Q&A - Ask Doubts and Get Answers

#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.

View Full Answer(1)

Posted by

infoexpert27

#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 $\angle XPY= 60^{\circ}$

View Full Answer(1)

Posted by

infoexpert27

#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, $XY\parallel BC$
$\therefore \bigtriangleup AXY\sim \bigtriangleup ABC$
$\frac{AX}{AB}= \frac{AY}{AC}= \frac{YX}{BC}= \frac{5}{7}$

View Full Answer(1)

Posted by

infoexpert27

#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
$\because B_{4}D\parallel B_{3}C$
$\therefore \frac{BC}{CD}= \frac{3}{1}\Rightarrow \frac{CD}{BC}= \frac{1}{3}$
$Now\, \therefore \frac{BD}{BC}= \frac{BC+CD}{BC}= 1+\frac{CD}{BC}= 1+\frac{1}{3}= \frac{4}{3}$
$Also\, DE\parallel CA$
$\therefore \bigtriangleup ABC\sim \bigtriangleup DBE$
$\frac{EB}{AB}= \frac{DE}{CA}= \frac{BD}{BC}= \frac{4}{3}$

View Full Answer(1)

Posted by

infoexpert27

#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.

View Full Answer(1)

Posted by

infoexpert27

#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 ${BD}'$

View Full Answer(1)

Posted by

infoexpert27

#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.

View Full Answer(1)

Posted by

infoexpert27

#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°

View Full Answer(1)

Posted by

infoexpert24

#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°

View Full Answer(1)

Posted by

infoexpert24

#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.

View Full Answer(1)

Exams

Colleges

Predictors

Resources

Quick links

Quick Links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Resources

Diploma Colleges

Other Exams

Exams

Resources

Upcoming Events

Exams

Top Schools

Products & Resources

NCERT Study Material

Top Countries

Student Visas

Exams

Colleges

Exams

Colleges

Upcoming Events

Resources

Exams

Colleges & Courses

Predictors

Resources

Online Courses

Products

Engineering Preparation

Medical Preparation

Top Streams

Specializations

Resources

Top Providers

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors & E-Books

Exams

Predictors & Articles

Colleges

Resources

Exams

Colleges

Resources

Top Courses & Careers

Colleges

Exams

Resources

Get Answers to all your Questions

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

Posted by

infoexpert27

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

Posted by

infoexpert27

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

Posted by

infoexpert27

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.

Posted by

infoexpert27

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