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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 \(3+2\sqrt2:32\sqrt3 \).
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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.
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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,
Step1 Draw a line segment
Step2 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 midpoint 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.
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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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.