Filter By

All Questions

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 centres, draw 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 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.

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

Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

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

1. Draw 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 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 $B_{5}X\parallel B_{7}B\, \, BY$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, $XY\parallel BC$
$\therefore \bigtriangleup AXY\sim \bigtriangleup ABC$
$\frac{AX}{AB}= \frac{AY}{AC}= \frac{YX}{BC}= \frac{5}{7}$

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

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 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.

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 ${BD}'$

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
$\therefore PB= AB-AP= 5-\frac{15}{4}= \frac{20-15}{4}= \frac{5}{4}cm$
$\frac{AP}{AB}= \frac{15}{4}\times \frac{4}{5}= \frac{1}{3}$
$AP:AB= 1:3$
$\therefore$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.

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°

Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

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°

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.