Verified step-by-step solutions for your class notes.
In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?
Answer: Let half the chord length be $x$. Using Pythagoras: $x = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$.
The total chord length is $2 \times 12 = \mathbf{24\text{ cm}}$.
An arc of a circle subtends an angle of $70^\circ$ at the centre. What is the measure of the angle subtended by the arc at a point on the circle?
Answer: By Theorem 9, the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
So, the angle on the circle = $70^\circ / 2 = \mathbf{35^\circ}$.
The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.
Answer: Radius $r = 26 / 2 = 13\text{ cm}$. Half the chord length $x = 24 / 2 = 12\text{ cm}$.
Let the distance from the centre be $d$. Using Pythagoras: $d = \sqrt{r^2 - x^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = \mathbf{5\text{ cm}}$.
A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?
Answer: Half the chord length $x = \sqrt{r^2 - d^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12\text{ cm}$.
The total length of the chord is $2 \times 12 = \mathbf{24\text{ cm}}$.
The diameter of a circle is AB. Point C is on the circumference. What is the measure of the $\angle ACB$? Explain your reasoning.
Answer: $\angle ACB = \mathbf{90^\circ}$.
Reasoning: The angle subtended by an arc at the centre is double the angle subtended at any point on the circle. The arc $AB$ (being a semicircle formed by the diameter) subtends a straight angle of $180^\circ$ at the centre. Therefore, the angle $\angle ACB$ at the circumference must be half of $180^\circ$, which is $90^\circ$. This is also known as the corollary "the angle in a semicircle is a right angle".
ABCD is a cyclic quadrilateral inscribed in a circle. If $\angle A$ measures $75^\circ$, what is the measure of $\angle C$? If $\angle B$ measures $110^\circ$, what is the measure of $\angle D$?
Answer: By Theorem 11, the sum of opposite angles in a cyclic quadrilateral is $180^\circ$.
$\angle C = 180^\circ - \angle A = 180^\circ - 75^\circ = \mathbf{105^\circ}$.
$\angle D = 180^\circ - \angle B = 180^\circ - 110^\circ = \mathbf{70^\circ}$.
Quadrilateral PQRS is inscribed in a circle. If $\angle P = (2x + 10)^\circ$ and $\angle R = (3x - 20)^\circ$, find the value of x and the measures of $\angle P$ and $\angle R$.
Answer: Since it is a cyclic quadrilateral, opposite angles $\angle P$ and $\angle R$ sum to $180^\circ$.
$(2x + 10) + (3x - 20) = 180$
$5x - 10 = 180 \implies 5x = 190 \implies \mathbf{x = 38}$.
$\angle P = 2(38) + 10 = 76 + 10 = \mathbf{86^\circ}$.
$\angle R = 3(38) - 20 = 114 - 20 = \mathbf{94^\circ}$.
(Check: $86 + 94 = 180$).
The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
Answer: Half the chord length $x = 16 / 2 = 8\text{ cm}$. Distance $d = 6\text{ cm}$.
Using Pythagoras: $r^2 = x^2 + d^2 = 8^2 + 6^2 = 64 + 36 = 100$.
Radius $r = \sqrt{100} = \mathbf{10\text{ cm}}$.
A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
Answer: This is a kite shape inscribed in a circle. The sides can form two right-angled triangles because $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. By the converse of Pythagoras, the angle between the sides of 5 and 12 is a right angle ($90^\circ$). Because the opposite angles must sum to $180^\circ$, they are both $90^\circ$.
The area is simply the sum of the areas of these two right triangles.
Area = $2 \times \left( \frac{1}{2} \times 5 \times 12 \right) = 5 \times 12 = \mathbf{60\text{ square units}}$.
*(Alternatively, use Brahmagupta's formula: $s = (5+5+12+12)/2 = 17$. Area = $\sqrt{(17-5)(17-5)(17-12)(17-12)} = \sqrt{12 \times 12 \times 5 \times 5} = 12 \times 5 = 60$).*
Show that rectangle is the only parallelogram that can be inscribed in a circle.
Answer: Let $ABCD$ be a parallelogram inscribed in a circle. Because it is a parallelogram, opposite angles are equal ($\angle A = \angle C$). Because it is inscribed in a circle (cyclic), opposite angles sum to $180^\circ$ ($\angle A + \angle C = 180^\circ$).
Substituting gives $2\angle A = 180^\circ \implies \angle A = 90^\circ$. The same logic applies to $\angle B$ and $\angle D$. A parallelogram with all right angles is a rectangle. Therefore, only rectangles can be inscribed in a circle.
Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., $\angle CDE = \angle ABC$, where E is a point on the extension of side CD).
Answer: Let's look at vertex $D$. The interior angle is $\angle ADC$ and the exterior angle is $\angle CDE$. Since they form a linear pair on the straight line $CDE$, $\angle ADC + \angle CDE = 180^\circ$.
Because $ABCD$ is a cyclic quadrilateral, its opposite interior angles sum to $180^\circ$. Thus, $\angle ADC + \angle ABC = 180^\circ$.
Equating the two expressions gives: $\angle ADC + \angle CDE = \angle ADC + \angle ABC$. Subtracting $\angle ADC$ from both sides leaves $\angle CDE = \angle ABC$. The exterior angle is equal to the interior opposite angle.
"There is no chord of a circle that is longer than its diameter." How do you justify this statement?
Answer: Let the endpoints of any chord be $A$ and $B$, and let the centre of the circle be $C$ with radius $r$. Consider $\Delta CAB$. By the triangle inequality theorem, the length of any side of a triangle must be strictly less than the sum of the other two sides. So, $AB < CA + CB$.
Since $CA = r$ and $CB = r$, this means $AB < 2r$. Therefore, any chord that does not pass through the centre is strictly less than $2r$ (the diameter). If the chord passes through the centre, it *is* the diameter, and $A, C, B$ are collinear, so $AB = 2r$. Thus, no chord can be longer than the diameter.
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