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)
Equation of median AD is
(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.
