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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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False
Not always
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By geometrical construction, it is possible to divide a line segment in the ratio \(3+2\sqrt2:3-2\sqrt3 \).
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False
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We can draw a tangent to a circle from a point which lies in the interior of the circle.
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False
From interior point the no. of tangents that can be drawn is O
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At any point on a circle, we can draw only one tangent
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True
Only one tangents can be drawn to a circle from a point on the same circle.
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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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True
Exactly two tangents can be drawn from an exterior point to a given cirlce.
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A pair of tangents can be constructed to a circle inclined at an angle of \(105^o \).
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True
If the angle between the pair of tangents is > 0 or < then we can construct a pair of tangents.
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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False
r > d
Point P lies inside the circle so no tangent is possible.
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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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In the figure,
Step-1 Draw a line segment
Step-2 Mark 5 arcs of equal length of 1.5 cm
Hence,
Hence, this is the answer.
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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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Given:
Line segment
Construction: (1) In circle of ABC.
Steps of Construction:
(1) Draw a line segment
(2) Draw a ray AX making an acute angle with AB at A.
(3) Locate points
at equal distance on
. Join
.
(4) From point draw a line
to meet
at
.
I.e., .
Then
Hence, and
are the requires parts of
.
Proof: In and
,
by
, we get
But (By construction)
Hence, point P, divides AB into the ratio 5 : 8.
Measuring two parts, we get.
AP = 2.9 cm and PB = 4.7 cm
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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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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;
Join to
.
Draw a line from point so that this line is parallel to
and intersects
at point
Draw a line from point parallel to
so that this line intersects
at point
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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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In ,
To construct – A of scale factor
which is similar to
as well.
Construction:- is divided into ratio
at
.
drawn parallel to
.
It intersects at
.
is the desired
.
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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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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 each of whose sides is (1.5 times = 3/2) of the corresponding sides of
.
So, we draw a line on
from point
which is
and meeting
at
.
Step: IX – From point draw
and meeting
at
Thus, is the required triangle, each of whose sides is (1.5 times = 3/2) of the corresponding sides of
.
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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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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.
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At a point P on the circle, draw a tangent, without using the centre of the circle.
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here, p is the point on circle,and a tangent is passing through this circle ,and points are points which passing through tangent.
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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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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°.
∠AOB and ∠APB Supplementary angles.
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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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1. Draw a line segment .
2. Draw .
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 .
6. Join .
7. Draw meeting
at
8. Hence, we obtain, is the point on
such that
.
9. Next, draw a ray AY, such that it makes an acute ∠CAY
10. Divide into four parts, namely
.
11. Join .
12. Draw meeting
at
. We get,
is the point on
such that
.
13. Join PQ and measure it.
14. .
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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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From the figure
Also,
Therefore,
Thus, is a rhombus.
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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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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.
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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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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 . Mark the mid point of
as
Step 5: From draw circle with radius
Step 6: Mark the intersection points of the circle drawn from 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.