Maths Class X Question Bank

Question 1 of 19

1. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
We can always divide a line segment in the ratio $\sqrt2:\dfrac{1}{\sqrt2} $ by geometrical construction.
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Hint

False
Not always
Question 2 of 19

2. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
By geometrical construction, it is possible to divide a line segment in the ratio $3+2\sqrt2:3-2\sqrt3 $.
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Hint

False
$\dfrac{3+2\sqrt2}{3-2\sqrt2}\times\dfrac{3+2\sqrt2}{3+2\sqrt2}\\ \\ =\dfrac{(3+2\sqrt2)^2}{1}=9+8+12\sqrt2\\ \\ =17+12\sqrt2$
Question 3 of 19

3. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
We can draw a tangent to a circle from a point which lies in the interior of the circle.
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Hint

False
From interior point the no. of tangents that can be drawn is O
Question 4 of 19

4. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
At any point on a circle, we can draw only one tangent
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Hint

True
Only one tangents can be drawn to a circle from a point on the same circle.
Question 5 of 19

5. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
From a point P which lies in the exterior of the circle, we can draw exactly two tangents to the circle.
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Hint

True
Exactly two tangents can be drawn from an exterior point to a given cirlce.
Question 6 of 19

6. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
A pair of tangents can be constructed to a circle inclined at an angle of $105^o $.
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Hint

True
If the angle between the pair of tangents is > 0 or < $180^o$ then we can construct a pair of tangents.
Question 7 of 19

7. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
A pair of tangents can be constructed from a point P to a circle of radius 4 cm situated at a distance of 3.5 cm from the centre.
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Hint

False
r > d
Point P lies inside the circle so no tangent is possible.
Question 8 of 19

8. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
Draw a line segment AB = 7.5 cm. Find a point P on it which divides it in the ratio 2 : 3.
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Hint

In the figure,

Step-1 Draw a line segment $AB=7.5\ cm$
$2x+3x=7.5\ cm\\ \\ x=1.5$
Step-2 Mark 5 arcs of equal length of 1.5 cm
Hence,
$AC=2\times 1.5\Rightarrow 3.0\ cm\\ BC=3\times 1.5\Rightarrow 4.5\ cm$
Hence, this is the answer.
Question 9 of 19

9. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
Draw a line segment of length 7.6 cm and divide it into the ratio 5 : 8. Measure the two parts.
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Hint

Given:
Line segment $AB=7.6\ cm$
Construction: (1) In circle of $\triangle$ ABC.
Steps of Construction:
(1) Draw a line segment $AB=7.6\ cm$
(2) Draw a ray AX making an acute angle with AB at A.
(3) Locate $5+8=13$ points $A_1,A_2,A_3,.....A_{13}$ at equal distance on $AX$ . Join $A_{13}B$ .
(4) From point $A_5$ draw a line $AA_5P||AA_{13}B$ to meet $AB$ at $P$ .
I.e., $\angle BA_{13}A=\angle PA_5A$ .
Then $AP:PB=5:8$
Hence, $AP$ and $PB$ are the requires parts of $AB$ .
Proof: In $\triangle AA_5P$ and $\triangle AA_{13}B$ ,
$A_{13}B||A_5P$
$\therefore$ by $BPT$ , we get
$\dfrac{AA_5}{A_5A_{13}}=\dfrac{AP}{PB}$
But $\dfrac{AA_5}{A_5A_{13}}=\dfrac{5}{8}$ (By construction)
$\therefore\dfrac{AP}{PB}=\dfrc{AP}{PB}=\dfrac{5}{8}$
Hence, point P, divides AB into the ratio 5 : 8.
Measuring two parts, we get.
AP = 2.9 cm and PB = 4.7 cm
Question 10 of 19

10. Question
1 point(s)
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Three sides PQ, QR and PR of $\triangle $PQR are 5 cm, 6 cm and 7 cm respectively. Construct the $\triangle $PQR . Construct a $\triangle $PQ’R’ such that each of its sides is $\dfrac{2}{3} $ of corresponding sides of $\triangle $PQR
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Hint

Making the triangle
Draw a line segment AB=5 cm
Draw an arc at 6 cm from point A.
Draw an arc at 7 cm from point B so that it intersects the previous arc.
Join the point of intersection from A and B.
This gives the required triangle ABC.
Dividing the base:
Draw a ray AX at an acute angle from AB
Plot three points on AX so that; $AA_1=A_1A_2=A_2A_3$
Join $A_3$ to $B$ .
Draw a line from point $A_2$ so that this line is parallel to $A_3B$ and intersects $AB$ at point $B'$
Draw a line from point $B'$ parallel to $BC$ so that this line intersects $AC$ at point $C'$
Question 11 of 19

11. Question
1 point(s)
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Draw a right triangle $ABC $ in which $BC=12\ cm,\ AB=5\ cm $ and $\angle B=90^o $. Construct a triangle similar to it and of scale factor $\dfrac{2}{3} $. Is the new triangle also a right triangle ?
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Hint

In $\triangle ABC$ ,
$AB=5\ cm,\ BC=12\ cm,\ \angle B=90^o$
To construct – A $\triangle$ of scale factor $\dfrac{2}{3}$ which is similar to $\triangle ABC$ as well.
Construction:- $BC$ is divided into ratio $2:3$ at $D$ .
$DE$ drawn parallel to $AB$ .
It intersects $AC$ at $E$ .
$\triangle DEC$ is the desired $\triangle$ .
Question 12 of 19

12. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
Construct a $\triangle ABC$ in which $AB = 5\ cm, \angle B = 60^o$ , altitude $CD = 3\ cm$. Constuct a $\triangle AQR $ similar to $\triangle ABC $ such that each side of $\triangle AQR $ is 1.5 times that of the corresponding side of $\triangle ABC $
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Hint

Step: I – First of all we draw a line segment $\mathrm{AB}=5 \mathrm{~cm}$.
Step: II – With $B$ as centre and draw an angle $\angle B=60^{\circ}$.
Step: III – From point $A$ and $B$ construct altitude $C D=$ $3 \mathrm{~cm}$, which cut the line $\mathrm{BS}$ at point $C$
Step: IV -Join $A C$ to obtain $\triangle \mathrm{ABC}$.
Step: V – Below $A B$, makes an acute angle $\angle B A X=60^{\circ}$.
Step: VI – Along $A X$, mark off five points $A_{1}, A_{2}$ and $A_{3}$ such that $\mathrm{AA}_{1}=\mathrm{A}_{1} \mathrm{~A}_{2}=\mathrm{A}_{2} \mathrm{~A}_{3}$
Step: VII – Join $\mathrm{A}_{2} \mathrm{~B}$.
Step: VIII -Since we have to construct a triangle $\triangle AQR$ each of whose sides is (1.5 times = 3/2) of the corresponding sides of $\triangle ABC$ .
So, we draw a line $A_3Q$ on $AX$ from point $A_3$ which is $A_3Q||A_2B$ and meeting $AB$ at $Q$ .
Step: IX – From $Q$ point draw $QR||BC$ and meeting $AC$ at $R$
Thus, $\triangle AQR$ is the required triangle, each of whose sides is (1.5 times = 3/2) of the corresponding sides of $\triangle ABC$ .
Question 13 of 19

13. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre
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Hint

The steps of construction are as follows:
1. Taking any point O as centre, draw a circle of 4 cm radius. Locate a point P which is 6 cm away from O. Join OP.
2. Bisect OP. Let M be the mid-point of P O .
3. Taking M as centre and MO as radius, draw a circle.
4. Let this circle intersect the previous circle at point Q and R.
5. Join PQ and PR. PQ and PR are the required tangents.
Question 14 of 19

14. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
At a point P on the circle, draw a tangent, without using the centre of the circle.
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Hint

here, p is the point on circle,and a tangent is passing through this circle ,and points $(x_1,y_1)$ are points which passing through tangent.
Question 15 of 19

15. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of $60^o $.
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Hint

Steps of Construction:
1. A circle is drawn with centre ‘O’ and radius 5 cm. Draw OA and OB radii and ∠AOB = 120°
2. Tangents are drawn at A and B and these intersects at P.
3. By measurement, ∠APB = 60°.
Angle between tangents PA and PB is 60°.
$\therefore$ ∠AOB and ∠APB Supplementary angles.
Question 16 of 19

16. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
Two line segments AB and AC include an angle of $60^o $,where $AB = 5\ cm$ and $AC = 7\ cm$. Locate points P and Q on AB and AC respectively such that $AP=\dfrac{3}{4}AB $ and $AQ=\dfrac{1}{4}\ AC $. Join P and Q and measure the length of PQ.
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Hint

1. Draw a line segment $AB = 5\ cm$ .
2. Draw $\angle BAZ=60^o$ .
3. With centre A and radius 7 cm, draw an arc cutting the line AZ at C.
4. Draw a ray AX, making an acute ∠BAX.
5. Divide AX into four equal parts, namely $AA_1=A_1A_2=A_2A_3=A_3A_4$ .
6. Join $A_4B$ .
7. Draw $A_3P||A_4B$ meeting $AB$ at $P$
8. Hence, we obtain, $P$ is the point on $AB$ such that $AB=\dfrac{3}{4}AB$ .
9. Next, draw a ray AY, such that it makes an acute ∠CAY
10. Divide $AY$ into four parts, namely $AB_1=B_1B_2=B_2B_3=B_3B_4$ .
11. Join $B_4C$ .
12. Draw $B_1Q||B_4C$ meeting $AC$ at $Q$ . We get, $Q$ is the point on $AC$ such that $AQ=\dfrac{1}{4} AC$ .
13. Join PQ and measure it.
14. $PQ=3.25\ cm$ .
Question 17 of 19

17. Question
1 point(s)
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Given a rhombus $ ABCD$ in which $AB=4\ cm $ and $\angle ABC=60^o $, divide it into two triangles say $ABC $ and $ADC $ by the diagonal $AC $. Construct the $\triangle AB’C’ $ similar to $\triangle ABC $ with scale factor $\dfrac{3}{5} $ Draw a line segment $C’D’ $ parallel to $CD $ where $D’ $ lies on $AD $. In $AB’C’D’ $ a rhombus?
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Hint

From the figure
$\dfrac{AB'}{AB}=\dfrac{2}{3}=\dfrac{A'C'}{AC}$
Also, $\dfrac{AC'}{AC}=\dfrac{C'D'}{CD}=\dfrac{AD'}{AD}=\dfrac{2}{3}$
Therefore, $AB' =B'C' =C'D' =AD' =\dfrac{2}{3}AB$
Thus, $AB'C'D'$ is a rhombus.
Question 18 of 19

18. Question
1 point(s)
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Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on outer circle construct the pair of tangents to the other
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Hint

Steps of construction :
1. Draw a circle with radius 3 cm and centre O.
2. Draw another circle with radius 5 cm and same centre O.
3. Take a point P on the circumference of larger circle and join O to p.
4. Taking OP as diameter draw another circle which intersects the smallest circle at A and B.
5. Join A to P and B to P.
Hence AP and BP are the required tangents.
Question 19 of 19

19. Question
1 point(s)
Write ‘T’ for true and ‘F’ for false statement. In each case, give reason for your answer.
Draw a circle of radius 3 cm. Take two point P and Q on one of its extended diameter each at a distance of 7 cm on opposites of its centre. Draw tangents to the circle from these two points P and Q.
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Hint

Step 1: Draw a circle with centre O and radius 3 cm using a compass.
Step 2: Draw a secant passing through the centre. Mark points P and Q on opposite sides of the centre at a distance of 7 cm from O.
Step 3: Place the compass on P and draw two arcs on opposite sides of OP. Now place the compass on O and draw two arcs intersecting the arcs drawn from point P.
Step 4: Join the intersection points of the arcs to obtain the perpendicular bisector of $OP$ . Mark the mid point of $OP$ as $M_1$
Step 5: From $M_1$ draw circle with radius $=M_1P=M_1O$
Step 6: Mark the intersection points of the circle drawn from $M_1$ with the circle drawn from O as A and B.
Step 7: Join P − A and P − B
Step 8: Repeat steps 3 to 7 for point Q and obtain tangents QC and QD
PA, PB, QC and QD are the required tangents.