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#### Question 1:

Mark (âœ“) 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) $\frac{-3}{5}$

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

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

#### Question 2:

Mark (âœ“) 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

(a) 3

Let f(x) = x3− 4x2 + 2x + k.
3 is the zero of f(x).
Therefore, f(3) = 0

#### Question 3:

Mark (âœ“) 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

(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 + 3x2x2 −3x −2x −6 = 0
x2(x + 3) −x(x + 3) −2(x + 3) = 0
(x + 3)(x2x − 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.

#### Question 4:

Mark (âœ“) 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)

(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)

#### Question 5:

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

(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 180o.

#### Question 6:

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

(d) 105°

#### Question 7:

Mark (âœ“) 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

(c) 60 cm2

#### Question 8:

Mark (âœ“) against the correct answer in each of the following:
In an isosceles right triangle, the length of the hypotenuse is .
(a)
(b) 6 cm
(c) 5 cm
(d) 4 cm
Figure

(d) 4 cm

In a right-angled triangle,

#### Question 9:

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

#### Question 10:

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

#### Question 11:

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

Let f(x) = a2x3− 4ax + 4a − 1.
(x 1) is a factor of f(x)
Therefore, f(1) = 0

#### Question 12:

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

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.

#### Question 13:

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

In a âˆ†ABC, 2$\angle$A = 3$\angle$B = 6$\angle$C = xo (say)

#### Question 14:

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

Side BC of $∆$ABC is produced to D.

Side CE of $∆$CED is produced to A.

#### 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: $\frac{\left(7+3\sqrt{5}\right)}{\left(3+\sqrt{5}\right)}-\frac{\left(7-3\sqrt{5}\right)}{\left(3-\sqrt{5}\right)}$

$\text{Given},\phantom{\rule{0ex}{0ex}}x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}},y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}{x}^{2}={\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\right)}^{2}=\frac{5+3+2\sqrt{15}}{5+3-2\sqrt{15}}=\frac{8+2\sqrt{15}}{8-2\sqrt{15}}\phantom{\rule{0ex}{0ex}}\mathrm{And}\phantom{\rule{0ex}{0ex}}{y}^{2}={\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)}^{2}=\frac{5+3-2\sqrt{15}}{5+3+2\sqrt{15}}=\frac{8-2\sqrt{15}}{8+2\sqrt{15}}$

OR

#### 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.

#### Question 17:

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

By the remainder theorem, we know that when p(x) is divided by (2x - 1), the remainder is $p\left(\frac{1}{2}\right)$.
Now,

#### Question 18:

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

(p q)3 + (q r)3 + (r p)
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)

#### Question 19:

In the given figure, in âˆ†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

#### Question 20:

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

#### Question 21:

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

#### Question 22:

In âˆ†ABC, ALBC and AM is the bisector of ∠A. Show that $\angle LAM=\frac{1}{2}\left(\angle B-\angle C\right)$.
Figure

#### Question 23:

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

#### Question 24:

Factorise: a3b3 + 1 + 3ab

#### Question 25:

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

OR

#### Question 26:

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

#### Question 27:

If (x3 + mx2x + 6) has (x − 2) as a factor and leaves a remainder r when divided by (x − 3), find the values of m and r.

#### 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.

#### Question 29:

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

#### 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?

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 .

#### Question 31:

In the given figure, in a âˆ†ABC, BEAC, ∠EBC = 40° and ∠DAC = 30°. Find the values of x, y and z.
Figure

#### Question 32:

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

From (1), (2) and (3), we have:

#### 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

#### Question 34:

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

Draw $EF\parallel AB\parallel CD$.
Then, $\angle AEF+\angle CEF=x°$
Now, $EF\parallel AB$ and AE is the transversal.

Again, $EF\parallel CD$ and CE is the transversal.

Therefore,

#### Question 1:

Mark (âœ“) 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) $2.\overline{)34}$

(c) $\sqrt{5}$

An irrational number between a and b is $\sqrt{ab}$.

#### Question 2:

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

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

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

#### Question 3:

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

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

#### Question 4:

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

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

#### Question 5:

Mark (âœ“) 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

(d) 3

p(x) = x3 + x2 + ax + 115 is exactly divisible by (x + 5).
Therefore, p($-$5) = 0

#### Question 6:

Mark (âœ“) 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

(b) x = 0

The equation of y-axis is x = 0.

#### Question 7:

Mark (âœ“) 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

(d) 20

#### Question 8:

Mark (âœ“) 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

(d) 105

#### Question 9:

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

#### Question 10:

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

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.

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.

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.

#### Question 11:

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

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

#### Question 12:

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

#### Question 13:

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

#### Question 14:

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

Draw $EF\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
$EF\parallel CD$ and FD is the transversal.

Therefore,

#### 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).

#### Question 16:

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

#### Question 17:

Factorise: (2x3 + 54).

#### Question 18:

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

#### Question 19:

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

#### Question 20:

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

#### Question 21:

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

Draw $OE\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
is the transversal.

#### Question 22:

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

#### Question 23:

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

Join BD to produce it to E.
Side BD of triangle ABD is produced to E.

Side BD of triangle BCD is produced to E.

#### Question 24:

Find six rational numbers between 3 and 4.

#### Question 25:

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

OR

#### Question 26:

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

#### 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.

#### Question 28:

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

#### Question 29:

Factorise: a3b3 + 1 + 3ab.

#### Question 30:

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

Draw $EF\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
is the transversal.

Now,

#### Question 31:

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

#### Question 32:

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

In the figure,

Now, divide $72°$ in the ratio 1 : 3.

Hence, the angles are 18o and 54o.

Given,
$AD=DB\phantom{\rule{0ex}{0ex}}⇒\angle DAB=\angle DBA=18°$

In $∆ABC$, we have:

#### Question 33:

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

#### Question 34:

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