Samacheer Kalvi Class 9 Maths Solutions Chapter 4 Geometry More Ques

I. Multiple Choice Questions.

Question 1. The angle sum of a convex polygon with number of sides 7 is .........
(a) 900°
(b) 1080°
(c) 1440°
(d) 720°
Answer: (a) 900°
In simple words: The sum of interior angles of a polygon with 'n' sides is found by the formula \( (n-2) \times 180° \). For a polygon with 7 sides, this means \( (7-2) \times 180° = 5 \times 180° = 900° \). This formula helps us calculate the total degrees inside any polygon.

🎯 Exam Tip: Remember the formula \( (n-2) \times 180° \) for the sum of interior angles of an n-sided polygon; it's a fundamental concept.

Question 2. What is the name of a regular polygon of six sides?
(a) Square
(b) Equilateral triangle
(c) Regular hexagon
(d) Regular octagon
Answer: (c) Regular hexagon
In simple words: A shape with six equal sides and six equal angles is called a regular hexagon. Each number of sides has a special name for its polygon.

🎯 Exam Tip: Familiarize yourself with the names of polygons based on their number of sides, especially for regular polygons where all sides and angles are equal.

Question 3. One angle of a parallelogram is a right angle. The name of the quadrilateral is .........
(a) square
(b) rectangle
(c) rhombus
(d) kite
Answer: (b) rectangle
In simple words: If a parallelogram has one angle that is 90 degrees, then all its angles must be 90 degrees, making it a rectangle. This is a special characteristic that defines a rectangle.

🎯 Exam Tip: Understand the properties of different quadrilaterals; for a parallelogram, if one angle is 90°, all are 90°, turning it into a rectangle.

Question 4. If all the four sides of a parallelogram are equal and the adjacent angles are of 120° and 60°, then the name of the quadrilateral is .........
(a) rectangle
(b) square
(c) rhombus
(d) kite
Answer: (c) rhombus
In simple words: A parallelogram with all four sides equal is always called a rhombus. The angles (like 120° and 60°) confirm it's not a square, which would have all 90° angles.

🎯 Exam Tip: Differentiate between special parallelograms: a rhombus has all sides equal, while a square has all sides equal and all angles 90°.

Question 5. In a parallelogram ∠A : ∠B = 1 : 2. Then ∠A .........
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Answer: (b) 60°
In simple words: In a parallelogram, two adjacent angles (like ∠A and ∠B) always add up to 180°. If their ratio is 1:2, it means one is \( 1x \) and the other is \( 2x \), so \( 1x + 2x = 180° \), which makes \( 3x = 180° \), so \( x = 60° \). Thus, ∠A is 60°.

🎯 Exam Tip: Remember that adjacent angles in a parallelogram are supplementary (sum to 180°); this is key for solving angle ratio problems.

Question 6. Which of the following is a formula to find the sum of interior angles of a quadrilateral of w-sides?
(a) \( \frac{n}{2} \times 180° \)
(b) \( (\frac{n+1}{2}) \times 180° \)
(c) \( (\frac{n-1}{2}) \times 180° \)
(d) \( (n - 2) 180° \)
Answer: (d) \( (n - 2) 180° \)
In simple words: To find the total degrees inside any polygon, subtract 2 from the number of sides (n), and then multiply that result by 180 degrees. This formula always works, regardless of the polygon's shape.

🎯 Exam Tip: The formula \( (n-2) \times 180° \) is universal for the sum of interior angles of any n-sided polygon, not just quadrilaterals (where n=4, so \( (4-2) \times 180° = 360° \)).

Question 7. Diagonal of which of the following quadrilaterals do not bisect it into two congruent triangles?
(a) rhombus
(b) trapezium
(c) square
(d) rectangle
Answer: (b) trapezium
In simple words: A trapezium has only one pair of parallel sides, which means its diagonal does not always divide it into two identical (congruent) triangles. In most other special quadrilaterals, a diagonal does create two congruent triangles.

🎯 Exam Tip: Understand the properties of quadrilaterals; only in a trapezium do the diagonals not necessarily bisect it into congruent triangles, unlike parallelograms, rhombuses, squares, and rectangles.

Question 8. The point of concurrency of the medians of a triangle is known as .........
(a) circumcentre
(b) incentre
(c) orthocentre
(d) centroid
Answer: (d) centroid
In simple words: When you draw all three medians (lines from a vertex to the midpoint of the opposite side) of a triangle, they all meet at a single point called the centroid. This point is also the center of gravity for the triangle.

🎯 Exam Tip: Memorize the names and definitions of the different points of concurrency in a triangle: centroid (medians), incentre (angle bisectors), circumcentre (perpendicular bisectors), and orthocentre (altitudes).

Question 9. Orthocentre of a triangle is the point of concurrency of .........
(a) medians
(b) altitudes
(c) angle bisectors
(d) perpendicular bisectors of side
Answer: (b) altitudes
In simple words: The orthocentre is the special point where all three altitudes of a triangle meet. An altitude is a line drawn from a vertex straight down to the opposite side, making a 90-degree angle.

🎯 Exam Tip: Distinguish between the types of concurrent points in a triangle; the orthocentre is where altitudes intersect, while the centroid is for medians.

Question 10. ABCD is a parallelogram as shown. Find x and y.
(a) 1, 7
(b) 2, 6
(c) 3, 5
(d) 4, 4
Answer: (c) 3, 5
In simple words: In a parallelogram, the diagonals cut each other in half. So, we can set up two equations using the parts of the diagonals. Solving these equations helps us find the values of x and y.

🎯 Exam Tip: Remember that the diagonals of a parallelogram bisect each other, meaning they cut each other into two equal parts. This property is crucial for solving problems involving diagonal lengths.

Question 11. A circle divides the plane into part .........
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
In simple words: A circle separates the flat surface it's drawn on into three parts: the area inside the circle, the area on the circle's boundary itself, and the area outside the circle. It creates three distinct regions.

🎯 Exam Tip: Understand the three distinct regions created by a circle in a plane: interior, exterior, and on the circumference itself.

Question 12. The longest chord of a circle is a of the circle..........
(a) radius
(b) diameter
(c) chord
(d) secant
Answer: (b) diameter
In simple words: The longest line segment that can connect two points on a circle's edge is the diameter. It always passes right through the center of the circle.

🎯 Exam Tip: A diameter is the longest possible chord in any circle, always passing through the center.

Question 13. Opposite angles of a cyclic quadrilateral are .........
(a) supplementary
(b) complementary
(c) equal
(d) none of these
Answer: (a) supplementary
In simple words: In a cyclic quadrilateral (a four-sided shape where all corners touch a circle), the angles that are opposite each other always add up to 180 degrees. This is a special rule for these shapes.

🎯 Exam Tip: Remember the key property of cyclic quadrilaterals: their opposite angles are supplementary (add up to 180°).

Question 14. The value of x from figure is if 'O' is the centre of the circle .........
(a) 20 cm
(b) 15 cm
(c) 12 cm
(d) 5 cm
Answer: (d) 5 cm
In simple words: When a line from the center of a circle is perpendicular to a chord, it cuts the chord in half. We can use the Pythagorean theorem with the radius (hypotenuse), half the chord length, and the distance from the center to the chord to find the missing value.

🎯 Exam Tip: Apply the property that a perpendicular from the center to a chord bisects the chord. Then use the Pythagorean theorem on the right-angled triangle formed by the radius, half-chord, and distance from center to chord.

Question 15. If PQ = x and 'O' is the centre of the circle, then x= .........
(a) 7 cm
(b) 14 cm
(c) 8 cm
(d) 13 cm
Answer: (b) 14 cm
In simple words: We can use the Pythagorean theorem here. If the radius is 25 cm and the perpendicular distance from the center to the chord is 24 cm, then half the chord length is \( \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7 \) cm. The full chord length (x) is double this value. This allows us to find the chord's length.

🎯 Exam Tip: Always remember that a perpendicular from the center of a circle to a chord bisects the chord. This forms a right-angled triangle allowing use of the Pythagorean theorem.

Question 16. In figure OM = ON = 8cm and AB = 30 cm, then CD = .........
(a) 15 cm
(b) 30 cm
(c) 40 cm
(d) 10 cm
Answer: (b) 30 cm
In simple words: In any circle, if two chords are equally far away from the center, then those two chords must be equal in length. Since OM and ON are both 8 cm, chords AB and CD must be the same length.

🎯 Exam Tip: Remember the theorem: chords equidistant from the center of a circle are equal in length.

Question 17. O is the centre of a circle, ∠AOB = 100°. Then angle ∠ ACB = .........
(a) 80°
(b) 40°
(c) 50°
(d) 60°
Answer: (c) 50°
In simple words: The angle that an arc makes at the center of a circle is always twice as big as the angle it makes at any point on the circle's edge. So, if the angle at the center is 100°, the angle at the circumference will be half of that.

🎯 Exam Tip: Recall the circle theorem: the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Question 18. In a circle with center O, ∠AOB = 20°, ∠BOC = 40°, if arc BC = 4 cm. Then length of arc AB is .........
(a) 8 cm
(b) 6 cm
(c) 2 cm
(d) 1 cm
Answer: (c) 2 cm
In simple words: The length of an arc on a circle is directly related to the size of the angle it makes at the center. If one arc has twice the central angle of another, it will have twice the length. Since angle AOB (20°) is half of angle BOC (40°), arc AB will be half the length of arc BC.

🎯 Exam Tip: Remember that arc length is proportional to the central angle it subtends. Use a ratio to find unknown arc lengths when angles are given.

Question 19. In the figure, OC = 3cm and radius of circle is 5 cm. Then AB = .........
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Answer: (d) 8 cm
In simple words: When a line from the center of a circle meets a chord at a 90-degree angle, it splits the chord into two equal parts. We can form a right-angled triangle with the radius as the hypotenuse, the distance OC as one side, and half the chord (AC) as the other side. Using the Pythagorean theorem, we find AC, then double it to get AB.

🎯 Exam Tip: Remember the property that a perpendicular from the center to a chord bisects the chord. This allows the use of the Pythagorean theorem to find unknown lengths in the right triangle formed.

Question 20. O is the centre of the circle. The value of x in the given diagram is .........
(a) 100°
(b) 160°
(c) 200°
(d) 80°
Answer: (d) 80°
In simple words: The angle at the center of a circle is double the angle at the circumference when both angles are made by the same arc. If the reflex angle at the center is 200°, the non-reflex angle (the smaller one) is \( 360° - 200° = 160° \). The angle x at the circumference is half of this interior central angle.

🎯 Exam Tip: Distinguish between the reflex and non-reflex central angles. The angle at the circumference is always half of the non-reflex central angle subtended by the same arc.

II. Answer the Following Questions.

Question 1. In the figure find x° and y°.
Answer: First, let's find x. The angle \( \angle ACD \) (120°) is an exterior angle of \( \triangle ABC \). It is equal to the sum of the two opposite interior angles, \( \angle A \) and \( \angle B \). So, \( \angle ACD = \angle A + \angle B \) \( 120° = 50° + x° \) \( x° = 120° - 50° \) \( \implies x° = 70° \)
Next, let's find y. In \( \triangle ABC \), the sum of all interior angles is 180°. \( \angle A + \angle B + \angle ACB = 180° \) \( 50° + x° + y° = 180° \) We already found \( x° = 70° \). \( 50° + 70° + y° = 180° \) \( 120° + y° = 180° \) \( y° = 180° - 120° \) \( \implies y° = 60° \) Therefore, the values are \( x = 70° \) and \( y = 60° \).
In simple words: First, use the rule that an outside angle of a triangle equals the sum of its two opposite inside angles to find x. Then, use the rule that all three inside angles of a triangle add up to 180 degrees to find y.

🎯 Exam Tip: Clearly state the geometric reasons (exterior angle theorem, sum of angles in a triangle) for each step to ensure full marks.

Question 2. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Answer: Let the angles of the quadrilateral be \( 3x, 5x, 9x \), and \( 13x \). We know that the sum of all interior angles of a quadrilateral is \( 360° \). So, we can set up the equation: \( 3x + 5x + 9x + 13x = 360° \) Combine the terms with x: \( 30x = 360° \) Now, solve for x: \( x = \frac{360°}{30} \) \( \implies x = 12° \)
Now substitute the value of x back into each ratio to find the angles: First angle: \( 3x = 3 \times 12° = 36° \) Second angle: \( 5x = 5 \times 12° = 60° \) Third angle: \( 9x = 9 \times 12° = 108° \) Fourth angle: \( 13x = 13 \times 12° = 156° \) The required angles of the quadrilateral are \( 36°, 60°, 108° \), and \( 156° \). To check, add them up: \( 36 + 60 + 108 + 156 = 360° \).
In simple words: Since all angles in a four-sided shape add up to 360 degrees, we can add the parts of the ratio together and find how much one part (x) is worth. Then, multiply x by each number in the ratio to get the actual size of each angle.

🎯 Exam Tip: When given angles in a ratio, represent them as multiples of 'x' and use the property that the sum of angles in a quadrilateral is 360°.

Question 3. Diagonal AC of a parallelogram ABCD bisects ∠A. Show that (i) it bisects ∠C also (ii) ABCD is a rhombus.
Answer:
(i) To prove that AC bisects ∠C:
We are given that ABCD is a parallelogram, and AC is its diagonal. Since ABCD is a parallelogram, AB is parallel to DC, and AC is a transversal. This means that \( \angle BAC = \angle ACD \) (These are alternate interior angles). Let's call this Equation (1).
Similarly, BC is parallel to AD, and AC is a transversal. This means that \( \angle BCA = \angle CAD \) (These are also alternate interior angles). Let's call this Equation (2).
We are given that the diagonal AC bisects ∠A, which means it divides ∠A into two equal parts. So, \( \angle BAC = \angle CAD \). Let's call this Equation (3).
From Equation (1) and Equation (3), we have \( \angle ACD = \angle CAD \). From Equation (2) and Equation (3), we have \( \angle BCA = \angle BAC \).
Since \( \angle BAC = \angle CAD \) (given), and \( \angle BAC = \angle ACD \) (from 1), and \( \angle CAD = \angle BCA \) (from 2), it follows that \( \angle ACD = \angle BCA \). This means that AC divides ∠C into two equal parts, so AC bisects ∠C.
(ii) To prove that ABCD is a rhombus:
From part (i), we found that \( \angle BAC = \angle CAD \) and \( \angle CAD = \angle BCA \). Therefore, \( \angle BAC = \angle BCA \).
In \( \triangle ABC \), if two angles are equal, then the sides opposite to those angles are also equal. Since \( \angle BAC = \angle BCA \), it means that the side opposite \( \angle BCA \) (which is AB) is equal to the side opposite \( \angle BAC \) (which is BC). So, \( AB = BC \). Let's call this Equation (4).
We know that ABCD is a parallelogram. In a parallelogram, opposite sides are equal. So, \( AB = DC \) and \( BC = AD \).
From Equation (4) and the properties of a parallelogram, we have: \( AB = BC \) (from our finding) \( AB = DC \) (opposite sides) \( BC = AD \) (opposite sides) Putting these together, we get \( AB = BC = CD = DA \). Since all four sides of the parallelogram ABCD are equal, it is a rhombus.
In simple words: (i) Because AC cuts angle A in half, and due to parallel lines, the angles it makes with C are also equal, meaning AC also cuts angle C in half. (ii) Since angle A is cut in half and opposite angles are equal, it turns out that all four sides of the parallelogram must be equal, which is what makes it a rhombus.

🎯 Exam Tip: When proving properties of parallelograms, utilize alternate interior angles, opposite sides being equal, and the property that sides opposite equal angles in a triangle are equal.

Question 4. ABCD is a parallelogram and AP and CQ are perpendiculars from vertex A and C on diagonal BD. Show that (i) \( \triangle APB \cong \triangle CQD \) (ii) AP = CQ.
Answer:
(i) To show that \( \triangle APB \cong \triangle CQD \):
Consider \( \triangle APB \) and \( \triangle CQD \). We are given that AP is perpendicular to BD and CQ is perpendicular to BD. So, \( \angle APB = 90° \) and \( \angle CQD = 90° \). Therefore, \( \angle APB = \angle CQD \) (Both are 90 degrees). Let's call this Equation (1).
Since ABCD is a parallelogram, its opposite sides are equal. So, \( AB = CD \). Let's call this Equation (2).
Also, in a parallelogram, opposite sides are parallel. So, \( AB || DC \). When BD is a transversal cutting these parallel lines, the alternate interior angles are equal. Thus, \( \angle ABP = \angle CDQ \) (Alternate interior angles). Let's call this Equation (3).
From Equations (1), (2), and (3), by the Angle-Side-Angle (ASA) congruence criterion, we can conclude that: \( \triangle APB \cong \triangle CQD \) (ASA Congruence).
(ii) To show that AP = CQ:
Since we have proved that \( \triangle APB \cong \triangle CQD \) in part (i), their corresponding parts must be equal. This is known as Corresponding Parts of Congruent Triangles (CPCTC). Therefore, the side AP in \( \triangle APB \) is equal to the corresponding side CQ in \( \triangle CQD \). \( AP = CQ \).
In simple words: (i) We can show the two triangles are exactly the same by checking their angles and sides. They both have a right angle, the same length for one side (because it's a parallelogram), and another pair of angles are equal because of parallel lines. (ii) Since the triangles are identical, their matching parts, like AP and CQ, must also be equal.

🎯 Exam Tip: For congruence proofs, clearly identify the three pairs of equal parts (angles or sides) and state the correct congruence criterion (e.g., ASA, SAS, SSS, RHS).

Question 5. ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Answer:
Let's consider \( \triangle ABC \). P is the midpoint of AB and Q is the midpoint of BC. By the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. So, \( PQ = \frac{1}{2} AC \) and \( PQ || AC \). Let's call this Equation (1).
Now consider \( \triangle ACD \). R is the midpoint of CD and S is the midpoint of DA. Again, by the Midpoint Theorem: \( SR = \frac{1}{2} AC \) and \( SR || AC \). Let's call this Equation (2).
From Equation (1) and Equation (2), we can conclude that: \( PQ = SR \) and \( PQ || SR \). This means that one pair of opposite sides of quadrilateral PQRS are equal and parallel, which makes PQRS a parallelogram.
Now, let's connect the midpoints of the other two sides. Consider \( \triangle ABD \). P is the midpoint of AB and S is the midpoint of AD. By the Midpoint Theorem: \( PS = \frac{1}{2} BD \) and \( PS || BD \). Let's call this Equation (3).
Consider \( \triangle BCD \). Q is the midpoint of BC and R is the midpoint of CD. By the Midpoint Theorem: \( QR = \frac{1}{2} BD \) and \( QR || BD \). Let's call this Equation (4).
From Equation (3) and Equation (4), we can conclude that: \( PS = QR \) and \( PS || QR \).
Now, we know that ABCD is a rectangle. A key property of a rectangle is that its diagonals are equal in length. So, \( AC = BD \).
Since \( PQ = \frac{1}{2} AC \) (from Equation 1) and \( PS = \frac{1}{2} BD \) (from Equation 3), and we know \( AC = BD \), it must be true that \( PQ = PS \).
We have already established that PQRS is a parallelogram. A parallelogram in which adjacent sides are equal (like PQ = PS) is defined as a rhombus. Therefore, the quadrilateral PQRS is a rhombus. This shows how midpoints connect to properties of polygons.
In simple words: By using the midpoint theorem on the smaller triangles formed by the diagonals, we show that the inside shape (PQRS) is a parallelogram. Since the original shape is a rectangle, its diagonals are equal, which makes the sides of our inside parallelogram also equal. A parallelogram with equal sides is a rhombus.

🎯 Exam Tip: The Midpoint Theorem is crucial for these types of proofs. Remember that in a rectangle, diagonals are equal, which is key to proving the inner quadrilateral is a rhombus.

Question 6. In the figure A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Answer:
We are given the central angles: \( \angle AOB = 60° \) \( \angle BOC = 30° \)
To find \( \angle AOC \), which is the total angle subtended by arc ABC at the center O: \( \angle AOC = \angle AOB + \angle BOC \) \( \angle AOC = 60° + 30° \) \( \implies \angle AOC = 90° \)
Now, D is a point on the circle other than arc ABC. We need to find \( \angle ADC \). According to the circle theorem, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Here, arc AC subtends \( \angle AOC \) at the center and \( \angle ADC \) at point D on the circumference. So, \( \angle AOC = 2 \times \angle ADC \) \( 90° = 2 \times \angle ADC \) To find \( \angle ADC \), divide 90° by 2: \( \angle ADC = \frac{90°}{2} \) \( \implies \angle ADC = 45° \)
In simple words: First, add the two given angles at the center (AOB and BOC) to find the total angle AOC. Then, use the rule that an angle at the center is twice the angle at the edge of the circle if they come from the same arc. Divide the central angle by two to get the angle at the edge.

🎯 Exam Tip: Clearly identify the arc that subtends both the central angle and the angle at the circumference to correctly apply the theorem about angles in a circle.

Question 7. In the given figure A, B, C and D are four points on a circle, AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
Answer:
First, let's look at \( \triangle CDE \). We are given \( \angle BEC = 130° \) and \( \angle ECD = 20° \). The angle \( \angle BEC \) is an exterior angle to \( \triangle CDE \). According to the exterior angle theorem, an exterior angle of a triangle is equal to the sum of its two opposite interior angles. So, \( \angle BEC = \angle ECD + \angle EDC \) \( 130° = 20° + \angle EDC \) To find \( \angle EDC \), subtract 20° from 130°: \( \angle EDC = 130° - 20° \) \( \implies \angle EDC = 110° \)
Now, we need to find \( \angle BAC \). Both \( \angle BAC \) and \( \angle BDC \) (which is the same as \( \angle EDC \)) are angles subtended by the same arc BC on the circumference of the circle. According to the circle theorem, angles subtended by the same arc in the same segment of a circle are equal. Therefore, \( \angle BAC = \angle BDC \) Since \( \angle EDC = 110° \), then \( \angle BDC = 110° \). Thus, \( \angle BAC = 110° \). This rule simplifies finding angles on a circle.
In simple words: First, use the rule that an outside angle of a triangle equals the sum of the two opposite inside angles to find angle EDC. Then, since angles from the same arc on the circle's edge are equal, angle BAC will be the same as angle EDC.

🎯 Exam Tip: When dealing with circles and intersecting chords, look for exterior angles of triangles and angles subtended by the same arc in the same segment. These theorems are very common.

Question 8. In the given figure KLMN is a cyclic quadrilateral. KD is the tangent at K. If LN is a diameter ∠NLK = 40° and ∠LNM = 50°. Find ∠MLN and ∠DKL.
Answer:
Given that LN is a diameter of the circle. Any angle inscribed in a semi-circle is a right angle (90°). Therefore, \( \angle LMN = 90° \) and \( \angle LKN = 90° \).
To find \( \angle MLN \): Consider \( \triangle LMN \). The sum of angles in a triangle is 180°. We have \( \angle LMN = 90° \) and we are given \( \angle LNM = 50° \). So, \( \angle MLN + \angle LMN + \angle LNM = 180° \) \( \angle MLN + 90° + 50° = 180° \) \( \angle MLN + 140° = 180° \) \( \angle MLN = 180° - 140° \) \( \implies \angle MLN = 40° \)
To find \( \angle DKL \): We are given \( \angle NLK = 40° \). In \( \triangle LKN \), \( \angle LKN = 90° \) (angle in a semi-circle). \( \angle LNK + \angle NLK + \angle LKN = 180° \) \( \angle LNK + 40° + 90° = 180° \) \( \angle LNK + 130° = 180° \) \( \implies \angle LNK = 50° \)
Now, KD is a tangent to the circle at K. KL is a chord. According to the Alternate Segment Theorem, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Here, \( \angle DKL \) is the angle between tangent KD and chord KL. The angle subtended by chord KL in the alternate segment is \( \angle KNL \) (which is \( \angle LNK \)). So, \( \angle DKL = \angle LNK \) Since \( \angle LNK = 50° \), then \( \angle DKL = 50° \). Thus, \( \angle MLN = 40° \) and \( \angle DKL = 50° \).
In simple words: First, use the rule that an angle in a semicircle is 90 degrees to find LMN. Then, use the sum of angles in a triangle to find MLN. For DKL, use the alternate segment theorem, which says the angle between the tangent and a chord is equal to the angle in the opposite part of the circle.

🎯 Exam Tip: This question combines several key circle theorems: angle in a semicircle, sum of angles in a triangle, and the Alternate Segment Theorem. Be sure to identify which theorem applies to each part of the problem.

Question 9. In the given figure ∠PQR = 100°, where P, Q and R are points on a circle with centre 'O'. Find ∠OPR.
Answer: We are given that \( \angle PQR = 100^\circ \). This angle is made by the chord PR in the smaller section of the circle.
The reflex angle \( \angle POR \) (the larger angle at the center) is twice the angle \( \angle PQR \) at the circumference.
\( \implies \) Reflex \( \angle POR = 2 \times \angle PQR = 2 \times 100^\circ = 200^\circ \).
The total angle around the center is \( 360^\circ \). So, the smaller angle \( \angle POR \) can be found by subtracting the reflex angle from \( 360^\circ \).
\( \angle POR = 360^\circ - \text{Reflex } \angle POR = 360^\circ - 200^\circ = 160^\circ \).
Now, consider triangle OPR. Here, OP and OR are both radii of the same circle, which means they have equal length. This makes triangle OPR an isosceles triangle.
In an isosceles triangle, the angles opposite the equal sides are also equal. Therefore, \( \angle OPR = \angle ORP \).
The sum of all angles in any triangle is \( 180^\circ \). So, in \( \triangle OPR \):
\( \angle OPR + \angle ORP + \angle POR = 180^\circ \)
Since \( \angle OPR = \angle ORP \), we can write:
\( 2 \angle OPR + 160^\circ = 180^\circ \)
\( 2 \angle OPR = 180^\circ - 160^\circ \)
\( 2 \angle OPR = 20^\circ \)
\( \angle OPR = \frac{20^\circ}{2} \)
\( \angle OPR = 10^\circ \). This property, where the angle at the center is twice the angle at any point on the remaining part of the circle, is a fundamental theorem in circle geometry.
In simple words: The angle at the center is twice the angle at the circumference. We used this to find the smaller angle at the center. Then, knowing the triangle has two equal sides (radii), we could find the missing angle.

🎯 Exam Tip: Remember that the angle subtended by an arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle. This is key for such problems.

Question 10. AB and CD are two parallel chords of a circle which are on either sides of the centre such that AB = 10 cm and CD = 24 cm. Find the radius if the distance between AB and CD is 17 cm.
Answer: We are given that two parallel chords, AB and CD, are on opposite sides of the circle's center. The length of chord AB is 10 cm, chord CD is 24 cm, and the perpendicular distance between them (PQ) is 17 cm.
A line drawn from the center of a circle perpendicular to a chord bisects the chord. So, for chord CD:
\( PC = PD = \frac{CD}{2} = \frac{24}{2} = 12 \) cm.
Similarly, for chord AB:
\( AQ = QB = \frac{AB}{2} = \frac{10}{2} = 5 \) cm.
Let the distance from the center O to chord CD be \( OP = x \).
Since the total distance between the chords is 17 cm, the distance from the center O to chord AB will be \( OQ = (17 - x) \) cm.
Now, we will use the Pythagorean theorem in the right-angled triangles formed.
In \( \triangle OPC \):
\( OC^2 = OP^2 + PC^2 \)
\( OC^2 = x^2 + 12^2 \)
\( OC^2 = x^2 + 144 \) (Equation 1)
In \( \triangle OAQ \):
\( OA^2 = AQ^2 + OQ^2 \)
\( OA^2 = 5^2 + (17 - x)^2 \)
\( OA^2 = 25 + (289 - 34x + x^2) \)
\( OA^2 = 314 - 34x + x^2 \) (Equation 2)
Both OC and OA are radii of the same circle, so their lengths (and their squares) must be equal.
Equating Equation 1 and Equation 2:
\( x^2 + 144 = 314 - 34x + x^2 \)
Subtract \( x^2 \) from both sides:
\( 144 = 314 - 34x \)
Move 314 to the left side:
\( 144 - 314 = -34x \)
\( -170 = -34x \)
Divide both sides by -34:
\( x = \frac{-170}{-34} \)
\( x = 5 \)
Now substitute the value of \( x \) back into Equation 1 to find the radius:
\( OC^2 = x^2 + 144 \)
\( OC^2 = 5^2 + 144 \)
\( OC^2 = 25 + 144 \)
\( OC^2 = 169 \)
To find OC, take the square root of 169:
\( OC = \sqrt{169} \)
\( OC = 13 \) cm.
Therefore, the radius of the circle is 13 cm. This problem uses the Pythagorean theorem, which is very useful for solving geometry problems involving right-angled triangles.
In simple words: First, we cut the chords in half. Then we used the Pythagorean theorem for two triangles involving the radius and the distances from the center. By making the radius squared equal in both cases, we found the unknown distance from the center, and then the actual radius.

🎯 Exam Tip: Always draw a clear diagram when solving chord problems. Remember that a perpendicular from the center to a chord bisects the chord, and the Pythagorean theorem is frequently used.