Date & Time: 14th March 2020, 11:00 am

Duration: 2h

- All questions are compulsory.
- Use of calculator is not allowed.
- The numbers to the right of the questions indicate full marks.
- In the case of MCQ’s Q. No. 1(A) only the first attempt will be evaluated and will be given credit.
- For every MCQ, the correct alternative (A), (B), (C) or (D) in front of subquestion number is to be written as an answer.

Some question and their alternative answer are given. Select the correct alternative.

Out of the following which is the Pythagorean triplet?

(1, 5, 10)

(3, 4, 5)

(2, 2, 2)

(5, 5, 2)

Chapter: [0.02] Pythagoras Theorem

Two circles of radii 5.5 cm and 3.3 cm respectively touch each other externally. What is the distance between their centres?

4.4 cm

2.2 cm

8.8 cm

8.9 cm

Chapter: [0.03] Circle

Fill in the blank using correct alternative.

Distance of point (–3, 4) from the origin is ___________

7

1

5

−5

Chapter: [0.04] Co-ordinate Geometry

Find the volume of a cube of side 3 cm:

27 cm^{3}

9 cm^{3}

81 cm^{3}

3 cm^{3}

Chapter: [0.07] Mensuration

The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.

Chapter: [0.01] Similarity

Find the diagonal of a square whose side is 10 cm.

Chapter: [0.07] Mensuration [0.07] Mensuration

`square`ABCD is cyclic. If ∠B = 110°, then find measure of ∠D.

Chapter: [0.03] Circle

Find the slope of the lines passing through the given point.

A(2, 3), B(4, 7)

Chapter: [0.04] Co-ordinate Geometry

In the figure given above, ‘O’ is the centre of the circle, seg PS is a tangent segment and S is the point of contact. Line PR is a secant.

If PQ = 3.6, QR = 6.4, find PS.

**Solution:**

PS^{2} = PQ × `square` ......(tangent secant segments theorem)

= PQ × (PQ + `square`)

= 3.6 × (3.6 + 6.4)

= 3.6 × `square`

= 36

∴ PS = `square` .....(by taking square roots)

Chapter: [0.03] Circle

If sec θ = `25/7`, find the value of tan θ.

**Solution:**

1 + tan^{2} θ = sec^{2} θ

∴ 1 + tan^{2} θ = `(25/7)^square`

∴ tan^{2} θ = `625/49 - square`

= `(625 - 49)/49`

= `square/49`

∴ tan θ = `square/7` ........(by taking square roots)

Chapter: [0.06] Trigonometry

In the figure given above, O is the centre of the circle. Using given information complete the following table:

Type of arc |
Name of the arc |
Measure of the arc |

Minor arc | `square` | `square` |

Major arc | `square` | `square` |

Chapter: [0.03] Circle

In ΔPQR, NM || RQ. If PM = 15, MQ = 10, NR = 8, then find PN.

Chapter: [0.01] Similarity

In ΔMNP, ∠MNP = 90˚, seg NQ ⊥ seg MP, MQ = 9, QP = 4, find NQ.

Chapter: [0.02] Pythagoras Theorem

In the given figure, M is the centre of the circle and seg KL is a tangent segment. L is a point of contact. If MK = 12, KL = `6sqrt3`, then find the radius of the circle.

Chapter: [0.03] Circle

Find the coordinates of midpoint of the segment joining the points (22, 20) and (0, 16).

Chapter: [0.04] Co-ordinate Geometry

A person is standing at a distance of 80 m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.

Chapter: [0.06] Trigonometry

In the given figure, X is any point in the interior of the triangle. Point X is joined to the vertices of triangle. seg PQ || seg DE, seg QR || seg EF. Complete the activity and prove that seg PR || seg DF.

**Proof:**

In ΔXDE, PQ || DE …(Given)

∴ `"XP"/"PD" = square/"QE"` …(Basic proportionality theorem)…(i)

In ΔXEF, QR || EF …(Given)

∴ `"XQ"/square = "XR"/square` ..........(`square`)....(ii)

∴ `"XP"/"PD" = square/square` ..…[From (i) and (ii)]

∴ seg PR || seg DF …(By converse of basic proportionality theorem

Chapter: [0.01] Similarity

If A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of `square`ABCD, show that `square`ABCD is a parallelogram.

**Solution:**

Slope of line = `("y"_2 - "y"_1)/("x"_2 - "x"_1)`

∴ Slope of line AB = `(2 - 1)/(8 - 6) = square` .......(i)

∴ Slope of line BC = `(4 - 2)/(9 - 8) = square` .....(ii)

∴ Slope of line CD = `(3 - 4)/(7 - 9) = square` .....(iii)

∴ Slope of line DA = `(3 - 1)/(7 - 6) = square` .....(iv)

∴ Slope of line AB = `square` ......…[From (i) and (iii)]

∴ line AB || line CD

∴ Slope of line BC = `square` …[From (ii) and (iv)]

∴ line BC || line DA

Both the pairs of opposite sides of the quadrilateral are parallel.

∴ `square`ABCD is a parallelogram.

Chapter: [0.04] Co-ordinate Geometry

In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.

Chapter: [0.02] Pythagoras Theorem

**Prove the following theorems:** Tangent segments drawn from an external point to the circle are congruent

Chapter: [0.03] Circle

Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.

Chapter: [0.05] Geometric Constructions

A metal cuboid of measures 16 cm × 11 cm × 10 cm was melted to make coins. How many coins were made, if the thickness and diameter of each coin was 2 mm and 2 cm respectively? (π = 3.14)

Chapter: [0.07] Mensuration

In ΔABC, PQ is a line segment intersecting AB at P and AC at Q such that seg PQ || seg BC. If PQ divides ΔABC into two equal parts having equal areas, find `"BP"/"AB"`.

Chapter: [0.01] Similarity

Draw a circle of radius 2.7 cm and draw a chord PQ of length 4.5 cm. Draw tangents at points P and Q without using centre.

Chapter: [0.03] Circle

In the given figure `square`ABCD is a square of side 50 m. Points P, Q, R, S are midpoints of side AB, side BC, side CD, side AD respectively. Find area of shaded region

Chapter: [0.07] Mensuration

Circles with centres A, B and C touch each other externally. If AB = 3 cm, BC = 3 cm, CA = 4 cm, then find the radii of each circle.

Chapter: [0.03] Circle

If sin θ + sin^{2} θ = 1 show that: cos^{2} θ + cos^{4} θ = 1

Chapter: [0.06] Trigonometry

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