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#10.4

Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to $\bigtriangleup ABC$ with scale factor $\frac{3}{2}$ . Justify the construction. Are the two triangles congruent? Note that all the three angles and two sides of the two triangles are equal

Solution

Steps of construction

Draw a line segment BC = 6 cm
Taking B and C as a centres, draw arc of radius AB= 4cm and AC=9cm
Join AB and AC
DABC is required triangle From B draw ray BM with acute angle $\angle XBM$
Make 3 points $B_{1},B_{2},B_{3}$ on BM with equal distance
Join $B_{2}C$ and $B_{3}$ draw $B_{3}X\parallel B_{2}C$ intersecting BC at X From point X draw XY||CA intersecting the extended line segment BA to Y Then $\bigtriangleup BXY$ is the required triangle whose sides are equal to $\frac{3}{2}$ of the $\bigtriangleup ABC$
Justification :
Here $B_{3}X\parallel B_{2}C$
$\therefore \frac{BC}{CX}= \frac{2}{1}$
$\therefore \frac{BX}{BC}= \frac{BC+CX}{BC}= 1+\frac{1}{2}= \frac{3}{2}$
Also $XY\parallel CA$
$\bigtriangleup ABC\sim \bigtriangleup YBX$
$\therefore \frac{YB}{AB}= \frac{YX}{AC}= \frac{BX}{BC}= \frac{3}{2}$
Here all the three angles are same but three sides are not same.
$\therefore$ The two triangles are not congruent because, if two triangles are congruent, then they have same shape and same size.

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#10.4

Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is $60^{\circ}$ . Also justify the
construction. Measure the distance between the centre of the circle and the point of intersection of tangents

Solution

Steps of construction

1. Draw a circle of radius OA = 4 cm with centre O
2. Produce OA to P such that $OA= AP= 4cm$
3. Draw a perpendicular bisector of OP=8cm
4.Now taking A as Centre draw circle of radius AP = OA = 4 cm
5.Which intersect the circle at x and y
6.Join PX and PY
7.PX and PY is the tangent of the circle
Justification
In $\bigtriangleup OAX$ we have
$OA= OP= 4\, cm$ (Radius)
$AX= 4\, cm$ (Radius of circle with centre A)
$\therefore OAX$ is equilateral triangle
$\angle OAX= 60^{\circ}$
$\Rightarrow \angle XAP= 120^{\circ}$
In $\bigtriangleup PAX$ we have
$PA= AX= 4cm$
$\angle XAP= 120^{\circ}$
$\angle APX= 30^{\circ}$
$\Rightarrow APY= 30^{\circ}$
Hence $\angle XPY= 60^{\circ}$

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#10.4

Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and $ABC= 60^{\circ}$ .Construct a triangle similar to ABC with scale factor $\frac{5}{7}$ . Justify the construction

Solution
Given : AB = 5 cm, BC = 6 cm

Steps of construction

Draw line segment AB = 5 cm
draw $< ABO= 60^{\circ}$ B taking as a centre draw an arc of radius BC=6cm
Join AC, DABC is the required triangle
From point A draw any ray $A{A}'$ with acute angle $\angle BA{A}'$
Mark 7 points $B_{1},B_{2},B_{3},B_{4},B_{5},B_{6},B_{7}$ with equal distance.
Join $B_{7}B$ and form $B_{5}$ draw $B_{5}X\parallel B_{7}B\, \, BY$ making the angle equal From point X draw $XY\parallel BC$ intersecting AC at Y. Then, DAMN is the required triangle whose sides are equal to $\frac{5}{7}$ of the corresponding sides of the $\bigtriangleup ABC$ ..

Justification: Here, $B_{5X}\parallel B_{7}B$ [by construction]
$\therefore \frac{AX}{XB}= \frac{5}{2}\Rightarrow \frac{XB}{AX}= \frac{2}{5}$
Now $\frac{AB}{AX}= \frac{AX+XB}{AX}$
$1+\frac{XB}{AX}= 1+\frac{2}{5}= \frac{7}{5}$
Also, $XY\parallel BC$
$\therefore \bigtriangleup AXY\sim \bigtriangleup ABC$
$\frac{AX}{AB}= \frac{AY}{AC}= \frac{YX}{BC}= \frac{5}{7}$

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#10.4

Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.

Solution

Steps of construction
1. Draw line BC = 5 cm
2.   Taking B and C as centres, draw two arcs of equal radius 6 cm intersecting each other at point A.
3.   Join AB and AC DABC is required isosceles triangle
4    From B draw ray   $B_{X}$ with an acute angle $CB{B}'$
6.   draw $B_{1},B_{2},B_{3},B_{4}$ at $BX$ with equal distance
7.   Join $B_{3}C$ and from $B_{4}$ draw line $B_{4}D\parallel B_{3}C,$ , intersect extended segment BC at point D.
8.   From point D draw $DE\parallel CA$ meting BA produced at E.
Then EBD is required triangle. We can name it PQR.
Justification
$\because B_{4}D\parallel B_{3}C$
$\therefore \frac{BC}{CD}= \frac{3}{1}\Rightarrow \frac{CD}{BC}= \frac{1}{3}$
$Now\, \therefore \frac{BD}{BC}= \frac{BC+CD}{BC}= 1+\frac{CD}{BC}= 1+\frac{1}{3}= \frac{4}{3}$
$Also\, DE\parallel CA$
$\therefore \bigtriangleup ABC\sim \bigtriangleup DBE$
$\frac{EB}{AB}= \frac{DE}{CA}= \frac{BD}{BC}= \frac{4}{3}$

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#10.4

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. Measure the length of a tangent and verify it by actual calculation.

Solution

Steps of construction
1. Draw two concentric circles with center O and radii 3 cm and 5 cm
2.   Taking any point P on outer circle, Join P and O
3.   Draw perpendicular bisector of OP let M be the mid point of OP
4.   Taking M as centre and OM as radius draw a circle which cuts inner circle at Q and R
5.   Join PQ and PR. Thus PQ and PR are required tangents
On measuring PQ and PR we find that PQ = PR = 4 cm
Calculations
$\bigtriangleup OQP, \angle OQP= 90^{\circ}$
$OP^{2}= OQ^{2}+PQ^{2}\left$ [using pythagoras theorem]
$\left ( 5 \right )^{2}= \left ( 3 \right )^{2}+\left ( PQ \right )^{2}$
$25-9= PQ^{2}$
$16= PQ^{2}$
$\sqrt{16}= PQ$
$4cm= PQ$
Hence the length of both tangents is 4 cm.

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#10.4

Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and $ABC= 60^{\circ}$ , divide it into triangles BCD and ABD
by the diagonal BD. Construct the triangle BD'C' similar to DBDC with scale factor $\frac{4}{3}$ . Draw the line segment D'A' parallel to DA where A' lies on extended side BA. Is A'BC'D' a parallelogram?

Solution

Steps of construction
1. Draw, A line AB = 3 cm
2 Draw a ray by making $\angle ABP= 60^{\circ}$
3.   Taking B as centre and radius equal to 5 cm. Draw an arc which cut BP at point C
4.   Again draw ray AX making $\angle {Q}'AX= 60^{\circ}$
5.   With A as centre and radius equal to 5 cm draw an arc which cut AX at point D
6.   Join C and D Here ABCD is a parallelogram
7.   Join BD , BD is a diagonal of parallelogram ABCD
8.   From B draw a ray BQ with any acute angle at point B i.e., $\angle CBQ$ is acute angle
9.   Locate 4 points $B_{1},B_{2},B_{3},B_{4}$ on BQ with equal distance.
10. Join $B_{3}C$ and from   $B_{4},{C}'$ parallel to $B_{3}C$ which intersect at point ${C}'$
11. From point   ${C}'$ draw line ${C}'{D}'$ which is parallel to CD
12. Now draw a line segment ${D}'{A}'$ parallel to DA
Note : Here ${A}',{C}'$ and ${D}'$ are the extended sides.
13.   ${A}'B{C}'{D}'$ is a parallelogram in which ${A}'{D}'= 6\cdot 5\, cm$ and ${A}'{B}= 4\, cm$ and $< {A}'B{D}'= 60^{\circ}$ divide it into triangles $B{C}'{D}'$ and ${A}'{BD}'$ by the diagonal ${BD}'$

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#10.4

Two line segments AB and AC include an angle of $60^{\circ}$ where AB = 5 cm and AC = 7 cm. Locate points P and Q on AB and AC, respectively such that $AP= \frac{3}{4}AB$ and $AQ= \frac{1}{4}AC$ . Join P and Q and measure the length PQ.

Solution
Given :
AB = 5 cm and AC = 7 cm
$AP= \frac{3}{4}AB\: \: \cdots 1$
$AQ= \frac{1}{4}AC\: \: \cdots 2$
From equation 1
$AP= \frac{3}{4}AB$
$AP= \frac{3}{4}\times 5= \frac{15}{4}\: \: \left [ \because AB= 5cm \right ]$
P is any point on B
$\therefore PB= AB-AP= 5-\frac{15}{4}= \frac{20-15}{4}= \frac{5}{4}cm$
$\frac{AP}{AB}= \frac{15}{4}\times \frac{4}{5}= \frac{1}{3}$
$AP:AB= 1:3$
$\therefore$ scale of a line segment AB is $\frac{1}{3}$

Steps of construction
1. Draw line segment AB = 5 cm
2.   Now draw ray AO which makes an angle i.e., $< BAO= 60^{\circ}$
3.   Which A as center and radius equal to 7 cm draw an arc cutting line AO at C
4.   Draw ray AP with acute angle BAP
5.   Along AP make 4 points $A_{1},A_{2},A_{3},A_{4}$ with equal distance.
6.   Join $A_{4}B$
7.   From   $A_{3}$ draw $A_{3}P$ which is parallel to $A_{4}B$ which meet AB at point P.
Then P is point which divides AB in ratio 3 : 1
   AP : PB = 3 : 1
8.   Now draw ray AQ, with an acute angle CAQ.
9.   Along AQ mark 4 points   $B_{1},B_{2},B_{3},B_{4}$ with equal distance.
10. Join $B_{4}C$
11. From $B_{1}$ draw $B_{1}Q$ which is parallel to $B_{4}C$ which meet AC at point Q.
Then Q is point which divides AC in ratio 1 : 3
   AQ : QC = 1 : 3
12. Finally join PQ and its measurement is 3.25 cm.

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#10.4

In Fig. 10.21, O is the centre of the circle, BD = OD and CD $\perp$ AB. Find $\angle$ CAB.

30°

Solution:

Given: In the figure BD = OD, CD $\perp$ AB

In $\triangle$ OBD,

BD = OD (given)

OD = OB (radius of the same circle)

OB = OD = BD

Hence the triangle is equilateral

$\angle$ BOD = $\angle$ OBD = $\angle$ ODB = 60°

Consider $\triangle$ MBC and MBD

MB = MB (common)

$\angle$ CMB = $\angle$ BMD = 90° (given)

CM = MD (perpendicular from the centre on the chord bisects the chord]

$\triangle$ MBC = $\triangle$ MBD (SAS rule)

$\angle$ MBC = $\angle$ MBD (CPCT)

$\angle$ MBC = $\angle$ OBD = 60° ( $\because$ OBD = 60°)

Since AB is the diameter of the circle

$\angle$ ACB = 90° (angle is a semi-circle)

$\angle$ CAB + $\angle$ CBA + $\angle$ ACB = 180° (By angle sum property of a triangle)

Putting the values,

$\angle$ CAB + 60° + 90° = 180°

$\angle$ CAB = 180° – (60° + 90°) = 30°

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#10.4

In Fig. 10.20, O is the centre of the circle, $\angle$ BCO = 30°. Find x and y.

x = 30° and y = 15°

Solution:

Given: O is the centre of the circle

$\angle$ BCO = 30°

Join OB and AC.

In $\triangle$ BOC,

CO = BO (Radius of the same circle)

$\angle$ OBC = $\angle$ OCB = 30° (angles opposite to equal sides in a triangle are equal)

In $\triangle$ OBC,

$\angle$ OBC + $\angle$ OCB + $\angle$ BOC = 180° (angle sum property of a triangle)

$\angle$ BOC = 180° – (30° + 30°) = 120°

$\angle$ BOC = 2 $\angle$ BAC (Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)

$\angle BAC = \frac{120^{\circ}}{2}=60^{\circ}$

Consider $\triangle$ AEB, $\triangle$ AEC

AE = AE (common)

$\angle$ AEB = $\angle$ AEC = 90°

BE = EC (given)

$\triangle$ AEB $\cong$ $\triangle$ AEC (SAS congruence)

$\angle$ BAE = $\angle$ CAE (CPCT)

$\angle$ BAE + $\angle$ CAE = $\angle$ BAC = 60°

2 $\angle$ BAE = 60°

$\angle$ BAE = 30° = x

Now, OD is perpendicular to AE

and, CE is perpendicular to AE

Two lines perpendicular to the same line are parallel to each other

So, OD || CE and OC is transversal

$\angle$ DOC = $\angle$ ECO = 30° (alternate interior angles)

Now, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle.

Consider the arc CD,

$\angle$ DOC = 2 $\angle$ DBC = 2y

30° = y

y = 15°

Hence, x = 30° and y = 15°

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#10.4

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q² = p² + 3r²

AB and AC are two chords of a circle of radius r such that AB = 2AC.

The distance of AB and AC from the centre are p and q, respectively

To Prove: 4q²= p² + 3r²

Proof:

Let AB = 2x then AC = x (Given, AB = 2AC)

Draw ON perpendicular to AB and OM perpendicular to AC

AM = MC = x/2 (perpendicular from the centre bisects the chord)

AN = NB = x (perpendicular from the centre bisects the chord)

In DOAM, applying Pythagoras theorem

AO² = AM²+ MO²

AO² = (x/2)² + q² ….(i)

In $\triangle$ OAN, applying Pythagoras theorem

AO² = (AN)² + (NO)²

AO² = (x)² + p² ….(ii)

From equation (i) and (ii)

$\\\left ( \frac{x}{2} \right )^{2}+\left ( q \right )^{2}=x^{2}+\left ( p \right )^{2}\\ \frac{x^{2}}{4}+q^{2}=x^{2}+p^{2}$

x² + 4q² = 4x² + 4p² [Multiply both sides by 4]

4q² = 3x² + 4p²

4q²= p² + 3 (x² + p²)

4q²= p²+ 3r² (In right angle $\triangle$ OAN, r² = x² + p²)

Hence Proved.

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All Questions

Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to with scale factor . Justify the construction. Are the two triangles congruent? Note that all the three angles and two sides of the two triangles are equal

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Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is . Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents

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Crack CUET with india's "Best Teachers"

Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and .Construct a triangle similar to ABC with scale factor . Justify the construction

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Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.

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JEE Main high-scoring chapters and topics

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. Measure the length of a tangent and verify it by actual calculation.

Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to $\bigtriangleup ABC$ with scale factor $\frac{3}{2}$ . Justify the construction. Are the two triangles congruent? Note that all the three angles and two sides of the two triangles are equal

Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is $60^{\circ}$ . Also justify the
construction. Measure the distance between the centre of the circle and the point of intersection of tangents

Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and $ABC= 60^{\circ}$ .Construct a triangle similar to ABC with scale factor $\frac{5}{7}$ . Justify the construction

Two line segments AB and AC include an angle of $60^{\circ}$ where AB = 5 cm and AC = 7 cm. Locate points P and Q on AB and AC, respectively such that $AP= \frac{3}{4}AB$ and $AQ= \frac{1}{4}AC$ . Join P and Q and measure the length PQ.

In Fig. 10.21, O is the centre of the circle, BD = OD and CD $\perp$ AB. Find $\angle$ CAB.

In Fig. 10.20, O is the centre of the circle, $\angle$ BCO = 30°. Find x and y.

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q² = p² + 3r²