Rs Aggarwal 2017 for Class 9 Math Chapter 1 - Real Numbers

Share with your friends

Rs Aggarwal 2017 Solutions for Class 9 Math Chapter 1 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 9 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 9 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 252:

Question 1:

Mark (✓) against the correct answer in each of the following:
Which of the following is a rational number?
(a) $\frac{2}{\sqrt{3}}$
(b) $\frac{\sqrt{2}}{3}$
(c) $3 \sqrt{5}$
(d) $\frac{- 3}{5}$

Answer:

(d) $\frac{- 3}{5}$

The number of the form $\frac{p}{q}$ , where p and q are integers and $q \neq 0$ , are known as rational numbers.
Here $- 3$ and 5 are integers and $5 \neq 0$ .
Therefore, $\frac{- 3}{5}$ is a rational number.

Page No 252:

Question 2:

Mark (✓) against the correct answer in each of the following:
The value of k, for which the polynomial x³ − 4x² + 2x + k has 3 as its zero, is
(a) 3
(b) −3
(c) 6
(d) −6

Answer:

(a) 3

Let f(x) = x³ − 4x² + 2x + k.
3 is the zero of f(x).
Therefore, f(3) = 0
$\Rightarrow 3^{3} - 4 \times 3^{2} + 2 \times 3 + k = 0 \Rightarrow 27 - 36 + 6 + k = 0 \Rightarrow - 3 + k = 0 ∴ k = 3$

Page No 252:

Question 3:

Mark (✓) against the correct answer in each of the following:
Which of the following are zeros of the polynomial x³ + 2x² − 5x − 6?
(a) −2
(b) 2
(c) −3
(d) 3

Answer:

(b) 2 and (c) −3
Let f(x) = x³ + 2x² − 5x − 6.
We have to find the zeros of the polynomial.
⇒ f(x) = 0
⇒ x³ + 2x² − 5x − 6 = 0
⇒ x³ + 3x² −x² −3x −2x −6 = 0
⇒ x²(x + 3) −x(x + 3) −2(x + 3) = 0
⇒ (x + 3)(x² − x − 2)
⇒ (x + 3)(x² − 2x + x − 2)
⇒ (x + 3){x(x − 2) +1(x − 2)} = 0
⇒ (x + 3)(x − 2)(x + 1) = 0
Either x + 3 = 0 or x − 2 = 0 or x + 1 = 0
⇒ x = − 3 or 2 or −1

Disclaimer: Here both 2 and − 3 are correct answers.

Page No 252:

Question 4:

Mark (✓) against the correct answer in each of the following:
The factorisation of −x² + 7x − 12 yields
(a) (x − 3) (x − 4)
(b) (3 + x) (4 − x)
(c) (x − 4) (3 − x)
(d) (4 − x) (3 − x)

Answer:

(c) (x − 4) (3 − x)

−x² + 7x − 12 = − (x² − 7x + 12)
                       = − (x² − 4x − 3x + 12)
                       = − { x(x − 4) −3 (x − 4)}
                       = − (x − 4)(x − 3)
                       = (x − 4)(3 − x)

Page No 253:

Question 5:

Mark (✓) against the correct answer in each of the following:
In the given figure, ∠BOC = ?
(a) 45°
(b) 60°
(c) 75°
(d) 56°
Figure

Answer:

(b) 60°

Since, OAB is a straight line, the sum of all the angles
on the same side of AOB, at point O on it, is 180^o.
$∴ 3 x + 5 x + 4 x = 180 \Rightarrow 12 x = 180 \Rightarrow x = 15 ∴ ∠ B O C = 4 \times 15 = 60 °$

Page No 253:

Question 6:

Mark (✓) against the correct answer in each of the following:
In the given figure, ∆ABC is an equilateral triangle and ∆BDC is an isosceles right triangle, right-angled at D. Then ∠ACD = ?
(a) 60°
(b) 90°
(c) 120°
(d) 105°
Figure

Answer:

(d) 105°

$∆ A B C is an equilateral triangle . So, every angle of ∆ A B C is 60 ° . Therefore, ∠ A C B = 60 ° Now, ∆ B C D is an isosceles triangle and ∠ B D C = 90 ° . Here, ∠ B C D = ∠ C B D Now, ∠ B C D + ∠ C B D + ∠ B D C = 180 ° (Angle sum property) \Rightarrow ∠ B C D + ∠ C B D + 90 ° = 180 ° \Rightarrow 2 ∠ B C D = 90 ° \Rightarrow ∠ B C D = 45 ° From the figure, ∠ A C D = ∠ A C B + ∠ B C D = 60 ° + 45 ° ∴ ∠ A C D = 105 °$

Page No 253:

Question 7:

Mark (✓) against the correct answer in each of the following:
Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is
(a) 30 cm²
(b) 45 cm²
(c) 60 cm²
(d) 78 cm²
Figure

Answer:

(c) 60 cm²

^{$Area of an isosceles triangle = \frac{1}{4} b \sqrt{4 a^{2} - b^{2}} H ere, a = equal sides and b = third side Here, a = 13 cm and b = 24 cm Therefore, a rea = \frac{1}{4} \times 24 \times \sqrt{4 \times 13^{2} - 12^{2}} = 6 \times \sqrt{4 \times 13^{2} - 24^{2}} = 6 \times \sqrt{4 \times 169 - 576} = 6 \times \sqrt{676 - 576} = 6 \times \sqrt{100} = 6 \times 10 = 60 {cm}^{2}$}

Page No 253:

Question 8:

Mark (✓) against the correct answer in each of the following:
In an isosceles right triangle, the length of the hypotenuse is $4 \sqrt{2} cm$ .
(a) $4 \sqrt{3} cm$
(b) 6 cm
(c) 5 cm
(d) 4 cm
Figure

Answer:

(d) 4 cm

In a right-angled triangle,
${(Hypotenuse)}^{2} = {(B ase)}^{2} + {(P erpendicular)}^{2} It is g iven that the triangle is isosceles . Therefore, base = perpendicular = x cm (say) Given : Hypotenuse = 4 \sqrt{2} cm Therefore, {(4 \sqrt{2})}^{2} = x^{2} + x^{2} \Rightarrow 32 = 2 x^{2} \Rightarrow x^{2} = 16 \Rightarrow x = 4 Therefore, length of each equal side = 4 cm$

Page No 253:

Question 9:

If $x = 7 + 4 \sqrt{3}$ , find the value of $(\sqrt{x} + \frac{1}{\sqrt{x}})$ .

Answer:

$Given, x = 7 + 4 \sqrt{3} \Rightarrow x = 4 + 4 \sqrt{3} + 3 \Rightarrow x = {(2)}^{2} + 2 \times 2 \times \sqrt{3} + {(\sqrt{3})}^{2} \Rightarrow x = {(2 + \sqrt{3})}^{2} \Rightarrow \sqrt{x} = \pm (2 + \sqrt{3}) When \sqrt{x} = 2 + \sqrt{3}, \frac{1}{\sqrt{x}} = \frac{1}{2 + \sqrt{3}} \Rightarrow \frac{1}{\sqrt{x}} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} \Rightarrow \frac{1}{\sqrt{x}} = \frac{2 - \sqrt{3}}{{(2)}^{2} - {(\sqrt{3})}^{2}} \Rightarrow \frac{1}{\sqrt{x}} = \frac{2 - \sqrt{3}}{4 - 3} \Rightarrow \frac{1}{\sqrt{x}} = 2 - \sqrt{3} Therefore, when \sqrt{x} = - (2 + \sqrt{3}), \frac{1}{\sqrt{x}} = - (2 - \sqrt{3}) = - 2 + \sqrt{3} Therefore, \sqrt{x} + \frac{1}{\sqrt{x}} = 2 + \sqrt{3} + 2 - \sqrt{3} or - (2 + \sqrt{3}) + (- 2 + \sqrt{3}) = 4 or - 4$

Page No 253:

Question 10:

Factorise: $(7 a^{3} + 56 b^{3})$ .

Answer:

$7 a^{3} + 56 b^{3} = 7 (a^{3} + 8 b^{3}) = 7 [{(a)}^{3} + {(2 b)}^{3}] = 7 (a + 2 b) [{(a)}^{2} + {(2 b)}^{2} - a \times 2 b] = 7 (a + 2 b) (a^{2} + 4 b^{2} - 2 a b)$

Page No 253:

Question 11:

Find the value of a for which (x − 1) is a factor of the polynomial (a²x³ − 4ax + 4a − 1).

Answer:

Let f(x) = a²x³ − 4ax + 4a − 1.
(x − 1) is a factor of f(x)
Therefore, f(1) = 0
$a^{2} \times 1^{3} - 4 \times a \times 1 + 4 a - 1 = 0 \Rightarrow a^{2} - 1 = 0 \Rightarrow a^{2} = 1 Therefore, a = \pm 1$

Page No 253:

Question 12:

In the given figure, if AC = BD, show that AB = CD. State Euclid's axiom used for it.
Figure

Answer:

From the given figure,
AC = AB + BC
BD = BC + CD
It is given that AC = BD.
Putting the values of AC and BD, we get:
AB + BC = BC + CD..................(1)
According to Euclid's axiom, when equals are subtracted from equals, the remainders are also equal.
Subtracting BC from both sides in equation (1), we get:
AB + BC $-$ BC = BC + CD $-$ BC
AB = CD
Hence, proved.

Page No 253:

Question 13:

In ∆ABC, if 2∠A = 3∠B = 6∠C, calculate the measure of ∠B.

Answer:

In a ∆ABC, 2 $∠$ A = 3 $∠$ B = 6 $∠$ C = x^o (say)

$Therefore, ∠ A = \frac{x °}{2} ∠ B = \frac{x °}{3} ∠ C = \frac{x °}{6} By angle sum property in ∆ A B C, ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow \frac{x °}{2} + \frac{x °}{3} + \frac{x °}{6} = 180 ° \Rightarrow x ° = 180 ° Therefore, ∠ B = \frac{x °}{3} = \frac{180 °}{3} = 60 ° .$

Page No 253:

Question 14:

In the given figure, ∠BAC = 30°, ∠ABC = 50° and ∠CDE = 40°. Find ∠AED.
Figure

Answer:

Side BC of $∆$ ABC is produced to D.
$∴ ∠ A C D = ∠ C A B + ∠ A B C [E xterior angle property] \Rightarrow ∠ A C D = (30 + 50) ° \Rightarrow ∠ A C D = 80 °$

Side CE of $∆$ CED is produced to A.
$∴ ∠ A E D = ∠ E C D + ∠ E D C [Exterior angle property] \Rightarrow ∠ A E D = (80 + 40) ° \Rightarrow ∠ A E D = 120 °$

Page No 254:

Question 15:

If $x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$ and $y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ , find the value of (x² + y²).
or
Simplify: $\frac{(7 + 3 \sqrt{5})}{(3 + \sqrt{5})} - \frac{(7 - 3 \sqrt{5})}{(3 - \sqrt{5})}$

Answer:

$Given, x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}, y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} Therefore, x^{2} = {(\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}})}^{2} = \frac{5 + 3 + 2 \sqrt{15}}{5 + 3 - 2 \sqrt{15}} = \frac{8 + 2 \sqrt{15}}{8 - 2 \sqrt{15}} And y^{2} = {(\frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}})}^{2} = \frac{5 + 3 - 2 \sqrt{15}}{5 + 3 + 2 \sqrt{15}} = \frac{8 - 2 \sqrt{15}}{8 + 2 \sqrt{15}}$

$Now, x^{2} + y^{2} = \frac{8 + 2 \sqrt{15}}{8 - 2 \sqrt{15}} + \frac{8 - 2 \sqrt{15}}{8 + 2 \sqrt{15}} = \frac{{(8 + 2 \sqrt{15})}^{2} + {(8 - 2 \sqrt{15})}^{2}}{(8 - 2 \sqrt{15}) (8 + 2 \sqrt{15})} = \frac{{(8)}^{2} + {(2 \sqrt{15})}^{2} + 2 \times 8 \times 2 \sqrt{15} + {(8)}^{2} + {(2 \sqrt{15})}^{2} - 2 \times 8 \times 2 \sqrt{15}}{{(8)}^{2} - {(2 \sqrt{15})}^{2}} = \frac{64 + 60 + 64 + 60}{64 - 60} = \frac{248}{4} = 62$

OR

$\frac{(7 + 3 \sqrt{5})}{(3 + \sqrt{5})} - \frac{(7 - 3 \sqrt{5})}{(3 - \sqrt{5})} = \frac{(7 + 3 \sqrt{5}) (3 - \sqrt{5}) - (7 - 3 \sqrt{5}) (3 + \sqrt{5})}{(3 + \sqrt{5}) (3 - \sqrt{5})} = \frac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15 - 21 - 7 \sqrt{5} + 9 \sqrt{5} + 15}{{(3)}^{2} - {(\sqrt{5})}^{2}} = \frac{4 \sqrt{5}}{9 - 5} = \frac{4 \sqrt{5}}{4} = \sqrt{5}$

Page No 254:

Question 16:

If 2 and $\frac{- 1}{3}$ are the zeros of the polynomial 3x³ − 2x² − 7x − 2, find the third zero of the polynomial.

Answer:

$To find the zeros of a polynomial, we equate th e polynomial to 0 . 3 x^{3} - 2 x^{2} - 7 x - 2 = 0 \Rightarrow 3 x^{3} - 6 x^{2} + 4 x^{2} - 8 x + x - 2 = 0 \Rightarrow 3 x^{2} (x - 2) + 4 x (x - 2) + 1 (x - 2) = 0 \Rightarrow (x - 2) (3 x^{2} + 4 x + 1) = 0 \Rightarrow (x - 2) (3 x^{2} + 3 x + x + 1) = 0 \Rightarrow (x - 2) [3 x (x + 1) + 1 (x + 1)] = 0 \Rightarrow (x - 2) (x + 1) (3 x + 1) = 0 \Rightarrow x - 2 = 0 or x + 1 = 0 or 3 x + 1 = 0 \Rightarrow x = 2 or - 1 or \frac{- 1}{3} As 2 and \frac{- 1}{3} are the given zeros, the third zero is - 1 .$

Page No 254:

Question 17:

Find the remainder when the polynomial f(x) = 4x³ − 12x² + 14x − 3 is divided by (2x − 1).

Answer:

$Let p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3 .$

By the remainder theorem, we know that when p(x) is divided by (2x - 1), the remainder is $p (\frac{1}{2})$ .
Now,
$p (\frac{1}{2}) = 4 \times {(\frac{1}{2})}^{3} - 12 \times {(\frac{1}{2})}^{2} + 14 \times \frac{1}{2} - 3 = \frac{1}{2} - 3 + 7 - 3 = \frac{3}{2}$

Page No 254:

Question 18:

Factorise: (p − q)³ + (q − r)³ + (r− p)³

Answer:

(p − q)³ + (q − r)³ + (r− p)³
Let p − q = x
q − r = y
r− p = z
Therefore, x + y+ z = p − q + q − r + r − p = 0
So,
(p − q)³ + (q − r)³ + (r− p)³= x³ + y³ + z³
= 3xyz (when x + y + z = 0, then x³ + y³ + z³ = 3xyz )
= 3(p − q)(q − r)(r− p)

Page No 254:

Question 19:

In the given figure, in ∆ABC, it is given that ∠B = 40° and ∠C = 50°. DE || BC and EF || AB. Find: (i) ∠ADE + ∠MEN (ii) ∠BDE and (iii) ∠BFE.
Figure

Answer:

$(i) D E ∥ B C Therefore, ∠ A D E = ∠ A B C = 40 ° (A lternate angle s) E F ∥ A B and M F is the produced line of E F . Therefore, M F ∥ A B . \Rightarrow ∠ M E N = ∠ A D E = 40 ° (A lternate angle s) ∠ A D E + ∠ M E N = 40 ° + 40 ° = 80 ° (ii) D E ∥ B C and A B is the transversal . Interior angles on the same side of a transversal are supplementary . ∴ ∠ B D E = 180 ° - ∠ A B C = 180 ° - 40 ° = 140 ° (iii) E F ∥ A B and B C is the transversal . Interior angles on the same side of a transversal are supplementary . ∴ ∠ B F E = 180 ° - ∠ A B C = 180 ° - 40 ° = 140 °$

Page No 254:

Question 20:

In the given figure, ∆ABC and ∆ABD are such that AD = BC, ∠1 = ∠2 and ∠3 = ∠4. Prove that BD = AC.
Figure

Answer:

$In ∆ A B C and ∆ A B D, A D = B C (Given) A B = A B (C ommon side) Given : ∠ 1 = ∠ 2 and ∠ 3 = ∠ 4 Therefore, ∠ 1 + ∠ 3 = ∠ 2 + ∠ 4 \Rightarrow ∠ D A B = ∠ A B C ∴ By S A S, ∆ D A B ≅ ∆ C B A Therefore, B D = A C (C orresponding sides of congruent triangles are equal)$

Page No 254:

Question 21:

In the given figure, C is the mid-point of AB. If ∠DCA = ∠ECB and ∠DBC = ∠EAC, prove that DC = EC.
Figure

Answer:

$In ∆ E C A and ∆ D C B, ∠ E A C = ∠ D B C (given) A C = B C (C is the mid point of A B) Now, ∠ D C A = ∠ E C B \Rightarrow ∠ D C A + ∠ D C E = ∠ E C B + ∠ D C E \Rightarrow ∠ A C E = ∠ B C D Therefore, by A S A congruency criteria, ∆ E C A ≅ ∆ D C B . \Rightarrow D C = E C (C orresponding si d es of congruent triangles are equal)$

Page No 254:

Question 22:

In ∆ABC, AL ⊥ BC and AM is the bisector of ∠A. Show that $∠ L A M = \frac{1}{2} (∠ B - ∠ C)$ .
Figure

Answer:

$In ∆ A B C, ∠ A = 180 ° - (∠ B + ∠ C) i . e . ∠ B A C = 180 ° - (∠ B + ∠ C) Now, A M is the angle bisector of ∠ A . ∴ ∠ B A M = \frac{1}{2} [180 ° - (∠ B + ∠ C)] ∴ ∠ B A M = 90 ° - \frac{1}{2} (∠ B + ∠ C) - - - - - (i) In ∆ A B L, ∠ A L B = 90 ° ∴ ∠ B A L = 180 ° - (∠ B + ∠ A L B) \Rightarrow ∠ B A L = 180 ° - (∠ B + 90 °) ∴ ∠ B A L = 90 ° - ∠ B - - - - - - - - - - (ii) Now, ∠ L A M = ∠ B A M - ∠ B A L Using (i) and (ii), we have : ∠ L A M = [90 ° - \frac{1}{2} (∠ B + ∠ C)] - [90 ° - ∠ B] = 90 ° - \frac{1}{2} ∠ B - \frac{1}{2} ∠ C - 90 ° + ∠ B ∴ ∠ L A M = \frac{1}{2} ∠ B - \frac{1}{2} ∠ C Hence, proved .$

Page No 255:

Question 23:

In the given figure, AB || CD, ∠BAE = 100° and ∠AEC = 30°. Find ∠DCE.
Figure

Answer:

$Construction : D raw G H ∥ F C as shown in the book by dotted line G H, which meets A B at A and C D at H . Now, ∠ G A E = ∠ A E C = 30 ° (alternate angle s)$
$And ∠ G A B = ∠ B A E - ∠ G A E = 100 ° - 30 ° = 70 ° Now, A B ∥ C D and G H is the transversal . ∠ D H A = ∠ G A B = 70 ° (corresponding angles) Again G H ∥ F C and D C is the transversal . ∴ ∠ D C E = ∠ D H A = 70 ° (corresponding angles)$

Page No 255:

Question 24:

Factorise: a³ − b³ + 1 + 3ab

Answer:

$a^{3} - b^{3} + 1 + 3 a b = {(a)}^{3} + {(- b)}^{3} + {(1)}^{3} - 3 (a) (- b) (1) = [a + (- b) + 1] [a^{2} + {(- b)}^{2} + 1^{2} - a \times (- b) - (- b) \times 1 - a \times 1] = (a - b + 1) (a^{2} + b^{2} + 1 + a b + b - a)$

Page No 255:

Question 25:

If $x = \frac{1}{(2 - \sqrt{3})}$ , show that the value of (x³ − 2x² − 7x + 5) is 3.
Or
Simplify: $\frac{1}{(1 + \sqrt{2})} + \frac{1}{(\sqrt{2} + \sqrt{3})} + \frac{1}{(\sqrt{3} + \sqrt{4})} + . . . . . . + \frac{1}{(\sqrt{8} + \sqrt{9})}$

Answer:

$x = \frac{1}{2 - \sqrt{3}} By rationalising we get : x = \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} \Rightarrow x = \frac{2 + \sqrt{3}}{4 - 3} \Rightarrow x = 2 + \sqrt{3} \Rightarrow x - 2 = \sqrt{3} Squaring both sides, we get : {(x - 2)}^{2} = 3 \Rightarrow x^{2} + 4 - 4 x = 3 \Rightarrow x^{2} - 4 x + 1 = 0 So, we can write x^{3} - 2 x^{2} - 7 x + 5 as x^{3} - 4 x^{2} + x + 2 x^{2} - 8 x + 2 + 3 . Hence, x^{3} - 2 x^{2} - 7 x + 5 = x^{3} - 4 x^{2} + x + 2 x^{2} - 8 x + 2 + 3 = x (x^{2} - 4 x + 1) + 2 (x^{2} - 4 x + 1) + 3 = x (0) + 2 (0) + 3 = 3$

OR

$\frac{1}{(1 + \sqrt{2})} + \frac{1}{(\sqrt{2} + \sqrt{3})} + \frac{1}{(\sqrt{3} + \sqrt{4})} + . . . . . . + \frac{1}{(\sqrt{8} + \sqrt{9})} B y rationalising we get : \frac{1}{(1 + \sqrt{2})} = \frac{1}{(\sqrt{2} + 1)} \times \frac{(\sqrt{2} - 1)}{(\sqrt{2} - 1)} = \frac{(\sqrt{2} - 1)}{2 - 1} = \sqrt{2} - 1 \frac{1}{(\sqrt{2} + \sqrt{3})} = \frac{1}{(\sqrt{3} + \sqrt{2})} \times \frac{(\sqrt{3} - \sqrt{2})}{(\sqrt{3} - \sqrt{2})} = \frac{(\sqrt{3} - \sqrt{2})}{3 - 2}$

$= \sqrt{3} - \sqrt{2} Similarly, \frac{1}{\sqrt{3} + \sqrt{4}} = \sqrt{4} - \sqrt{3} \frac{1}{\sqrt{4} + \sqrt{5}} = \sqrt{5} - \sqrt{4} \frac{1}{\sqrt{5} + \sqrt{6}} = \sqrt{6} - \sqrt{5} \frac{1}{\sqrt{6} + \sqrt{7}} = \sqrt{7} - \sqrt{6} \frac{1}{\sqrt{7} + \sqrt{8}} = \sqrt{8} - \sqrt{7} \frac{1}{\sqrt{8} + \sqrt{9}} = \sqrt{9} - \sqrt{8} Adding all the terms of the left side and right side, we get : \frac{1}{(1 + \sqrt{2})} + \frac{1}{(\sqrt{2} + \sqrt{3})} + \frac{1}{(\sqrt{3} + \sqrt{4})} + . . . . . + \frac{1}{(\sqrt{8} + \sqrt{9})} = \sqrt{9} - 1 = 3 - 1 = 2$

Page No 255:

Question 26:

If $x = \frac{\sqrt{a + 2 b} + \sqrt{a - 2 b}}{\sqrt{a + 2 b} - \sqrt{a - 2 b}}$ , then show that $b x^{2} - a x + b = 0$ .

Answer:

$Given, x = \frac{\sqrt{a + 2 b} + \sqrt{a - 2 b}}{\sqrt{a + 2 b} - \sqrt{a - 2 b}} Rationalising the denominator, we get : x = \frac{(\sqrt{a + 2 b} + \sqrt{a - 2 b})}{(\sqrt{a + 2 b} - \sqrt{a - 2 b})} \times \frac{(\sqrt{a + 2 b} + \sqrt{a - 2 b})}{(\sqrt{a + 2 b} + \sqrt{a - 2 b})} \Rightarrow x = \frac{{(\sqrt{a + 2 b} + \sqrt{a - 2 b})}^{2}}{{(\sqrt{a + 2 b})}^{2} - {(\sqrt{a - 2 b})}^{2}} \Rightarrow x = \frac{{(\sqrt{a + 2 b})}^{2} + {(\sqrt{a - 2 b})}^{2} + 2 \sqrt{a + 2 b} \sqrt{a - 2 b}}{(a + 2 b) - (a - 2 b)} \Rightarrow x = \frac{a + 2 b + a - 2 b + 2 \sqrt{a^{2} - {(2 b)}^{2}}}{4 b} \Rightarrow x = \frac{2 a + 2 \sqrt{a^{2} - 4 b^{2}}}{4 b} \Rightarrow x = \frac{a + \sqrt{a^{2} - 4 b^{2}}}{2 b} \Rightarrow 2 b x = a + \sqrt{a^{2} - 4 b^{2}} \Rightarrow 2 b x - a = \sqrt{a^{2} - 4 b^{2}} Squaring both sides we get : {(2 b x - a)}^{2} = {(\sqrt{a^{2} - 4 b^{2}})}^{2} \Rightarrow 4 b^{2} x^{2} + a^{2} - 4 a b x = a^{2} - 4 b^{2} \Rightarrow 4 b^{2} x^{2} - 4 a b x + 4 b^{2} = 0 \Rightarrow 4 b (b x^{2} - a x + b) = 0 Therefore, b x^{2} - a x + b = 0 Hence, proved .$

Page No 255:

Question 27:

If (x³ + mx² − x + 6) has (x − 2) as a factor and leaves a remainder r when divided by (x − 3), find the values of m and r.

Answer:

$Let f (x) = x^{3} + m x^{2} - x + 6 . Here, (x - 2) is a factor of f (x) . Therefore, f (2) = 0 \Rightarrow 2^{3} + m \times 2^{2} - 2 + 6 = 0 \Rightarrow 8 + 4 m - 2 + 6 = 0 \Rightarrow 4 m + 12 = 0 \Rightarrow 4 m = - 12 \Rightarrow m = - 3 Now, using m = - 3 i n f (x), we get : f (x) = x^{3} - 3 x^{2} - x + 6 When f (x) is divided by (x - 3), the remainder will be f (3), and the remainder is given as r . Now, f (3) = 3^{3} - 3 \times 3^{2} - 3 + 6 or, r = 27 - 27 - 3 + 6 or, r = 3 Therefore, m = - 3 and r = 3 .$

Page No 255:

Question 28:

If r and s be the remainders when the polynomials (x³ + 2x² − 5ax − 7) and (x³ + ax² − 12ax + 6) are divided by (x + 1) and (x − 2), respectively and 2r + s = 6, find the value of a.

Answer:

$Let f (x) = x^{3} + 2 x^{2} - 5 a x - 7 . When f (x) is divided by (x + 1), the remainder is r . Therefore, f (- 1) = r \Rightarrow {(- 1)}^{3} + 2 {(- 1)}^{2} - 5 a (- 1) - 7 = r \Rightarrow - 1 + 2 + 5 a - 7 = r \Rightarrow 5 a - 6 = r \Rightarrow r = 5 a - 6 - - - - - - - (1) Again, let g (x) = x^{3} + a x^{2} - 12 a x + 6 . When g (x) is divided by (x - 2), the remainder is s . Therefore, g (2) = s \Rightarrow {(2)}^{3} + a {(2)}^{2} - 12 a (2) + 6 = s \Rightarrow 8 + 4 a - 24 a + 6 = s \Rightarrow s = - 20 a + 14 - - - - - (2) From (1) \times 2 + (2), we have : 2 r + s = 2 (5 a - 6) + (- 20 a + 14) \Rightarrow 6 = 10 a - 12 - 20 a + 14 \Rightarrow 6 = - 10 a + 2 \Rightarrow - 10 a = 4 \Rightarrow a = \frac{- 2}{5}$

Page No 255:

Question 29:

Prove that:
$(a + b)^{3} (b + c)^{3} + (c + a)^{3} - 3 (a + b) (b + c) (c + a) = 2 (a^{3} + b^{3} + c^{3} - 3 a b c)$ .

Answer:

$L . H . S . = {(a + b)}^{3} + {(b + c)}^{3} + {(c + a)}^{3} - 3 (a + b) (b + c) (c + a) = (a + b + b + c + c + c + a) [{(a + b)}^{2} + {(b + c)}^{2} + {(c + a)}^{2} - (a + b) (b + c) - (b + c) (c + a) - (c + a) (a + b)] = (2 a + 2 b + 2 c) (a^{2} + b^{2} + 2 a b + b^{2} + c^{2} + 2 b c + c^{2} + a^{2} + 2 c a - a b - a c - b^{2} - b c - b c - a b - c^{2} - c a - c a - b c - a^{2} - a b) = 2 (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = 2 (a^{3} + b^{3} + c^{3} - 3 a b c) = R . H . S . Hence, proved .$

Page No 255:

Question 30:

On a graph paper plot the following points:
A(3,3), B(2, 4), C(5, 5), D(0, 2), E(3, −3) and F(−5,−5).
Which of these points are the mirror images on (i) x-axis (ii) y-axis?

Answer:

Points A(3,3), B(2,4), C(5,5), D(0,2), E(3,-3) and F(-5,-5)
are shown on the graph paper below.

(3,-3) is the mirror image of (3,3) on the x-axis .

Page No 255:

Question 31:

In the given figure, in a ∆ABC, BE ⊥ AC, ∠EBC = 40° and ∠DAC = 30°. Find the values of x, y and z.
Figure

Answer:

$In ∆ BCE, ∠ CBE + ∠ CEB + ∠ BCE = 180 ° (Angle sum property) \Rightarrow 40 ° + 90 ° + x ° = 180 ° \Rightarrow x ° + 130 ° = 180 ° \Rightarrow x ° = 180 ° - 130 ° \Rightarrow x ° = 50 ° Therefore, x = 50 I n ∆ A D C, ∠ A D C + ∠ A C D + ∠ C A D = 180 ° (Angle sum property) \Rightarrow ∠ A D C + 50 ° + 30 ° = 180 ° \Rightarrow ∠ A D C = 180 ° - 80 ° \Rightarrow ∠ A D C = 100 ° Now, ∠ A D B + ∠ A D C = 180 ° (Linear pair) \Rightarrow y ° + 100 ° = 180 ° \Rightarrow y ° = 180 ° - 100 ° \Rightarrow y ° = 80 ° Therefore, y = 80$

$Let, l ine segment s A D and B E intersect at F . Therefore, s ide E F of △ A E F is produced to B . ∠ A F B = ∠ F A E + ∠ A E F (Exterior angle property) \Rightarrow z ° = 30 ° + 90 ° \Rightarrow z ° = 120 ° Therefore, z = 120 .$

Page No 256:

Question 32:

In the given figure, ABC is a triangle in which AB = AC. D is a point in the interior of ∆ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC.
Figure

Answer:

$In ∆ B C D, ∠ D B C = ∠ D C B Therefore, ∆ D B C is an isosceles triangle and the opposite sides of equal angles are equal . ∴ D B = D C - - - - (1)$
$A B = A C So, ∆ A B C is isosceles triangle and the opposite angles of the equal sides are equal . ∴ ∠ A B C = ∠ A C B and ∠ D B C = ∠ D C B ∴ ∠ A B C - ∠ D B C = ∠ A C B - ∠ D C B ∴ ∠ A B D = ∠ A C D - - - - - - - - - (2)$
$And A B = A C - - - - - - (3) (given)$
From (1), (2) and (3), we have:

$∆ A D B ≅ ∆ A D C [By SAS congruency criteria] ∴ ∠ B A D = ∠ C A D Therefore, A D bisects ∠ B A C .$

Page No 256:

Question 33:

In the given figure, ABCD is a square and EF is parallel to diagonal DB and EM = FM. Prove that: (i) BF = DE (ii) AM bisects ∠BAD.
Figure

Answer:

$(i) We have, C D = B C - - - - - - - (1) (S ides of a square are equal) and ∠ B D C = ∠ D B C = 45 ° Given : E F ∥ D B ∴ ∠ B D C = ∠ F E C (C orresponding angles) ∴ ∠ F E C = 45 ° Similarly, ∠ E F C = 45 ° ∴ △ C E F is an isosceles triangle . and C E = C F - - - - - - - (2) From (1) - (2), we get : C D - C E = B C - C F \Rightarrow D E = B F ∴ B F = D E$

$(ii) In △ A B F and △ A D E, A B = A D (All sides of a square are equal) ∠ A B F = ∠ A D E (Both angles are equal to 90 °) B F = D E (Proved above) ∴ △ A B F ≅ △ A D E (by S A S congruency criteria) ∴ ∠ B A F = ∠ D A E - - - - (3) And A F = A E - - - - - - (4) Now, In △ A M E and △ A M F, A M = A M (C ommon side) E M = F M (G iven) A E = A F [from (4)] ∴ △ A M E ≅ △ A M F (By SSS congruency criteria) ∴ ∠ E A M = ∠ F A M ∴ ∠ F A M = ∠ E A M - - - - - (5) From (3) + (5), we get : ∠ B A F + ∠ F A M = ∠ D A E + ∠ E A M ∠ B A M = ∠ D A M Therefore, A M bisects ∠ B A D .$

Page No 256:

Question 34:

In the given figure, AB || CD. If ∠BAE = 100° and ∠ECD = 120° then x = ?
Figure

Answer:

Draw $E F ∥ A B ∥ C D$ .
Then, $∠ A E F + ∠ C E F = x °$
Now, $E F ∥ A B$ and AE is the transversal.
$∴ ∠ A E F + ∠ B A E = 180 ° [Sum of consecutive interior angles is supplementary] \Rightarrow ∠ A E F + 100 ° = 180 ° \Rightarrow ∠ A E F = 80 °$
Again, $E F ∥ C D$ and CE is the transversal.
$∠ C E F + ∠ E C D = 180 ° [Sum of consecutive interior angles is supplementary] \Rightarrow ∠ C E F + 120 ° = 180 ° \Rightarrow ∠ C E F = 60 °$
Therefore,
$x ° = ∠ A E F + ∠ C E F = (80 + 60) ° = 140 ° Therefore, x = 140$

Page No 257:

Question 1:

Mark (✓) against the correct answer in each of the following:
An irrational number between 2 and 2.5 is
(a) $\sqrt{3}$
(b) 2.3
(c) $\sqrt{5}$
(d) $2.34$

Answer:

Page No 257:

Question 2:

Mark (✓) against the correct answer in each of the following:
Which of the following is a polynomial in one variable?
(a) x² + x⁻²
(b) $\sqrt{3 x} + 9$
(c) $x^{2} + 2 x - \sqrt{x} + 3$
(d) $\sqrt{3} + 2 x - x^{2}$

Answer:

(d) $\sqrt{3} + 2 x - x^{2}$

Here, the powers of x in all the terms are non-negative integers.
Therefore, it is a polynomial in one variable.

Page No 257:

Question 3:

Mark (✓) against the correct answer in each of the following:
$\frac{1}{(\sqrt{18} - \sqrt{32})} = ?$
(a) $\sqrt{2}$
(b) $\frac{1}{\sqrt{2}}$
(c) $- \sqrt{2}$
(d) $\frac{- 1}{\sqrt{2}}$

Answer:

(d) $\frac{- 1}{\sqrt{2}}$

$\frac{1}{\sqrt{18} - \sqrt{32}} = \frac{1}{3 \sqrt{2} - 4 \sqrt{2}} = \frac{1}{- \sqrt{2}} = \frac{- 1}{\sqrt{2}}$

Page No 257:

Question 4:

Mark (✓) against the correct answer in each of the following:
If $p (x) = (x^{4} - x^{2} + x)$ , then $p (\frac{1}{2}) = ?$
(a) $\frac{1}{16}$
(b) $\frac{3}{16}$
(c) $\frac{5}{16}$
(d) $\frac{7}{16}$

Answer:

(c) $\frac{5}{16}$

$p (x) = x^{4} - x^{2} + x Therefore, p (\frac{1}{2}) = {(\frac{1}{2})}^{4} - {(\frac{1}{2})}^{2} + \frac{1}{2} \Rightarrow p (\frac{1}{2}) = \frac{1}{16} - \frac{1}{4} + \frac{1}{2} \Rightarrow p (\frac{1}{2}) = \frac{1 - 4 + 8}{16} \Rightarrow p (\frac{1}{2}) = \frac{5}{16}$

Page No 258:

Question 5:

Mark (✓) against the correct answer in each of the following:
If p(x) = x³ + x² + ax + 115 is exactly divisible by (x + 5) then a = ?
(a) 8
(b) 6
(c) 5
(d) 3

Answer:

(d) 3

p(x) = x³ + x² + ax + 115 is exactly divisible by (x + 5).
Therefore, p( $-$ 5) = 0
$\Rightarrow {(- 5)}^{3} + {(- 5)}^{2} + a \times (- 5) + 115 = 0 \Rightarrow - 125 + 25 - 5 a + 115 = 0 \Rightarrow - 5 a + 15 = 0 \Rightarrow 5 a = 15 ∴ a = 3$

Page No 258:

Question 6:

Mark (✓) against the correct answer in each of the following:
The equation of y-axis is
(a) y = 0
(b) x = 0
(c) y = 0
(d) y = constant

Answer:

(b) x = 0

The equation of y-axis is x = 0.

Page No 258:

Question 7:

Mark (✓) against the correct answer in each of the following:
In the given figure, the value of x is
(a) 10
(b) 12
(c) 15
(d) 20
Figure

Answer:

(d) 20

$∠ A C D + ∠ B C D = 180 ° (Linear pair) \Rightarrow 4 x ° + 5 x ° = 180 ° \Rightarrow 9 x ° = 180 ° \Rightarrow x ° = 20 ° Therefore, x = 20$

Page No 258:

Question 8:

Mark (✓) against the correct answer.
In the given figure, CE || BA and EF || CD. If ∠BAC = 40°, ∠ACB = 65° and ∠CEF = x° then the value of x is
(a) 40
(b) 65
(c) 75
(d) 105
Figure

Answer:

(d) 105

$C E ∥ A B ∴ ∠ A C E = ∠ B A C = 40 ° (Alternate angles) ∴ ∠ B C E = ∠ B C A + ∠ A C E = 65 ° + 40 ° = 105 ° Again E F ∥ C D and line segment E C intersect s them . ∴ ∠ C E F = ∠ B C E (alternate angle) \Rightarrow x ° = 105 ° Therefore, x = 105 .$

Page No 258:

Question 9:

Factorise: $\sqrt{2} x^{2} + 3 x + \sqrt{2}$ .

Answer:

$\sqrt{2} x^{2} + 3 x + \sqrt{2} = \sqrt{2} x^{2} + 2 x + x + \sqrt{2} = \sqrt{2} x (x + \sqrt{2}) + 1 (x + \sqrt{2}) = (x + \sqrt{2}) (\sqrt{2} x + 1)$

Page No 258:

Question 10:

Prove that $\sqrt{5}$ is an irrational number.

Answer:

We will show that $\sqrt{5}$ is irrational by contradiction.
Let us assume that $\sqrt{5}$ is rational.

Therefore we can represent it in the form of $\frac{a}{b}$ .
$\sqrt{5}$ = $\frac{a}{b}$
Let us assume that $\frac{a}{b}$ is in its lowest form, i.e. a and b are co-prime.
$\sqrt{5} = \frac{a}{b} \Rightarrow a = \sqrt{5} b \Rightarrow a^{2} = 5 b^{2} - - - - - (1)$

Therefore, 5 is the factor of $a^{2}$ . If 5 is a factor of $a^{2}$ , 5 will also be the factor of a.

So ,we can write a = 5c.

${(5 c)}^{2} = 5 b^{2} \Rightarrow 25 c^{2} = 5 b^{2} \Rightarrow b^{2} = 5 c^{2}$

Therefore b² is divisible by 5.
Since, it is divisible by 5, b will also be divisible by 5.

Therefore, a and b have at least 5 as a common factor.

But this contradicts that a and b are co-prime.

This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.

Therefore, $\sqrt{5}$ is irrational.

Page No 258:

Question 11:

Draw the graph of the equation y = 2x + 3.

Answer:

To draw the graph of y = 2x + 3, make a table of values of x and y satisfying the given equation.
Now, we have

x	y
0	3
$- \frac{3}{2}$	0

Page No 258:

Question 12:

If $x = (3 + \sqrt{8})$ , find the value of $(x^{2} + \frac{1}{x^{2}})$ .

Answer:

$x = 3 + \sqrt{8} \Rightarrow \frac{1}{x} = \frac{1}{3 + \sqrt{8}} \Rightarrow \frac{1}{x} = \frac{1}{3 + \sqrt{8}} \times \frac{3 - \sqrt{8}}{3 - \sqrt{8}} \Rightarrow \frac{1}{x} = \frac{3 - \sqrt{8}}{9 - 8} \Rightarrow \frac{1}{x} = 3 - \sqrt{8}$
$and x + \frac{1}{x} = 3 + \sqrt{8} + 3 - \sqrt{8} = 6 So, x + \frac{1}{x} = 6 Squaring both sides, we get : \Rightarrow {(x + \frac{1}{x})}^{2} = 6^{2} \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 \times x \times \frac{1}{x} = 36 \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 36 \Rightarrow x^{2} + \frac{1}{x^{2}} = 36 - 2 ∴ x^{2} + \frac{1}{x^{2}} = 34$

Page No 258:

Question 13:

Find the area of the triangle whose sides measure 52 cm, 56 cm and 60 cm.

Answer:

$Let the s ides of the triangle be a = 52 cm b = 56 cm c = 60 cm ∴ s = \frac{a + b + c}{2} = \frac{52 + 56 + 60}{2} = 84 cm$

$Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{84 (84 - 52) (84 - 56) (84 - 60)} = \sqrt{84 \times 32 \times 28 \times 24} = \sqrt{3 \times 28 \times 4 \times 4 \times 2 \times 28 \times 2 \times 2 \times 2 \times 3} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (4 \times 4) \times (28 \times 28)} = 2 \times 2 \times 3 \times 4 \times 28 = 1344 {cm}^{2}$

Page No 258:

Question 14:

In the given figure, AB || CD. Find the value of x.
Figure

Answer:

Draw $E F ∥ A B ∥ C D$ .
Now, $E F ∥ A B and B F$ is the transversal.
$∴ ∠ A B F + ∠ E F B = 180 ° [Sum of consecutive interior angles is supplementary] \Rightarrow 40 ° + ∠ E F B = 180 ° \Rightarrow ∠ E F B = 140 °$
Also,
$E F ∥ C D$ and FD is the transversal.
$∴ ∠ C D F + ∠ E F D = 180 ° [Sum of consecutive interior angles is supplementary] \Rightarrow 35 ° + ∠ E F D = 180 ° \Rightarrow ∠ E F D = 145 °$
Therefore,
$x ° = ∠ E F B + ∠ E F D \Rightarrow x ° = (140 + 145) ° \Rightarrow x ° = 285 ° Therefore, x = 285$

Page No 258:

Question 15:

Find the values of a and b so that the polynomial (x⁴ + ax³ − 7x² + 8x + b) is exactly divisible by (x + 2) as well as (x + 3).

Answer:

$Let p (x) = x^{4} + a x^{3} - 7 x^{2} + 8 x + b Now as p (x) is exactly divisible by (x + 2) and (x + 3), p (- 2) = 0 and p (- 3) = 0 For p (- 2) = 0 \Rightarrow {(- 2)}^{4} + a {(- 2)}^{3} - 7 {(- 2)}^{2} + 8 (- 2) + b = 0 \Rightarrow 16 - 8 a - 28 - 16 + b = 0 \Rightarrow - 8 a + b = 28 - - - - - - - - - (1) For p (- 3) = 0 \Rightarrow {(- 3)}^{4} + a {(- 3)}^{3} - 7 {(- 3)}^{2} + 8 (- 3) + b = 0 \Rightarrow 81 - 27 a - 63 - 24 + b = 0 \Rightarrow - 27 a + b = 6 - - - - - - - - - (2) Subtracting (2) from (1), we get : 19 a = 22 Therefore, a = \frac{22}{19} Putt ing a = \frac{22}{19} in (1), we get : - 8 \times \frac{22}{19} + b = 28 \Rightarrow b = 28 + \frac{176}{19} = \frac{708}{19}$

Page No 258:

Question 16:

Using remainder theorem, find the remainder when p(x) = x³ − 3x² + 4x + 50 is divided by (x + 3).

Answer:

$p (x) = x^{3} - 3 x^{2} + 4 x + 50 According to remainder theorem, w hen p (x) is divided by (x + 3) the remainder is p (- 3) . Now, p (- 3) = {(- 3)}^{3} - 3 {(- 3)}^{2} + 4 (- 3) + 50 = - 27 - 27 - 12 + 50 = - 16 Therefore, remainder = - 16$

Page No 259:

Question 17:

Factorise: (2x³ + 54).

Answer:

$2 x^{3} + 54 = 2 (x^{3} + 27) = 2 (x^{3} + 3^{3}) = 2 (x + 3) (x^{2} - x \times 3 + 3^{2}) = 2 (x + 3) (x^{2} - 3 x + 9)$

Page No 259:

Question 18:

Find the product (a − b − c) (a² + b² + c² + ab + ac − bc).

Answer:

$(a - b - c) (a^{2} + b^{2} + c^{2} + a b + a c - b c) = [a + (- b) + (- c)] [{(a)}^{2} + (- b^{2}) + {(- c)}^{2} - (a) (- b) - (a) (- c) - (- b) (- c)] = {(a)}^{3} + {(- b)}^{3} + {(- c)}^{3} - 3 (a) (- b) (- c) = a^{3} - b^{3} - c^{3} - 3 a b c$

Page No 259:

Question 19:

In ∆ABC, if ∠A − ∠B = 33° and ∠B − ∠C = 18°, find the measure of each angle of the triangle.

Answer:

$∠ A - ∠ B = 33 ° - - - - - - (1) and ∠ B - ∠ C = 18 ° - - - - - - (2) From (1) - (2), we g e t : (∠ A - ∠ B) - (∠ B - ∠ C) = 33 ° - 18 ° \Rightarrow ∠ A - ∠ B - ∠ B + ∠ C = 15 ° \Rightarrow ∠ A - 2 ∠ B + ∠ C = 15 ° - - - - - (3) By the angle sum property, we know that ∠ A + ∠ B + ∠ C = 180 ° - - - - - - (4) From (4) - (3), we get : (∠ A + ∠ B + ∠ C) - (∠ A - 2 ∠ B + ∠ C) = 180 ° - 15 ° \Rightarrow ∠ A + ∠ B + ∠ C - ∠ A + 2 ∠ B - ∠ C = 165 ° \Rightarrow 3 ∠ B = 165 ° \Rightarrow ∠ B = 55 ° P u t t i n g ∠ B = 55 ° in (1), we get : ∠ A - 55 ° = 33 ° \Rightarrow ∠ A = 88 ° Putting ∠ B = 55 ° in (2), we get : 55 ° - ∠ C = 18 ° \Rightarrow ∠ C = 55 ° - 18 ° = 37 ° Therefore, ∠ A = 88 °, ∠ B = 55 ° and ∠ C = 37 °$

Page No 259:

Question 20:

In the figure, in ∆ABC, the angle bisectors of ∠B and ∠C meet at a point O. Find the measure of ∠BOC.
Figure

Answer:

$In ∆ A B C, using angle sum property, we have : ∠ C A B + ∠ A B C + ∠ B C A = 180 ° \Rightarrow 70 ° + ∠ A B C + ∠ B C A = 180 ° ∴ ∠ A B C + ∠ B C A = 110 ° - - - - - (1) Now, B O is the angle bisector of ∠ A B C . \Rightarrow ∠ O B C = \frac{1}{2} ∠ A B C and C O is the angle bisector of ∠ B C A . \Rightarrow ∠ O C B = \frac{1}{2} ∠ B C A ∴ ∠ O B C + ∠ O C B = \frac{1}{2} (∠ A B C + ∠ B C A) - - - - - (2) In ∆ O B C, using angle sum property, we get : ∠ B O C + ∠ O B C + ∠ O C B = 180 ° Using (2) we get : ∠ B O C + \frac{1}{2} (∠ A B C + ∠ B C A) = 180 ° Now, using (1) we get : ∠ B O C + \frac{1}{2} (110 °) = 180 ° \Rightarrow ∠ B O C + 55 ° = 180 ° \Rightarrow ∠ B O C = 180 ° - 55 ° = 125 ° Therefore, ∠ B O C = 125 °$

Page No 259:

Question 21:

In the given figure, AB || CD. If ∠BAO = 110°, ∠AOC = 20° and ∠OCD = x°, find the value of x.
Figure

Answer:

Draw $O E ∥ A B ∥ C D$ .
Now, $O E ∥ A B and O A$ is the transversal.
$∴ ∠ O A B + ∠ A O E = 180 ° [Sum of consecutive interior angles is supplementary] \Rightarrow ∠ O A B + ∠ A O C + ∠ C O E = 180 ° \Rightarrow 110 ° + 20 ° + ∠ C O E = 180 ° \Rightarrow ∠ C O E = 50 °$
Also,
$O E ∥ C D and O C$ is the transversal.
$∴ ∠ O C D + ∠ C O E = 180 ° [Sum of consecutive interior angles is supplementary] \Rightarrow x ° + 50 ° = 180 ° \Rightarrow x ° = 130 ° Therefore, x = 130$

Page No 259:

Question 22:

In a right-angled triangle, prove that the hypotenuse is the longest side.

Answer:

$Let ∆ A B C be a right - angle d triangle, in which ∠ C is the right angle . In ∆ A B C, ∠ A C B = 90 ° ∠ A C B + ∠ C B A + ∠ B A C = 180 ° (angle sum property) \Rightarrow 90 ° + ∠ C B A + ∠ B A C = 180 ° \Rightarrow ∠ C B A + ∠ B A C = 180 ° - 90 ° \Rightarrow ∠ C B A + ∠ B A C = 90 ° So, ∠ C B A and ∠ B A C are acute angles . \Rightarrow ∠ C B A < 90 ° and ∠ B A C < 90 ° \Rightarrow ∠ C B A < ∠ A C B and ∠ B A C < ∠ A C B \Rightarrow A B > A C and A B > B C (side opposite to greater angle is greater) Therefore, hypotenuse A B is the longest side . Hence, proved .$

Page No 259:

Question 23:

In the given figure, prove that:
x = α + β + γ.
Figure

Answer:

Join BD to produce it to E.
Side BD of triangle ABD is produced to E.
$∴ ∠ A D E = ∠ B A D + ∠ A B D - - - - (1) (Exterior angle property)$
Side BD of triangle BCD is produced to E.
$∴ ∠ C D E = ∠ C B D + ∠ B C D - - - - (2) (Exterior angle property)$
$Adding (1) and (2), we get: ∠ A D E + ∠ C D E = ∠ B A D + ∠ A B D + ∠ C B D + ∠ B C D \Rightarrow ∠ C D A = ∠ B A D + ∠ B C D + ∠ A B D + ∠ C B D \Rightarrow x ° = ∠ B A D + ∠ B C D + ∠ C B A \Rightarrow x ° = α ° + γ ° + β ° ∴ x = α + β + γ .$

Page No 259:

Question 24:

Find six rational numbers between 3 and 4.

Answer:

$Let x = 3, y = 4, n = 6 d = \frac{y - x}{n + 1} = \frac{4 - 3}{6 + 1} = \frac{1}{7} Then, six rational numbers are : x + d, x + 2 d, x + 3 d, x + 4 d, x + 5 d, x + 6 d = 3 + \frac{1}{7}, 3 + \frac{2}{7}, 3 + \frac{3}{7}, 3 + \frac{4}{7}, 3 + \frac{5}{7}, 3 + \frac{6}{7} = \frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}$

Page No 259:

Question 25:

If $\frac{(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})} = a + \sqrt{15} b$ , find the values of a and b.
OR
Factorise: $(5 a - 7 b)^{3} + (9 c - 5 a)^{3} + (7 b - 9 c)^{3}$ .

Answer:

$\frac{(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})} = a + \sqrt{15} b Rationalising the denomiantor of the left side, we get: \frac{(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})} \times \frac{(\sqrt{5} + \sqrt{3})}{(\sqrt{5} + \sqrt{3})} = a + \sqrt{15} b \Rightarrow \frac{{(\sqrt{5} + \sqrt{3})}^{2}}{{(\sqrt{5})}^{2} - {(\sqrt{3})}^{2}} = a + \sqrt{15} b \Rightarrow \frac{{(\sqrt{5})}^{2} + {(\sqrt{3})}^{2} + 2 \sqrt{5} \times \sqrt{3}}{5 - 3} = a + \sqrt{15} b \Rightarrow \frac{5 + 3 + 2 \sqrt{15}}{2} = a + \sqrt{15} b \Rightarrow \frac{8 + 2 \sqrt{15}}{2} = a + \sqrt{15} b \Rightarrow \frac{2 (4 + \sqrt{15})}{2} = a + \sqrt{15} b \Rightarrow 4 + \sqrt{15} = a + \sqrt{15} b Comparing the left and right side s, we get a = 4, b = 1 .$

OR

${(5 a - 7 b)}^{3} + {(9 c - 5 a)}^{3} + {(7 b - 9 c)}^{3} L e t x = 5 a - 7 b y = 9 c - 5 a z = 7 b - 9 c ∴ x + y + z = 0 If x + y + z = 0, t hen x^{3} + y^{3} + z^{3} = 3 x y z Therefore, {(5 a - 7 b)}^{3} + {(9 c - 5 a)}^{3} + {(7 b - 9 c)}^{3} = 3 (5 a - 7 b) (9 c - 5 a) (7 b - 9 c)$

Page No 259:

Question 26:

Factorise: $12 (x^{2} + 7 x)^{2} - 8 (x^{2} + 7 x) (2 x - 1) - 15 (2 x - 1)^{2}$ .

Answer:

$12 {(x^{2} + 7 x)}^{2} - 8 (x^{2} + 7 x) (2 x - 1) - 15 {(2 x - 1)}^{2} = 12 {(x^{2} + 7 x)}^{2} - 18 (x^{2} + 7 x) (2 x - 1) + 10 (x^{2} + 7 x) (2 x - 1) - 15 {(2 x - 1)}^{2} = 6 (x^{2} + 7 x) [2 (x^{2} + 7 x) - 3 (2 x - 1)] + 5 (2 x - 1) [2 (x^{2} + 7 x) - 3 (2 x - 1)] = [2 (x^{2} + 7 x) - 3 (2 x - 1)] [6 (x^{2} + 7 x) + 5 (2 x - 1)] = (2 x^{2} + 14 x - 6 x + 3) (6 x^{2} + 42 x + 10 x - 5) = (2 x^{2} + 8 x + 3) (6 x^{2} + 52 x - 5)$

Page No 259:

Question 27:

If (x³ + ax² + bx + 6) has (x − 2) as a factor and leaves a remainder 3 when divided by (x − 3), find the values of a and b.

Answer:

$Let f (x) = x^{3} + a x^{2} + b x + 6 Here, (x - 2) is a factor of f (x) Therefore, f (2) = 0 \Rightarrow 2^{3} + a \times 2^{2} + 2 b + 6 = 0 \Rightarrow 8 + 4 a + 2 b + 6 = 0 \Rightarrow 4 a + 2 b = - 14 \Rightarrow 2 a + b = - 7 - - - - - (1) Therefore, when f (x) is divided by (x - 3) the remainder will be f (3), which is given to be 3 . \Rightarrow f (3) = 3 \Rightarrow 3^{3} + a \times 3^{2} + 3 b + 6 = 3 \Rightarrow 27 + 9 a + 3 b + 6 = 3 \Rightarrow 9 a + 3 b + 33 = 3 \Rightarrow 9 a + 3 b = - 30 \Rightarrow 3 a + b = - 10 Subtracting (1) from (2), we get : a = - 3 Putting a = - 3 in (1), we get: 2 \times (- 3) + b = - 7 \Rightarrow - 6 + b = - 7 \Rightarrow b = - 1 Therefore, a = - 3 and b = - 1 .$

Page No 259:

Question 28:

Without actual division, show that (x³ − 3x² − 13x + 15) is exactly divisible by (x² + 2x − 3).

Answer:

$x^{3} - 3 x^{2} - 13 x + 15 = x^{3} - 5 x^{2} + 2 x^{2} - 10 x - 3 x + 15 = x^{2} (x - 5) + 2 x (x - 5) - 3 (x - 5) = (x - 5) (x^{2} + 2 x - 3) Here, x^{2} + 2 x - 3 is a factor of x^{3} - 3 x^{2} - 13 x + 15 . Therefore, x^{3} - 3 x^{2} - 13 x + 15 is exactly divisible by x^{2} + 2 x - 3 .$

Page No 260:

Question 29:

Factorise: a³ − b³ + 1 + 3ab.

Answer:

$a^{3} - b^{3} + 1 + 3 a b = a^{3} + {(- b)}^{3} + 1^{3} - 3 a (- b) 1 = [a + (- b) + 1] [a^{2} + {(- b)}^{2} + 1^{2} - a (- b) - (- b) 1 - 1 a] = (a - b + 1) (a^{2} + b^{2} + 1 + a b + b - a)$

Page No 260:

Question 30:

In the given figure, AB || CD, ∠ECD = 100°, ∠EAB = 50° and ∠AEC = x°. Find the value of x.
Figure

Answer:

Draw $E F ∥ A B ∥ C D$ .
Now, $E F ∥ A B and E A$ is the transversal.
$∴ ∠ E A B + ∠ A E F = 180 ° [Sum of consecutive interior angles is supplementary] \Rightarrow 50 ° + ∠ A E F = 180 ° \Rightarrow ∠ A E F = 180 ° - 50 ° = 130 °$
Also,
$E F ∥ C D and E C$ is the transversal.
$∴ ∠ E C D + ∠ C E F = 180 ° [Sum of consecutive interior angles is supplementary] \Rightarrow 100 ° + ∠ C E F = 180 ° \Rightarrow ∠ C E F = 80 °$
Now,
$∠ A E C + ∠ C E F = ∠ A E F \Rightarrow x ° + 80 ° = 130 ° \Rightarrow x ° = 130 ° - 80 ° = 50 ° Therefore, x = 50$

Page No 260:

Question 31:

Prove that the bisectors of the angles of a linear pair are at right angles.

Answer:

$In the figure, ∠ A C D and ∠ B C D form a linear pair . \Rightarrow ∠ A C D + ∠ B C D = 180 ° C E and C F bisect ∠ A C D and ∠ B C D, respectively . ∠ A C D + ∠ B C D = 180 ° \Rightarrow \frac{∠ A C D}{2} + \frac{∠ B C D}{2} = 90 ° \Rightarrow ∠ E C D + ∠ D C F = 90 ° (C E and C F bisect ∠ A C D and ∠ B C D respectivel y) \Rightarrow ∠ E C F = 90 ° ∠ E C F is the angle between C E and C F, which bisect the linear pair of ∠ A C D and ∠ B C D . Hence, it is proved that the angle bisectors of a linear pair are at right angles to each other .$

Page No 260:

Question 32:

In the given figure, AD bisects ∠BAC in the ratio 1 : 3 and AD = DB. Determine the value of x.
Figure

Answer:

In the figure,

$∠ B A C + 108 ° = 180 ° [Linear pair] \Rightarrow ∠ B A C = 72 °$

Now, divide $72 °$ in the ratio 1 : 3.
$∴ a + 3 a = 72 ° \Rightarrow a = 18 ° ∴ a = 18 ° and 3 a = 54 °$
Hence, the angles are 18^o and 54^o.
$∴ ∠ B A D = 18 ° and ∠ D A C = 54 °$

Given,
$A D = D B \Rightarrow ∠ D A B = ∠ D B A = 18 °$

In $∆ A B C$ , we have:
$∠ B A C + ∠ A B C + ∠ A C B = 180 ° [angle sum property] \Rightarrow 72 ° + 18 ° + x ° = 180 ° \Rightarrow x ° = 90 ° Therefore, x = 90$

Page No 260:

Question 33:

In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, find ∠MAN.
Figure

Answer:

$In ∆ A B C, using angle - sum property, we have : ∠ A B C + ∠ B C A + ∠ C A B = 180 ° \Rightarrow 70 ° + 20 ° + ∠ C A B = 180 ° \Rightarrow 90 ° + ∠ C A B = 180 ° \Rightarrow ∠ C A B = 180 ° - 90 ° \Rightarrow ∠ C A B = 90 ° Now, A N is the angle bisector of ∠ A . ∴ ∠ B A N = \frac{1}{2} ∠ C A B = \frac{1}{2} \times 90 ° = 45 ° In ∆ A B M, using angle - sum property, we have : ∠ A B M + ∠ B M A + ∠ M A B = 180 ° \Rightarrow 70 ° + 90 ° + ∠ M A B = 180 ° \Rightarrow 160 ° + ∠ M A B = 180 ° \Rightarrow ∠ M A B = 180 ° - 160 ° = 20 ° From the figure, ∠ M A N = ∠ B A N - ∠ M A B = 45 ° - 20 ° = 25 ° Therefore, ∠ M A N = 25 °$

Page No 260:

Question 34:

If the bisector of the vertical angle of a triangle bisects the base, prove that the triangle is isosceles.

Answer:

$In ∆ A B C, ∠ B A D = ∠ C A D and B D = C D Extend A D to point E so that C E is parallel to A B . Join C E . Taking transversal A E intersecting A B and C E, ∠ B A D = ∠ D E C (interior alternate angle) We know that ∠ B A D = ∠ C A D = ∠ D E C In △ A C E, ∠ C A D = ∠ D E C ∴ A C = C E In △ A B D and △ E C D, ∠ A D B = ∠ E D C (vertically - opposite angle) ∠ B A D = ∠ C E D (interior alternate angle) B D = C D (given) By A A S, ∆ A B D ≅ △ E C D Therefore, A B = E C (by cpct) We know E C = A C (already proven) Therefore, A B = A C and △ A B C is isosceles .$

View NCERT Solutions for all chapters of Class 9