SSC BOARD PAPERS IMPORTANT TOPICS COVERED FOR BOARD EXAM 2024

### March 2016 Maths Board paper Sum No 26

PART – A

1. Solve any five sub-questions: [5]

(i) ∆DEF ~ ∆MNK. If DE = 2, MN = 5, then find the value of A(∆ DEF)/ A(∆ MNK).

Solution:

Given,

∆DEF ~ ∆MNK

DE = 2, MN = 5

We know that the ratio of areas of similar triangles is equal to squares of the ratio of their corresponding sides.

A(ΔDEF)/ A(ΔMNK) = DE2/MN2

= (2)2/(5)2

= 4/25

(ii) In the following figure, in ∆ABC, ∠B = 90°, ∠C = 60°, ∠A = 30°, AC = 16 cm. Find BC.

Solution:

Given,

ΔABC is a 30° – 60° – 90° triangle.

Side opposite to 30° = BC

BC = (1/2) × Hypotenuse

= (1/2) × AC

= (1/2) × 16

= 8 cm

Therefore, BC = 8 cm

(iii) In the following figure, m(arc PMQ) = 110°, find ∠PQS.

Solution:

Given,

m(arc PMQ) = 110°

We know that the measure of the angle formed by the intersection of a chord and tangent of the circle equal to the half the angle made by the arc with the chord.

∠PQS = (½) × m(arc PMQ)

= (1/2) × 110°

= 55°

(iv) If the angle θ = -30°, find the value of cos θ.

Solution:

Given,

θ = -30°

We know that,

cos (-θ) = cos θ

cos θ = cos (-30°)

= cos 30°

= √3/2

(v) Find the slope of the line with inclination 60°.

Solution:

Given,

Inclination of line = θ = 60°

The slope of the line = tan θ

= tan 60°

= √3

(vi) Using Euler’s formula, find V if E = 10, F = 6.

Solution:

Given,

E = 10, F = 6

Using Euler’s formula,

F + V = E + 2

6 + V = 10 + 2

V = 12 – 6

V = 6

2. Solve any four sub-questions: [8]

(i) In the following figure, in ∆PQR, seg RS is the bisector of ∠PRQ. If PS = 9, SQ = 6, PR = 18, find QR.

Solution:

Given that, in triangle PQR, seg RS is the bisector of ∠PRQ.

PS = 9, SQ = 6, PR = 18

By the angle bisector property,

PR/QR = PS/SQ

18/QR = 9/6

⇒ QR = (18 × 6)/9

⇒ QR = 12

(ii) In the following figure, a tangent segment PA touching a circle in A and a secant PBC are shown. If AP = 12, BP = 9, find BC.

Solution:

Given,

PA = tangent

PBC = secant

By the tangent – secant theorem,

PB × PC = PA2

9 × PC = (12)2 [given AP = 12, BP = 9]

PC = 144/9

PC = 16

Now,

PC = PB + BC

16 = 9 + BC

⇒ BC = 16 – 9

⇒ BC = 7

(iii) Draw an equilateral ∆ ABC with side 6.4 cm and construct its circumcircle.

Solution:

(iv) For the angle in standard position if the initial arm rotates 130° in an anticlockwise direction, then state the quadrant in which the terminal arm lies. (Draw the Figure and write the answer.)

Solution:

Given,

The initial arm rotates 130° in the anticlockwise direction from the standard position.

The measure of angle 130° lies between 90° and 180°.

Hence, the terminal arm lies in quadrant II.

(v) Find the area of the sector whose arc length and radius are 16 cm and 9 cm, respectively.

Solution:

Given,

Length of the arc = 16 cm

Radius = r = 9 cm

Area of sector = (r/2) × Length of arc

= (9/2) × 16

= 9 × 8

= 72 cm2

(vi) Find the surface area of a sphere of radius 1.4 cm. (π = 22/7)

Solution:

Given,

Radius of the sphere = r = 1.4 cm

The surface area of the sphere = 4πr2

= 4 × (22/7) × 1.4 × 1.4

= 24.64 cm2

Therefore, the surface area of the sphere is 24.64 cm2.

3. Solve any three sub-questions: [9]

(i) Adjacent sides of a parallelogram are 11 cm and 17 cm. If the length of one of its diagonals is 26 cm, find the length of the other.

Solution:

Given,

Adjacent sides of a parallelogram are 11 cm and 17 cm.

The length of one diagonal is 26 cm.

AB = CD = 17 cm

BC = AD = 11 cm

BD = 26 cm

We know that,

Sum of squares of sides of a parallelogram = Sum of squares of its diagonals

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

(17)2 + (11)2 + (17)2 + (11)2 = AC2 + (26)2

289 + 121 + 289 + 121 = AC2 + 676

AC2 = 820 – 676

AC2 = 144

AC = 12 cm

Therefore, the length of the other diagonal is 12 cm.

(ii) In the following figure, secants containing chords RS and PQ of a circle intersect each other in point A in the exterior of a circle. If m(arc PCR) = 26°, m(arc QDS) = 48°, then find:

a. m ∠PQR

b. m ∠SPQ

c. m ∠RAQ

Solution:

Given,

m(arc PCR) = 26°

m(arc QDS) = 48°

By the inscribed angle theorem,

a. ∠PQR = 1/2 × m(arc PCR)

= (1/2) × 26°

= 13°

∠PQR = ∠AQR = 13….(i)

b. ∠SPQ = 1/2 × m(arc QDS)

= (1/2) × 48°

= 24°

∠SPQ = 24°

c. In triangle AQR,

∠RAQ + ∠AQR = ∠SRQ (remote interior angle theorem)

∠SRQ = ∠SPQ (angles subtended by the same arc)

Therefore,

∠RAQ + ∠AQR = ∠SPQ

m∠RAQ = m∠SPQ – m∠AQR

= 24° – 13° [From (i) and (ii)]

= 11°

m∠RAQ = 11°

(iii) Draw a circle of radius 3.5 cm. Take any point K on it. Draw a tangent to the circle at K without using the centre of the circle.

Solution:

Therefore, XKX’ is the required tangent to the circle with a radius of 3.5 cm.

(iv) If sec α = 2/√3, then find the value of (1 – cosec α)/ (1 + cosec α), where α is in IV quadrant.

Solution:

Given,

sec α = 2/√3

Thus, sec α = r/x = 2/√3

Let r = 2k and x = √3k

We know that,

r2 = x2 + y2

(2k)2 = (√3k)2 + y2

y2 = 4k2 – 3k2

y2 = k2

y = ±k

Given that α lies in quadrant IV.

Therefore, y = -k

cosec α = r/y = 2k/-k = -2

Now,

(1 – cosec α)/ (1 + cosec α)

= [1 – (-2)]/ [1 + (-2)]

= (1 + 2)/ (1 – 2)

= -3

(v) Write the equation of the line passing through the pair of points (2, 3) and (4, 7) in the form of y = mx + c.

Solution:

Let the given points be:

A(2, 3) = (x1, y1)

B(4, 7) = (x2, y2)

Equation of the line passing through the points (x1, y1) and (x2, y2) is:

(x – x1)/ (x2 – x1) = (y – y1)/ (y2 – y1)

(x – 2)/ (4 – 2) = (y – 3)/ (7 – 3)

(x – 2)/ 2 = (y – 3)/ 4

4(x – 2) = 2(y – 3)

4x – 8 = 2y – 6

4x – 8 – 2y + 6 = 0

4x – 2y – 2 = 0

2(2x – y – 1) = 0

2x – y – 1 = 0

y = 2x – 1

This is of the form y = mx + c [m = 2, c = -1]

Hence, the required equation of line is y = 2x – 1.

4. Solve any two sub-questions: [8]

(i) Prove that “The length of the two tangent segments to a circle drawn from an external point are equal”.

Solution:

Given,

PQ and PR are the tangents to the circle with centre O from an external point P.

To prove: PQ = PR

Construction:

Join OQ, OR and OP.

Proof:

We know that the radius is perpendicular to the tangent through the point of contact.

∠OQP = ∠ORP = 90°

In right ΔOQP and ORP,

OQ = OR (radii of the same circle)

OP = OP (common)

By RHS congruence criterion,

ΔOQP ≅ ΔORP

By CPCT,

PQ = PR

Hence proved.

(ii) A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and width of the river. (√3 = 1.73)

Solution:

Let AB be the tree and BC be the width of the river.

CD = 40 m

BC = x and AB = h

In right triangle ABC,

tan 60° = AB/BC

√3 = h/x

h = √3x….(i)

In right triangle ABD,

tan 30° = AB/BD

1/√3 = h/(x + 40)

x + 40 = h√3

x + 40 = (√3x)√3 [From (i)]

x + 40 = 3x

2x = 40

x = 20 m

Substituting x = 20 in (i),

h = 20√3

= 20 × 1.73

= 34.6 m

Therefore, the height of the tree is 34.6 m and the width of the river is 20 m.

(iii) A(5, 4), B(-3, -2) and C(1, -8) are the vertices of a triangle ABC. Find the equations of median AD and the line parallel to AC passing through point B.

Solution:

Given,

Vertices of a triangle ABC are A(5, 4), B(-3, -2) and C(1,-8).

A(5, 4) = (x1, y1)

B(-3, -2) = (x2, y2)

C(1, -8) =(x3, y3)

Let D(x, y) be the median of triangle ABC.

D is the midpoint of BC.

D(x, y) = [(x2 + x3)/2, (y2 + y3)/2]

= [(-3 + 1)/2, (-2 – 8)/2]

= (-2/2, -10/2)

= (-1, -5)

D(-1, -5) = (x4, y4)

(x – x1)/ (x4 – x1) = (y – y1)/(y4 – y1)

(x – 5)/ (-1 – 5) = (y – 4)/ (-5 – 4)

(x – 5)/(-6) = (y – 4)(-9)

-9(x – 5) = -6(y – 4)

-9x + 45 = -6y + 24

9x – 45 – 6y + 24 = 0

9x – 6y – 21 = 0

3(3x – 2y – 7) = 0

3x – 2y – 7 = 0

Hence, the required equation of median AD is 3x – 2y – 7 = 0.

We know that the line parallel to AC = Slope of AC

Slope of AB = (y3 – y1)/ (x3 – x1)

= (-8 – 4)/ (1 – 5)

= -12/-4

= 3

Thus, m = 3

Equation of the line parallel to AC and passing through the point B(-3, -2) is

y – y2 = m(x – x2)

y – (-2) = 3[x – (-3)]

y + 2 = 3x + 9

3x + 9 – y – 2 = 0

3x – y + 7 = 0

5. Solve any two sub-questions: [10]

(i) In the following figure, AE = EF = AF = BE = CF = a, AT ⊥ BC. Show that AB = AC = √3 × a

Solution:

Given,

AE = EF = AF = BE = CF

AT ⊥ EF

ΔAEF is an equilateral triangle.

ET = TF = a/2

BT = CT = a + (a/2)….(i)

In right triangle ATB and ATC,

AT = AT (common)

∠ATB = ∠ATC (right angles)

BT = CT [From (i)]

By SAS congruence criterion,

ΔATB ≅ ΔATC

BY CPCT,

AB = AC

In triangle AEF,

AE = AF = EF (given)

ΔAEF is an equilateral triangle.

Altitude of an equilateral triangle = AT = (√3/2)a

In right triangle ATB,

AB2 = AT2 + BT2

= [(√3/2)a]2 + [a + (a/2)]2

= (3a2/4) + (3a/2)2

= (3a2/4) + (9a2/4)

= 12a2/4

= 3a2

AB = √3a

Therefore, AB = AC = √3 × a

(ii) ∆ SHR ~ ∆ SVU. In ∆ SHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and SH/SV = 3/5.

Construct ∆ SVU.

Solution:

(iii) Water flows at the rate of 15 m per minute through a cylindrical pipe, having the diameter 20 mm. How much time will it take to fill a conical vessel of base diameter 40 cm and depth 45 cm?

Solution:

Given,

Diameter of cylindrical pipe = 20 mm

Radius of the cylindrical pipe = r = 20/2 = 10 mm = 1 cm

Speed of water = h = 15 m per minute = 1500 cm/min

Volume of cylindrical pipe = Volume of water flowing per minute

= πr2h

= (22/7) × 1 × 1 × 1500

= 33000/7 cm3

Also,

Diameter of conical vessel = 40 cm

Radius of conical vessel = R = 40/2 = 20 cm

Depth = H = 45

Capacity of the conical vessel = Volume of cone

= (1/3) πR2H

= (1/3) × (22/7) × 20 × 20 × 45

= 396000/21 cm3

Time required to fill the vessel = Volume of conical vessel/ Volume of water flowing per minute

= (396000/21)/ (33000/7)

= (396000 × 7)/ (21 × 33000)

= 4

Hence, the time required to fill the conical vessel is 4 min.