Math Ncert Exemplar Solutions for Class 10 Maths Chapter 10 - Constructions

Answer:

Given, a line segment AB in the ratio 5 : 7.

Now, draw a ray AX which makes an acute angle ∠BAX, then mark A + B points at equal distance.

Here, A = 5 and B = 7.

Therefore, the minimum number of these points = A + B
                                                                              = 5 + 7
                                                                              = 12
Hence, the correct answer is option D.

Answer:

Given, a line segment AB in the ratio 4 : 7
$\Rightarrow$A : B = 4 : 7
Minimum number of points located at equal distances on the ray AX = A + B
                                                                                                               = 4 + 7
                                                                                                               = 11
Here, A₁, A₂, A₃, ... are located at equal distances on the ray AX.
Point B is joined to the last point is A₁₁.

Hence, the correct answer is option B.
 

Answer:

Given, a line segment AB in the ratio 5 : 6.
⇒ A : B = 5 : 6

To divide a line segment AB in the ratio 5 : 6, the fifth division of AX and the sixth division of BY are joined. Thus, A₅ and B₆ are joined. 
Hence, the correct answer is option A.

 

Answer:

Steps:
1. Locate points B₁, B₂, B₃, B₄, B₅, B₆ and B₇ on BX at equal distance.


2. Join the last point B₇ to C.


Hence, the correct answer is option C.

Answer:

When a triangle similar to the given triangle with its sides $\frac{m}{n}$ of the corresponding sides of given triangle is to be constructed, the minimum number of points to be located at an equal distance is equal to the greater of m and n.
Here, $\frac{m}{n}=\frac{8}{5}$.

So, the minimum number of point to be located at equal distance on ray BX is 8.
Hence, the correct answer is option C.

Answer:

First, make the figure as per the given information.

Here,
OP and OQ are the tangents to the circle. Thus, they intersect the circle at right angles.

Now, the sum of angles of a quadrilateral, here PRQO, is 360°.
∴ ∠POQ + ∠ORP + ∠PRQ + ∠OQR = 360°
⇒ 60° + 90° + θ + 90° = 180°
⇒ θ = 120°
The angle between them should be 120°.
Hence, the correct answer is option D.

Answer:

The ratio in which the line segment is to be divided is $\sqrt{3}:\frac{1}{\sqrt{3}}$.
$\therefore \mathrm{Required} \mathrm{ratio}=3:1 [ \mathrm{multiplying} \sqrt{3} \mathrm{in} \mathrm{each} \mathrm{term}]$
So, $\sqrt{3}:\frac{1}{\sqrt{3}}$ can be simplified as 3 : 1 and 3 as well as 1 both are positive integers.
Hence, by geometrical construction is possible to divide a line segment in the ratio 3 : 1.

Answer:

False

To draw a triangle similar to a given $△$ABC with its sides $\frac{7}{3}$ of the corresponding sides of $△$ABC, the points B₁, B₂, ...., B₇ are located at equal distances on BX, B₃ is joined to C and then a line segment B₇C' is drawn parallel to B₃C where C' lies on BC produced.

So, B₇C’ is parallel to B₃C.

Answer:

False

Let r be radius of circle and d be the distance of the point from the centre.
Here, r = 3.5 cm and d = 3 cm.
⇒ r > d
Point P lies in inside the circle, as shown below:

Thus, it is not possible to make tangents to the circle from point P which lies inside the circle.

Answer:

True

The angle between the pair of tangents should always be greater than 0° and less than 180 °.

Hence, we can draw a pair of tangents to a circle inclined at an angle of 170°, as shown below:

Answer:

Steps of construction:
1. Draw a line segment AB = 7 cm.



2. Draw a ray AX, making an acute ∠BAX.



3. Along AX, mark 3 + 5 = 8 points namely A₁, A₂, A₃, A₄, A₅, A₆, A₇, A₈ such that
AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇ = A₇A₈.


4. Join A₈B.


5. From A₃, draw A₃P  ∥ A₈B meeting AB at P. (By making an angle equal to ∠BA₈ at A₃).
Then, P is the point on AB which divides it in the ratio 3 : 5. Thus, AP : PB = 3 : 5.


Justification:
Let AA₁ = A₁A₂ = A₂A₃ = A₃A₄ =........= A₇A₈ = x
In $△$ABA₈, A₃P âˆ¥ A₈B.
$\therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{{\mathrm{AA}}_3}{{\mathrm{A}}_3{\mathrm{A}}_8}=\frac{3x}{5x}=\frac{3}{5}$
Hence, AP : PB = 3 : 5.

 

Answer:

Steps of construction:
1. Draw a line segment BC = 12 cm.



2. From B, draw a line which makes a right angle.

a. Now as point B is the initial point as the centre, draw an arc of any radius such that the arc meets BC at point D.



b. With D as centre and with the same radius as before, draw another arc cutting the previous one at point E.



c. Now with E as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point F.



d. With E and F as centres and with a radius more than half the length of FE, draw two arcs intersecting at point G.



e. Join points B and G. The angle formed is a right angle i.e., ∠GBC = 90°.



3. From B as centre, draw an arc of 5 cm which intersects GB at A.



4. Join AC to obtain the right angled triangle ABC.


5. From B, draw an acute ∠CBH downwards.



6. On ray BH, mark three points B₁, B₂ and B₃ such that BB₁ = B₁B₂ = B₂B₃.



7. Join B₃C.



8. From point B₂, draw B₂N âˆ¥ B₃C intersecting BC at N.



9. From point N, draw NM ∥ CA intersect BA at M. $△$MBN is the required triangle, right angled at B.



As per the construction, $△$MNB is the required triangle
Justification:
Let BB₁ = B₁B₂ = B₂B₃ = x.

Considering $△$BNB₂ and $△$BCB₃, 
∠B = ∠B                                      (common)
∠BNB₂ =∠BCB₃                         (corresponding angles of the same transverse as B₃C âˆ¥ B₂N)
∴ $△$BNB₂ $~$$△$BCB₃                 (by AA criteria)

 $\Rightarrow \frac{\mathrm{BN}}{\mathrm{BC}}=\frac{{\mathrm{BB}}_2}{{\mathrm{BB}}_3}=\frac{2x}{3x}=\frac{2}{3}$          .....(1)

Similarly, in $△$MBN and $△$ABC,
∠ A  = ∠ A                                    (common)
∠BAC = ∠BMN                           (corresponding angles of the same transverse as AC âˆ¥ MN)
∴ $△$MBN $~$ $△$ABC                    (by AA criteria)

$\Rightarrow \frac{\mathrm{MB}}{\mathrm{AB}}=\frac{\mathrm{MN}}{\mathrm{AC}}=\frac{\mathrm{BN}}{\mathrm{BC}}=\frac{2}{3}$          .....(2)

∴ the constructed triangle $△$MNB is of scale $\frac{2}{3}$ times of the $△$ABC.

Also, as we can clearly see, ∠B = 90° is common in both the triangles $△$ABC and $△$MNB. Hence, the constructed triangle $△$MNB is also a right angle triangle.


 

Answer:

Steps of construction:

1. Draw the line segment BC = 6 cm.




2. Taking B and C as centres, draw two arcs of radii 4 cm and 5 cm respectively intersecting each other at A.



3. Join BA and CA. $△$ABC is the required triangle.



4. From B, draw any ray BD downwards making at acute angle.



5. Mark five points B₁, B₂, B₃, B₄ and B₅ on BD, such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.


6. Join B₃C and from B₅, draw B₅M âˆ¥ B₃C intersecting the extended line segment BC at M.


7. From point M, draw MN âˆ¥ CA intersecting the extended line segment BA at N.



8. Then, $△$NBM is the required triangle whose sides are equal to $\frac{5}{3}$ of the corresponding sides of the $△$ABC.

Justification:
As per the construction, ∆MNB is the required triangle
Let BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = x.

In triangles $△$BCB₃ and $△$BMB₅, 
∠B = ∠B                                   (common)
∠BCB₃ = ∠BMB₅                    (corresponding angles of the same transverse as B₅D âˆ¥ B₃C)
∴ $△$BCB₃ $~$ $△$BMB₅             (by AA congruency criteria)

$\Rightarrow \frac{\mathrm{BM}}{\mathrm{BC}}=\frac{{\mathrm{BB}}_5}{{\mathrm{BB}}_3}=\frac{5x}{3x}=\frac{5}{3}$                           .....(1)

Similarly, in $△$MBN and $△$CBA,
∠B = ∠B                                         (common)
∠BAC = ∠BNM                             (corresponding angles of the same transverse as AC âˆ¥ MN)
∴ $△$MBN $~$ $△$CBA                      (by AA congruency criteria)

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{NB}}=\frac{\mathrm{AC}}{\mathrm{NM}}=\frac{\mathrm{BC}}{\mathrm{BM}}=\frac{5}{3}$                           .....(2)

From (1) and (2), the constructed triangle $△$NBM is of scale $\frac{5}{3}$ times of the $△$ABC.

 

Answer:

Steps of construction:

1. Draw a circle of radius 4 cm. Let centre of this circle be O.



2. Take a point M at 6 cm away from the radius.



3. Join OM and bisect it. Now, with M and O as centres and with radius more than half of  OM, draw two arcs on the either sides of the line. Let the arcs meet at A and B and M₁ be mid-point of OM.



4. Taking M₁ as centre and M₁O as radius, draw a circle to intersect the previous circle with radius 4 and centre O at two points P and Q.


5. Join PM and MQ. PM and MQ are the required tangents from the point M to circle with centre O and radius 4 cm.



Hence, obtain the tangents PM and QM to the circle with centre O and radius 4 cm.
 

Answer:

Thinking process:
I. Firstly, we find the ratio of AB in which P divides it with the help of the relation AP = $\frac{3}{4}$AB.
II. Secondly, we find the ratio of AC in which Q divides it with the help of the relation AQ = $\frac{1}{4}$AC.
III. Now, construct the line segment AB and AC in which P and Q respectively divides it in the ratio from step I and II respectively.
IV. Finally, get the points P and Q. After that, join PQ and get the required measurement of PQ.

Given that, AB = 5 cm and AC = 7 cm.
Also, $\mathrm{AP}=\frac{3}{4}\mathrm{AB} \mathrm{and} \mathrm{AQ}=\frac{1}{4}\mathrm{AC}$                        .....(1)
$\Rightarrow \mathrm{AP}=\frac{3}{4}\times 5=\frac{15}{4} \mathrm{cm}$
Then, PB = AB − AP
$\Rightarrow \mathrm{PB}=5-\frac{15}{4}=\frac{5}{4} \mathrm{cm}$                   
$\Rightarrow \mathrm{AP}:\mathrm{PB}=\frac{15}{4}:\frac{5}{4}$
Hence, AP : PB = 3 : 1 i.e., scale factor of line segment AB is $\frac{3}{1}$.

Again from (1),
$\mathrm{AQ}=\frac{1}{4}\mathrm{AC}=\frac{1}{4}\times 7=\frac{7}{4} \mathrm{cm}$
Then,
$\mathrm{QC}=\mathrm{AC}-\mathrm{AQ}=7-\frac{7}{4}=\frac{21}{4} \mathrm{cm}$
$\Rightarrow \mathrm{AQ} : \mathrm{QC}=\frac{7}{4}:\frac{21}{4}$
Hence, AQ : QC = 1 : 3 i.e., scale factor of line segment AQ is $\frac{1}{3}$.

Steps of construction:
1. Draw a line segment AB = 5 cm.



2. Now, draw a ray AZ making an acute ∠BAZ = 60°.



3. With A as centre and radius 7 cm, draw an arc intersecting AZ at C.


4. Draw a ray AX making an acute ∠BAX.



5. Along AX, mark 1 + 3 = 4 points A₁, A₂, A₃ and A₄ such that A₁A₂ = A₂A₃ = A₃A₄.


6. Join A₄B.


7. From A₃, draw A₃P âˆ¥ A₄B meeting AB at P. [by making an angle equal to ∠AA₄B]
Then, P is the point on AB which divides it in the ratio 3 : 1.
So, AP : PB = 3 : 1

8. Draw a ray AY making an acute ∠CAY.



9. Along AY, mark 3 + 1 = 4 points B_1, B₂, B₃ and B₄ such that AB₁ = B₁B₂ = B₂B₃ = B₃B₄.


10. Join B₄C.


11. From B₁, draw B₁Q âˆ¥ B₄C meeting AC at Q.   [by making an angle equal to ∠AB₄C]
Then, Q is the point on AC which divides it in the ratio 1 : 3.
So, AQ : QC = 1 : 3.


12. Finally, join PQ and its measurement is 3.25 cm.

 

Answer:

Thinking process:
I. Firstly we draw a line segment, then either of one end of the line segment with length 5 cm and making an angle 60° with this end. We know that is parallelogram both opposite sides are equal and parallel, then again draw a line with 5 cm making an angle with 60° from other end of line segment. Now, join both parallel line by a line segment whose measurement is 3 cm, we get a parallelogram. After that we draw a diagonal and get a triangle $△$BOC.
II. Now, we construct $△$BD'C' similar to $△$BDC with scale factor $\frac{4}{3}$.
III. Now, draw the line segment D'A' parallel to DA.
IV. Finally, we get the required parallelogram A'BC'D'.

Steps of construction:
1. Draw a line segment AB = 3 cm.



2. Now, draw a ray BY making an acute ∠ABY = 60°.



3. With B as centre and radius equal to 5 cm, draw an arc cutting BY at point C.



4. Again draw a ray AZ making an acute ∠ZAX₁ = 60°.


5. With A as centre and radius equal to 5 cm, draw an arc on AZ cutting it at point D.



6. Now, join CD and finally make a parallelogram ABCD.



7. Join BD, which is a diagonal of parallelogram ABCD and from B draw any ray BX making an acute angle ∠CBX.


8. Locate 4 points B₁, B₂, B₃, B₄ on BX, such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.


9. Join B₃C and from B₄, draw a line B₄C' âˆ¥ B₃C intersecting the extended line segment BC at C'.


11. From point C', draw C'D' ∥ CD intersecting the extended line segment BD at D'. Then, $△$D'BC' is the required triangle whose sides are $\frac{4}{3}$ of the corresponding sides of $△$DBC.

12. Now draw a line segment D'A' parallel to DA, where A' lies on the extended side BA i.e., a ray BX₁.


13. Finally, we observe that A'BC'D' is a parallelogram in which A'D' = 6.5 cm, A'B = 4 cm and ∠A'BD' = 60° divide it into triangles BC'D' and A'BD' by the diagonal BD'.



Justification that A'BC'D' is a parallelogram:
As per the construction,
CD âˆ¥ C'D' and BC' âˆ¥ A'D'                                                                           .....(1)

Also,
As per the given data, ∠ ABC = 60°.

Consider the parallelogram ABCD, the sum of complementary angles is 180°.
So, ∠ABC + ∠BCD = 180°
⇒ ∠BCD = 180° − ∠ ABC
                = 180° − 60°
                = 120°
Therefore, ∠BCD = 120°.

Now, as CD âˆ¥ C'D',
⇒ ∠BCD = ∠BC'D'                                          (as corresponding angles)
Hence, ∠BC'D' = 120°                                                                               .....(2)

As BC âˆ¥ AD and AD âˆ¥ A'D', hence BC âˆ¥ A'D',
∠ABC = ∠D'A'X₁                                          (corresponding angles)
So, ∠D'A'X₁ = ∠ABC = 60°
Now, consider ∠D'A'X₁ + ∠D'A'B = 180°           (linear angle)
60° + ∠D’A’B = 180°
⇒ ∠D’A’B = 180° − 60° = 120°                                                               .....(3)

From (1), (2) and (3), we can say that A’BC'D' is a parallelogram as opposite sides and opposite angles are equal.
 

Answer:

Given, two concentric circles of radii 3 cm and 5 cm with centre O. We have to draw pair of tangents from point P on outer circle to the other.
Steps of construction:
1. Draw two concentric circles with centre O and radii 3cm and 5cm.



2. Taking any point P on outer circle. Join OP.



3. Bisect OP, let M’ be the mid-point of OP.



4. Taking M’ as centre and OM’ as radius, draw a dotted circle cutting the inner circle at points A and B.


5. Join PA and PB. Thus, PA and PB are required tangents.



6. On measuring PA and PB, we find that PA = PB = 4 cm.



Actual calculation:
In the right angle $△$OAP,
∠PAO = 90°

According to Pythagoras theorem,
(hypotenuse)² = (base)² + (perpendicular)²
⇒ PA² = (5)²$-$ (3)²
            = 25 $-$ 9
            = 16 cm
⇒ PA = 4 cm         (∵ side cannot be negative)
Hence, the length of both tangents is 4 cm.
Therefore, PA = PB = 4 cm.

Answer:

Steps of construction:
1. Draw a line segment BC = 5 cm.



2. Construct MN, the perpendicular bisector of line segment BC meeting BC at P'.



3. Taking B and C as centres, draw two arcs of equal radius 6 cm intersecting each other at A.

4. Join BA and CA. So, $△$ABC is the required isosceles triangle and from B, draw any ray BX making an acute angle, ∠CBX.


5. Locate four points B₁, B₂, B₃ and B₄ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.

6. Join B₃C and from B₄, draw a line B₄R ∥ B₃C intersecting the extended line segment BC at R.

7. From point R, draw RP âˆ¥ CA meeting the extended line BA at P.


Then, $△$PBR is the required triangle.

Justification:
Let BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = x.
As per the construction, B₄R âˆ¥ B₃C.
Now,
$\frac{\mathrm{BC}}{\mathrm{CR}}=\frac{{\mathrm{BB}}_3}{{\mathrm{B}}_3{\mathrm{B}}_4}=\frac{3x}{x}=\frac{3}{1}$
$\therefore \frac{\mathrm{BC}}{\mathrm{CR}}=\frac{3}{1}$

$\Rightarrow \frac{\mathrm{BR}}{\mathrm{BC}}=\frac{\mathrm{BC}+\mathrm{CR}}{\mathrm{BC}}=\frac{\mathrm{BC}}{\mathrm{BC}}+\frac{\mathrm{CR}}{\mathrm{BC}}=1+\frac{1}{3}=\frac{4}{3}$

Also, from the construction, RP âˆ¥ CA.

In $△\mathrm{PBR} \mathrm{and} △\mathrm{ABC}$,
$\angle \mathrm{PBR} = \angle \mathrm{ABC}$
$\angle \mathrm{PRB}=\angle \mathrm{ACB}$                                            [Corresponding Angles]
$\Rightarrow △\mathrm{PBR}~△\mathrm{ABC}$                                        [By AA criteria]
and $\frac{\mathrm{PB}}{\mathrm{AB}}=\frac{\mathrm{RP}}{\mathrm{CA}}=\frac{\mathrm{BR}}{\mathrm{BC}}=\frac{4}{3}$

Hence, the new triangle is similar to the given triangle whose sides are $\frac{4}{3}$ times of the corresponding sides of the isosceles $△$ABC.

Answer:

Steps of construction:
1. Draw a line segment AB = 5 cm.
2. From point B, draw ∠ABY = 60° and mark an arc on it such that BC = 6 cm.
3. Join AC, $△$ABC is the required triangle.
4. From A, draw any ray AX downwards making an acute angle ∠BAX.
5. Mark 7 points B₁, B₂, B₃, B₄, B₅, B₆ and B₇ on AX, such that AB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.
6. Join B₇B and from B₅, draw B₅M âˆ¥ B₇B intersecting AB at point M.
7. From point M, draw MN âˆ¥ BC intersecting AC at N. Then, $△$AMN is the required triangle whose sides are equal to $\frac{5}{7}$ of the corresponding sides of the $△$ABC.

Justification:
Here, B₅M âˆ¥ B₇B (by construction).
$$\begin{gathered} \therefore \frac{\mathrm{AM}}{\mathrm{MB}}=\frac{5}{2} \\ \Rightarrow \frac{\mathrm{MB}}{\mathrm{AM}}=\frac{2}{5} \end{gathered}$$

Now,
$\begin{array}{rcl}\frac{\mathrm{AB}}{\mathrm{AM}}& =& \frac{\mathrm{AM}+\mathrm{MB}}{\mathrm{AM}}\\ & =& 1+\frac{\mathrm{MB}}{\mathrm{AM}}\\ & =& 1+\frac{2}{5}\\ & =& \frac{7}{5}\end{array}$

Also, MN âˆ¥ BC.
$$\begin{gathered} \therefore △\mathrm{AMN}~△\mathrm{ABC} \\ \Rightarrow \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AC}}=\frac{\mathrm{NM}}{\mathrm{BC}}=\frac{5}{7} \end{gathered}$$
 

Answer:

Steps of Construction:
Step 1:  A circle with center O and radius 4 cm is drawn.
Step 2:  OQ is joined when Q is any point on the circle.
Step 3: ∠OQR = 30$°$ is drawn and QR intersects the circle at R.
Step 4: Centering Q and R and with radius QR two arcs are drawn, they intersect at P.
Step 5: PQ and PR are joined.
PQ and PR are the required tangents.

Answer:

Steps of construction:
Step 1: Construct a triangle ABC with AB = 4 cm, BC = 6 cm and AC = 9 cm.
Step 2: Draw a ray AX making an acute angle with the base AC and mark 3 points A₁​, A₂​, A₃​ such that AA₁​ = A₁​A₂ â€‹= A₂​A₃.
Step 3: Join A₂C and draw a line A₃​C′ such that A₂C is parallel to A_3​C′ where C′ lies on the produced AC.
Step 4: Now draw another line parallel to BC at C′ such that it meets the produced AB at B′.

Here, $△$AB′C′ is the required triangle similar to $△$ABC with scale factor $\frac{3}{2}$.