Download HOTS Level 1 And Level2 MATHS QUESTIONS

HOTS Level 1 And Level2
2 Mark Questions
Q1 In the given figure PQ,PR and AB are tangents at points Q,R and S respectively of a circle.
If PQ =8 cm .Find the Perimeter of triangle
P
A
Q
B
RRr
S
R
Sol. AQ=AS
BR=BS
PQ=PR=8cm
Perimeter of ∆ APB =AP+AB+PB
= PQ-AQ+AS+BS+PR-BR
=PQ+PR
=8+8=16cm
Q2. PT is a Tangent to circle with centre O. OT=56cm,TP=90 cm Find OP
Sol. A Tangnt to the Circle is perpendicularto the radius at the point of contact
So, OT
TP
Implies ∆OTP is a rt angle ∆
Therefore Op2=OT2+ TP2
562+902= 3136+8100
=11236 = √11236
OP=106cm
Q3. L and M are two Tangents to a circle. How could the centre of Circle is Ascertained
L
M
Sol. Draw the perpendiculars on the point of contact of both Tangents .Let these
perpendiculars meet at a point O then O will be the Centre of Circle
3 Mark Questions
Q1.Theorem: The tangent at any point on a circle is perpendicular to the radius drawn to
the point of contact.
Given: -A tangent AB with point of contact P.
To prove: OP ⊥ AB
Proof: Consider point C on AB other than P.
C must lie outside the circle. (∵A tangent can have only one point of contact with the
circle)
OC > OP (∵ C lies outside the circle)
This is true for all positions of C on AB.
Thus, OP is the shortest distance between point P and line segment AB.
Hence, OP ⊥ AB.
Q2. Theorem :Tangents drawn to a circle from an external point are equal in length.
Given:- Two tangents AB and AC from an external point A to points B and C on a circle.
To prove: AB = AC
Construction: Join OA, OB and OC.
Proof:In triangles OAB and OAC,
∠OBA = 90⁰ (Radius OB ⊥ Tangent AB at B)
∠OCA = 90⁰ (Radius OC ⊥ Tangent AC at C)
In triangles OBA and OCA,
∠OBA = ∠OCA = 90⁰
OB = OC (Radii of the same circle)
OA = OA (Common side)
C (RHS congr ence r le)
Hence, AB = AC (By C.P.C.T)
Q3. In figure 5, the common tangents AB and CD to two circles with centres O and O’ intersect in E.
Prove that the points O , E and O’ are collinear.
In figure 5, the common tangents AB and CD to two circles with centres O and O’ intersect in E. Prove
that the points O , E and O’ are collinear.
Sol. By the property of tangents drawn to a circle from an external point, we have
∠1= ∠2
……………….(i)
∠3= ∠4
……………….(ii)
Also ∠AED = ∠CEB (Vertically opposite angles) …………(iii)
Adding (i),(ii) and (iii), we get
∠1+ ∠3+ ∠AED= ∠2+ ∠4+ ∠CEB
But (∠1+ ∠3+ ∠AED) + (∠2+ ∠4+ ∠CEB) = 360° , (angles at a point)
∴ we must have
∠1+ ∠3+ ∠AED= ½ (360°) = 180°
⇒ EO and EO’ are collinear
⇒ O, E and O’ lie in the same line
Q4.From the given fig. A Circle touches all four sides of Quadrilateral ABCD. Prove that
AB+CD=BC+DA
A
S
P
D
R
Q
C
Sol. From the fig. AS=AP, SD=DR , PB=BQ, CR=CQ(Tangents)
AS+SD+BQ+CQ=AP+PB+CR+DR
AD+BC=AB+CD
Q5. Prove that the ||gm circumscribing a circle is a rhombu
A
M
B
Q
D
P
C
Sol. Given: ABCD ||gm touching the circle at M,N,P,Q
To prove: ABCD is a rhombus
Proof: AQ=AM
DQ=DP
BN=MB
NC=PC
Adding the above we get
AD+BC=AB+CD
AD=BC and AB=CD
AD=AB=BC=CD
It is a rhombus
Hot questions for bright students
Area related to circles
2 marks Questions
Q.1.
A race track is in the form of a ring whose circumference is 352m, and the outer
circumference is 396m. Find the width of the track?
oR
Rr
Sol.
Let the outer and inner radii of the ring be R and r metres respectively .
Then
2R = 396 and 2r = 352
 2 x 22/7 x R = 396 and 2 x 22/7 x r = 352
 R= 396 x 7/22 x 1/2 and r= 352 x t/22 x 1/2
 R = 63m and r = 56 m
Hence , width of the track
= (R – r ) metres,
=(63 – 56 ) metres
Q.2.
A wheel has diameter 84cm. Find how many complete revolution must it take to cover
792 meters.
Sol.
Let r be the radius of the wheel.
Then , Diameter = 84 cm
 2r = 84
 r = 42cm
 Circumference of the wheel = 2r cm
= 2 x 22/7 x 42 cm
= 264 cm
= 2.64m
So, the wheel comes 2.64 metres in one complete revolutions.
 Total number of revolutions in covering 792 metres = 792/2.64 = 300
Hence , the wheel takes 300 revolution in covering 792 metres.
Q.3.
A chord AB of a circle of radius 15cm makes and angle of 600 at the center of the circle .
Find the area of the minor segment.
(take =3.14) 3=1.73)
Sol.
We know that area of a minor segment of angle  in a circle of radius r is given by:
A={
} r2
A={
A = {
} (15)2
} x 225 cm2
O
 A = (0 . 5233 – 0.4330) 225 cm2
15cm
= 225 x 0.902 cm2
= 20.295 cm2
A
B
3 marks Questions
Q1.
In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is
drawn with BC as diameter. Find the area of the shaded region.
(Answer = 98 cm2)
Solution: BC2=AB2+AC2
=142+142
BC= 14 2 cm.
Area of shaded portion = (Area of semi circle BC as diameter ) – [Area of quadrant with AB as
radius – Area of triangle ABC]
=
= 154-154+98
=98 cm2
Q2.
The area of an equilateral triangle is 17320.5 cm2. Taking each angular point as center a
circle is described with radius equal to half the length of the side of triangle as shown in
figure. Find the area of triangle not included in circle.( 3  1.73205 )
Solution: AREA =1732.05cm2
3 /4 a2 =1732.05
a2/4= 1732.05/ 3
A=3{60/360x3.14x(a2/2)} cm2
A=1570.05cm2
Required Area =Area of triangle-Area of 3 sectors each of angle 600= 162.01cm2
Q3.
Find the area of the shaded region in the figure given here, if BC = BD = 8 cm, AC = AD
= 15 cm, and O is the centre of the circle.
Solution: AB=
( AC 2  BC 2 )
Radius= 17/2 cm.
Area of Shaded Region =
r2 – 2xArea of triangle
= 3.14 x17/2x17/2- 2x1/2x8x15
= 106.87 cm2
Q4.
Find the area of the region shown shaded in the given figure.
Solution: Area of 1 semicircle =
1
2
r2
1 22 2
= x
x2
2 7
= 44/7 cm2
Area of 4 semi circles= 4x44/7 =176/7 cm2
Area of Small Square =4x4 =16 cm2
Area of Shaded Region = 14x14 –[16+176/7]
= 196-[288/7]
=196-41.1
= 154.9 cm2
Q5.
In the given figure, find the area of the shaded design, where ABCD is a square of side 1cm and semi-circle are drawn with each side of the square as diameter (use   3.14 )
Solution
A
I
B
Let us mark the four unshaded regions as I,II,
III, and IV
IV
D
II
III
C
Area of I + Area of III = Area of ABCD – Area of two semicircles of each of radius 5 cm.
1
= (10x10-2x x x52) cm2 = (100 – 3.14x25) cm2
2
=(100 – 78.5) cm2 = 21.5 cm2
Similarly, Area of II + Area of IV = 21.5 cm2
So, Area of shaded design = Area of ABCD – Area of (I+II+III+IV)
= (100-2x21.5) cm2 =(100-43) cm2 = 57 cm2.
HOTS QUESTIONS
LEVEL - 1
1.
A vessel is in the form of hollow hemisphere mounted by a hollow cylinder .The
diameter of
hemisphere is 14 cm and total height of vessel is 13 cm. Find the inner surface area of
vessel.
Sol. Here the radius of hemisphere = radius of cylinder = r cm =7 cm
And height of cylinder, h= 13-7=6 cm
Now, inner S.A of vessel = curved surface area of cylindrical part + curved surface area
of
cylindrical part
= (2
2
)=2
=2x
x 7 x (6 +7)
=2x22x13 =572cm2
2.
The largest possible sphere is curved out of a
wooden solid cube of side 7 cm. Find the volume of wood left.
Sol. Diameter of sphere carved out = side of cube = 7 cm
R= 3.5 cm
3
3
3
Volume of cube = a =7 =343 cm
Volume of sphere carved out = 4/3
=179.66 cm3
Volume of wood left = 343-179.66
3
=163.34 cm3
3. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of
uniform thickness. Find the thickness of wire.
Sol.
The volume of the rod = x(1/2)2x 8 cm3= 2 cm3
The length of new wire of the same volume= 18 m=1800 cm
If r is the radius of cross-section of wire its volume= xr2x1800
xr2x1800=2
R=1/30
The thickness of wire of cross section is 1/15 cm.
=0.67 mm
LEVEL - 2
1. A right triangle, whose side are 15cm and 20cm (other than hypotenuse) is made to revolve
about
it hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value
of pie as found appropriate).
Sol.
Let ABC be the right angled triangle such that AB=15cm and AC=20 cm.
Using Pythagoras theorem, we have
BC2=AB2+AC2
BC2 =152+202
BC2 =225 + 400 = 625
BC=25cm
Let OB=x and OA=y
Applying Pythagoras theorem in triangles OAB and OAC, we have
AB2=OB2 +OA2 and AC2=OA2 +OC2
152 =x2+y2 and 202=y2+ (25-x2)
x2+y2= 225 and (25-x) 2 +y2= 400
{(25-x) 2+y2} – { x2+y2} = 400 – 225
(25-x) 2 -x2= 175
(25-x-x) (25-x+x)= 175
(25-2x) X 25= 175
25-2x=7
2x=18
X=9
Putting x=9 in x2+y2=225, we get
81+y2=225
y2=144
y=12
Thus, we have OA=12cm and OB=9cm
Volume of double cone = vol. of cone CAA’+ vol. of cone BAA’
=
(OA2) × OC +
(OA2) × OB
=
× 122 × 16+
=
× 144 (16+9) =
× 122 × 9
× 3.14 × 144 × 25cm3
=3768 cm3
Surface area of double cone = curved surface area of cone CAA’ +curved surface area
of cone
BAA’
=
× OA × AC +
=
× 12 × 20 +
=420
× OA × AC
× 12 × 25
cm3
=420 × 3.14 = 1318.8 cm2
2.
Sol.
A cistern, internally measuring 150cm X 120cm X 110cm, has 129600cm3 of water in it.
Porous bricks are placed in the water until the cistern is full to the brim. Each brick
absorbs one-seventeenth of its own volume of water. How many bricks can be put in
without overflowing the water, each brick being 22.5cm X 7.5cm X 6.5cm?
We have
Volume of cistern =150 X 120 X 110 = 1980000cm3
Volume of water in cistern = 129600cm3
Volume of one brick = 22.5 x 7.5 x 6.5 = 1096.875cm3
Volume of water absorb by one brick =
X 1096.875 cm3
Let n be the total no, of brick which can be put in the cistern without water overflowing
then,
Volume of water absorbed by n bricks = n x
Volume of water left in cistern = (129600-
X 1096.875 cm3
X 1096.875) cm3
Since the cistern is filled up to the brim
Therefore volume of water left in cistern + volume of brick = volume of cistern
= 129600n x 1096.875-
X 1096.875 = 1980000 – 129600
1096.875 x (n -
) = 1850400
1096.875 x
17550 x
n=
3.
X 1096.875 + n x 1096.875 = 1980000
= 1850400
= 1850400
= 1792.41 is congruent =1792
The height of a right circular cone is trisected by two planes drawn parallel to the base.
Show that the volumes of the three portions from the top are in ratio 1:7:19.
Sol. Let VAB a right circular cone of height 3h and base radius r.
This cone is cut by planes parallel to its base at points O- and L, such that VL = LO-=h
Since triangles VOA and VO-A- are similar
V
=
=
D
C
L
r1 =
Also,
B’
A’
O’
=
r
A
0
O
=
r2 =
Let v1 be the volume of cone VCD
r22h =
Then, v1 =
2
r2h
h=
Let v2 be t e vol me of fr t m ’ ’DC.
V2 =
(r21+ r22+ r1r2)h
V2 =
(
+
2
V2=
+
)h
h
Let V3 be t e vol me of fr t m
V3 =
(r2+ r21+ r1r)h
V3 =
(r2+
V3=
2
+
’ ’.
en
)h
h
Required ratio = V1:V2:V3=
4.
en
r2h:
2
h:
2
h=1:7:19.
The radius of the base of a right circular cone is r. It is cut by a plane parallel to the base at the
height h from the base. The distance of the boundary of the upper surface from the centre of
the
frustum is
. Show that the volume of the frustum is
2
h.
B
Sol We have,
O’
r
’2
and
’=
A’
.
B’
In
’2
’2+ ’ ’2
h
h2+
=h2 + ’ ’2
’ ’
A
2
Therefore Volume of frustum =
Volume of frustum =
5.
{r2+(
r
2
{r +( ) +r x }h
)+
}h=
2
B
O
h.
Water in a canal, 6m wide and 1.5m deep , is flowing with a speed of 10km/h. How much Area
will
it irrigate in 30 min if 8 cm of standing water is needed?
Solution: We have, width of the canal =6m
Depth of the canal =1.5m
Now, length of water column per hour =10km
Therefore length of water column in half hour = 5km= 5000m
Therefore volume of water flow in 30min= 1.5x6x5000 = 45000m3
Here, standing water needed in 8cm = 0.08m
Therefore area irrigated in 30 min =
=
2
=562500 m (1 hectare= 10000 m2)
=56.25 hectare
HOTS ON PROBABILITY
(2 marks questions)
1. Two customers Shyam and Ekta are visiting a particular shop in the same week
(Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another
day. What is the probability that both will visit the shop on (i) the same day? (ii)
Consecutive days?
Solution: The total number of days is 5 and hence both of them can reach the shop in 5
ways.
Hence, total number of outcomes = 5 x 5 = 25
They can reach on the same day in 5 ways, i.e. (Tue, Tue), (Wed ,Wed), (Thur,Thur),
(Fri ,Fri) and (Sat ,Sat)
(i)P(Reaching on same day)
(ii) They can reach on consecutive days in following 8 ways: (tue ,wed), (wed, tue),
(wed ,thur), (thur wed), (thurfri), (frithu), (fri,sat), (sat ,fri)
P(Reaching on consecutive days)
2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is
thrown two times and the total score in two throws is noted. Complete the following
table which gives a few values of the total score on the two throws:
What is the probability that total score is (i) even? (ii) 6?
Solution:
Following table shows the sample space:
+ 1 2 2 3 3 6
1 2 3 3 4 4 7
2 3 4 4 5 5 8
2 3 4 4 5 5 8
3 4 5 5 6 6 9
3 4 5 5 6 6 9
6 7 8 8 9 9 12
Total number of outcomes = 36
(i)Number of even score = 18
Hence, P (even score)
(ii)Number of times 6 comes = 4
Hence, P (score of 6)
3. A game consists of tossing a one rupee coin 3 times and noting its outcome each
time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and
loses otherwise. Calculate the probability that Hanif will lose the game.
Solution: Possible outcomes of 3 tosses of coin
1st = HHH
2nd = HHT or HTH or THH
3rd = TTH or THT or HTT
4th = TTT
Total number of events = 8
Number of favorable events = 6 (refer to 2nd and 3rd case)
(3 marks questions)
1.Find the probability of having 53 Sundays in
(i) a leap year
2 1
,
(ii) a non leap year (Ans: 7 7 )
Ans:
An ordinary year has 365 days i.e. 52 weeks and 1 day
This day can be any one of the 7 days of the week.
1
 P(that this day is Sunday) = 7
1
Hence, P(an ordinary year has 53 Sunday) = 7
A leap year 366 days i.e. 52 weeks and 2 days
This day can be any one of the 7 days of the week
2
 P (that this day is Sunday) = 7
2
Hence, P(a leap year has 53 Sunday) =
7
2. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue
ball is double that of a red ball , determine the number of blue balls in the bag.
(Ans:10)
Ans: Let the number of blue balls is the bag be x
Then total number of balls is the bag = 5 + x
 Number of all possible outcomes = 5 + x
Number of outcomes favourable to the event of drawing a blue ball = x
( there are x blue balls)
 Probability of drawing a blue ball
x
5 x
5
Similarly, probability of drawing a red ball = 5  x
According to the answer
x
5
5 x = 2 ( 5 x )
x = 10
3. A number x is selected from the numbers 1,2,3 and then a second number y is
randomly selected from the numbers 1,4,9. What is the probability that the product xy of
the two numbers will be less than 9?
Ans:Numberx can be selected in three ways and corresponding to each such
way there are three ways of selecting number y. Therefore, two numbers can be
selected in 9 ways as listed below: (1,1), (1,4), (2,1), (2,4), (3,1)
Favorable number of elementary events = 5
Hence, required probability = 5/9
4. Suppose you drop a die at random on the rectangular region shown below. What is
the probability that it will land inside the circle with diameter 1m?
Solution: Area of rectangle = 6 sq m
Area of circle
Area of rectangle gives the total number of events and area of circle gives the number
of favourable events.
5. A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
Solution: Possible outcomes of 2 throws of a die can be shown by following table:
1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3, 1 3, 2 3, 3 3, 4 3, 5 3, 6
4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
Total number of events = 36
Number of times when 5 does not come up either of times = 25
(ii) 5 will come up at least once?
Solution: Number of times 5 comes at least once = 11