Grade 10 jeopardy

Indices

100

What is 2³x2⁴? simplified

2⁷

200

Simplify 5⁶/5²?

5⁴

300

What is (3²)⁴ equal to

3⁸

400

Express 8^−2/3 as a fraction

1/4

500

Solve for x in 3^x=81.

x=4

Proportonality

100

If y is directly proportional to x and y=10 when x=2, find y when x=5.

y=25

200

If a is inversely proportional to b and a = 4 when b = 6, find a when b = 2.

a = 12

300

The area of a circle is proportional to the square of its radius. If the radius doubles, by what factor does the area increase?

By a factor of 4

400

If p varies as the cube of q, and p = 16 when q = 2, find p when q = 3.

p = 54

500

The intensity of light I at a certain point is inversely proportional to the square of the distance d from the light source.

If the intensity of a light bulb is 90 lumens at 2 meters, what will the intensity be at 5 meters?

I=14.4 lumens

Congruence

100

Two triangles ABC and DEF have sides:

• AB = 6 cm, BC = 8 cm, AC = 10 cm
• DE = 8 cm, EF = 10 cm, DF = 6 cm
• Are these triangles congruent? If so, by which rule?

Yes, by SSS (Side-Side-Side) Congruence (order of sides doesn't matter as long as they match).

200

Two triangles XYZ and PQR have:

• XY = 9 cm, YZ = 12 cm, and ∠XYZ = 50°
• PQ = 9 cm, QR = 12 cm, and ∠PQR = 50°

Are these triangles congruent? If so, by which rule?

Yes, by SAS (Side-Angle-Side) Congruence.

300

Triangle JKL and Triangle MNO both have angles 40°, 70°, and 70°.
However, their sides are different.

Are these triangles congruent? If so by which rule?

No, because AAA (Angle-Angle-Angle) does not prove congruence—it only proves similarity. The triangles could have different sizes.

400

Two right-angled triangles ABC and DEF are given:

• Both triangles have a right angle at C and F.
• AC = 6 cm, BC = 8 cm, and the hypotenuse AB = 10 cm in Triangle ABC.
• DF = 8 cm, EF = 6 cm, and the hypotenuse DE = 10 cm in Triangle DEF.

Are these triangles congruent? If so, by which rule?

Yes, by RHS (Right-Angle Hypotenuse Side) Congruence. The hypotenuses and one leg in each triangle are equal, so by RHS, the triangles are congruent.

500

• Circle O has radius r = 10 cm.
• Two tangents, PA and PB, are drawn from an external point P to the circle, meeting the circle at points A and B, respectively.
• The angle ∠APB = 60°.

Prove that triangle OPA ≅ triangle OPB.

• Step 1: Use properties of tangents.

  • PA = PB because tangents from the same external point to a circle are equal in length.

• Step 2: Use the radii.

  • OA = OB = 10 cm because they are both radii of the same circle.

• Step 3: Show that the angles are equal.

  • Since PA = PB and OA = OB, and ∠APB = 60°, we can use the fact that the angle between two tangents is the same for both triangles.

• Step 4: Apply the SAS (Side-Angle-Side) congruence criterion.
  We have:

  • PA = PB
  • OA = OB
  • ∠APB = ∠BPA = 60°

• By SAS, we conclude that triangle OPA ≅ triangle OPB.

Circle Theorems

100

Solve 3/x = 6/(x+2).

Answer: x = 6

200

Express (1/x) + (1/y) as a single fraction.

Answer: (x + y) / (xy)

300

Simplify (x² - 9) / (x - 3).

Answer: x + 3 (for x ≠ 3)

400

Solve (2x+3) / (x+1) = 4.

Answer: x = 1/2

500

Factorize and simplify (x² - 4) / (x² - x - 6).

Answer: (x - 2) / (x - 3) (for x ≠ -2, 3)

Algebraic Fractions

100

In a circle, ABC is a triangle inscribed in the circle with AC as the diameter.
Find the measure of ∠ABC.

90^o

200

In a circle with center O, ∠AOB = 120°.
Find the angle ∠ACB, where C is any point on the circumference on the same arc AB.

60^o

300

In a cyclic quadrilateral ABCD, it is given that ∠ABC = 75° and ∠BAD = 120°.
Find <ADC and ∠BCD.

105^o , 60^o

400

A tangent PT to a circle at point P forms an angle of 60° with a chord AP.
If ∠PAQ = 90° (where Q is a point on the circumference and A is on the circle), find the measure of ∠PQA.

30^o

500

In a circle with center O, the chord AB is perpendicular to the diameter CD at point P. The angle ∠BOC = 120°.

Prove that △OBC is an equilateral triangle. Find the value of ∠BAP.

1. Proving △OBC is equilateral:

   1. Use the fact that AB⊥CD, so P is the midpoint of AB.
   2. The radius OB and OC are equal (since both are radii of the same circle).
   3. Use the angle ∠BOC = 120°. By symmetry, the distance between points B and C should be equal to the radius of the circle.
   4. Therefore, △OBC is equilateral, as all the angles and sides are equal.

2. Finding ∠BAP:

   1. Apply the Alternate Segment Theorem or the Angles in the Same Segment Theorem to find ∠BAP.
   2. Since AB is a chord and AB⊥CD, the angle between the chord and the tangent at P will relate to the angles subtended by the same chord.