Question Set D

CBSE 9th Standard Mathematics Introduction to Euclid's Geometry Important Questions And Answers Study Material - QB365

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Questions:

Generic 4

8 x 4 = 32

Draw the graph of the linear equation x+y=7. Verify from the graph that (8,-1) is a solution of the equation x+y=7

In a rectangle ABCD, E is a point which bisects BC Prove that AE = ED

Prove that if one triangle is equal to the sum of the other two angles the triangle is right angled.

In figure, C is the mid-point of AB and Dis the mid-point of AC. Prove that AD = \(1\over2\) AB.

Plot the points A, B, C, D from the table:

Points	A	B	C	D
x	8	-5	13	-4
y	10	13	-5	-16

and answer the following:
(a) Write the coordinates of A, B, C, D.
(b) Shade the triangle ABC.

If x+2y=10, xy=15, find \(x^3+8y^3\).

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

Express \(1.\overline { 32 } +0.\overline { 35 } \) in the form \(\frac { p }{ q } \) , where p and q are integers and \(q\neq 0\)

Multiple Choice Question

14 x 1 = 14

In ΔABC, if ㄥA > ㄥB > ㄥC then:

(a)

AB > AC

(b)

AC < BC

(c)

AB > BC

(d)

AC > BC

10.

Two triangles are congruent, if any two pairs of angles and one pair of corresponding sides are equal. This rule is known as

(a)

SAS congruence rule

(b)

ASA congruence rule

(c)

AAS congruence rule

(d)

SSS congruence rule

11.

In \(\triangle \) ABC, the bisectors of \(\angle ABC\) and \(\angle BCA\) intersect each other at O. The measure of \(\angle BOC\) is:

(a)

\(90^{ 0 }+\angle A\)

(b)

\(90^{ 0 }+\frac { \angle A }{ 2 } \)

(c)

\(180^{ 0 }-\angle A\)

(d)

\(90^{ 0 }-\frac { \angle A }{ 2 } \)

12.

The angle complementary to \(90^{ 0 }\)-\(9^{ 0 }\) is

(a)

\(90^{ 0 }\)+\(9^{ 0 }\)

(b)

\(9^{ 0 }\)

(c)

\(180^{ 0 }\)-\(9^{ 0 }\)

(d)

\(360^{ 0 }\)-\(9^{ 0 }\)

13.

John Playfair was a

(a)

french mathematician

(b)

Scottish mathematician

(c)

Indian mathematician

(d)

Egyptian mathematician

14.

The number of lines that can pass through a given point is:

(a)

two

(b)

none

(c)

only one

(d)

infinite many

15.

The point of the from \(\left({p\over2},p\right)\) always lies on the graph of the equation:

(a)

2x=y

(b)

x=2y

(c)

x=y+2

(d)

x+2=y

16.

Which of the following ordered pairs is a solution of the equation x-2y=6?

(a)

(2, 4)

(b)

(0,3)

(c)

(-4, 1)

(d)

(4, -1)

17.

By plotting the points O(0,0), A(1,0), B(1,1), C(0,1) and joining OA, AB, BC and CO, the figure we obtain is:

(a)

Square

(b)

Rectangle

(c)

Trapezium

(d)

Rhombus

18.

In fourth quadrant, x is

(a)

+ve

(b)

-ve

(c)

(d)

None of these

19.

One of the factors of \(42+y-y^2\) is:

(a)

\((7+y)\)

(b)

\((6-y)\)

(c)

\((7-y)\)

(d)

\((-6+y)\)

20.

If (x+3) is the factor of polynomial \(x^3+ax^2+x+3\) then the value of a is:

(a)

(b)

(c)

(d)

-3

21.

\(\left( -2-\sqrt { 3 } \right) \left( -2+\sqrt { 3 } \right) \) when simplified is:

(a)

positive and irrational

(b)

positive and rational

(c)

negative and irrational

(d)

negative and rational

22.

The rational number between -1/5 and -2/5 is

(a)

(b)

-1/4

(c)

-3/10

(d)

-7/25

Generic 2

7 x 2 = 14

23.

In the figure, \(\triangle\)ABC and \(\triangle\)DBC are two isosceles triangles on the same base BC Prove that \(\angle\) ABD = \(\angle\)ACD.

24.

In the given figure, if AB = CD, then prove that AC = BD. Also write the Euclid's axiom used for proving it.

25.

If the point (2k-3, k+2) lies on the graph of the equation 2x+3y+15=0, find value of k.

26.

In the figure PQ \(\bot \) S find 'x' and 'y' where \(\angle PRS=120^{ 0 }\quad and\quad \angle PSR\quad =30^{ 0 }\)

27.

Write the coordinates of A, B, C and D from the figure given below:

28.

Find the degree of the polynomials given below:
\(2-{ y }^{ 2 }-{ y }^{ 3 }+{ 2y }^{ 8 }\)

29.

Find the value of \(\frac { 4 }{ { \left( 216 \right) }^{ \frac { 2 }{ 3 } } } -\frac { 4 }{ { \left( 256 \right) }^{ \frac { 3 }{ 4 } } } \)

Generic 3

5 x 3 = 15

30.

If \((\frac{a}{b})^{x-1}=(\frac{b}{a})^{2x-8}\) , then find the value of x.

31.

Write three solutions of the equation 3x=y+3. Draw its graph and find the points where the graph intersects the axes.

32.

(i) Why is Axiom 5, in the list of Euclid's axioms, considered a 'universal truth'? (Note that the question is not about the fifth postulate).
(ii) How would you rewrite Eulid's fifth postulate so that it would be easier to understand?

33.

(i) Plot the points M(5,-3) and N(-3,-3).
(ii) What is the length of MN?
(iii) Find the coordinates of points A, B and C lying on Mn such that MA=AB=BC=CN

34.

In which quadrant or on which axes the following points lie?
P(9-2,4), Q(3,-1), R(-1,0) and S(0,-4)

1 Marks

6 x 1 = 6

35.

Given\(\triangle OAP\cong \triangle OBP\) in the figure below. Prove the criteria by which the triangles are congruent.

36.

In the figure below, AOB is a straight line. Calculate the measure of \(\angle COD\).

37.

What is a straight line?

38.

Calculate the value of \(\frac{16^{3/4}}{16^{-1/4}}\)

39.

In the expression x²+\(\frac { \pi }{ 2 } x-7\), what is the co-efficient of x?

40.

In a one day cricket match, Raina and Dhoni scored 198 runs. Express this as a linear equation in two carables.

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Answers Key:

Generic 4

8 x 4 = 32

x+y=7
y=7-x

x	5	7	4
y	2	0	3

From graph it is clear that (8,-1) lies on the line AB.
Hence (8,-1) is a solution of the given equation.

Given: In a rectangle ABCD, E is a point which bisects BC
To Prove: AE = ED

Proof: In \(\triangle EBA\) and \(\triangle ECD\)
EB = EC
\(\angle EBA=\angle ECD\) | Each 90⁰ (ABCD is a rectangle)
BA = CD | Opposite sides of rectangle ABCD
\(\triangle EBA\cong \triangle ECD\) | SAS congruenece rule
AE = DE | C.P.C.T
AE = ED

Let in \(\triangle \)ABC \(\angle A=\angle B+\angle C\)
We know that \(\angle A=\angle B+\angle C=180^{ 0 }\)
The sum of the three angles of a triangle is \(180^{ 0 }\)
\(\Rightarrow \angle A+\angle A=180^{ 0 }\)
\(\Rightarrow 2\angle A=180^{ 0 }\)
\(\Rightarrow \angle A=\frac { 180^{ 0 } }{ 2 } =90^{ 0 }\)
Hence Triangle ABC is a right-angled triangle.

\(\because\) C is the midpoint of AB
\(\therefore \) AC = CB
AC + AC = CB + AC
| If equals are added to equals, then the wholes are equal (Euclid's Axiom (ii))]
\(\Rightarrow \) 2AC = AB I CB + AC coincides with AB
\(\Rightarrow \) \(1\over2\)(2AC) = \(1\over2\) AB
| Things which are halves of the same thing are equal (Euclid's Axiom (vii»]
\(\Rightarrow \) AC =\(1\over2\)AB
\(\Rightarrow \) \(1\over2\)AC = \(1\over2\)(\(1\over2\)AB)
| Things which are halves of the same thing are equal to one another (Euclid's Axiom (vii))]
\(1\over2\)AC = \(1\over2\)AB
AD = \(1\over4\)AB
\(\because\) D is the mid-point of AC
\(\therefore \)AD = DC =\(1\over2\)AC (as above)

(a) The coordinates A, B, C, D are (8,10), (-5,13), (13,5), (-4,-16) respectively.
(b) The triangle ABC has been shaded.

We know that
\((x+2y)^3=(x)^3+(2y)^3+3(x)(2y)(x+2y)\) Using Identity VI
\(\Rightarrow(x+2y)^3=x^3+8y^3+6xy(x+2y)\)
\(\Rightarrow (10)^3=x^3+8y^3+6(15)(10)\)
\(\Rightarrow 1000=x^3+8y^3+900\)
\(\Rightarrow x^3+8y^3=1000-900=100\)

\(\frac { 1 }{ x } =\frac { 1 }{ 3+2\sqrt { 2 } } =\frac { 1 }{ 3+2\sqrt { 2 } } \times \frac { 3-2\sqrt { 2 } }{ 3-2\sqrt { 2 } } \)
\(\\ =\frac { 3-2\sqrt { 2 } }{ { \left( 3 \right) }^{ 2 }-{ \left( 2\sqrt { 2 } \right) }^{ 2 } } =\frac { 3-2\sqrt { 2 } }{ 9-8 } \)
\(\\ =\frac { 3-2\sqrt { 2 } }{ 1 } =3-2\sqrt { 2 } \)
\(\\ x+\frac { 1 }{ x } =\left( 3+2\sqrt { 2 } \right) +\left( 3-2\sqrt { 2 } \right) =6\)
\(\\ { \left( x+\frac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3(x)\left( \frac { 1 }{ x } \right) \left( x+\frac { 1 }{ x } \right)\)
\( \\ { \left( 6 \right) }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3(6)\)
\(\\ 216={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +18\)
\(\\ { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } =198\)

Let x = \(1.\overline { 32 } \)
x = 1.3222...
10x = 13.222... ...(1)
100x = 132.222... ....(2)
Subtracting (1) from (2), we get
90x = 119
x = \(\frac { 119 }{ 90 } \) ....(3)
Let x = \(0.\overline { 35 } \)
Then x = 0.35 35 35...(4)
100x = 35.35 35 35...(5)
Subtracting (4) from (5), we get
99x = 35
x = 35/99
\(1.\overline { 32 } +0.\overline { 35 } =\frac { 119 }{ 90 } +\frac { 35 }{ 99 } \)
\(\\ =\frac { 1309+350 }{ 990 } =\frac { 1659 }{ 990 } =\frac { 553 }{ 330 } \)
Here, p = 553, q =3 30(\(\neq 0\))

Multiple Choice Question

14 x 1 = 14

ㄥA > ㄥB > ㄥC
BC > AC

10.

Theorem

11.

\(\angle A+\angle B+\angle C=180^{ 0 }\)
\(\Rightarrow \angle B+\angle C=180^{ 0 }-\angle A\)
\(\angle BOC+\frac { \angle B }{ 2 } +\frac { \angle C }{ 2 } =180^{ 0 }\)
\(\Rightarrow \angle BOC+\frac { 180^{ 0 }-\angle A }{ 2 } =180^{ 0 }\)
\(\angle BOC=90^{ 0 }+\frac { \angle A }{ 2 } \)

12.

Required angle=\(180^{ 0 }\)-(\(90^{ 0 }\)+\(9^{ 0 }\)) = \(90^{ 0 }\)+\(9^{ 0 }\)

13.

(b)
Scottish mathematician

14.

(d)
infinite many

15.

\(\left({p\over2},p\right)\) satisfies 2x=y

16.

(4, -1) satisfies x-2y=6

17.

(a)
Square

18.

(a)
+ve

19.

\(42+y-y^2=42+7y-6y-y^2\)
\(=7(6+y)-y(6+y)\)
\(=(6+y)(7-y)\)

20.

\(f(x)=x^3+ax^2+x+3\)
\(x+3=0\Rightarrow \ x=-3\)
\(f(-3)=0\)
\(\Rightarrow \ (-3)^3+a(-3)^2+(-3)+3=0\)
\(\Rightarrow \ a=0\)

21.

(b)
positive and rational

22.

Generic 2

7 x 2 = 14

23.

Join AD.
In \(\triangle\)ABC and \(\triangle\)ACD,
AB = AC (Given)
BD = CD (Given)
AD = AD (Common)
By using SSS Congruency Rule,
\(\triangle ABD\cong \triangle ACD\)
\(\therefore \angle ABD=\angle ACD (By c.p.c.t)\)

24.

AB = CD(Given)
\(\Rightarrow\)AB + BC = BC + CD
\(\Rightarrow\)AC = BD
Euclid's axiom used: If equals are added to equals, the wholes are equal.

25.

Putting x=2k-3, y=k+2 in 2x+3y=15=0, we get
2(2k-3)+3(k+2)+15=0
\(\Rightarrow 4k-6+3k+6+15=0\)
\(\Rightarrow k=\frac{-15}{7}\)

26.

\(30^{ 0 },30^{ 0 }\)

27.

\(A\rightarrow \left( 5,0 \right)\)
\(\\ B\rightarrow \left( 0,3 \right) \)
\(\\ C\rightarrow \left( -5,0 \right)\)
\(\\ D\rightarrow \left( 0,-2 \right)\)

28.

29.

Generic 3

5 x 3 = 15

30.

\((\frac{a}{b})^{x-1}=(\frac{b}{a})^{2x-8}\)
\((\frac{a}{b})^{x-1}=(\frac{a}{b})^{[-2x-8]}\)
⇒ x - 1 = -[2x - 8]
3x = 9
x = 3

31.

3x=y+3; three solutions are x=1, y=0; x=2, y=3 and x=0, y=-3

From graph it is clear that line meets x-axis at(1,0) and y-axis at (0,-3)

32.

(i) Since this is true for anything in any part of the world, this is a universal truth.
(ii) If the sum of the cointerior angles made by a transversal intersect two straight lines at distinct points is less than 180^o, then the lines cannot be parallel.

33.

(i)

(ii) 8 units
(iii) \(A\rightarrow \left( 3,-3 \right)\)
\(\\ B\rightarrow \left( 1,-3 \right)\)
\(\\ C\rightarrow \left( -1,-3 \right) \)

34.

\(P\rightarrow \left( II \right) \)
\(\\ Q\rightarrow \left( IV \right)\)
\(\\ R\rightarrow \left( x-axis \right)\)
\( \\ S\rightarrow \left( y-axis \right) \)

1 Marks

6 x 1 = 6

35.

()
OA = OB (Given)
OP = OP (Common)
\(\angle\)AOP = \(\angle\)BOP (Given)
\(\triangle\)OAP\(\cong\)\(\triangle\)OBP (By SAS)

36.

()
\(\because\) x+20^o+2x-20^o+60^o=180^o (\(\because\) Straight line makes an angle of 180^o)
\(\Rightarrow\) 3x=180^o-60^o=120^o
\(\Rightarrow\) x=40^o
Thus, \(\angle COD\)=2x-20^o=80^o-20^o=60^o