Triangle ABC is given, with altitudes CD and BE from vertices C and B to opposite sides AB and AC respectively being equal. It is proved that triangle ABC must be isosceles by showing that triangles CBD and BCE are congruent by the right angle-hypotenuse (RHS) criterion, implying corresponding angles are equal, and then using corresponding parts of congruent triangles to show sides AB and AC are equal, making triangle ABC isosceles.

1. Mathematics IX (Term - I) 1
SECTIONA
(Question numbers 1 to 8 carry 1 mark each. For each question, four alternative choices
have been provided of which only one is correct. You have to select the correct choice).
1. When x3
– 6x2
+ 9x + 3 is divided by (x – 1), the remainder is :
(a) 20 (b) 7 (c) 13 (d) –7
Sol. (b) Let p(x) = x3
– 6x2
+ 9x + 3
∴ p(1) = 1 – 6 + 9 + 3 = 13 – 6 = 7
2. The coefficient of y in (x + y + z)2
is :
(a) 2x (b) 2z (c) x + z (d) 2x + 2z
Sol. (d) (x + y + z)2
= x2
+ y2
+ z2
+ 2xy + 2yz + 2zx
= x2
+ y2
+ z2
+ y (2x + 2z) + 2zx
Thus, the coefficient of y is (2x + 2z).
3. On factorising x2
+ 8x + 15, we get :
(a) (x + 3) (x – 5) (b) (x – 3) (x + 5) (c) (x + 3) (x + 5) (d) (x – 3) (x – 5)
Sol. (c) x2
+ 8x + 15 = x2
+ 3x + 5x + 15
= x (x + 3) + 5 (x + 3) = (x + 3) (x + 5)
4. The length of the sides of a triangle are 5 cm, 7 cm and 8 cm. Area of the triangle is :
(a) 300 cm2
(b) 100 2
3 cm (c) 50 2
3 cm (d) 10 2
3 cm
Sol. (d) s =
5 7 8
cm =10 cm
2
+ +
∴ Area of the triangle = 2
10 (10 – 5) (10 – 7) (10 – 8) cm
= 2 2
10 5 3 2 cm 10 3 cm× × × = .
5. To locate n , where n is a positive integer, on the number line, we first locate :
(a) n (b) 1n − (c) 1n + (d)
2
n
Sol. (b) We can locate n , for any positive integer n, after 1n − has been located.
6. Lines AB and CD intersect each other at O. If ∠AOC : ∠BOC = 2 : 7, then ∠BOD
equals :
(a) 20° (b) 40° (c) 60° (d) 80°
Sol. (b) Let ∠AOC = 2x and ∠BOC = 7x, then
2x + 7x = 180° [Linear pair]
⇒ 9x = 180°
⇒ x = 20°
MODEL TEST PAPER – 2 (SOLVED)
Maximum Marks : 90 Maximum Time : 3 hours
General Instructions : Same as in CBSE Sample Question Paper.

2. 2 Mathematics IX (Term - I)
∴ ∠AOC = 2 × 20° = 40°
Also, ∠BOD = ∠AOC = 40° [Vertically opposite angles]
7. In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be
congruent by SAS axiom if :
(a) BC = EF
(b) AC = DE
(c) BC = DE
(d) AC = EF
Sol. (b) In ∆ABC and ∆DEF
AB = FD [Given]
∠A = ∠D [Given]
The two triangles will be congruent by SAS rule, if AC = DE.
8. An isosceles right triangle has area 8 cm2
.The length of its hypotenuse is :
(a) 32 cm (b) 16 cm (c) 48 cm (d) 24 cm
Sol. (a) We have,
1
2
× x × x = 8 ⇒ x = 4
∴ Hypotenuse of the triangle = 2 2
4 4 cm = 32 cm+
SECTION B
(Question numbers 9 to 14 carry 2 marks each)
9. Using remainder theorem, find the value of k so that (4x2
+ kx – 1) leaves the remainder
2 when divided by (x – 3).
Sol. Let f(x) = 4x2
+ kx – 1
⇒ f(3) = 4(3)2
+ k(3) – 1 = 2
⇒ 4 × 9 + 3k – 1 = 2 ⇒ 3k = 2 – 35 ⇒ 3k = –33 ⇒ k = – 11
10. Evaluate 103 × 107, without multiplying directly.
Sol. We have, 103 × 107 = (100 + 3) (100 + 7)
= (100)2
+ (3 + 7) × 100 + 3 × 7
[Using (x + a) (x + b) = x2
+ (a + b) x + ab]
= 10000 + 1000 + 21
= 11021
11. Angles X, Y, and Z of a triangle are equal. Prove that ∆XYZ is equilateral.
Sol. In ∆XYZ, ∠X = ∠Y ⇒ YZ = ZX ... (i)
[Sides opposite to equal angles are equal]
Also, ∠Y = ∠Z ⇒ ZX = XY ... (ii)
From (i) and (ii), we have XY = YZ = ZX
So, triangle XYZ is equilateral. Proved.

3. Mathematics IX (Term - I) 3
OR
In the figure, PQ = PR and ∠Q = ∠R. Prove that ∆PQS ≅ ∆PRT.
Sol. In ∆PQS and ∆PRT,
∠Q = ∠R [Given]
PQ = PR [Given]
∠P = ∠P [Common]
∴ By ASA axiom, ∆PQS ≅ ∆PRT. Proved.
12. Where will you find all points with negative abscissa and positive ordinate?
Sol. In the II quadrant we can find all points with negative abscissa and positive ordinate
because the sign of all points in the II quadrant are (–, +).
13. Find the value of (13
+ 23
+ 33
)–3/2
Sol. We have, (13
+ 23
+ 33
)–3/2
= (1 + 8 + 27)–3/2
= (36)–3/2
= (62
)–3/2
= 62×(–3/2)
=
3
3
1 1
6
6 216
−
= = .
14. Two adjacent angles on a straight line are in the ratio 2 : 7. Find the measure of the
greater angle.
Sol. Let the angles are 2x and 7x.
∴ 2x + 7x = 180° ⇒ 9x = 180° ⇒ x = 20°
∴ Greater angle = 7 × 20° = 140°.
SECTION C
(Question numbers 15 to 24 carry 3 marks each)
15. Simplify
( )
2
2 3 1
3 1 4
−
− −
and then express it with a rational denominator...
Sol. We have,
( )
2
2 3 1
3 1 4
−
− −
=
2 3 1
3 1 2 3 4
−
+ − −
=
2 3 1
2 3
−
−
=
( ) 3
3
2 3 1
2 3
−
×
−
[Rationalising the denominator]
=
6 3
6
−
−
=
3 6
6
−
OR
Show how 5 can be represented on the number line.
Sol. Draw a number line as shown in the given figure. Let the point O represent 0 (zero)
and point A represent 2. Draw perpendicular AX at A on the number line and cut-off
AB = 1 unit. Join OB.
We have, OA = 2 units and AB = 1 unit
Using Pythagoras theorem, we have

4. 4 Mathematics IX (Term - I)
OB2
= OA2
+ AB2
⇒ OB2
= (2)2
+ 12
= 5 ⇒ OB = 5
Taking O as the centre and OB = 5 as radius, draw an arc cutting the number line
at C. Clearly, OC = OB = 5 .
Hence, C represents 5 on the number line.
16. Express
3
3 2 2 5+
with a rational denominator...
Sol. We have,
3
3 2 2 5+
Here, rationalising factor of denominator is 3 2 2 5− . So, multiply both numerator
and denominator by 3 2 2 5−
3
3 2 2 5+
=
( )
( ) ( )
( )
( ) ( )
2 2
3 3 2 2 5 3 3 2 2 5
3 2 2 5 3 2 2 5 3 2 2 5
× − −
=
+ × − −
=
( ) ( )3 3 2 2 5 3 3 2 2 5
18 20 2
× − − −
=
−
.
17. In the figure, in ∆ABC, ∠DAC = ∠ECA and AB = BC.
Prove that ∆ABD ≅ ∆CBE.
Sol. In ∆ABC, AB = BC
⇒ ∠C = ∠A
⇒ ∠BCE + ∠ECA = ∠BAD + ∠DAC
⇒ ∠BCE = ∠BAD ... (i) [∵ ∠ECA = ∠DAC]
In ∆ABD and ∆CBE, we have
∠BAD = ∠BCE [From (i)]
AB = CB [Given]
∠ABD = ∠CBE [Common]
∴ By ASA axiom, ∆ABD ≅ ∆CBE. Proved.
18. Two lines AB and CD intersect at a point O such that ∠BOC + ∠AOD = 280°, as
shown in the given figure. Find all the four angles.
Sol. We have, ∠BOC + ∠AOD = 280°
But, ∠BOC = ∠AOD [Vertically opposite angles]

5. Mathematics IX (Term - I) 5
∴ 2∠BOC = 280° ⇒ ∠BOC = 140°
Now, ∠COA + ∠BOD = 360° – 280° = 80°
But, ∠COA = ∠BOD [Vertically opposite angles]
∴ 2∠COA = 80° ⇒ ∠COA = 40°
So, measures of all 4 angles are 140°, 40°, 140°, 40°.
19. Find the remainder when the polynomial f(x) = x4
+ 2x3
– 3x2
+ x – 1 is divided by
(x – 2).
Sol. x – 2 = 0 ⇒ x = 2
By the remainder theorem, we know that when f(x) is divided by (x – 2), the remainder
is f(2).
Now, f(2) = 24
+ 2 × 23
– 3 × 22
+ 2 – 1 = 16 + 16 – 12 + 2 – 1 = 21
Hence, the required remainder is 21.
20. Factorise : 27p3
–
1
216
9
2
1
4
2
− +p p
Sol. We have, 27p3
–
1
216
9
2
1
4
2
− +p p = (3p)3
–
1
6
3
⎛
⎝⎜
⎞
⎠⎟ – 3 × (3p)2
×
1
6
⎛
⎝⎜
⎞
⎠⎟ + 3 × (3p)
1
6
2
⎛
⎝⎜
⎞
⎠⎟
= 3
1
6
3
p −
⎛
⎝⎜
⎞
⎠⎟ [∵ (a – b)3
= a3
– b3
– 3a2
b + 3ab2
]
= 3
1
6
p −
⎛
⎝⎜
⎞
⎠⎟ 3
1
6
p −
⎛
⎝⎜
⎞
⎠⎟ 3
1
6
p −
⎛
⎝⎜
⎞
⎠⎟ .
OR
Factorise : a b3 3
2 2−( )
Sol. We have, a b3 3
2 2−( ) = ( )a b3
3
2− ( )⎡
⎣⎢
⎤
⎦⎥
= a b a a b b−( ) + ( )+ ( )⎡
⎣⎢
⎤
⎦⎥
2 2 22
2
= a b a ab b−( ) + +( )2 2 22 2
∴ a b3 3
2 2−( ) = a b a ab b−( ) + +( )2 2 22 2
21. In the given figure, AB ⊥ BC, AD || BC,
∠CDE = 75° and ∠ACD = 22°. Find
the values of x and y.
Sol. In the given figure, AB ⊥ BC,
AD || BC, ∠CDE = 75° and ∠ACD = 22°.
Now, ∠BCD = ∠CDE [Alternate angles]
⇒ y + 22° = 75°
⇒ y = 75° – 22° = 53°
In ∆ABC, x = 180° – (90° + 53°) [Angle sum property of a triangle]
= 180° – 143° = 37°
Hence, x = 37° and y = 53°.
C B
A D
O

6. 6 Mathematics IX (Term - I)
22. A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that
∠BAL = ∠ACB.
Sol. In right ∆BAC, we have
⇒ ∠BAC = ∠ABC + ∠ACB ...(i)
Again, in right ∆ALB, we have
⇒ ∠ALC = ∠ABL + ∠BAL
[Exterior angle property]
⇒ ∠ALC = ∠ABC + ∠BAL ...(ii)
[∵ ∠ABL = ∠ABC]
From (i) and (ii), we get
∠ABC + ∠ACB = ∠ABC + ∠BAL [∵ ∠BAC = ∠ALC = 90°]
⇒ ∠ACB = ∠BAL
Or, ∠BAL = ∠ACB. Proved.
OR
If the altitudes from two vertices of a triangle to opposite sides are equal, prove that
triangle is isosceles.
Sol. ABC is a triangle in which altitude CD = altitude BE.
In ∆CBD and ∆BCE, we have
Hypotenuse BC = Hypotenuse BC [Common]
CD = BE [Given]
∠BDC = ∠CEB = 90°
∴ ∆CBD ≅ ∆BCE [RHS]
⇒ ∠CBD = ∠BCE [CPCT]
⇒ AB = AC [Side opposite to equal angles are equal]
Hence, the triangle is isosceles. Proved.
23. The difference between the legs of a right-angled triangle is 14 cm. The area of the
triangle is 120 cm2
. Calculate the perimeter of the triangle.
Sol. Let the sides containing the right angle be x cm and (x – 14) cm.
Then, the area of the triangle =
1
2
× x × (x – 14) cm2
But, area = 120 cm2
[Given]
∴
1
2
× x (x – 14) = 120 ⇒ x2
– 14x – 240 = 0
⇒ x2
– 24x + 10x – 240 = 0 ⇒ x(x – 24) + 10(x – 24) = 0
⇒ (x – 24) (x + 10) = 0 ⇒ x = 24 [Neglecting x = – 10]
∴ One side = 24 cm, other side = (24 – 14) cm = 10 cm.
Hypotenuse = ( ) ( )
2 2
24 10+ cm = 576 100+ cm = 676 cm = 26 cm
∴ Perimeter of the triangle = (24 + 10 + 26) cm = 60 cm
24. In the figure, ∠Q > ∠R. If QS and RS are bisectors of ∠Q and ∠R respectively, then
show that SR > SQ.
Sol. We have, ∠Q > ∠R

7. Mathematics IX (Term - I) 7
⇒
1
2
∠Q >
1
2
∠R ⇒ ∠SQR > ∠SRQ
⇒ SR >SQ [Side opposite to greater angle is greater]
Proved.
SECTION D
(Question numbers 25 to 34 carry 4 marks each)
25. Factorise :
2 5 1
2
6 12
x x− +
Sol. We have,
2 5 1
2
6 12
x x− + =
( )2
24 10 1
12
− +x x
=
1
12
(24x2
– 10x + 1)
=
1
12
(24x2
– 6x – 4x + 1) =
1
12
[6x(4x – 1) – 1(4x – 1)]
=
1
12
(4x – 1) (6x – 1)
Hence, 2 5 1
2
6 12
x x− + =
1
12
(4x – 1) (6x – 1)
OR
If p = 2 – a, prove that a3
+ 6ap + p3
– 8 = 0.
Sol. We have, p = 2 – a ⇒ a + p – 2 = 0
Now, a3
+ 6ap + p3
– 8 = a3
+ p3
+ (–2)3
– 3ap(–2)
= {a + p + (–2)}{a2
+ p2
+ (–2)2
– ap – p(–2) – a(–2)}
= (a + p – 2) (a2
+ p2
+ 4 – ap + 2p + 2a)
= 0 × (a2
+ p2
+ 4 – ap + 2p + 2a) = 0.
26. If a + b = 10 and a2
+ b2
= 58, find the value of a3
+ b3
.
Sol. We know that (a + b)2
= a2
+ b2
+ 2ab
Putting a + b = 10 and a2
+ b2
= 58, we get
102
= 58 + 2ab ⇒ 100 = 58 + 2ab ⇒ 2ab = 100 – 58 = 42 ⇒ ab = 21
Thus, we have a + b = 10 and ab = 21
Now, (a + b)3
= a3
+ b3
+ 3ab (a + b)
⇒ (10)3
= a3
+ b3
+ 3 × 21 × 10 [Putting a + b = 10 and ab = 21]
⇒ 1000 = a3
+ b3
+ 630
⇒ a3
+ b3
= 1000 – 630 = 370.
27. Factorise : x8
– y8
Sol. We have, x8
– y8
= {(x4
)2
– (y4
)2
} = (x4
– y4
)(x4
+ y4
)
= {(x2
)2
– (y2
)2
} (x4
+ y4
)
= (x2
– y2
) (x2
+ y2
) (x4
+ y4
)
= (x – y)(x + y) (x2
+ y2
) (x4
+ y4
)
= (x – y)(x + y) (x2
+ y2
){(x2
)2
+ (y2
)2
+ 2x2
y2
– 2x2
y2
}
= (x – y) (x + y)(x2
+ y2
) {(x2
+ y2
)2
– ( 2 xy)2
}
= (x – y) (x + y)(x2
+ y2
)(x2
+ y2
– 2 xy)(x2
+ y2
+ 2 xy)

8. 8 Mathematics IX (Term - I)
28. (i) Plot each of the points
A (–2, 4), B (–2, –3),
C(4, – 3) and D(4, 4).
(ii) Draw the segments AB, BC,
CD and DA. What is the name
of the figure ABCD?
(iii) What are the coordinates of the
point where the segment AD
cuts the y-axis?
(iv) What are the coordinates of the
points where the segment CD
cuts the x-axis?
Sol. (i) Points A(–2, 4), B(–2, –3),
C(4, –3) and D(4, 4) are plotted
in the adjoining figure.
(ii) From the graph, we observe that
AB = CD = 7 units and BC = AD = 6 units.
Therefore, the figure ABCD is a rectangle.
(iii) From the graph, it is clear that the segment AD cuts the y-axis at (0, 4).
(iv) From the graph, it is clear that segment CD cuts the x-axis at (4, 0)
29. ∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that
AD = AB (see figure). Show that ∠BCD is a right angle.
Sol. We have, AB = AC [Given]
∠ACB = ∠ABC ...(i)
[Angles opposite to equal sides are equal]
Also, AB = AD [Given]
∴ AD = AC [∵ AB = AC]
∴ ∠ACD = ∠ADC ...(ii)
[Angles opposite to equal sides are equal]
Adding (i) and (ii), we get
∠ACB + ∠ACD = ∠ABC + ∠ADC
⇒ ∠BCD = ∠ABC + ∠ADC ...(iii)
Now, in ∆BCD, we have
∠BCD + ∠DBC + ∠BDC = 180° [Angle sum property of a triangle]
∴ ∠BCD + ∠BCD = 180° ⇒ 2∠BCD = 180° ⇒ ∠BCD = 90°. Proved.
30. Prove that the perimeter of a triangle is greater than the sum of its three altitudes.
Sol. We have, ∆ABC in which AD, BE and CF are its altitudes.
In ∆ABD, AD < AB ...(i)
[Perpendicular line segment is the shortest]
Similarly, in ∆ADC, AD < AC ...(ii)
Adding (i) and (ii), we get
2AD < AB + AC ...(iii)
Similarly, we can prove

9. Mathematics IX (Term - I) 9
2BE < BC + AB ...(iv)
And, 2CF < AC + BC ...(v)
Adding (iii), (iv) and (v), we get
2(AD + BE + CF) < 2 (AB + BC + CA)
⇒ AB + BC + CA > AD + BE + CF. Proved.
31. In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray
lying between rays OP and OR. Prove that
∠ROS =
1
2
(∠QOS – ∠POS)
Sol. We have, ∠ROS = ∠ROP – ∠POS ...(i)
And ∠ROS = ∠QOS – ∠QOR ...(ii)
Adding (i) and (ii),
∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠ROP – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS
[∵ ∠QOR = ∠ROP = 90°]
⇒ ∠ROS =
1
2
(∠QOS – ∠POS). Proved.
32. In the given figure, AB || CD. Find the value of x
Sol. Through E, draw a line GEH || AB || CD.
Now, GE || AB and EA is a transversal.
∴ ∠GEA = ∠EAB = 50° [Alt. Int. ∠s]
Again, EH || CD and EC is a transversal.
∴ ∠HEC + ∠ECD = 180° [Co. Int. ∠s]
⇒ ∠HEC + 100° = 180° ⇒ ∠HEC = 80°
Now, GEH is a straight line.
∴ ∠GEA + ∠AEC + ∠HEC = 180° [Straight angle]
⇒ 50° + x° + 80° = 180° ⇒ x° = 50°
Hence, x = 50.
33. If x =
5 21
2
−
, find the value of
2
2
1
x
x
+ .
Sol. We have, x +
1
x
=
5 21 2
2 5 21
−
+
−
=
( )
( )( )
2 5 215 21
2 5 21 5 21
+−
+
− +

10. 10 Mathematics IX (Term - I)
=
( )2 5 215 21
2 25 21
+−
+
−
=
5 21 5 21
2 2
− +
+ = 5
Now,
2
2
2
1 1
x x
x x
 
+ = + 
 
– 2 = 52
– 2 = 25 – 2 = 23.
OR
Simplify :
3 2 3 2
3 2 3 2
 + −
+  − + 
Sol. We have,
( ) ( )
( )
( )
( ) ( ){ }
( )
2 2
2 2
3 2 3 2 3 2 3 23 2
3 23 2 3 2 3 2 3 2
+ + + ++
= × = =
−− − + −
= ( ) ( ) ( )
2 2 2
3 2 3 2 2 3 2+ = + + × × = ( )3 2 2 6 5 2 6+ + = +
And,
( )
( )
( )
( )
3 2 3 23 2
3 2 3 2 3 2
− −−
= ×
+ + −
=
( )
( )
2
23 2
3 2
3 2
−
= −
−
= ( ) ( )
2 2
3 2 2 3 2+ − × × = 3 + 2 – 2 6 = ( )5 2 6−
∴
3 2 3 2
3 2 3 2
+ −
+
− +
= ( ) ( )5 2 6 5 2 6+ + − = 10.
34. Prove that :
–1 –1 2
–1 –1 –1 –1 2 2
2
– –
a a b
a b a b b a
+ =
+
Sol. We have,
–1 –1
–1 –1 –1 –1
–
a a
a b a b
+
+
=
1/ 1/ 1/ 1/
1 1 1 1 –
–
+ = +
+
+
a a a a
b a b a
a b a b ab ab
=
1 1
. .
– –
ab ab b b
a b a a b a b a b a
+ = +
+ +
=
2 2
2 2
( – ) ( ) –
( )( – ) –
b b a b b a b ba b ab
b a b a b a
+ + + +
=
+
=
2
2 2
2
–
b
b a
. Proved.