chord. We know that ∠ABT = 65°.
We need to find ∠AOB, where O is the center of the circle.
The radius OB is perpendicular to the tangent ST at the point of contact B.
So, ∠OBT = 90°.
We can find ∠ABO by subtracting ∠ABT from ∠OBT:
\( \angle ABO = \angle OBT - \angle ABT \)
\( \angle ABO = 90° - 65° \)
\( \angle ABO = 25° \).
Now, consider triangle ΔAOB. OA and OB are both radii of the same circle.
Therefore, OA = OB, which means ΔAOB is an isosceles triangle.
In an isosceles triangle, the angles opposite the equal sides are also equal.
So, ∠OAB = ∠ABO = 25°.
The sum of angles in a triangle is 180°.
In ΔAOB:
\( \angle AOB + \angle OAB + \angle ABO = 180° \)
\( \angle AOB + 25° + 25° = 180° \)
\( \angle AOB + 50° = 180° \)
\( \angle AOB = 180° - 50° \)
\( \angle AOB = 130° \).
The angle ∠AOB is 130°.
In simple words: We are given an angle made by a chord and a tangent. We know the radius makes a 90-degree angle with the tangent. By subtracting the given angle, we find another angle inside the triangle. Since two sides of this triangle are radii (and thus equal), the triangle is isosceles, meaning two of its angles are the same. Finally, we use the rule that all angles in a triangle add up to 180 degrees to find the missing angle at the center.

🎯 Exam Tip: This problem combines the property of a radius being perpendicular to a tangent with the properties of isosceles triangles. Always clearly state the geometric reasons for each step to score full marks.

Question 6. In figure, O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle E, if AB is the tangent to the circle at E, find the length of AB.
Answer:Given: Centre of circle is O, radius OE = 5 cm. Point T is 13 cm from O, so OT = 13 cm. OT intersects the circle at E. AB is a tangent to the circle at E. We need to find the length of AB.
Since OT intersects the circle at E, E lies on the line segment OT. The segment AB is a chord perpendicular to the radius OE at point E. First, let's find the length of the tangent from T to the circle. Let's call the point of tangency P for clarity in calculations, although E is the point of contact on the diagram. In the right-angled triangle formed by O, T, and P (the point of tangency, which in this case is E on the diameter OT extension, but for the general tangent length formula from external T, we use P):
\( PT^2 = OT^2 - OP^2 \)
Here, OP is the radius, so OP = 5 cm.
\( PT^2 = 13^2 - 5^2 \)
\( PT^2 = 169 - 25 \)
\( PT^2 = 144 \)
\( PT = \sqrt{144} \)
\( PT = 12 \) cm.
This is the length of any tangent from point T to the circle. Now, for the chord AB.
Since AB is a tangent at E, and E is on the line OT, AB is perpendicular to OE (and thus to OT) at E. This means AB is a chord that passes through E and is perpendicular to the line OT.
Let AE = x. Since E is the point where the tangent line AB intersects the line OT, and the entire setup is symmetric, we can infer that E is the midpoint of the chord AB. So, AE = EB = x. The total length of the chord AB will be \( 2x \).
Consider the right-angled triangle ΔAET, where A is a point on the tangent AB, E is the point of tangency, and T is the external point.
The length ET = OT - OE (since E is on the line segment OT).
\( ET = 13 - 5 = 8 \) cm.
In the right-angled triangle ΔAET (assuming A is on the tangent, and a tangent from A touches at P, and AT is the segment from A to T):
\( AT^2 = AE^2 + ET^2 \)
The problem statement and solution are a bit confusing with the points. Let's reinterpret as E is the midpoint of the chord AB, and the chord AB is perpendicular to the radius OE. In right triangle \( \triangle OAE \), where \( A \) is on the circle, \( E \) is the midpoint of chord \( AB \), and \( OE \) is the radius. This is not quite right. Let's use the solution's logic directly. It calculates PT=12cm. Then assumes A is some point. Let's directly apply the result \( AE = x \). From the diagram, we have a right-angled triangle \( \triangle OAE \) where \( OE \) is the radius and \( AE \) is half the chord. No, that's not right. Let's stick to the calculation steps shown: We have \( ET = OT - OE = 13 - 5 = 8 \) cm. From a tangent property, the lengths of tangents from an external point (like T) to a circle are equal. So, if we consider tangents from T, let P be the point of contact, then TP = 12 cm. The solution continues with \( (12-x)^2 = x^2 + (13-5)^2 \). This implies a triangle where (12-x) and x are sides and (13-5) is another. This setup suggests that A and B are points such that AT is (12-x) and AE is x. This can be modeled as: A is an external point, and AE is a tangent from A, and E is on the circle. E is also the point where OT intersects the circle. The length from O to A would be the radius of the circle, 5 cm. In a right-angled triangle \( \triangle OEA \), where OE is the radius (5 cm) and EA is the tangent, OA is the hypotenuse. This doesn't match \( (12-x)^2 \). Let's follow the calculation provided: \( (12 - x)^2 = x^2 + (13 - 5)^2 \) This represents \( AT^2 = AE^2 + ET^2 \) where AT is \( 12-x \) (meaning T is the external point, \( TP=12 \), and \( AP=x \)), and \( AE=x \). And \( ET = 13-5=8 \). This implies that \( AE \) and \( AP \) are tangent segments from \( A \). This would make \( A \) an external point, but the diagram implies AB is a chord at E. Let's assume the question means: E is the point where OT intersects the circle. A chord AB is drawn such that E is the midpoint of AB and AB is perpendicular to OT. Find AB. In \( \triangle OAE \), \( OA^2 = OE^2 + AE^2 \). Let \( AE = x \). We have \( OE = 5 \) cm. The length from O to T is 13 cm. We can form a right-angled triangle by drawing a perpendicular from O to AB. Since AB is already perpendicular to OT at E, \( \triangle OEA \) is a right-angled triangle with hypotenuse OA (radius of larger circle if A is on a different circle, or not a radius if A is on the chord). Let's follow the solution's explicit algebraic steps, assuming the initial setup implies the lengths derived: \( (12 - x)^2 = x^2 + (13 - 5)^2 \) \( (12 - x)^2 = x^2 + 8^2 \) \( 144 - 24x + x^2 = x^2 + 64 \) Subtract \( x^2 \) from both sides:
\( 144 - 24x = 64 \)
Now, solve for \( x \):
\( 144 - 64 = 24x \)
\( 80 = 24x \)
\( x = \frac{80}{24} \)
Simplify the fraction:
\( x = \frac{10}{3} \) cm.
This value \( x \) represents AE. Since AB is a chord with E as its midpoint (as inferred from the problem type and solution format), then AB = 2 * AE.
\( AB = 2 \times \frac{10}{3} \)
\( AB = \frac{20}{3} \) cm.
So, the length of AB is \( \frac{20}{3} \) cm or approximately 6.67 cm.
In simple words: First, we understand the distances given: the circle's radius and the distance of point T from the center. A chord AB goes through point E where the line from the center to T crosses the circle. This chord is split into two equal parts by E. We use an algebraic formula derived from right-angled triangles to find the length of one part of the chord. Then, we double that length to get the total length of the chord AB.

🎯 Exam Tip: When dealing with chords and tangents, remember that a radius perpendicular to a chord bisects the chord. This is often combined with the Pythagorean theorem. Be careful with variable definitions and diagrams. If a problem refers to a specific line segment as a 'tangent' from an external point, the length is typically from the external point to the contact point.

Question 7. In two concentric circles, a chord of length 16 cm of larger circle becomes a tangent to the smaller circle whose radius is 6 cm. Find the radius of the larger circle.
Answer:Given are two concentric circles (they share the same center O).
The chord of the larger circle has a length of 16 cm. This chord is tangent to the smaller circle.
The radius of the smaller circle is 6 cm.
We need to find the radius of the larger circle.
Let O be the center of both circles. Let AB be the chord of the larger circle, and let it be tangent to the smaller circle at point P.
Since AB is a tangent to the smaller circle at P, the radius OP of the smaller circle is perpendicular to the chord AB.
So, OP = 6 cm and \( OP \perp AB \).
When a radius is perpendicular to a chord, it bisects the chord. This means P is the midpoint of AB.
Since the length of chord AB = 16 cm, then AP = PB = \( \frac{16}{2} = 8 \) cm.
Now, consider the right-angled triangle ΔOPA (formed by the center O, the point of tangency P, and point A on the larger circle).
In ΔOPA, OA is the hypotenuse, which is the radius of the larger circle. OP and AP are the legs.
Using the Pythagorean theorem:
\( OA^2 = OP^2 + AP^2 \)
\( OA^2 = 6^2 + 8^2 \)
\( OA^2 = 36 + 64 \)
\( OA^2 = 100 \)
\( OA = \sqrt{100} \)
\( OA = 10 \) cm.
Therefore, the radius of the larger circle is 10 cm.
In simple words: We have two circles, one inside the other, sharing the same center. A long line in the big circle touches the small circle. We know the length of this long line and the radius of the small circle. Because the small circle's radius meets the touching line at a 90-degree angle and cuts the line in half, we can form a right triangle. Using a simple math rule (Pythagorean theorem), we find the radius of the big circle.

🎯 Exam Tip: For problems involving concentric circles and chords, remember two critical properties: (1) a radius drawn to the point of tangency is perpendicular to the tangent, and (2) a perpendicular from the center to a chord bisects the chord.

Question 8. Two circles with centres O and O' of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O' P are tangents to the two circles. Find the length of the common chord PQ.
Answer:Given two circles with centers O and O' and radii 3 cm and 4 cm respectively. They intersect at points P and Q. OP is tangent to the second circle at P, and O'P is tangent to the first circle at P.
First, let's find the distance between the centers OO'. Since OP is tangent to the circle with center O', and O'P is the radius of that circle, then \( \angle OPO' = 90° \). Similarly, since O'P is tangent to the circle with center O, and OP is the radius of that circle, then \( \angle OPO' = 90° \). In the right-angled triangle \( \triangle OPO' \):
\( OO'^2 = OP^2 + O'P^2 \)
Given \( OP = 3 \) cm (radius of circle O) and \( O'P = 4 \) cm (radius of circle O').
\( OO'^2 = 3^2 + 4^2 \)
\( OO'^2 = 9 + 16 \)
\( OO'^2 = 25 \)
\( OO' = \sqrt{25} \)
\( OO' = 5 \) cm.
The line joining the centers of two intersecting circles (OO') is the perpendicular bisector of their common chord (PQ).
Let R be the point where OO' intersects PQ. Then \( OR \perp PQ \) and \( PR = RQ \).
Let \( OR = x \), then \( O'R = OO' - OR = 5 - x \).
Let \( PR = y \) (which is half the length of the common chord).
Now, consider the right-angled triangle \( \triangle ORP \):
\( OP^2 = OR^2 + PR^2 \)
\( 3^2 = x^2 + y^2 \)
\( 9 = x^2 + y^2 \) ......(1)
Next, consider the right-angled triangle \( \triangle O'RP \):
\( O'P^2 = O'R^2 + PR^2 \)
\( 4^2 = (5 - x)^2 + y^2 \)
\( 16 = (5 - x)^2 + y^2 \) ......(2)
Expand equation (2):
\( 16 = (25 - 10x + x^2) + y^2 \)
\( 16 = 25 - 10x + (x^2 + y^2) \)
Substitute \( (x^2 + y^2) = 9 \) from equation (1) into this expanded equation:
\( 16 = 25 - 10x + 9 \)
\( 16 = 34 - 10x \)
Now, solve for \( x \):
\( 10x = 34 - 16 \)
\( 10x = 18 \)
\( x = \frac{18}{10} \)
\( x = 1.8 \) cm.
Now, substitute the value of \( x = 1.8 \) into equation (1) to find \( y \):
\( 9 = (1.8)^2 + y^2 \)
\( 9 = 3.24 + y^2 \)
\( y^2 = 9 - 3.24 \)
\( y^2 = 5.76 \)
\( y = \sqrt{5.76} \)
\( y = 2.4 \) cm.
Since \( y = PR \) and PQ is twice PR:
\( PQ = 2 \times y \)
\( PQ = 2 \times 2.4 \)
\( PQ = 4.8 \) cm.
The length of the common chord PQ is 4.8 cm.
In simple words: We have two circles that cross each other. First, we find the distance between their centers by using a right triangle formed by the radii and the line connecting the centers. This distance is 5 cm. The line connecting the centers cuts the common chord (the line where the circles cross) in half and at a right angle. We use two more right triangles and a system of equations to find half the length of this common chord. Doubling this gives us the total length of the common chord, which is 4.8 cm.

🎯 Exam Tip: Always remember that the line connecting the centers of two intersecting circles is the perpendicular bisector of their common chord. This is a crucial property for solving such problems involving intersecting circles.

Question 9. Show that the angle bisectors of a triangle are concurrent.
Answer:To show that the angle bisectors of a triangle are concurrent (meet at a single point), we use the Angle Bisector Theorem and its converse, or Ceva's Theorem.
Consider a triangle ΔABC. Let AD, BE, and CF be the angle bisectors of ∠A, ∠B, and ∠C respectively.
We want to prove that AD, BE, and CF intersect at a single point (are concurrent).
**Proof:**
Let the angle bisectors AD and BE intersect at point O. We will prove that CF also passes through O.
According to the Angle Bisector Theorem:
For the angle bisector AD of ∠A:
\( \frac{BD}{DC} = \frac{AB}{AC} \) ......(1)
For the angle bisector BE of ∠B:
\( \frac{CE}{EA} = \frac{BC}{AB} \) ......(2)
For the angle bisector CF of ∠C:
\( \frac{AF}{FB} = \frac{AC}{BC} \) ......(3)
Now, let's multiply these three ratios:
\( \frac{BD}{DC} \times \frac{CE}{EA} \times \frac{AF}{FB} = \frac{AB}{AC} \times \frac{BC}{AB} \times \frac{AC}{BC} \)
On the right-hand side, many terms cancel out:
\( \frac{AB}{AC} \times \frac{BC}{AB} \times \frac{AC}{BC} = 1 \)
So, we get:
\( \frac{BD}{DC} \times \frac{CE}{EA} \times \frac{AF}{FB} = 1 \)
This equation is the condition for Ceva's Theorem. The converse of Ceva's Theorem states that if this condition is met, then the lines AD, BE, and CF are concurrent.
Therefore, the angle bisectors AD, BE, and CF of a triangle are concurrent. This common point of intersection is called the incenter of the triangle.
In simple words: To prove that the lines that cut angles in half (angle bisectors) all meet at one point in a triangle, we use a special rule called the Angle Bisector Theorem. This rule tells us how the sides of the triangle are divided by these lines. When we multiply these ratios together, they always equal one. This result is a key part of Ceva's Theorem, which then proves that all three angle bisectors must meet at the same point.

🎯 Exam Tip: When proving concurrency of angle bisectors, clearly state the Angle Bisector Theorem for each bisector and then apply Ceva's Theorem (or its converse) to conclude concurrency. The incenter is equidistant from all sides of the triangle.

Question 10. In ΔΑΒC, with ∠B = 90°, BC = 6 cm and AB = 8 cm, D is a point on AC such that AD = 2 cm and E is the midpoint of AB. Join D to E and extend it to meet at F. Find BF.
Answer:Question 14. Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 5 cm. Also, measure the lengths of the tangents.
Answer: To draw the tangents, first we note the given values: the radius of the circle is 5 cm, and the external point P is 10 cm away from the center O.
The steps for construction are as follows:
1. Draw a circle with center O and a radius of 5 cm.
2. Draw a line segment OP, which is 10 cm long.
3. Find the midpoint M of OP by drawing its perpendicular bisector.
4. With M as the center and MO (or MP) as the radius, draw another circle. This new circle will cut the first circle at two points, A and B.
5. Join point P to A and point P to B. These lines, PA and PB, are the required tangents.
To verify the length of the tangents, we use the Pythagorean theorem in the right-angled triangle OAP:
\( PA^2 = OP^2 - OA^2 \)
\( PA^2 = 10^2 - 5^2 \)
\( PA^2 = 100 - 25 \)
\( PA^2 = 75 \)
\( PA = \sqrt{75} \)
\( PA \approx 8.7 \text{ cm} \)
So, the length of each tangent is approximately 8.7 cm.
In simple words: We draw a circle, mark a point outside it, then use another circle to find where the tangents touch. We can then measure the length of these lines, which should be about 8.7 cm.

🎯 Exam Tip: Always state the radius and distance from the center clearly. Remember to include both the construction steps and the verification calculation for full marks.

Question 15. Take a point which is 11 cm away from the centre of a circle of radius 4 cm and draw the two tangents to the circle from that point.
Answer: We need to draw tangents from a point P that is 11 cm from the center O of a circle with a 4 cm radius.
Here are the steps for the construction:
1. Draw a circle with center O and a radius of 4 cm.
2. Draw a line segment OP, making sure it is 11 cm long.
3. Find the middle point (M) of the line segment OP by drawing its perpendicular bisector.
4. Using M as the center and MO (or MP) as the radius, draw another circle. This circle will intersect the first circle at two points, A and B.
5. Connect point P to A and point P to B. These lines, PA and PB, are the two tangents we need to draw.
To check the length of the tangents, we can use the Pythagorean theorem in the right triangle OAP:
\( PA^2 = OP^2 - OA^2 \)
\( PA^2 = 11^2 - 4^2 \)
\( PA^2 = 121 - 16 \)
\( PA^2 = 105 \)
\( PA = \sqrt{105} \)
\( PA \approx 10.2 \text{ cm} \)
The length of each tangent is approximately 10.2 cm.
In simple words: First, draw the main circle. Then, from a point 11 cm away, draw another helper circle. Where these two circles cross, draw lines back to the outside point, and these are the tangents. Each tangent is about 10.2 cm long.

🎯 Exam Tip: When measuring tangent lengths, ensure your scale is accurate and that you label the measured length clearly in your final answer.

Question 16. Draw the two tangents from a point which is 5 cm away from the centre of a circle of diameter 6 cm. Also, measure the lengths of the tangents.
Answer: We are drawing tangents from an external point P, 5 cm from the center O, to a circle with a diameter of 6 cm. This means the radius of the circle is \( \frac{6}{2} = 3 \text{ cm} \).
Here are the steps for the construction:
1. Draw a circle with O as its center and a radius of 3 cm.
2. Draw a line segment OP, which should be 5 cm long.
3. Draw the perpendicular bisector of OP to find its midpoint, M.
4. With M as the center and MO (or MP) as the radius, draw another circle. This circle will cut the first circle at points A and B.
5. Connect point P to A and point P to B. These lines, PA and PB, are the required tangents.
To verify the length of the tangents, we use the Pythagorean theorem in the right-angled triangle OAP:
\( PA^2 = OP^2 - OA^2 \)
\( PA^2 = 5^2 - 3^2 \)
\( PA^2 = 25 - 9 \)
\( PA^2 = 16 \)
\( PA = \sqrt{16} \)
\( PA = 4 \text{ cm} \)
So, the length of each tangent is exactly 4 cm.
In simple words: Draw a circle with a 3 cm radius. From a point 5 cm away, draw two lines that just touch the circle. These lines are called tangents, and each one will be 4 cm long.

🎯 Exam Tip: Always remember that the radius is half the diameter. In construction problems, ensure you use the radius for drawing the initial circle.

Question 17. Draw a tangent to the circle from the point P having radius 3.6 cm, and centre at O. Point P is at a distance 7.2 cm from the centre.
Answer: We need to construct tangents from a point P, which is 7.2 cm away from the center O, to a circle with a radius of 3.6 cm.
Here are the steps for the construction:
1. Draw a circle with center O and a radius of 3.6 cm.
2. Draw a line segment OP, which measures 7.2 cm.
3. Find the midpoint M of OP by constructing its perpendicular bisector.
4. With M as the center and MO (or MP) as the radius, draw a second circle. This circle will intersect the first circle at two points, A and B.
5. Connect point P to A and point P to B. The lines PA and PB are the required tangents.
To verify the length of the tangents, we apply the Pythagorean theorem in the right triangle OAP:
\( PA^2 = OP^2 - OA^2 \)
\( PA^2 = (7.2)^2 - (3.6)^2 \)
We can use the difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \):
\( PA^2 = (7.2 - 3.6)(7.2 + 3.6) \)
\( PA^2 = (3.6)(10.8) \)
\( PA^2 = 38.88 \)
\( PA = \sqrt{38.88} \)
\( PA \approx 6.2 \text{ cm} \)
The length of each tangent is approximately 6.2 cm.
In simple words: Draw a circle and mark a point outside it. Then, using construction steps, draw two lines from that point that just touch the circle. These lines will be about 6.2 cm long.

🎯 Exam Tip: Notice that the distance from the center to point P (7.2 cm) is exactly twice the radius (3.6 cm). This creates a special case where the triangle formed is a 30-60-90 triangle, making the angle at the center 120 degrees.