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Page No 252:
Question 1:
Mark (✓) against the correct answer in each of the following:
Which of the following is a rational number?
(a)
(b)
(c)
(d)
Answer:
(d)
The number of the form , where p and q are integers and , are known as rational numbers.
Here and 5 are integers and .
Therefore, 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
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 180o.
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°
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
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 .
(a)
(b) 6 cm
(c) 5 cm
(d) 4 cm
Figure
Answer:
(d) 4 cm
In a right-angled triangle,
Page No 253:
Question 9:
If , find the value of .
Answer:
Page No 253:
Question 10:
Factorise: .
Answer:
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
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, 2A = 3B = 6C = xo (say)
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.
Side CE of CED is produced to A.
Page No 254:
Question 15:
If and , find the value of (x2 + y2).
or
Simplify:
Answer:
OR
Page No 254:
Question 16:
If 2 and are the zeros of the polynomial 3x3 − 2x2 − 7x − 2, find the third zero of the polynomial.
Answer:
Page No 254:
Question 17:
Find the remainder when the polynomial f(x) = 4x3 − 12x2 + 14x − 3 is divided by (2x − 1).
Answer:
By the remainder theorem, we know that when p(x) is divided by (2x - 1), the remainder is .
Now,
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:
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:
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:
Page No 254:
Question 22:
In ∆ABC, AL ⊥ BC and AM is the bisector of ∠A. Show that .
Figure
Answer:
Page No 255:
Question 23:
In the given figure, AB || CD, ∠BAE = 100° and ∠AEC = 30°. Find ∠DCE.
Figure
Answer:
Page No 255:
Question 24:
Factorise: a3 − b3 + 1 + 3ab
Answer:
Page No 255:
Question 25:
If , show that the value of (x3 − 2x2 − 7x + 5) is 3.
Or
Simplify:
Answer:
OR
Page No 255:
Question 26:
If , then show that .
Answer:
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:
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:
Page No 255:
Question 29:
Prove that:
.
Answer:
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:
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:
From (1), (2) and (3), we have:
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:
Page No 256:
Question 34:
In the given figure, AB || CD. If ∠BAE = 100° and ∠ECD = 120° then x = ?
Figure
Answer:
Draw .
Then,
Now, and AE is the transversal.
Again, and CE is the transversal.
Therefore,
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)
(b) 2.3
(c)
(d)
Answer:
(c)
An irrational number between a and b is .
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)
(c)
(d)
Answer:
(d)
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:
(a)
(b)
(c)
(d)
Answer:
(d)
Page No 257:
Question 4:
Mark (✓) against the correct answer in each of the following:
If , then
(a)
(b)
(c)
(d)
Answer:
(c)
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
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
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
Page No 258:
Question 9:
Factorise: .
Answer:
Page No 258:
Question 10:
Prove that is an irrational number.
Answer:
We will show that is irrational by contradiction.
Let us assume that is rational.
Therefore we can represent it in the form of .
=
Let us assume that is in its lowest form, i.e. a and b are co-prime.
Therefore, 5 is the factor of . If 5 is a factor of , 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 is rational.
Therefore, 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 |
0 |
Page No 258:
Question 12:
If , find the value of .
Answer:
Page No 258:
Question 13:
Find the area of the triangle whose sides measure 52 cm, 56 cm and 60 cm.
Answer:
Page No 258:
Question 14:
In the given figure, AB || CD. Find the value of x.
Figure
Answer:
Draw .
Now, is the transversal.
Also,
and FD is the transversal.
Therefore,
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:
Page No 258:
Question 16:
Using remainder theorem, find the remainder when p(x) = x3 − 3x2 + 4x + 50 is divided by (x + 3).
Answer:
Page No 259:
Question 17:
Factorise: (2x3 + 54).
Answer:
Page No 259:
Question 18:
Find the product (a − b − c) (a2 + b2 + c2 + ab + ac − bc).
Answer:
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:
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:
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 .
Now, is the transversal.
Also,
is the transversal.
Page No 259:
Question 22:
In a right-angled triangle, prove that the hypotenuse is the longest side.
Answer:
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.
Side BD of triangle BCD is produced to E.
Page No 259:
Question 24:
Find six rational numbers between 3 and 4.
Answer:
Page No 259:
Question 25:
If , find the values of a and b.
OR
Factorise: .
Answer:
OR
Page No 259:
Question 26:
Factorise: .
Answer:
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:
Page No 259:
Question 28:
Without actual division, show that (x3 − 3x2 − 13x + 15) is exactly divisible by (x2 + 2x − 3).
Answer:
Page No 260:
Question 29:
Factorise: a3 − b3 + 1 + 3ab.
Answer:
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 .
Now, is the transversal.
Also,
is the transversal.
Now,
Page No 260:
Question 31:
Prove that the bisectors of the angles of a linear pair are at right angles.
Answer:
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,
Now, divide in the ratio 1 : 3.
Hence, the angles are 18o and 54o.
Given,
In , we have:
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:
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:
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