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#### Question 1:

If the sum of the roots of the equation $3{x}^{2}-\left(3k-2\right)x-\left(k-6\right)=0$ is equal to the product of its roots, then k = ?

(a) 1
(b) −1
(c) 0
(d) 2

#### Answer:

(d) 2

Let the roots of the equation be $\alpha \mathrm{and}\beta$.
Then,
$\alpha +\beta =\frac{-\left[-\left(3k-2\right)\right]}{3}=\frac{3k-2}{3}$ and $\alpha \beta =\frac{-\left(k-6\right)}{3}=\frac{-k+6}{3}$
Therefore,
$\frac{3k-2}{3}=\frac{-k+6}{3}\phantom{\rule{0ex}{0ex}}$
$⇒3k-2=-k+6$

#### Question 2:

The number of all-digit numbers divisible by 6 is

(a) 12
(b) 15
(c) 16
(d) 18

#### Answer:

(b) 15
All 2 − digit numbers divisible by 6, are
12, 18, 24, 30........96
This is an AP, in which a = 12, d = (18 − 12) = 6 and  l = 96
Let the number of these terms be n.
Then, Tn = 96
a + (n − 1)d = 96
⇒ 12 + (n − 1) × 6 = 96
⇒ (n − 1) × 6 = 84
⇒ (n − 1) = 14 ⇒ n = 15

#### Question 3:

A fair die is thrown once. The probability of getting a composite number is

(a) $\frac{1}{3}$
(b) $\frac{1}{6}$
(c) $\frac{2}{3}$
(d) 0

#### Answer:

(a) $\frac{1}{3}$
In a single throw of a die, the possible outcomes are
1 , 2, 3, 4, 5 and 6
The total number of possible outcomes = 6
Let E be the event of getting a composite number.
Then, the favourable outcomes are 4 and 6.
Number of favourable outcomes = 2
∴ P(getting a composite number) = P(E) = $\frac{2}{6}=\frac{1}{3}$

#### Question 4:

Which of the following statements is true?

(a) The tangents drawn at the end points of a chord of a circle are parallel.
(b) From a point P on the exterior of a circle, only two secants can be drawn through P to the circle.
(c) From a point P in the plane of a circle, two tangents can be drawn to the circle.
(d) The perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

#### Answer:

(d) The perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

Let O be the centre of the given circle.

A tangent PR has been drawn touching the circle at point P.
Draw QP ⊥ RP at point P, such that point Q lies on the circle. $\angle$OPR = 90° (Since radius ⊥ tangent)
Also, $\angle$QPR = 90° (Given)
$\angle$OPR = $\angle$QPR
Now, the above case is possible only when the centre O lies on the line QP.
Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

#### Question 5:

In the given figure, PA and PB are tangents to a circle, such that PA = 8 cm and ∠APB = 60°. The length of the chord AB is
Figure

(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 6 cm

#### Answer:

(a) 8 cm

Since the lengths of tangents drawn from a point to a circle are equal, PA = PB.
Therefore, $\angle$PAB = $\angle$PBA = x° (say)
Then, x° + x° + 60° = 180°
⇒2x° = 120°
x° = 60°
∴ Each angle of ΔPAB is 60°; therefore, it is an equilateral triangle.
AB = PA = PB = 8 cm

#### Question 6:

The angle of depression of an object from a 60-m-high tower is 30°. The distance of the object from the tower is

(a) $20\sqrt{3}$ m
(b) $60\sqrt{3}$ m
(c) $40\sqrt{3}$ m
(d) 120 m

#### Answer:

(b) $60\sqrt{3}$ m

Let the height of the tower be AB = 60 m.
​Let the angle of depression of the point C be 30°.

$\angle ACB=\angle PAC=30°\phantom{\rule{0ex}{0ex}}$
∴ In the right-angled ΔABC,
$\mathrm{tan}30°=\frac{AB}{BC}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{1}{\sqrt{3}}=\frac{60}{BC}\phantom{\rule{0ex}{0ex}}$
$⇒BC=60\sqrt{3}\mathrm{m}$
Hence, the distance of the object from the tower is $60\sqrt{3}$ m.

#### Question 7:

In what ratio does the point P(2, −5) divide the line segment joining A(−3, 5) and B(4, −9)?

(a) 3 : 2
(b) 2 : 1
(c) 5 : 2
(d) 5 : 3

#### Answer:

(c) 5 : 2

Let the required ratio be k : 1.
Then, by section formula, the co-ordinates of P are
$P\left(\frac{4k-3}{k+1},\frac{-9k+5}{k+1}\right)$
[ Since P(2, -5) is given]
⇒ 4k - 3 = 2k + 2 and  -9k + 5 = - 5k - 5
⇒ 2k = 5 and 4k = 10
$k=\frac{5}{2}$ in each case.
So, the required ratio is $\frac{5}{2}:1$, which is 5 : 2.

#### Question 8:

Three solid spheres of radii 6 cm, 8 cm and 10 cm are melted to form a sphere. The radius of the sphere so formed is

(a) 24 cm
(b) 16 cm
(c) 14 cm
(d) 12 cm

#### Answer:

(d) 12 cm

Radius (r1) of 1st sphere = 6 cm
Radius (r2) of 2nd sphere = 8 cm
Radius (r3) of 3rd sphere = 10 cm
Let the radius of the resulting sphere be r.
We know that the object formed by recasting these spheres will be same in volume as the sum of the volumes of these spheres. Volume of 3 spheres = Volume of resulting sphere
$\frac{4}{3}\mathrm{\pi }\left[{{r}_{1}}^{3}+{{r}_{2}}^{3}+{{r}_{3}}^{3}\right]=\frac{4}{3}{\mathrm{\pi r}}^{3}$
$⇒\frac{4}{3}\mathrm{\pi }\left[{6}^{3}+{8}^{3}+{10}^{3}\right]=\frac{4}{3}{\mathrm{\pi r}}^{3}\phantom{\rule{0ex}{0ex}}$
r3 = (216 + 512 + 1000) = 1728
r = 12 cm
Hence, the radius of the sphere so formed is 12 cm.

#### Question 9:

Find the value of p for which the quadratic equation ${x}^{2}-2px+1=0$ has no real roots.

#### Answer:

Given equation: x2 − 2px + 1 = 0
Here, a = 1, b  = − 2p , c = 1
For no real roots, b2 − 4ac < 0
b2 − 4ac = (−2p)2 − (4 ×1 × 1)
⇒ 4p2 − 4 < 0
⇒ 4(p2 − 1) < 0
⇒ (p2 − 1) < 0    [ Since 4 < 0, which is not possible]
⇒ −1 < p < 1

#### Question 10:

Find the 10th term from the end of the AP 4, 9, 14, ..., 254.
Or, which term of the AP 24, 21, 18, 15, ... is the first negative term?

#### Answer:

Here, a = 4, d = (9 - 4) = 5, l = 254 and n = 10
Now, nth term from the end = {l - (n - 1)d}
∴ 10th term from the end = {254 - (10 - 1) × 5}
= {254 - (9 × 5)} = (254 - 45) = 209
Hence, the 10th term from the end is 209.

OR
Here, a = 24, d = (21 - 24) = - 3
Let the nth term of the given AP be the first negative term.
Then, Tn < 0 ⇒ {a+ (n - 1)d} < 0
⇒ { 24 + (n - 1) × ( - 3) < 0
⇒ (27 - 3n) < 0
⇒ 27 < 3n
⇒ 3n > 27
⇒ n > 9
n = 10
Hence, the 10th term is the first negative term of the given AP.

#### Question 11:

A circle is touching the side BC of a ∆ABC at P and is touching AB and AC when produced to Q and R, respectively.
Prove that (perimeter of ∆ABC).
Figure

#### Answer:

We know that the lengths of tangents drawn from an external point to a circle are equal.

AQ = AR ..........(i)         [tangent from A]
BP = BQ ...........(ii)        [tangent from B]
CP =  CR ...........(iii)      [tangent from C]
Perimeter of ΔABC
= AB  + BC + AC
= AB + BP + CP + AC
= AB + BQ + CR + AC            [using (ii) and (iii)]
= AQ + AR
= 2AQ     [ using (i)]

#### Question 12:

Two vertices of ∆ABC are A(6, 4) and B(−2, 2) and its centroid is G(3, 4). Find the coordinates of its third vertex C.

#### Answer:

Two vertices of ΔABC are A(6, 4) and B(− 2, 2). Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
$G\left(\frac{6-2+a}{3},\frac{4+2+b}{3}\right)\phantom{\rule{0ex}{0ex}}$
$⇒G\left(\frac{4+a}{3},\frac{6+b}{3}\right)$
But it is given that the centroid is G(3,4).
$\frac{4+a}{3}=3$ and $\frac{6+b}{3}=4$
⇒ 4 + a = 9  and 6 + b = 12
a = 5 and b = 6
Hence, the third vertex of ΔABC is C(5, 6).

#### Question 13:

A box contains 150 oranges. If one orange is taken out from the box at random and the probability of its being rotten is 0.06, then find the number of good oranges in the box.

#### Answer:

Total number of oranges = 150
Probability of it being good =1 - 0.06 = 0.94
Let the number of good oranges be x.
P(getting good oranges) =
$0.94=\frac{x}{150}\phantom{\rule{0ex}{0ex}}$

Hence, the required number of good oranges in the box is 141.

#### Question 14:

A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.

#### Answer:

Given that the diameter of the common base of the cone and hemisphere is 7 cm.
∴ Radius of the cone (r ) = 7 cm
The total height of the toy is 31 cm.

∴
The height of the cone (h) = 31 - 7 = 24 cm
Surface area of the hemisphere = $2{\mathrm{\pi r}}^{2}$
= $\left(2×\frac{22}{7}×7×7\right)$ = 308 cm2
Slant height of the cone (l) = $\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{\left(7\right)}^{2}+{\left(24\right)}^{2}}=\sqrt{625}$  = 25 cm

∴  Curved surface area of the cone = $\mathrm{\pi rl}=\left(\frac{22}{7}×7×25\right)$ = 550 cm2
Hence,
Total surface area of the toy = Surface area of the hemisphere + Curved surface area of the cone
= (308 + 550) cm2= 858 cm2

#### Question 15:

Solve: ${a}^{2}{b}^{2}{x}^{2}-\left(4{b}^{4}-3{a}^{4}\right)x-12{a}^{2}{b}^{2}=0.$

#### Answer:

The given equation is:
a2b2x2 − (4b4 − 3a4)x − 12a2b2 = 0
Comparing it with Ax2 + Bx + C = 0, we get:
A = a2b2 , B = − (4b4 − 3a4) and C = − 12a2b2
D = (B2 − 4AC)
= (3a4 − 4b4 )2 − 4(a2b2)( − 12a2b2 )
= 9a8 + 16b8 − 24a4b4 + 48a4b4
= (3a4)2 + (4b4)2 + 2 × 3a4 × 4b4
= (3a4 + 4b4)2 ≥ 0
Hence, the given equation has real roots. These are
$\alpha =\frac{-B+\sqrt{D}}{2A}=\frac{\left(4{b}^{4}-3{a}^{4}\right)+\left(3{a}^{4}+4{b}^{4}\right)}{2{a}^{2}{b}^{2}}=\frac{8{b}^{4}}{2{a}^{2}{b}^{2}}=\frac{4{b}^{2}}{{a}^{2}}\phantom{\rule{0ex}{0ex}}$
$\beta =\frac{-B-\sqrt{D}}{2A}=\frac{\left(4{b}^{4}-3{a}^{4}\right)-\left(3{a}^{4}+4{b}^{4}\right)}{2{a}^{2}{b}^{2}}=\frac{-6{a}^{4}}{2{a}^{2}{b}^{2}}=\frac{-3{a}^{2}}{{b}^{2}}$
Hence, $\frac{4{b}^{2}}{{a}^{2}}\mathrm{and}\frac{-3{\mathrm{a}}^{2}}{{\mathrm{b}}^{2}}$ are the roots of the given equation.

#### Question 16:

If the 8th term of an AP is 31 and its 15th term is 16 more than the 11 the term, find AP.
Or, find the sum of all two-digit odd positive numbers.

#### Answer:

Here,
a8 = 31 and a15 = 16 + a11
⇒ 31 = a + 7d .............(i)
and a + 14d = 16 + a + 10d .............(ii)
By solving equation (ii), we get:
4d = 16 ⇒ d = 4
On substituting d = 4 in (i), we get:
31 = a + 7 × 4 ⇒ a = 3
Since a = 3 and d = 4, required AP = 3, 7 , 11, 15 .....

OR
All two - digit odd positive numbers are 11, 13, 15, 17 ......99.
This is an AP in which a = 11, d = (13 - 11) = 2 and  l = 99.
Let the number of terms be n. Then,
Tn = 99
a + (n - 1) × d = 99
⇒ 99 = 11 + (n - 1) × 2
⇒ 99 = 11 + 2n - 2
⇒ 90 = 2n
n = 45
Required sum = $\frac{n}{2}\left(a+l\right)$
= $\frac{45}{2}\left(11+99\right)=\left(\frac{45}{2}×110\right)$
= (45 × 55) = 2475

#### Question 17:

In the adjoining figure, PA and PB are tangents drawn from an external point P to a circle with centre O.
Prove that ∠APB = 2∠OAB
Figure

Or, in the adjoining figure, quadrilateral ABCD is circumscribed. If the radius of the incircle with centre O is 10 cm and AD ⊥ DC, find the value of x.
Figure

#### Answer:

Given:  PA and PB are the tangents to a circle with centre O from a point P outside it.
To prove: $\angle$APB = 2$\angle$OAB
Proof : Let $\angle$APB = x°
We know that the tangents to a circle from an external point are equal. So, PA =  PB
Since, the angles opposite to the equal sides of a triangle are equal,
PA = PB ⇒ $\angle$PBA = $\angle$PAB

Also, the sum of the angles of a triangle is 180°.
$\angle$APB + $\angle$PAB +$\angle$PBA = 180°
x° + 2$\angle$PAB = 180°        [Since ∠PBA = ∠PAB]
$⇒\angle \mathrm{PAB}=\frac{1}{2}\left(180°-\mathrm{x}°\right)=\left(90°-\frac{1}{2}\mathrm{x}°\right)$
But PA is a tangent and OA is the radius of the given circle.
$\angle$OAB + $\angle$PAB = 90°
$\angle \mathrm{OAB}+\left(90°-\frac{1}{2}\mathrm{x}°\right)=90°\phantom{\rule{0ex}{0ex}}$
$⇒\angle \mathrm{OAB}=\frac{1}{2}\mathrm{x}°=\frac{1}{2}\angle \mathrm{APB}$
Hence, $\angle$APB = 2$\angle$OAB

OR
Disclaimer :  In the figure side AB is given 27 cm, instead of AB , BP should be 27 cm.
Since the lengths of tangents from an external point to a circle are equal, we have:
BP =  BQ =  27 cm
∴ QC = (BC - BQ) = (38 - 27) cm = 11 cm
⇒ CQ =  CR = 11 cm
Join OR. Then, SORD is a rectangle.
∴ OS =  DR = 10 cm
Hence, DC = (DR + RC) = (10 + 11) cm = 21 cm

#### Question 18:

Draw a circle of radius 6 cm. From a point 10 cm away from its centre construct a pair of tangents to the circle. Measure the length of each of the tangent segments.

#### Answer:

Given that tangents PQ and PR are drawn from a point P to a circle with centre O.
Here, OQ = OR = 6 cm and OP = 10 cm.

Now,
PQ is tangent at Q and OQ is the radius through Q.
∴ OQ ⊥ QP
Similarly,
OR ⊥ PR
In the right-angled  ΔOQP and ΔORP, we get:
OQ = OR     [ radii of the same circle]
OP = OP      [ common side]
∴ Δ OQP ≅ Δ ORP    [ By RHS congruence]
Hence QP = PR
Again,
By Pythagoras' theorem, we get:
⇒ OP2 = OQ2 + QP2
⇒ (10)2 = (6)2 + QP2
⇒ QP2 = [100 - 36] = 64 ⇒ QP = 8 cm
Thus, QP = RP = 8 cm

#### Question 19:

The three vertices of a parallelogram ABCD, taken in order, are A(1, −2), B(3, 6) and C(5, 10). Find the coordinates of the fourth vertex D.

#### Answer:

Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and let its fourth vertex be D(a, b).
Join AC and BD.
Let AC and BD intersect at point O.

We know that the diagonals of a parallelogram bisect each other.
So, O is the mid-point of AC as well as of BD.
Mid-point of AC = $\left(\frac{1+5}{2},\frac{-2+10}{2}\right)$ ⇒ (3, 4)
Mid-point of BD = $\left(\frac{3+a}{2},\frac{6+b}{2}\right)$
$\frac{3+a}{2}=3$ and $\frac{6+b}{2}=4$
⇒ 3 + a = 6 and 6 + b = 8
a = 3 and b = 2
Hence, the fourth vertex is D(3, 2).

#### Question 20:

Find the third vertex a ∆ABC if two of its vertices are B(−3, 1) and C(0, −2) and its centroid is at the origin.

#### Answer:

Two vertices of ABC are B(- 3, 1) and C(0, -2). Let the third vertex be A(a, b).
Then, the coordinates of its centroid are
$G\left(\frac{-3+0+a}{3},\frac{1-2+b}{3}\right)\phantom{\rule{0ex}{0ex}}$
$=G\left(\frac{-3+a}{3},\frac{-1+b}{3}\right)$
But is is given that the centroid is G(0,0).
Therefore, $\frac{-3+a}{3}=0$ and $\frac{-1+b}{3}=0$
⇒ - 3+ a = 0 and - 1+ b = 0
a = 3 and b = 1
Hence, the third vertex of ∆ABC is A(3 ,1).

#### Question 21:

Cards marked with all 2-digit numbers are placed in a box and are mixed thoroughly. One card is drawn at random. Find the probability that the number on the card is

(a) divisible by 10
(b) a perfect square
(c) a prime number less than 25

#### Answer:

Cards are marked with numbers 10, 11, 12, …, 99.
Total number of cards = 90

(a) The number of cards divisible by 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90.
The number of favourable cases = 9
Probability of getting a card with a number divisible by 10 = $\frac{9}{90}=\frac{1}{10}$

(b) The numbers of cards which are perfect squares are 16, 25, 36, 49, 64, 81
The number of favourable cases = 6
Probability of getting a card that is a perfect square = $\frac{6}{90}=\frac{1}{15}$

(c) Prime numbers less than 25 are 2, 3, 5, 7, 11 , 13, 17, 19, 23.
The number of favourable cases = 9
Probability of getting a card with a prime number less than 25 = $\frac{9}{90}=\frac{1}{10}$

#### Question 22:

A road 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.

#### Answer:

Let the radius of the park be r metres.

Then, its circumference = $2\mathrm{\pi r}$
Therefore,
$2\mathrm{\pi r}=352⇒2×\frac{22}{7}×\mathrm{r}=352\phantom{\rule{0ex}{0ex}}$
$⇒\mathrm{r}=\left(352×\frac{7}{44}\right)=56$
Thus, the inner radius = 56 m
Outer radius = (56 + 7) m = 63 m
Area of the road = $\mathrm{\pi }\left[{\left(63\right)}^{2}-{\left(56\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}$ m2
$=\frac{22}{7}×\left(63+56\right)\left(63-56\right)\phantom{\rule{0ex}{0ex}}$ m2
$=\left(\frac{22}{7}×119×7\right)$ m2 = 2618 m2

#### Question 23:

A round table cover shown in the adjoining figure has six equal designs. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.50 per cm2.
Figure

Or, in an equilateral triangle of side 12 cm, a circle is inscribed touching its sides. Find the area of the portion of the triangle not included in the circle.

#### Answer:

Given: radius =  OA = OB = 28 cm

Area of the shaded region = (area of the circle - area of the hexagon)
=
$⇒\left[\left(\frac{22}{7}×28×28\right)-\left(6×\frac{\sqrt{3}}{4}×28×28\right)\right]\phantom{\rule{0ex}{0ex}}$cm2
$⇒\left(28×28\right)\left[\frac{22}{7}-\frac{3}{2}×1.73\right]\phantom{\rule{0ex}{0ex}}$cm2
$⇒784×\left(3.142-2.595\right)$ cm2 = (784 × 0.547) cm2 = 428.84 cm2
Rate of design =  Rs 0.50 per cm2
Hence, total cost of designing = Rs (429.84 × 0.50) = Rs 214.42
OR
Let ΔABC be an equilateral triangle of side 12 cm.
Let AD ⊥ BC. Then D is the mid-point of BC.

∴ BD = DC = 6 cm and AB = 12 cm
∴ AD = $\sqrt{A{B}^{2}-B{D}^{2}}\phantom{\rule{0ex}{0ex}}$ cm
$=\sqrt{{\left(12\right)}^{2}-{\left(6\right)}^{2}}\phantom{\rule{0ex}{0ex}}$ cm
$=\sqrt{144-36}$ cm = $\sqrt{108}$ cm = $6\sqrt{3}$cm
Let O be the centre of the inscribed circle.
Then, O is the centroid of ΔABC.
∴ AO : OD = 2 : 1 and OD = $\frac{1}{3}AD=\frac{1}{3}×6\sqrt{3}=2\sqrt{3}$
r = OD = $2\sqrt{3}$cm
Required area = (Area of ΔABC) - (Area of the incircle)
= $\left[\frac{\sqrt{3}}{4}×{\left(12\right)}^{2}-3.14×{\left(2\sqrt{3}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}$ cm2
$=\left(36\sqrt{3}-3.14×12\right)\phantom{\rule{0ex}{0ex}}$ cm2
$=\left(62.28-37.68\right)$ cm2 = 24.6 cm2

#### Question 24:

If a sphere has the same surface area as the total surface area of a circular cone of height 40 cm and radius 30 cm, find the radius of the sphere.

#### Answer:

Given: radius of the cone (r) = 30 cm and height of the cone (h) = 40 cm
Slant height of the cone (l) = $\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{\left(30\right)}^{2}+{\left(40\right)}^{2}}$
$⇒\sqrt{900+1600}⇒\sqrt{2500}=50$ cm
Total surface area of the circular cone = $\mathrm{\pi r}\left(\mathrm{l}+\mathrm{r}\right)$ sq. units
$=\mathrm{\pi }×30\left(50+30\right)$ cm2
TSA of a sphere is same as TSA of a circular cone.
Therefore,
Total surface area of a sphere = Total surface area of circular cone
$4{\mathrm{\pi r}}^{2}=\mathrm{\pi }×30×80\phantom{\rule{0ex}{0ex}}$
$⇒{\mathrm{r}}^{2}=\frac{\left(30×80\right)}{4}=600\phantom{\rule{0ex}{0ex}}$
$⇒\mathrm{r}=10\sqrt{6}$ cm
Hence, radius of the sphere will be $10\sqrt{6}$cm.

#### Question 25:

A two-digit number is such that product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number.

Or, two water taps together can fill a tank in $9\frac{3}{8}$ hours. The larger tap takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

#### Answer:

Let the tens digit be x and the units digit be y.
∴ Product of the digits = xy = 35............(i)
Original number is (10x + y).
If the digits interchange their position, the number becomes (10y + x).
Thus,
(10x + y) + 18 = (10y + x)
9x − 9y = − 18
xy = − 2 .............(ii)
From (i), we get:  $y=\frac{35}{x}$
On substituting the value of y in (ii), we get:
$x-\frac{35}{x}=-2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$⇒{x}^{2}+7x-5x-35=0\phantom{\rule{0ex}{0ex}}$
$⇒x\left(x+7\right)-5\left(x+7\right)=0\phantom{\rule{0ex}{0ex}}$
$⇒\left(x-5\right)\left(x+7\right)=0$
x = 5 or x = − 7
Since the digits cannot be negative, x = 5 and
Hence, the required number = 10(5) + 7 = 57

OR

Let the smaller tap fill the tank in x hours.
Then, the larger tap will fill it in (x − 10) hours.
Time taken by both together to fill the tank = $9\frac{3}{8}=\frac{75}{8}$ hours
Part filled by smaller tap in 1 hour = $\frac{1}{x}$
Part filled by larger tap in 1 hour = $\frac{1}{\left(x-10\right)}$
Part filled by both the taps in 1 hour = $\frac{8}{75}$
$\frac{1}{x}+\frac{1}{\left(x-10\right)}=\frac{8}{75}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\left(x-10\right)+x}{x\left(x-10\right)}=\frac{8}{75}⇒\frac{\left(2x-10\right)}{x\left(x-10\right)}=\frac{8}{75}$
75(2x −10) = 8x(x −10)         [By cross−multiplication]
⇒ 150x − 750 = 8x2 − 80x
⇒ 8x2 − 230x + 750 = 0
⇒ 4x2 − 115x + 375 = 0
⇒ 4x2 − 100x − 15x + 375 = 0 ⇒ 4x(x − 25) − 15(x − 25) = 0
⇒ (4x − 15)(x − 25) = 0 ⇒ (4x − 15) = 0 or (x − 25) = 0

x = 25
Hence, the time taken by the smaller tap to fill the tank = 25 hours.
And the time taken by the larger tap to fill the tank = (25 − 10) hours = 15 hours.

#### Question 26:

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact to the centre.

#### Answer:

Given:
PA and PB are the tangents drawn from a point P to a circle with centre O. Also, the line segments OA and OB are drawn.

To prove:$\angle$APB + $\angle$AOB = 180°
Proof:
We know that the tangent to a circle is perpendicular to the radius through the point of  contact.
∴ PA ⊥ OA ⇒ $\angle$OAP = 90°
PB ⊥ OB ⇒ $\angle$OBP = 90°
$\angle$OAP + $\angle$OBP = (90° + 90°)= 180° ......................(i)
But we know that the sum of all the angles of a quadrilateral is 360°.
$\angle$OAP + $\angle$OBP + $\angle$APB + $\angle$AOB = 360° .................(ii)
From (i) and (ii), we get:
$\angle$APB + $\angle$AOB = 180°

#### Question 27:

From the top of a 7-m-high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.

#### Answer:

Let AB be the building that is 7 metres high. AE ⊥ CD, where CD is the cable tower.
Let AE = BD = x m

In ΔAED,
$\angle$EAD = 45° = angle of depression
$\frac{\mathrm{AE}}{\mathrm{ED}}=\mathrm{cot}45°\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\mathrm{x}}{7}=1⇒\mathrm{x}=7$
In ΔACE,
$\angle$CAE = 60° = angle of elevation
$\angle$AEC = 90°
$\frac{\mathrm{CE}}{\mathrm{AE}}=\mathrm{tan}60°\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\mathrm{h}}{x}=\sqrt{3}$
$⇒h=x\sqrt{3}=\left(7×1.73\right)=12.11$
Height of the tower = CD = CE + ED = (12.11+ 7) m = 19.11 m

#### Question 28:

Puja works in a bank and she gets a monthly salary of Rs 35000 with annual increment of Rs 1500. What would be her monthly salary in the 10th year of her employment?

#### Answer:

Here a = 35000 , d = 1500  and T10 = ?
We know that
Tn = a + (n - 1)d
T10 = 35000 + (10 - 1) × 1500
= 35000 + (9 × 1500) = (35000 + 13500) = 48,500
Hence, Puja's monthly salary in the 10th year will be Rs. 48,500.

#### Question 29:

In the given figure, ABDCA represents the quadrant of a circle of radius 7 cm with centre A. Calculate the area of the shaded region.

#### Answer:

We have:

Shaded area = Area of quadrant - Area of ΔABE
= $\left[\frac{1}{4}{\mathrm{\pi r}}^{2}-\frac{1}{2}×\mathrm{h}×\mathrm{b}\right]\phantom{\rule{0ex}{0ex}}$sq. units
$⇒\left[\left\{\frac{1}{4}×\frac{22}{7}×7×7\right\}-\left\{\frac{1}{2}×7×2\right\}\right]\phantom{\rule{0ex}{0ex}}$ cm2
$⇒\left(\frac{77}{2}-7\right)=\frac{77-14}{2}=\frac{63}{2}=31.5$ cm2

#### Question 30:

The radii of the circular ends of a solid frustum of a cone are 33cm and 27 cm and its slant height is 10 cm. Find its capacity and total surface area.

#### Answer:

R = 33 cm, r = 27 cm and  l = 10 cm
l2 = h2 + (R - r)2
h2 = l2 - (R - r)2
= {(10)2 - (33 - 27)2} = (100 - 36) = 64
h = 8 cm
$\mathrm{Capacity}=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{R}}^{2}+{\mathrm{r}}^{2}+\mathrm{Rr}\right)\phantom{\rule{0ex}{0ex}}$ cm3
$\left\{\frac{1}{3}×\frac{22}{7}×8\left({33}^{2}+{27}^{2}+33×27\right)\right\}\phantom{\rule{0ex}{0ex}}$ cm3
$\left\{\frac{1}{3}×\frac{22}{7}×8\left(1089+729+891\right)\right\}=\left\{\frac{1}{3}×\frac{22}{7}×8×2709\right\}=22704$ cm3
Again,
sq. cm
$=\frac{22}{7}\left[{33}^{2}+{27}^{2}+10\left(33+27\right)\right]\phantom{\rule{0ex}{0ex}}$ cm2
$=\frac{22}{7}\left[1089+729+10\left(60\right)\right]\phantom{\rule{0ex}{0ex}}$ cm2
$=\frac{22}{7}\left[2418\right]=7599.43$ cm2
Hence, capacity of the frustum is 22704 cm3 and total surface area of frustum = 7599.43 cm2.

#### Question 31:

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at the point E and PA = 14 cm, find the perimeter of ∆PCD.
Figure

#### Answer:

Given : PA and PB are tangents to the circle with centre O.
CD is a tangent to the circle at E, which intersects PA and PB at C and D, respectively, and PA = 14 cm.

We know that tangents drawn from an external point to a circle are equal.
∴ PA = PB = 14 cm
CA = CE  and DB = DE
Now, perimeter of ΔPCD = PC + CD + PD
= PC + (CE + ED) + PD
= PC + (CA + DB) + PD
= (PC + CA) + (DB + PD)
= PA + PB
= 14 cm + 14 cm = 28 cm

#### Question 32:

Construct ∆ABC in which BC = 5.4 cm, AB = 4.5 cm and ∠ABC = 60°. Construct a triangle similar to this triangle, whose sides are $\frac{3}{4}$ the corresponding sides of ∆ABC.

#### Answer:

Steps of construction:
Step 1: Draw a line segment BC = 5.4 cm.
Step 2: Draw an angle of 60° at B, so that $\angle$XBC = 60°.
Step 3: With centre B and radius 4.5 cm, draw an arc intersecting XB at A.
Step 4: Join AC
Thus, ΔABC is the required triangle.
Step 5: Draw a line BY below BC.
Step 6: Cut off 4 equal distances from B, such that BB1 = B1B2 = B2B3 = B3B4.
Step 7: Join CB4
Step 8: Draw C'B3 parallel to CB4.
Step 9: Draw C'A' parallel to CA through C' intersecting BA produced to A'.
Thus, ΔA'BC' is the required similar triangle.

#### Question 33:

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag.

#### Answer:

Let the number of blue balls in the bag be x.
Then, total number of balls = ( 5 + x)
Given: P(a blue ball) = 3 × P(a red ball)

Hence, the number of blue balls in the bag is 15.

#### Question 34:

In what ratio is the line segment joining the points (−2, −3) and (3, 7) divided by the y-axis? Also, find the coordinates of the point of division.

#### Answer:

Let the y - axis cut the join A(-2, -3) and B(3, 7) at the point P in the ratio k : 1.
Then, by section formula, the co-ordinates of P are
$\left(\frac{3k-2}{k+1},\frac{7k-3}{k+1}\right)$
But P lies on the y - axis; so, its abscissa is 0.
$\frac{3k-2}{k+1}=0⇒3k-2=0⇒k=\frac{2}{3}$
So, the required ratio is $\frac{2}{3}:1$, which is 2 : 3.

$P\left(0,\frac{7×\frac{2}{3}-3}{\frac{2}{3}+1}\right)⇒P\left(0,1\right)$
Hence, the point of intersection of AB and the y - axis is  P(0, 1).

#### Question 1:

The value of k, for which the equation $2{x}^{2}kx+3=0$ has two real equal roots, are

(a) $±2\sqrt{3}$
(b) $±3\sqrt{2}$
(c) $±2\sqrt{6}$
(d) $±\sqrt{6}$

#### Answer:

(c) $±2\sqrt{6}$
The given equation is 2x2 + kx + 3 = 0.
This is of the form ax2 + bx + c = 0, where, a = 2, b = k and c = 3.
Therefore,
D = (b2 - 4ac) = (k2 - 4 × 2 × 3) = (k2 - 24)
For real and equal roots, we must have:
D = 0 ⇒ k2 - 24 = 0 ⇒k2 = 24 ⇒ $k=±\sqrt{24}=±2\sqrt{6}$
Hence, $2\sqrt{6}$ and $-2\sqrt{6}$ are the required values of k.

#### Question 2:

How many terms are there in the AP 7, 11, 15, ..., 139?

(a) 31
(b) 32
(c) 33
(d) 34

#### Answer:

(d) 34

In the given AP, we have:
a = 7 and d = (11 - 7) = 4
Suppose there are n terms in the given AP. Then,
Tn = 139 ⇒ a + (n - 1)d = 139
⇒ 7 + (n - 1) × 4 = 139
⇒  4n = 136
⇒ n = 34
Hence, there are 34 terms in the given AP.

#### Question 3:

One card is drawn from a well-shuffled deck of 52 cards. The probability of drawing a 10 of a black suit is

(a) $\frac{1}{13}$
(b) $\frac{1}{26}$
(c) $\frac{1}{52}$
(d) $\frac{3}{52}$

#### Answer:

(b) $\frac{1}{26}$

Number of possible outcomes = 52
Number of favourable outcomes = 2
Therefore,
P(drawing a 10 of a black suit) = $\frac{2}{52}=\frac{1}{26}$

#### Question 4:

In a circle of radius 7 cm, tangent PT is drawn from a point P, such that PT = 24 cm. If O is the centre of the circle, then OP = ?
Figure

(a) 30 cm
(b) 28 cm
(c) 25 cm
(d) 31 cm

#### Answer:

(c) 25 cm

OP2 = OT2 + PT2 = (7)2 + (24)2
OP2 = ( 49 + 576) = 625 ⇒ OP = $\sqrt{625}=25$ cm

#### Question 5:

The ratio in which the line segment joining the points A(−3, 2) and B(6, 1) is divided by the y-axis is

(a) 3 : 1
(b) 1 : 3
(c) 2 : 1
(d) 1 : 2

#### Answer:

(d) 1 : 2

Let the y - axis cut the line segment joining A(- 3, 2) and B(6, 1) at the point P in the ratio k : 1.
Then, by section formula, the co-ordinates of P are
$\left(\frac{6k-3}{k+1},\frac{k+2}{k+1}\right)$
But P lies on the y-axis. So, its abscissa is 0.
$\frac{6k-3}{k+1}=0\phantom{\rule{0ex}{0ex}}⇒6k-3=0⇒6k=3⇒k=\frac{3}{6}=\frac{1}{2}$
So, the required ratio is $\frac{1}{2}:1$, which is 1 : 2.

#### Question 6:

The distance of the point P(6, −6) from the origin is

(a) 6 units
(b) $\sqrt{6}$ units
(c) $3\sqrt{2}$ units
(d) $6\sqrt{2}$ units

#### Answer:

(d) $6\sqrt{2}$ units

Let P(6, - 6) be the given point and O(0, 0) be the origin.
Then, OP = $\sqrt{{\left(6-0\right)}^{2}+{\left(-6-0\right)}^{2}}=\sqrt{{6}^{2}+{\left(-6\right)}^{2}}$
= $\sqrt{36+36}=\sqrt{72}=\sqrt{36×2}=6\sqrt{2}$ units

#### Question 7:

A kite is flowing at a height of 75 cm from the level ground, attached to a string inclined at 60° to the horizontal. The length of the string, with no slack in it, is

(a) $50\sqrt{2}\mathrm{m}$
(b) $25\sqrt{3}\mathrm{m}$
(c) $50\sqrt{3}\mathrm{m}$
(d) 37.5 m

#### Answer:

(c) $50\sqrt{3}\mathrm{m}$

Let OB be the length of the string from the level ground and O be the point of the observer.
Then, AB = 75 m and $\angle$OAB = 90° and $\angle$AOB = 60°.
Let OB be l metres.
From the right-angled ΔOAB, we have:

$⇒\frac{l}{75}=\frac{2}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$
$⇒l=\left(75×\frac{2}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\right)=\left(25×2×\sqrt{3}\right)=50\sqrt{3}$ m

#### Question 8:

A solid metal cone with radius of base 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. The number of balls formed is

(a) 16
(b) 32
(c) 24
(d) 28

#### Answer:

(b) 32

Radius of the cone = 12 cm and its height = 24 cm
Volume of cone = $\frac{1}{3}\mathrm{\pi }{r}^{2}\mathrm{h}=\left(\frac{1}{3}×\mathrm{\pi }×12×12×24\right){\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}$

Volume of each ball = $\frac{4}{3}{\mathrm{\pi R}}^{3}=\frac{4}{3}\mathrm{\pi }×3×3×3$      [ Since diameter of each ball is 6 cm]
$=\left(36\mathrm{\pi }\right){\mathrm{cm}}^{3}$
Thus,
Number of balls formed =
$=\frac{\left(48×24\mathrm{\pi }\right)}{36\mathrm{\pi }}=32$

#### Question 9:

If the roots the equation $\left(a-b\right){x}^{2}+\left(b-c\right)x+\left(c-a\right)=0$ are equal, prove that b + c = 2a.

#### Answer:

Given equation is (a - b)x2 + (b - c)x + (c - a) = 0
It is of the form Ax2 + Bx + C = 0
We know that D = B2 - 4AC
D = (b - c)2 - 4(a - b)(c - a) = 0
For real and equal roots, D = 0
Now, D = 0
⇒ (b - c)2 - 4(a - b)(c - a) = 0
b2 - 2bc + c2 - 4(ac - a2  - bc + ab ) = 0
b2 - 2bc + c2 - 4ac + 4a2  + 4bc - 4ab = 0
⇒ 4a2 + b2 + c2 - 4ab + 2bc - 4ac = 0
⇒ (-2a)2 + b2 + c2 + 2(-2a)b + 2bc + 2c(- 2a) = 0
⇒ [ (- 2a ) + b + c]2 = 0
⇒ (- 2a + b + c)2 = 0
⇒ - 2a + b + c = 0
b + c = 2a

#### Question 10:

Find the 10th term from the end of the AP 4, 9, 14, ..., 254.
Or, Which term of the AP 3, 15, 27, 34, ... will be 132 more than its 54th term?

#### Answer:

Here, a = 4, d = (9 − 4) = 5, l = 254 and n = 10
Now, nth term from the end = {l − (n − 1)d}
∴ 10th term from the end = {254 − (10 − 1) × 5}
= {254 − (9 × 5)} = (254 − 45) = 209
Hence, the 10th term from the end is 209.

OR
Here, a = 3, d = (15 − 3) = 12
Tn = 132 + T54
a + (n − 1)d = 132 + {3 + (54 − 1) × 12}
⇒ 3 + (n − 1) × 12 = 132 + 639
⇒(n − 1) × 12 = (771 − 3) = 768
⇒ (n − 1) = $\frac{768}{12}=64$
n = (64 + 1) = 65
Hence, the required term is the 65th term.

#### Question 11:

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

#### Answer:

Given: CD and EF are the tangents at the end-points A and B of the diameter AB  of a circle with centre O.
To prove: $\mathrm{CD}\parallel \mathrm{EF}$

Proof: CD is the tangent to the circle at the point A.
$\angle$BAD = 90°
EF is the tangent to the circle at the point B.
$\angle$ABE = 90°
Thus,
$\angle$BAD = $\angle$ABE (each equal to 90°)
But these are alternate angles.
$\mathrm{CD}\parallel \mathrm{EF}$

#### Question 12:

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at the point E and PA = 14 cm, find the perimeter of ∆PCD.
Figure

#### Answer:

Given : PA and PB are tangents to the circle with centre O.
CD is a tangent to the circle at E, which intersects PA and PB at C and D, respectively and PA = 14 cm.

We know that tangents drawn from as external point to a circle are equal.
∴ PA = PB = 14 cm
CA = CE  and DB = DE
Now, perimeter of ΔPCD = PC + CD + PD
= PC + (CE + ED) + PD
= PC + (CA + DB) + PD
= (PC + CA) + (DB + PD)
= PA + PB
= 14 cm + 14 cm = 28 cm

#### Question 13:

The area of the circular base of a cone is 616 cm2 and its height is 48 cm. Find its whose surface area.

#### Answer:

Area of circular base = 616 cm2
Therefore,

And h = 48 cm
Then slant height, l =
Total surface are of cone =

#### Question 14:

In the adjoining figure, the area enclosed between two concentric circles is 770 cm2 and the radius of the outer circle is 21 cm. Find the radius of the inner circle.
Figure

#### Answer:

Let R and r be the outer and inner radii of a ring.
Then, area of the ring = $\mathrm{\pi }\left({\mathrm{R}}^{2}-{\mathrm{r}}^{2}\right)$
Given, R = 21 cm
Area of the ring = 770 cm2
Thus,
$770=\frac{22}{7}\left\{{\left(21\right)}^{2}-{r}^{2}\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{770×7}{22}=441-{r}^{2}$
⇒ 245 - 441 = - r2
⇒ - 196 = - r2
⇒ 196 = r2
r = 14 cm
Therefore, inner radius = 14 cm

#### Question 15:

Solve for x: $12ab{x}^{2}-\left(9{a}^{2}-8{b}^{2}\right)x-6ab=0.$

#### Answer:

2abx2 − (9a2 − 8b2)x − 6ab = 0
⇒ 12abx2 − 9a2x + 8b2x − 6ab = 0
⇒ 3ax(4bx − 3a) + 2b(4bx − 3a) = 0
⇒ (3ax + 2b)(4bx − 3a) = 0
⇒ (3ax + 2b) = 0 or (4bx − 3a) = 0
⇒ 3ax = − 2b or 4bx = 3a
⇒ $x=\frac{-2b}{3a}$ or $x=\frac{3a}{4b}$

#### Question 16:

If the 8th term of an AP is 31 and its 15th term is 16 more than the 11th term, find the AP.
Or, find  the sum of all odd integers between 2 and 100 that are divisible by 3.

#### Answer:

Here,
a8 = 31 and a15 = 16 + a11
⇒ 31 = a + 7d .............(i)
And a15 = 16 + a11
⇒  a + 14d = 16 + a + 10d .............(ii)
By solving equation (ii), we get:
4d = 16
d = 4
On substituting d = 4 in (i), we get:
31 = a + 7 × 4
a = 3
Since a = 3 and d = 4, required AP = 3, 7 , 11, 15 .....

OR

All odd integers between 2 and 100 divisible by 3 are
3, 9 , 15 , 21 ................ 99
Here a = 3 , d = 6 and nth term = 99
Let the number of terms be n. Then,
Tn = 99
a + (n - 1)× d
⇒ 99 = 3+ (n - 1) × 6
⇒ 99 = 3+ 6n - 6
⇒ 102 = 6n
n = 17
Required sum = $\frac{n}{2}\left(a+l\right)$
= $\frac{17}{2}×\left(3+99\right)=\left(17×51\right)=867$

#### Question 17:

Prove that the parallelogram circumscribing a circle is a rhombus.
Or, ∆ABC is drawn to circumscribe a circle of radius 4 cm, such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 8 cm and 6 cm respectively. Find AB and AC.
Figure

#### Answer:

Given: A parallelogram ABCD circumscribes a circle with centre O.
Thus, AB = CD and AD = BC
To Prove :  AB = BC = CD = AD
Proof :
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
∴    AP = AS .......(i)              [tangents from A]
BP = BQ .......(ii)             [tangents from B]
CR = CQ .......(iii)            [tangents from C]
DR = DS .......(iv)             [tangents from D]

∴ AB + CD = AP + BP + CR + DR
= AS + BQ + CQ + DS            [From (i), (ii), (iii) and (iv)]
= (AS + DS) + (BQ + CQ)
= AD + BC
Thus, AB + CD = AD + BC
⇒ AB + AB = AD+ AD
⇒ 2AB = 2AD
⇒ AB = AD
∴ CD = AB = AD = BC
Hence, ABCD is a rhombus.

OR
Let the given circle touch the sides AB and AC of the triangle at point F and E, respectively, and let the length of the line segment AF be x.

In ΔABC, BF = BD = 6 cm               (Tangents on the circle from point C)
CE = CD = 8 cm                               (Tangents on the circle from point B)
AE = AF = x cm                                (Tangents on the circle from point A)
AB = AF + FB = (x + 6) cm
BC = BD + DC = (8 + 6) = 14 cm
CA = CE + EA = (8 + x ) cm
∴ Perimeter of triangle (2s) = AB + BC + CA = (x + 8 + 14 + 6 + x) = (28 + 2x) cm
Semi-perimeter (s) = (14 + x) cm
Now,

$=\sqrt{\left\{14+x\right\}\left\{\left(14+x\right)-14\right\}\left\{\left(14+x\right)-\left(6+x\right)\right\}\left(14+x\right)-\left(8+x\right)}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{\left(14+x\right)\left(x\right)\left(8\right)\left(6\right)}=4\sqrt{3\left(14x+{x}^{2}\right)}$
Area of ΔOBC = $\frac{1}{2}×\mathrm{OD}×\mathrm{BC}=\frac{1}{2}×4×14=28\phantom{\rule{0ex}{0ex}}$
Area of ΔOCA = $\frac{1}{2}×OE×\mathrm{AC}=\frac{1}{2}×4×\left(8+\mathrm{x}\right)=16+2\mathrm{x}\phantom{\rule{0ex}{0ex}}$
Area of ΔOAB = $\frac{1}{2}×OF×\mathrm{AB}=\frac{1}{2}×4×\left(6+\mathrm{x}\right)=12+2\mathrm{x}$
Again,
Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB
$4\sqrt{3\left(14x+{x}^{2}\right)}=28+16+2x+12+2x\phantom{\rule{0ex}{0ex}}$
$⇒4\sqrt{3\left(14x+{x}^{2}\right)}=56+4x\phantom{\rule{0ex}{0ex}}$
$⇒4\sqrt{3\left(14x+{x}^{2}\right)}=4\left(14+x\right)\phantom{\rule{0ex}{0ex}}$
$⇒\sqrt{3\left(14x+{x}^{2}\right)}=14+x$
Squaring both the sides, we get:
3(14x + x2) = (14 + x)2
⇒ 42x + 3x2 = 196 + 28x + x2
⇒ 2x2 +14x -196 = 0
x2 + 7x - 98 = 0
x2 + 14x - 7x - 98 = 0
x (x + 14) - 7( x + 14) = 0
⇒ (x - 7)(x + 14) = 0
⇒ (x +14 ) = 0 or (x − 7) = 0
Therefore, x = − 14 or x =  7
However, x = − 14 is not possible as the length of the sides cannot be negative.
Therefore, x = 7
Hence, AB = x + 6 = 7 + 6 = 13 cm
AC = 8 + x = 8 + 7 = 15 cm

#### Question 18:

Draw a circle of diameter 12 cm. From a point 10 cm away from its centre, construct a pair of tangents to the circle. Measure the length of each tangent segment.

#### Answer:

Given that tangents PQ and PR are drawn from a point P to a circle with centre O.
Here, OQ = OR = 6 cm and OP = 10 cm.

Now,
PQ is a tangent at Q and OQ is the radius through Q.
∴ OQ ⊥ QP
Similarly,
OR ⊥ PR
In the right-angled  ΔOQP and ΔORP, we get:
OQ = OR     [ radii of the same circle]
OP = OP      [ common side]
∴ ΔOQP ≅ ΔORP    [ By RHS congruence]
Hence QP = PR
Again,
By Pythagoras' theorem, we get:
⇒ OP2 = OQ2 + QP2
⇒ (10)2 = (6)2 + QP2
⇒ QP2 = [100 - 36] = 64 ⇒ QP = 8 cm
Thus, QP = RP = 8 cm

#### Question 19:

Show that the points A(a, a), B(−a, −a) and $\mathrm{C}\left(-a\sqrt{3},a\sqrt{3}\right)$ are the vertices of an equilateral triangle.

#### Answer:

The given points are .

$\mathrm{AB}=\sqrt{{\left[a-\left(-a\right)\right]}^{2}+{\left[a-\left(-a\right)\right]}^{2}}=\sqrt{{\left(2a\right)}^{2}+{\left(2a\right)}^{2}}=\sqrt{4{a}^{2}+4{a}^{2}}=\sqrt{8{a}^{2}}=2\sqrt{2}a$ units
$\mathrm{BC}=\sqrt{{\left[-a-\left(-a\sqrt{3}\right)\right]}^{2}+{\left[-a-\left(a\sqrt{3}\right)\right]}^{2}}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{{\left[a\left(\sqrt{3}-1\right)\right]}^{2}+{\left[a\left(\sqrt{3}+1\right)\right]}^{2}}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{{a}^{2}\left(3+1-2\sqrt{3}\right)+{a}^{2}\left(3+1+2\sqrt{3}\right)}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{{a}^{2}\left(4-2\sqrt{3}\right)+{a}^{2}\left(4+2\sqrt{3}\right)}=\sqrt{{a}^{2}\left(4-2\sqrt{3}+4+2\sqrt{3}\right)}=\sqrt{8{a}^{2}}=2\sqrt{2}a$ units
$CA=\sqrt{{\left[a-\left(-a\sqrt{3}\right)\right]}^{2}+{\left[a-a\sqrt{3}\right]}^{2}}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{{\left[a\left(1+\sqrt{3}\right)\right]}^{2}+{\left[a\left(1-\sqrt{3}\right)\right]}^{2}}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{{a}^{2}\left(1+3+2\sqrt{3}\right)+{a}^{2}\left(1+3-2\sqrt{3}\right)}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{{a}^{2}\left(4+2\sqrt{3}\right)+{a}^{2}\left(4-2\sqrt{3}\right)}=\sqrt{{a}^{2}\left(4+2\sqrt{3}+4-2\sqrt{3}\right)}=\sqrt{8{a}^{2}}=2\sqrt{2}a$ units
∴ AB = BC = CA = $2\sqrt{2}a$ units
Hence, ΔABC is an equilateral triangle and each of its sides is $2\sqrt{2}a$ units.

#### Question 20:

Find the area of a rhombus if its vertices are A(3, 0) B(4, 5), C(−1, 4) and D(−2, −1).

#### Answer:

Let A (3, 0), B (4, 5), C (−1, 4) and D (−2, −1) be the vertices of a rhombus ABCD.

units
units
Hence, area of rhombus = $\frac{1}{2}×4\sqrt{2}×6\sqrt{2}$ = 24 sq. units

#### Question 21:

Cards marked with numbers 13, 14, 15, ..., 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card drawn is
(a) divisible by 5
(b) a perfect square

#### Answer:

Total numbers of cards = 60 - 13 + 1 = 48
(a) Cards with numbers divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60
Number of such cards = 10
∴ P(getting a number divisible by 5) = $\frac{10}{48}=\frac{5}{24}$

(b) Perfect squares are 16, 25, 36, 49.
Number of such cards = 4
∴ P(getting a perfect square) = $\frac{4}{48}=\frac{1}{12}$

#### Question 22:

A window in a building is at a height of 10 m from the ground. The angle of depression of a point P on the ground from the window is 30°. The angle of elevation of the top of the building from the point P is 60°. Find the height of the building.

Or, in a violent storm, a tree is bent by the wind. The top of the tree meets the ground at an angle of 30°, at a distance of 30 metres from the root. At what height from the bottom did the tree get bent? What was the original height of the tree?

#### Answer:

Given, height of the window from the ground DB = 10 m
Let H be the height of the building.
Let O be the point of observation. Then,

DB = 10 m, $\angle$DPB = 30° and $\angle$APB =  60°
BP = x m
$\mathrm{tan}30°=\frac{10}{x}$
$\frac{1}{\sqrt{3}}=\frac{10}{x}⇒x=10\sqrt{3}\mathrm{m}$ ...............(i)
Again,
$\mathrm{tan}60°=\frac{H}{x}\phantom{\rule{0ex}{0ex}}$
$⇒\sqrt{3}=\frac{H}{x}$
$⇒H=\sqrt{3}x\phantom{\rule{0ex}{0ex}}$

Hence, required height of the building is 30 m.

OR
Let AB be the original height of the tree, which gets bent at a point C. After being bent, let the part CB take the position CD, meeting the ground at D. Then AD = 30 m, ∠ADC = 30°, ∠DAC = 90° and CD = CB

Let AC = x metres and CD = CB = y metres
From right-angled ΔDAC, we have:
$\frac{AC}{AD}=\mathrm{tan}30°=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{x}{30}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$
$⇒x=\left(30×\frac{1}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\right)=10\sqrt{3}\mathrm{m}$
Also, from right-angled ΔDAC, we have:
$\frac{DC}{AD}=\mathrm{sec}30°=\frac{2}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{y}{30}=\frac{2}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$
$⇒y=\left(30×\frac{2}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\right)=20\sqrt{3}\mathrm{m}$
Thus AB = (x + y) metres = $\left(10\sqrt{3}+20\sqrt{3}\right)\mathrm{m}=30\sqrt{3}\mathrm{m}$
Hence, the tree got bent at a height of $10\sqrt{3}\mathrm{m}$ and the original height of the tree was $30\sqrt{3}\mathrm{m}$.

#### Question 23:

A wire bent in the form of a circle of radius 42 cm is cut and again bent in the form of a square. Find the ratio of the areas of the regions enclosed by the circle and the square.

#### Answer:

Length of the wire =  Circumference of the circle
= $\left(2×\frac{22}{7}×42\right)=264$ cm
Therefore,
Perimeter of the square = Length of the wire = 264 cm
Hence, the side of the square = $\left(\frac{264}{4}\right)\mathrm{cm}=66\mathrm{cm}$
Thus,
Area of the circle = ${\mathrm{\pi r}}^{2}=\left(\frac{22}{7}×42×42\right)$ cm2 = 5544 cm2
Area of the square = (Side)2 = (66 cm)2 = 4356 cm2

Therefore,

Hence,
Area of circle  :  Area of square = 14 : 11

#### Question 24:

A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?

#### Answer:

Radius of the sphere = 10.5 cm
Volume of the sphere = $\frac{4}{3}{\mathrm{\pi r}}^{3}=\left(\frac{4}{3}×\mathrm{\pi }×10.5×10.5×10.5\right)$ cm3 =$\left(\frac{4}{3}\mathrm{\pi }×\frac{21}{2}×\frac{21}{2}×\frac{21}{2}\right)$ cm3 = $\left(\frac{3087\mathrm{\pi }}{2}\right)$ cm3
Radius of each cone = 3.5 cm and height = 3 cm
Volume of each cone = $\left(\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\right)$ = $\left(\frac{1}{3}\mathrm{\pi }×3.5×3.5×3\right)$ cm3 =$\left(\frac{1}{3}\mathrm{\pi }×\frac{7}{2}×\frac{7}{2}×3\right)$ cm3 = $\left(\frac{49\mathrm{\pi }}{4}\right)$ cm3
Required number of cones =
Hence, required number of cones is 126.

#### Question 25:

In the given figure, ∆ABC is right-angled at A. Semi-circles are drawn on AB, AC and BC as diameters. It is given that AB = 3 cm and AC = 4 cm. Find the area of the shaded region.
Figure

#### Answer:

Given: AB = 3 cm and AC = 4
Thus, BC = 5 cm (By Pythagoras' theorem)

Area of shaded region = {ar (ΔABC + ar(semi-circle APB) + ar(semi-circle AQC)} - ar(semi-circle BAC)
= $\left[\left(\frac{1}{2}×3×4\right)+\left(\frac{1}{2}\mathrm{\pi }×\frac{3}{2}×\frac{3}{2}\right)+\left(\frac{1}{2}\mathrm{\pi }×2×2\right)-\left(\frac{1}{2}\mathrm{\pi }×\frac{5}{2}×\frac{5}{2}\right)\right]$ cm2
= $\left\{6+\frac{1}{2}\mathrm{\pi }\left(\frac{9}{4}+4-\frac{25}{4}\right)\right\}$ cm2
= (6 + 0) cm2 = 6 cm2

#### Question 26:

Rs 250 is divided equally among a certain number of children. If there were 25 more children, each would have received 50 paise less. Find the number of children.
Or, the hypotenuse of a right-angled triangle is 6 cm more than twice the shortest side. If the third side is 2 cm less than the hypotenuse, find the sides of the triangle.

#### Answer:

Let the required number of children be x.
And total amount = Rs 250 = 25000 paise
Then, share of each child = $\frac{25000}{x}$ paise
If there were (x + 25) children, share of each child = $\frac{25000}{\left(x+25\right)}$paise
$\frac{25000}{x}-\frac{25000}{\left(x+25\right)}=\frac{50}{1}$
$⇒25000\left(\frac{1}{x}-\frac{1}{x+25}\right)=\frac{50}{1}\phantom{\rule{0ex}{0ex}}$
$⇒25000\left[\frac{x+25-x}{x\left(x+25\right)}\right]=\frac{50}{1}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{25000×25}{50}={x}^{2}+25x\phantom{\rule{0ex}{0ex}}$
$⇒12500={x}^{2}+25x\phantom{\rule{0ex}{0ex}}$
$⇒{x}^{2}+25x-12500=0\phantom{\rule{0ex}{0ex}}$
$⇒{x}^{2}+125x-100x-12500=0\phantom{\rule{0ex}{0ex}}$
$⇒x\left(x+125\right)-100\left(x+125\right)=0\phantom{\rule{0ex}{0ex}}$
$⇒\left(x-100\right)\left(x+125\right)=0\phantom{\rule{0ex}{0ex}}$

Since the number of children cannot be negative, x = 100

OR

Let the length of the shortest side be x cm.
Then, hypotenuse = (2x + 6) cm
And the third side = (2x + 6 - 2) cm = (2x + 4) cm
By Pythagoras' theorem, we have:
(2x+ 6)2 = x2 + (2x + 4)2

x2 - 8x - 20 = 0
x2 - 10x + 2x - 20 = 0
x(x - 10) + 2(x - 10) = 0
⇒ (x + 2)(x - 10) = 0
x = - 2 or x = 10
Since, the side of a triangle is never negative, the length of the shortest side = 10 cm
Length of the hypotenuse = (2x + 6) cm = 26 cm
Length of the third side = ( 26 - 2) = 24 cm
Hence, the sides of the triangle are 10 cm, 26 cm and 24 cm.

#### Question 27:

If the sum of first n, 2n and 3n terms of an AP be S1, S2 and S3 respectively, then prove that S3 = 3 (S2 − S1).

#### Answer:

Let a be the first term and d the common difference of the given AP. Then,
S1 = sum of first n terms of the given AP
S2 = sum of first 2n terms of the given AP
S3 = sum of first 3n terms of the given AP
Therefore,
${S}_{1}=\frac{n}{2}\left\{2a+\left(n-1\right)d\right\}\phantom{\rule{0ex}{0ex}}$
${S}_{2}=\frac{2n}{2}\left\{2a+\left(2n-1\right)d\right\}\phantom{\rule{0ex}{0ex}}$
${S}_{3}=\frac{3n}{2}\left\{2a+\left(3n-1\right)d\right\}$
Thus,
$3\left({S}_{2}-{S}_{1}\right)=3\left[\left\{2na+n\left(2n-1\right)d\right\}-\left\{na+\frac{1}{2}n\left(n-1\right)d\right\}\right]\phantom{\rule{0ex}{0ex}}$
$=3\left[na+\frac{3}{2}{n}^{2}d-\frac{1}{2}nd\right]=\frac{3n}{2}\left[2a+3nd-d\right]\phantom{\rule{0ex}{0ex}}$
$=\frac{3n}{2}\left[2a+\left(3n-1\right)d\right]={S}_{3}$
Hence, S3 = 3 (S2 − S1)

#### Question 28:

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of $1500\sqrt{3}\mathrm{m}$, find its speed.

#### Answer:

Let A and B be the two positions of the jet plane and O be the point of observation. Let OX be the horizontal ground.

Then, $\angle$XOA = 60° and $\angle$XOB = 30°
Draw AL ⊥ OX and BM ⊥ OX.
Let AL = BM = h metres = $1500\sqrt{3}\mathrm{m}$
From right ΔOLA, we have:

$⇒\frac{\mathrm{OL}}{1500\sqrt{3}}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$
$⇒\mathrm{OL}=1500$
From right ΔOMB, we have:
$\frac{\mathrm{OM}}{\mathrm{MB}}=\mathrm{cot}30°=\sqrt{3}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\mathrm{OM}}{1500\sqrt{3}}=\sqrt{3}\phantom{\rule{0ex}{0ex}}$
$⇒\mathrm{OM}=4500$
Therefore,
LM = (OM - OL) = (4500 - 1500) = 3000 m
Thus, the jet plane covers 3000 m in 15 seconds.
Hence, the speed of the jet plane =

#### Question 29:

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

#### Answer:

Given:  A circle with centre O and a tangent AB at point P of the circle.
To prove: OP ⊥ AB
Construction: Take a point Q, other than P, on the tangent AB. Join OQ.

Proof:
Q is a point on the tangent AB other than the point of contact P.
∴ Q lies outside the circle.
Let OQ intersect the circle at R.
Then,
OR < OQ    [ a part is less than the whole]    ................(i)
But OP =  OR    [radii of the same circle]     .................(ii)
∴ OP < OQ      [ from (i) and (ii)]
Thus, OP is shorter than any other line segment joining O to any point of AB, other than P.
In other words, OP is the shortest distance between the point O and the line AB.
But the shortest distance between a point and a line is the perpendicular distance.
∴ OP ⊥ AB.

#### Question 30:

A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the given figure.
Prove that AB + CD = AD + BC.
Figure

#### Answer:

We know that the lengths of tangents drawn from an exterior point to a circle are equal.
∴    AP = AS .......(i)              [tangents from A]
BP = BQ .......(ii)             [tangents from B]
CR = CQ .......(iii)            [tangents from C]
DR = DS .......(iv)             [tangents from D]
∴ AB + CD = AP + BP + CR + DR
= AS + BQ + CQ + DS            [From (i), (ii), (iii) and (iv)]
= (AS + DS) + (BQ + CQ)
= AD + BC
Thus, AB + CD = AD + BC

#### Question 31:

A solid is made up of a cube and a hemisphere is attached on its top, as shown in the figure. Each edge of the cube measures 5 cm and the hemisphere has a diameter of 4.2 cm. Find the total area to be painted.
Or, the diameters of the lower and upper ends of a bucket, in the form of a frustum of a cone, are 10 cm and 30 cm respectively. If its heights is 24 cm, find
(i) the capacity of the bucket
(ii) the area of the metal sheet used to make the bucket

#### Answer:

Total surface area of the cube = 6 × (edge)2
= (6 × 5 × 5) cm2
Area to be painted = (Total surface area of the cube) - (Base area of the hemisphere) + (CSA of the hemisphere)
= $\left(150-{\mathrm{\pi r}}^{2}+2{\mathrm{\pi r}}^{2}\right)\phantom{\rule{0ex}{0ex}}$ cm2
$=\left(150+{\mathrm{\pi r}}^{2}\right)\phantom{\rule{0ex}{0ex}}$cm2
$=\left(150+\frac{22}{7}×2.1×2.1\right)\phantom{\rule{0ex}{0ex}}$ cm2
$=\left(150+\frac{693}{50}\right)$ cm2
= (150 + 13.86) cm2 = 163.86 cm2
Hence, required area to be painted is 163.86 cm2.

OR

We have:
R = 15 cm, r = 5 cm and  h = 24 cm
l2 = h2 + (R - r)2
= ((24)2 + (15 - 5)2} = (576 + 100) = 676
l = 26 cm

(i)

$=\frac{1}{3}\left\{3.14×24\left({15}^{2}+{5}^{2}+15×5\right)\right\}{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}$

= 8164 cm3

(ii)

#### Question 32:

Find the value of k for which the points A(−1, 3), B(2, k) and C(5, −1) are collinear.

#### Answer:

Let A(−1, 3), B(2, k) and C(5, − 1) be the given points. Then,
(x1 = − 1, y1 = 3), (x2 = 2 , y2 = k) and (x3 = 5, y3 = − 1)
Since the given points are collinear, the area of the triangle formed by them must be 0.
$\frac{1}{2}$[x1(y2y3) + x2(y3y1) + x3(y1y2)] = 0
⇒ − 1(k + 1) + 2( − 1 − 3) + 5(3 − k) = 0
⇒ − 1(k + 1) + 2( − 4) + 5(3 − k) = 0
⇒ − k − 1 − 8 + 15 − 5k = 0
⇒ − 6k = − 6
⇒ 6k = 6 ⇒ k = 1
Hence, the required value of k is 1.

#### Question 33:

Two dice are thrown at the same time. Find the probability that the sum of two numbers appearing on the top of the dice is more than 9.

#### Answer:

When two dice are thrown simultaneously, the possible outcomes are :
(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)
Number of possible outcomes = 36
Let E be the event that the sum of the numbers appearing on the top of the two dice is more than 9.
The favourable outcomes are:
(4, 6), (5, 5), (5, 6), (6, 4), (6, 5) ,(6, 6)
Number of favourable outcomes = 6
P(getting a sum more than 9) = P(E) = $\frac{6}{36}=\frac{1}{6}$

#### Question 34:

A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical portion is 53 m, find the area of canvas needed to make the tent.

#### Answer:

For cylinder:
Height of cylinder (h) = 3 m
Base radius of cylinder (r) = 52.5 m

For cone:
Slant height of cone (l) = 53 m
Base radius of cone = Base radius of cylinder (r) = 52.5 m

∴ Area of canvas required =
= $\mathrm{\pi r}\left(\mathrm{l}+2\mathrm{h}\right)$ sq. units
= $\frac{22}{7}×52.5\left(53+2×3\right)$ m2
= $\frac{22}{7}×52.5×59$ m2 = 9,735 m2
Hence, area of canvas needed to make the tent = 9,735 m2

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