Samacheer Kalvi Class 9 Maths Solutions Chapter 4 Geometry Exercise 4.4

Question 1. Find the value of x in the given figure.

Answer: First, we know that if an angle is formed in a semicircle, it is always \( 90^\circ \). So, \( \angle BCA = 90^\circ \). For any triangle, the sum of all its internal angles is \( 180^\circ \). In triangle ABC, we are given \( \angle ABC = 60^\circ \). We also have \( \angle BAC \) which is represented by x. So, we can set up the equation for the sum of angles in triangle ABC: \( \angle BAC + \angle ABC + \angle BCA = 180^\circ \) Substitute the known values: \( x + 60^\circ + 90^\circ = 180^\circ \) Add the constant angles together: \( x + 150^\circ = 180^\circ \) To find the value of x, subtract \( 150^\circ \) from \( 180^\circ \): \( x = 180^\circ - 150^\circ \) \( x = 30^\circ \) Therefore, the value of x is \( 30^\circ \).In simple words: An angle in a semicircle is \( 90^\circ \). Since all angles in a triangle add up to \( 180^\circ \), we can subtract the two known angles ( \( 60^\circ \) and \( 90^\circ \) ) from \( 180^\circ \) to find the missing angle x.

🎯 Exam Tip: Remember that the angle subtended by a diameter at any point on the circumference is always a right angle (\( 90^\circ \)). This is a key property of circles.

Question 2. In the given figure, AC is the diameter of the circle with centre O. If \( \angle ADE = 30^\circ \); \( \angle DAC = 35^\circ \) and \( \angle CAB = 40^\circ \). Find (i) \( \angle ACD \) (ii) \( \angle ACB \) (iii) \( \angle DAE \)

Answer:
(i) To find \( \angle ACD \):
Since AC is the diameter, the angle formed in the semicircle, \( \angle ADC \), is \( 90^\circ \). We know that the sum of angles in any triangle is \( 180^\circ \). In triangle ADC, we have \( \angle ADC = 90^\circ \) and \( \angle DAC = 35^\circ \). \( \angle ADC + \angle ACD + \angle DAC = 180^\circ \) \( 90^\circ + \angle ACD + 35^\circ = 180^\circ \) Combine the known angles: \( 125^\circ + \angle ACD = 180^\circ \) Subtract \( 125^\circ \) from \( 180^\circ \) to find \( \angle ACD \): \( \angle ACD = 180^\circ - 125^\circ \) \( \angle ACD = 55^\circ \) (ii) To find \( \angle ACB \):
Similarly, since AC is the diameter, the angle formed in the semicircle, \( \angle ABC \), is \( 90^\circ \). In triangle ABC, the sum of its angles is \( 180^\circ \). We have \( \angle ABC = 90^\circ \) and \( \angle CAB = 40^\circ \). \( \angle ACB + \angle CBA + \angle BAC = 180^\circ \) \( \angle ACB + 90^\circ + 40^\circ = 180^\circ \) Combine the known angles: \( \angle ACB + 130^\circ = 180^\circ \) Subtract \( 130^\circ \) from \( 180^\circ \) to find \( \angle ACB \): \( \angle ACB = 180^\circ - 130^\circ \) \( \angle ACB = 50^\circ \) (iii) To find \( \angle DAE \):
In the cyclic quadrilateral ACDE, the sum of opposite angles is \( 180^\circ \). We already found \( \angle ACD = 55^\circ \). Since \( \angle ACD \) and \( \angle AED \) are opposite angles in the cyclic quadrilateral ACDE, their sum is \( 180^\circ \). \( \angle AED + \angle ACD = 180^\circ \) \( \angle AED + 55^\circ = 180^\circ \) \( \angle AED = 180^\circ - 55^\circ \) \( \angle AED = 125^\circ \) Now, consider triangle DAE. The sum of its angles is \( 180^\circ \). We have \( \angle AED = 125^\circ \) and \( \angle EDA = 30^\circ \). \( \angle DAE + \angle AED + \angle EDA = 180^\circ \) \( \angle DAE + 125^\circ + 30^\circ = 180^\circ \) Combine the known angles: \( \angle DAE + 155^\circ = 180^\circ \) Subtract \( 155^\circ \) from \( 180^\circ \) to find \( \angle DAE \): \( \angle DAE = 180^\circ - 155^\circ \) \( \angle DAE = 25^\circ \).In simple words: We used the property that angles in a semicircle are \( 90^\circ \) and that angles in a triangle add up to \( 180^\circ \) to find the first two angles. For the third angle, we used the cyclic quadrilateral property (opposite angles add to \( 180^\circ \)) and then the triangle angle sum rule again.

🎯 Exam Tip: When dealing with circles, always look for diameters to identify \( 90^\circ \) angles, and remember the properties of cyclic quadrilaterals where opposite angles sum to \( 180^\circ \).

Question 3. Find all the angles of the given cyclic quadrilateral ABCD in the figure.

Answer: In a cyclic quadrilateral ABCD, the sum of opposite angles is always \( 180^\circ \). So, we can set up two equations: \( \angle B + \angle D = 180^\circ \) \( (6x - 4^\circ) + (7x + 2^\circ) = 180^\circ \) Combine like terms: \( 13x - 2^\circ = 180^\circ \) Add \( 2^\circ \) to both sides: \( 13x = 180^\circ + 2^\circ \) \( 13x = 182^\circ \) Divide by 13 to find x: \( x = \frac{182^\circ}{13} \) \( x = 14^\circ \) Now, let's find the measure of angle B and D using the value of x: \( \angle B = 6x - 4^\circ = 6(14^\circ) - 4^\circ = 84^\circ - 4^\circ = 80^\circ \) \( \angle D = 7x + 2^\circ = 7(14^\circ) + 2^\circ = 98^\circ + 2^\circ = 100^\circ \) Next, we use the second pair of opposite angles: \( \angle A + \angle C = 180^\circ \) \( (2y + 4^\circ) + (4y - 4^\circ) = 180^\circ \) Combine like terms: \( 6y = 180^\circ \) Divide by 6 to find y: \( y = \frac{180^\circ}{6} \) \( y = 30^\circ \) Now, let's find the measure of angle A and C using the value of y: \( \angle A = 2y + 4^\circ = 2(30^\circ) + 4^\circ = 60^\circ + 4^\circ = 64^\circ \) \( \angle C = 4y - 4^\circ = 4(30^\circ) - 4^\circ = 120^\circ - 4^\circ = 116^\circ \) So, all the angles of the cyclic quadrilateral ABCD are: \( \angle A = 64^\circ \), \( \angle B = 80^\circ \), \( \angle C = 116^\circ \), \( \angle D = 100^\circ \).In simple words: For a shape called a cyclic quadrilateral, opposite angles always add up to \( 180^\circ \). We used this rule twice, first with angles B and D to find the value of x, and then with angles A and C to find the value of y. Once we knew x and y, we could calculate the exact size of each angle.

🎯 Exam Tip: Always double-check your calculations by ensuring that \( \angle A + \angle C = 180^\circ \) and \( \angle B + \angle D = 180^\circ \) with your final angle values.

Question 4. In the given figure, ABCD is a cyclic quadrilateral where diagonals intersect at P and \( \angle BAC = 60^\circ \) find (i) \( \angle CAD \) (ii) \( \angle BCD \)

Answer:
(i) To find \( \angle CAD \):
Angles in the same segment of a circle are equal. This means angles subtended by the same arc at the circumference are equal. The arc CD subtends \( \angle CAD \) and \( \angle CBD \) (which is the same as \( \angle DBC \)) at the circumference. We are given \( \angle DBC = 40^\circ \). Therefore, \( \angle CAD = \angle DBC = 40^\circ \). (ii) To find \( \angle BCD \):
Again, using the property that angles in the same segment are equal: The arc BC subtends \( \angle BAC \) and \( \angle BDC \) at the circumference. We are given \( \angle BAC = 60^\circ \). Therefore, \( \angle BDC = \angle BAC = 60^\circ \). Now, consider the triangle BCD. The sum of angles in a triangle is \( 180^\circ \). We have \( \angle BCD + \angle BDC + \angle CBD = 180^\circ \). Substitute the known values: \( \angle BCD + 60^\circ + 40^\circ = 180^\circ \) Combine the known angles: \( \angle BCD + 100^\circ = 180^\circ \) Subtract \( 100^\circ \) from \( 180^\circ \) to find \( \angle BCD \): \( \angle BCD = 180^\circ - 100^\circ \) \( \angle BCD = 80^\circ \).In simple words: We used the rule that angles in the same part of a circle are equal. This helped us find the first angle directly. For the second angle, we used this rule again to find a missing angle in triangle BCD, then used the fact that all angles in a triangle add up to \( 180^\circ \).

🎯 Exam Tip: Always identify which arc subtends which angles. Angles from the same arc to different points on the circumference are always equal.

Question 5. In the given figure, AB and CD are the parallel chords of a circle with centre O. Such that AB = 8 cm and CD = 6 cm. If OM \( \perp \) AB and OL \( \perp \) CD distance between LM is 7cm. Find the radius of the circle?

Answer: Let OM be represented by x. Since LM is \( 7 \) cm and OM is x, then OL must be \( (7 - x) \) cm. When a perpendicular from the center of a circle meets a chord, it bisects the chord (divides it into two equal parts). For chord AB = \( 8 \) cm, AM = \( \frac{8}{2} = 4 \) cm. For chord CD = \( 6 \) cm, CL = \( \frac{6}{2} = 3 \) cm. Now, consider the right-angled triangle AOM. OA is the radius (r). Using the Pythagorean theorem ( \( \text{hypotenuse}^2 = \text{side1}^2 + \text{side2}^2 \) ): \( OA^2 = AM^2 + OM^2 \) \( r^2 = 4^2 + x^2 \) \( r^2 = 16 + x^2 \) ... (Equation 1) Next, consider the right-angled triangle OCL. OC is also the radius (r). Using the Pythagorean theorem: \( OC^2 = OL^2 + CL^2 \) \( r^2 = (7 - x)^2 + 3^2 \) \( r^2 = (49 - 14x + x^2) + 9 \) \( r^2 = 58 + x^2 - 14x \) ... (Equation 2) Since both equations equal \( r^2 \), we can set them equal to each other: \( 16 + x^2 = 58 + x^2 - 14x \) Subtract \( x^2 \) from both sides: \( 16 = 58 - 14x \) Add \( 14x \) to both sides and subtract \( 16 \): \( 14x = 58 - 16 \) \( 14x = 42 \) Divide by 14 to find x: \( x = \frac{42}{14} \) \( x = 3 \) cm Now that we have x, substitute it back into Equation 1 to find the radius (r): \( r^2 = 16 + x^2 \) \( r^2 = 16 + 3^2 \) \( r^2 = 16 + 9 \) \( r^2 = 25 \) Take the square root of both sides: \( r = \sqrt{25} \) \( r = 5 \) cm The radius of the circle is \( 5 \) cm.In simple words: We used the Pythagorean theorem twice to set up equations for the radius, considering two right triangles formed by the center, the midpoint of each chord, and an endpoint of each chord. By solving these equations, we found the distance of one chord from the center, and then used that to calculate the circle's radius.

🎯 Exam Tip: Remember that a perpendicular from the center to a chord always bisects the chord. This creates right-angled triangles crucial for applying the Pythagorean theorem.

Question 6. The arch of a bridge has dimensions as shown, where the arch measure 2 m at its highest point and its width is 6 m. What is the radius of the circle that contains the arch?

Answer: The width of the arch is AB = \( 6 \) m. D is the midpoint of AB, so AD = \( \frac{6}{2} = 3 \) m. The height of the arch at its highest point is CD = \( 2 \) m. We need to find the radius of the circle that contains this arch. Let this radius be r. Consider the right-angled triangle ADC. Using the Pythagorean theorem ( \( \text{hypotenuse}^2 = \text{side1}^2 + \text{side2}^2 \) ): \( AC^2 = AD^2 + DC^2 \) \( AC^2 = 3^2 + 2^2 \) \( AC^2 = 9 + 4 \) \( AC^2 = 13 \) \( AC = \sqrt{13} \) \( AC \approx 3.6 \) m Now, we use a property of intersecting chords. If CD is extended to meet the circle at a point (let's call it E), then \( AD \times DB = CD \times DE \). Since CD is part of the diameter that is perpendicular to the chord AB, it also bisects AB. If the full circle has center O and radius r, and D is the midpoint of AB, then CD is part of the perpendicular diameter. Let the center of the circle be O. Let C be the highest point on the arch, and D be the midpoint of the base AB. The radius is r. The distance from O to D is \( (r - CD) = (r - 2) \) m. In the right triangle formed by OA (radius), OD, and AD: \( OA^2 = OD^2 + AD^2 \) \( r^2 = (r - 2)^2 + 3^2 \) \( r^2 = (r^2 - 4r + 4) + 9 \) \( r^2 = r^2 - 4r + 13 \) Subtract \( r^2 \) from both sides: \( 0 = -4r + 13 \) Add \( 4r \) to both sides: \( 4r = 13 \) Divide by 4: \( r = \frac{13}{4} \) \( r = 3.25 \) m The radius of the circle that contains the arch is \( 3.25 \) m.In simple words: We pictured the bridge arch as part of a full circle. By using the arch's width and height, we formed a right-angled triangle. Then, using the Pythagorean theorem and the idea of the circle's radius and center, we set up an equation to find the radius.

🎯 Exam Tip: For problems involving segments of circles, draw the full circle and identify the center and radius. Often, the perpendicular from the center to a chord will be involved in a right-angled triangle, allowing for the application of the Pythagorean theorem.

Question 7. In figure \( \angle ABC = 120^\circ \), where A, B and C are points on the circle with centre O. Find \( \angle OAC \)?

Answer: 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. Here, arc AC subtends \( \angle AOC \) at the center O and \( \angle ABC \) at point B on the circumference. So, the reflex angle \( \angle AOC = 2 \times \angle ABC \). Given \( \angle ABC = 120^\circ \): Reflex \( \angle AOC = 2 \times 120^\circ = 240^\circ \). The sum of angles around a point is \( 360^\circ \). The smaller angle \( \angle AOC \) is: \( \angle AOC = 360^\circ - \text{Reflex } \angle AOC \) \( \angle AOC = 360^\circ - 240^\circ = 120^\circ \). Now, consider triangle OAC. OA and OC are both radii of the same circle, so OA = OC. This means triangle OAC is an isosceles triangle. In an isosceles triangle, the angles opposite to the equal sides are also equal. So, \( \angle OAC = \angle OCA \). The sum of angles in triangle OAC is \( 180^\circ \): \( \angle OAC + \angle OCA + \angle AOC = 180^\circ \) Substitute \( \angle OAC \) for \( \angle OCA \) and \( \angle AOC = 120^\circ \): \( \angle OAC + \angle OAC + 120^\circ = 180^\circ \) \( 2 \times \angle OAC = 180^\circ - 120^\circ \) \( 2 \times \angle OAC = 60^\circ \) Divide by 2: \( \angle OAC = \frac{60^\circ}{2} \) \( \angle OAC = 30^\circ \).In simple words: The angle at the center is twice the angle at the edge of the circle if they use the same arc. We used this to find the angle at the center. Then, because two sides of the triangle from the center are radii (and thus equal), the other two angles in that triangle must also be equal. We then used the \( 180^\circ \) rule for triangles to find the final angle.

🎯 Exam Tip: Remember the relationship between central angles and inscribed angles, and that triangles formed by two radii and a chord are always isosceles.

Question 8. A school wants to conduct tree plantation programme. For this a teacher allotted a circle of radius 6m ground to nineth standard students for planting sapplings. Four students plant trees at the points A, B, C and D as shown in figure. Here AB = 8m, CD = 10m and AB \( \perp \) CD. If another student places a flower pot at the point P, the intersection of AB and CD, then find the distance from the centre to P.

Answer: The radius of the circle, OA (or OD), is \( 6 \) m. Chord AB = \( 8 \) m. Chord CD = \( 10 \) m. When a perpendicular from the center O meets a chord, it bisects it. For chord AB, AM = \( \frac{8}{2} = 4 \) m. For chord CD, CN = \( \frac{10}{2} = 5 \) m. Consider the right-angled triangle AOM. OA is the radius (6 m). Using the Pythagorean theorem ( \( \text{hypotenuse}^2 = \text{side1}^2 + \text{side2}^2 \) ): \( OA^2 = AM^2 + OM^2 \) \( 6^2 = 4^2 + OM^2 \) \( 36 = 16 + OM^2 \) \( OM^2 = 36 - 16 \) \( OM^2 = 20 \) \( OM = \sqrt{20} \) m Now consider the right-angled triangle CON. OC is the radius (6 m). Using the Pythagorean theorem: \( OC^2 = CN^2 + ON^2 \) \( 6^2 = 5^2 + ON^2 \) \( 36 = 25 + ON^2 \) \( ON^2 = 36 - 25 \) \( ON^2 = 11 \) \( ON = \sqrt{11} \) m We are given that AB \( \perp \) CD, and O is the center. Since OM \( \perp \) AB and ON \( \perp \) CD, the quadrilateral formed by O, M, P, N (where P is the intersection of the chords) is a rectangle. This means that opposite sides are equal. So, MP = ON = \( \sqrt{11} \) m. And NP = OM = \( \sqrt{20} \) m. We need to find the distance from the center O to the intersection point P, which is the diagonal OP of the rectangle ONPM. In right-angled triangle ONP: \( OP^2 = ON^2 + NP^2 \) \( OP^2 = (\sqrt{11})^2 + (\sqrt{20})^2 \) \( OP^2 = 11 + 20 \) \( OP^2 = 31 \) \( OP = \sqrt{31} \) m \( OP \approx 5.56 \) m The distance from the center to P is approximately \( 5.56 \) m.In simple words: First, we found the distances from the center to each chord using the radius and half the chord length. Since the chords cross at a \( 90^\circ \) angle, the shape formed by the center and the points where the perpendiculars meet is a rectangle. We then used the Pythagorean theorem again to find the length of the diagonal of this rectangle, which is the distance from the center to the meeting point of the chords.

🎯 Exam Tip: When chords intersect perpendicularly, remember that the distances from the center to the chords, along with the segments of the chords, often form rectangles or right triangles, allowing for easy use of the Pythagorean theorem.

Question 9. In the given figure, \( \angle POQ = 100^\circ \) and \( \angle PQR = 30^\circ \), then find \( \angle RPO \).

Answer: The angle subtended by an arc at the circumference is half the angle subtended by the same arc at the center. Arc PQ subtends \( \angle POQ = 100^\circ \) at the center O and \( \angle PRQ \) at the circumference. So, \( \angle PRQ = \frac{1}{2} \times \angle POQ \). \( \angle PRQ = \frac{1}{2} \times 100^\circ = 50^\circ \). Now consider triangle PQR. The sum of its angles is \( 180^\circ \). \( \angle RPQ + \angle PQR + \angle PRQ = 180^\circ \) We know \( \angle PQR = 30^\circ \) and \( \angle PRQ = 50^\circ \). \( \angle RPQ + 30^\circ + 50^\circ = 180^\circ \) \( \angle RPQ + 80^\circ = 180^\circ \) \( \angle RPQ = 180^\circ - 80^\circ = 100^\circ \). In triangle OPQ, OP and OQ are both radii of the same circle, so OP = OQ. This means triangle OPQ is an isosceles triangle. The angles opposite to the equal sides are also equal. So, \( \angle OPQ = \angle OQP \). The sum of angles in triangle OPQ is \( 180^\circ \). We know \( \angle POQ = 100^\circ \). \( \angle OPQ + \angle OQP + \angle POQ = 180^\circ \) \( \angle OPQ + \angle OPQ + 100^\circ = 180^\circ \) \( 2 \times \angle OPQ = 180^\circ - 100^\circ \) \( 2 \times \angle OPQ = 80^\circ \) \( \angle OPQ = \frac{80^\circ}{2} \) \( \angle OPQ = 40^\circ \). Finally, we need to find \( \angle RPO \). We know \( \angle RPQ = 100^\circ \) and \( \angle OPQ = 40^\circ \). From the figure, \( \angle RPQ = \angle RPO + \angle OPQ \). So, \( \angle RPO = \angle RPQ - \angle OPQ \). \( \angle RPO = 100^\circ - 40^\circ \) \( \angle RPO = 60^\circ \).In simple words: First, we used the rule that the angle at the center of a circle is twice the angle at the edge for the same arc. This helped us find \( \angle PRQ \). Then, we used the sum of angles in triangle PQR to find \( \angle RPQ \). Next, because two sides of triangle OPQ are radii, it's an isosceles triangle, which allowed us to find \( \angle OPQ \). Finally, we subtracted \( \angle OPQ \) from \( \angle RPQ \) to get \( \angle RPO \).

🎯 Exam Tip: Break down complex geometry problems into smaller triangles. Apply angle properties of circles (central vs. inscribed angles) and triangles (sum of angles, isosceles triangle properties) step by step.