Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
Filter By
Clear Filter

All Questions

Q: 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

  1. Draw a line segment BC = 6 cm
  2. Taking B and C as a centre, draw an arc of radius AB= 4cm and AC=9cm
  3. Join AB and AC
  4. DABC is required triangle From B draw ray BM with acute angle \angle XBM
  5. Make 3 points B_{1},B_{2},B_{3}  on BM with equal distance
  6. 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 three angles are the same but the three sides are not the same.
    \thereforeThe two triangles are not congruent because, if two triangles are congruent, then they have the same shape and same size.

View Full Answer(1)
Posted by

infoexpert27

Q: 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. Construct a circle with $O$ as a centre and a 4 cm radius
2. Construct any diameter $A O B$
3. Construct an angle $\angle A O P=600$ where $O P$ is the radius which intersect the circle at the point $P$
4. Construct $P Q$ perpendicular to $O P$ and $B E$ perpendicular to $O B$
$P Q$ and $B E$ intersect at the point $R$
5. $R P$ and $R B$ are the required tangents
6. The measurement of $O R$ is 8 cm

Justification:
PR is the tangent to a circle

$$
\angle O P Q=900
$$

$B R$ is the tangent to a circle

$$
\angle O B R=900
$$
So we get

$$
\angle P O B=180-60=1200
$$
In BOPR

$$
\angle B R P=360-(120+90+90)=600
$$
Therefore, the distance between the centre of the circle and the point of intersection of tangents is 8 cm.

View Full Answer(1)
Posted by

infoexpert27

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

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

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

Steps of construction

  1. Draw a line segment AB = 5 cm
  2. draw < ABO= 60^{\circ}  B taking as a centre draw an arc of radius BC=6cm
  3. Join AC, DABC is the required triangle
  4.  From point A draw any ray A{A}'   with an acute angle \angle BA{A}'
  5.  Mark 7 points B_{1},B_{2},B_{3},B_{4},B_{5},B_{6},B_{7}  with equal distance.
  6. Join B_{7}B  and form B_{5}  draw B_{5}X\parallel B_{7}B\, \, BYmaking 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}  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

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 $B C=5 \mathrm{~cm}$
2. Taking $B$ and $C$ as centres, draw two arcs of equal radius 6 cm intersecting each other at point $A$.
3. Join $A B$ and $A C D A B C$ is required isosceles triangle

4 From B draw ray $B_X$ with an acute angle $C B B^{\prime}$
6. draw $B_1, B_2, B_3, B_4$ at $B X$ with equal distance
7. Join $B_3 C$ and from $B_4$ the draw line $B_4 D \| B_3 C$, intersect extended segment BC at point D .

Then EBD is required triangle. We can name it PQR.

$$
\begin{aligned}
& \text { Justification } \\
& \because B_4 D \| B_3 C \\
& \therefore \frac{B C}{C D}=\frac{3}{1} \Rightarrow \frac{C D}{B C}=\frac{1}{3} \\
& \text { Now } \therefore \frac{B D}{B C}=\frac{B C+C D}{B C}=1+\frac{C D}{B C}=1+\frac{1}{3}=\frac{4}{3} \\
& \text { Also } D E \| C A \\
& \therefore \triangle A B C \sim \triangle D B E \\
& \frac{E B}{A B}=\frac{D E}{C A}=\frac{B D}{B C}=\frac{4}{3}
\end{aligned}
$$

View Full Answer(1)
Posted by

infoexpert27

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on the 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 centre O and radii 3 cm and 5 cm
2.   Taking any point P on the outer circle, Join P and O
3.   Draw a perpendicular bisector of OP let M be the midpoint of OP
4.   Taking M as the centre and OM as the radius draw a circle which cuts the 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} [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

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 the 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 the centre and radius equal to 5 cm. Draw an arc which cuts BP at point C
4.   Again draw ray AX making  \angle {Q}'AX= 60^{\circ}
5.   With A as the centre and radius equal to 5 cm draw an arc which cuts 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 an 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 a line {C}'{D}'  which is parallel to CD
 12. Now draw a line segment {D}'{A}'  parallel to the 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

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

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
\thereforescale of a line segment AB is \frac{1}{3}


 Steps of construction
  1.   Draw a line segment AB = 5 cm
  2.   Now draw ray AO which makes an angle i.e., < BAO= 60^{\circ}
  3.   Which A as centre 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 a point which divides AB in a 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 a point which divides AC in a 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

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

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

In \triangleOBD,

BD = OD         (given)

OD = OB         (radius of the same circle)

OB = OD = BD

Hence the triangle is equilateral

\angleBOD = \angleOBD = \angleODB = 60°

Consider \triangleMBC and MBD

MB = MB                               (common)

\angleCMB = \angleBMD = 90°          (given)

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

\triangleMBC =\triangleMBD                     (SAS rule)

\angleMBC =\angleMBD                    (CPCT)

\angleMBC = \angleOBD = 60°           (\becauseOBD = 60°)

Since AB is the diameter of the circle

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

\angleCAB +\angleCBA + \angleACB = 180°       (By angle sum property of a triangle)

Putting the values,

\angleCAB + 60° + 90° = 180°

\angleCAB = 180° – (60° + 90°) = 30°

View Full Answer(1)
Posted by

infoexpert24

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

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

Given: O is the centre of the circle

\angleBCO = 30°

Join OB and AC.

In \triangleBOC,

CO = BO (Radius of the same circle)

\angleOBC =\angleOCB = 30°  (angles opposite to equal sides in a triangle are equal)

In \triangleOBC,

\angleOBC + \angleOCB + \angleBOC = 180°  (angle sum property of a triangle)

\angleBOC = 180° – (30° + 30°) = 120°

\angleBOC = 2\angleBAC (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 \triangleAEB, \triangleAEC

AE = AE (common)

\angleAEB =\angleAEC = 90°

BE = EC (given)

\triangleAEB \cong \triangleAEC  (SAS congruence)

\angleBAE =\angleCAE  (CPCT)

\angleBAE + \angleCAE = \angleBAC = 60°

2\angleBAE = 60°

\angleBAE = 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

\angleDOC = \angleECO = 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,

\angleDOC = 2\angleDBC = 2y

30° = y

y = 15°

Hence, x = 30° and y = 15°

View Full Answer(1)
Posted by

infoexpert24

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 = 2a then AC = a                   (Given, AB = 2AC)

Draw ON perpendicular to AB and OM perpendicular to AC

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

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

In DOAM, applying Pythagoras theorem

AO2 = AM2 + MO2

AO2 = (a/2)2 + q2                                 ….(i)    

In\triangleOAN, applying Pythagoras theorem

AO2 = (AN)2 + (NO)2

AO2 = (a)2 + p2                                      ….(ii)

From equation (i) and (ii)

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

a2 + 4q2 = 4a2 + 4p2                [Multiply both sides by 4]

4q2 = 3x2 + 4p2

4q2 = p2 + 3 (a2 + p2)

4q2 = p2 + 3r2                                           (In right angle \triangleOAN, r2 = x2 + p2)

Hence Proved.

View Full Answer(1)
Posted by

infoexpert24

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE
filter_img