Test: Properties of Equal Chords - Grade 9 MCQ

Given, PQ = 10 cm, XY = 24 cm and radius of circle = 13 cm

We need to find the distance between PQ and XY.

Let AB be the diameter of the circle passing through O.

Then, OB = OA = AB/2 = 13 cm

Let M be the midpoint of PQ and N be the midpoint of XY.

Then, OM is perpendicular to PQ and ON is perpendicular to XY.

Also, OM = ON = OB = 13 cm (radii of the same circle).

Now, consider the right-angled triangles OMP and ONX.

We have, OP = OX = PQ/2 = 5 cm and OY = OQ = XY/2 = 12 cm.

Using Pythagoras theorem, we can find

MP and NX as: MP = sqrt(OP^2 - OM^2) = sqrt(5^2 - 13^2) = sqrt(144) = 12 cm NX = sqrt(OX^2 - ON^2) = sqrt(12^2 - 13^2) = sqrt(23) cm

Therefore, the distance between PQ and XY is MN = MP + NX = 12 + sqrt(23) cm Hence, option (A) 17 cm is the correct answer.

The angle subtended by an arc at the center of the circle is twice the angle subtended by the same arc at any point on the circumference.

• ∠ABC=70^∘, subtended by the arc AC.
• Therefore, the angle subtended by the same arc AC at the circumference on the opposite side (∠BCD) can be calculated.

∠BCD = 180^∘ − (∠BAD + ∠ABC)

∠BCD = 180^∘ − (50^∘+70^∘)

∠BCD = 60^∘

∠SOR = 37.5° 

∠SQT = 1/2 ∠SOR (Angle at the circumference is half of the angle at the centre)

⇒ 37.5°/2

⇒ 18.75°                                                                                             

∠QSP = 90° (angle made from the diameter to the circumference is 90°)

Now,

∠PSQ + ∠QST = 180° (Linear pair angle)

∠QST = 180° - 90° = 90°

In ∆QST,

∠STQ = 180° - (90° + 18.75°)

⇒ ∠STQ = 71.25°

We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the
alternate segment.
Thus, ∠ AOC = 2 ∠ ADC
⇒ 100 ° = 2 ∠ ADC
∴ ∠ ADC = 50 °
The opposite angles of a cyclic quadrilateral are supplementary and ABCD is a cyclic quadrilateral.
Thus, ∠ ADC + ∠ ABC = 180 °
⇒ 50 ° + ∠ ABC = 180 °
⇒ ∠ ABC = (180 ° – 50 ° ) = 130 °
∴ ∠ ADC = 50 ° and ∠ ABC = 130 °

Given: ∠BOA=90^∘ and ∠AOC=110^∘

We know that angles around a point add up to 360^∘

∴∠BOC+∠BOA+∠AOC=360^∘

⇒∠BOC+90^∘+110^∘=360^∘

⇒∠BOC=360^∘−200^∘

∠BOC=160^∘

We know that 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.

Arc BC is subtending ∠BOC at the center and ∠BAC on the remaining part of the circle, so ∠BOC=2×∠BAC

Given : An arc ABC of a circle with center O , and a point C on the remaining part of the circumference.

To Prove : Angle AOB =  Twice angle ABC

Construction : Join OC and produce it to a suitable point D

Since AB and CD are equal chords, they are equidistant from the center O.

- Let's consider a point F on AB such that OF is perpendicular to AB.

- We can draw a perpendicular from O to AB, intersecting it at point G.

- Now, we have two right-angled triangles OGF and OFE.

- In triangle OGF, OG is the radius of the circle, and OF is the perpendicular from the center O to the chord AB.

- In triangle OFE, OE is the perpendicular from the center O to the chord CD, and OF is the perpendicular from the point F on AB to the chord CD.

- Since OE = 5 cm (given), OF = OF (common side), and OG = OG (radius of the circle), we have two right-angled triangles with a common side and a common hypotenuse.

- Therefore, triangle OGF and triangle OFE are congruent by the Hypotenuse-Leg congruence criterion.

- By this congruence, we can conclude that OG = OE = 5 cm.

- Since OG is the radius of the circle, it is equal to the radius of the circle, which means the length of OF is 5 cm.

Hence, the length of the perpendicular OF on AB is 5 cm.

Therefore, option A is the correct answer.