NCERT Exemplar Class 10 Maths Solutions Chapter 10 Constructions

Question:1

To divide a line segment AB in the ratio $5:7$, first, a ray AX is drawn so that$\mathrm{\angle }BAX$ is an acute angle, and then at equal distances, points are marked on the ray AX such that the minimum number of these points is
(A) 8 (B) 10 (C) 11 (D) 12

Answer(D) 12
Solution
Given: $\mathrm{\angle }BAX$ is an acute angle.
The required ratio is $5:7$
Let m = 5, n = 7

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 12 points on AX at equal distances. (Because here $m+n=12$)
3. Join $A_{12}B$
4. Through the point $A_5$ draw a line parallel to $A_{12}B$ intersecting AB at the point P.
Then $AP:PB=5:7$
$\because A_{5}P\parallel A_{12}B$
$\therefore \frac{A{A}_5}{A_5{A}_{12}}=\frac{AP}{PB}$ (By Basic Proportionality theorem)
By construction $\frac{A{A}_5}{A_5{A}_{12}}=\frac{5}{7}$
$\therefore \frac{AP}{PB}=\frac{5}{7}$
Hence, the number of points is 12.

Question:2

To divide a line segment AB in the ratio $4:7$, a ray AX is drawn first such that $\mathrm{\angle }BAX$ is an acute angle and then points $A_1,A_2,A_3\cdots$ are located at equal distances on the ray AX and the point B is joined to:
(A) A₁₂ (B) A₁₁ (C) A₁₀ (D) A₉

Answer(B) A₁₁
Solution
Given: $\mathrm{\angle }BAX$ is an acute angle
The required ratio is $4:7$
Let m = 4, n = 7
$m+n=4+7=11$

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 11 points on AX at equal distances (because m + n = 11)
3. Join $A_{11}B$
4. Through the point A4 draw a line parallel to $A_{11}B$ intersecting AB at the point P.
Then $AP:PB=4:7$
Hence, point B is joined to A₁₁.

Question:3

To divide a line segment AB in the ratio $5:6$, draw a ray AX such that $\mathrm{\angle }BAX$ is an acute angle, then draw a ray BY parallel to AX and the points $A_1,A_2,A_3\cdots$and $B_1,B_2,B_3\cdots$are located at equal distances on ray AX and BY, respectively. Then the points joined are
(A) A₅ and B₆ (B) A­ and B₅ (C) A₄ and B₅ (D) A₅ and B₄

Answer(A) A₅ and B₆

Solution
Given: $\mathrm{\angle }BAX$ and $\mathrm{\angle }ABY$ both are acute angles and AX parallel to BY
The required ratio is $5:6$
Let m = 5 , n = 6

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray $BY\parallel AX$.
3. Locate the points $A_1,A_2,A_3,A_4,A_5$ on AX at equal distances
4. Locate the points $B_1,B_2,B_3,B_4,B_5$ on BY at a distance equal to the distance between points on the AX line.
5. Join $A_5{B}_6$.
Let it intersect AB at a point C in the figure.
Then $AC:CB=5:6$
Here $△A{A}_{5}C$ is similar to $△B{B}_{6}C$
Then $\frac{A{A}_5}{B{B}_6}=\frac{5}{6}=\frac{AC}{BC}$
$\therefore$by construction $\frac{A{A}_5}{B{B}_6}=\frac{5}{6}$
$\therefore \frac{AC}{BC}=\frac{5}{6}$
$\therefore$ Points joined one A₅ and B₆.

Question:4

To construct a triangle similar to a given $△ABC$ with its sides of $\frac{3}{7}$ the corresponding sides of $△ABC$, first, draw a ray BX such that CBX is an acute angle and X lies on the opposite side of A Aconcerningo BC. Then locate points B₁, B₂, B₃, ... on BX at equal distances, and the next step is to join
(A) B₁₀ to C (B) B₃ to C (C) B₇ to C (D) B₄ to C

Answer(C) B₇ to C
Solution Given: $\mathrm{\angle }CBX$ is an acute angle.

Steps of construction
1. Draw any ray BX making an angle with BC on the side opposite vertex A.
2. Locate 7 points on BX in equidistant
3. Now join B₇ to C
4. Draw a line through B₃$C^{\prime }$ parallel to B₇C .

Question:5

To construct a triangle similar to a given $△ABC$ with its sides $\frac{8}{5}$ of the corresponding sides of $△ABC$ draw a ray BX such that $\mathrm{\angle }CBX$ is an acute angle and X is on the opposite side of A concerning BC. The minimum number of points to be located at equal distances on ray BX is
(A) 5 (B) 8 (C) 13 (D) 3

Answer (B) 8
Solution
To construct a triangle similar to a triangle, with its sides $\frac{8}{5}$ of the corresponding sides of a given triangle, the minimum number of points to be located at an equal distance is equal to the greater of 8 and 5 in .$\frac{8}{5}$ Here $8>5$
So, the minimum number of points to be located at equal distances on ray BX is 8.

Question:6

To draw a pair of tangents to a circle which are inclined to each other at an Angle of ${60}^{\circ }$, it is required to draw tangents at the endpoints of those two radii of the circle, and the angle between them should be
(A) ${135}^{\circ }$ (B) ${90}^{\circ }$ (C) ${60}^{\circ }$ (D) ${120}^{\circ }$

Answer(D) 120°
Solution
According to question:-

Given :$\mathrm{\angle }QPR={60}^{\circ }$
Let $\mathrm{\angle }QOR=x$
As we know, the angle between the agent and the radius of a circle is 90 degrees
$\mathrm{\angle }PQO=\mathrm{\angle }PRO={90}^{\circ }$
We know that $\mathrm{\angle }PQO+\mathrm{\angle }PRO+\mathrm{\angle }QPR+\mathrm{\angle }QOR={360}^{\circ }$
[$\because$ sum of interior angles of quadrilateral is ${360}^{\circ }$ ]
${90}^{\circ }+{90}^{\circ }+x+{60}^{\circ }={360}^{\circ }$|
$240+x={360}^{\circ }$
$x={120}^{\circ }$

Question:1

Write True or False and give reasons for your answer in each of the following:
By geometrical construction, it is possible to divide a line segment in the ratio $\sqrt{3}:\frac{1}{\sqrt{3}}$

Answer [True]
Solution
We need both positive integers to divide a line segment in the ratio.
So, we can simplify it by multiplying both the terms by .$\sqrt{3}$
We obtain $\sqrt{3}\times \sqrt{3}:\sqrt{3}\times \frac{1}{\sqrt{3}}$
$\Rightarrow 3:1$
So, the required ratio is $3:1$
In geometrical construction is possible to divide a line segment in the ratio $3:1$

Question:2

Write True or False and give reasons for your answer in each of the following:
To construct a triangle similar to a given $△ABC$ with its sides $\frac{7}{3}$ of the corresponding sides of $△ABC$, draw a ray BX making an acute angle with BCand X lies on the opposite side of A concerning BC. The points B₁, B₂,..., B₇ are located at equal distances on BX, B3 is joined to C and then a line segment $B_6{C}^{\prime }$ is drawn parallel to B₃C where $C^{\prime }$ lies on BC produced. Finally, line segment $A^{\prime }{C}^{\prime }$ is drawn parallel to AC.

Answer: False
Solution
According to the question:-
To construct a triangle similar to a given $△ABC$ with its sides $\frac{7}{3}$ of the corresponding sides of $△ABC$
1. Draw a line segment BC
2. Taking B and C as centres, draw two arcs of suitable radii intersecting each other at A.
3. Join BA and CA?ABC is the required triangle.
4. From B draw any ray BX downwards making an acute angle CBX.
5. Locate seven points B₁, B₂, b₃, …. B₇ on BX such that BB₁ = B₁B₂ = B₁B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.
6. Join B₃C and from B₇ draw a line B₇C’ ? B₃C intersecting the extended line segment BC at C’.
7. From point C’ draw C’A’? CAintersectsg the extended line segment BA at A

But as given, if we join B₃C and from B₆ draw a line B₆C’? B₃C intersecting the extended line segment BC at C’.
$B{B}_3/B{B}_6=BC/{BC}^{\prime }=3/6$
$BC/{BC}^{\prime }=1/2$
$BC:{BC}^{\prime }=1:2$
Hence, the sides are not in the ratio of 7:3
So, the required triangle can not be constructed in this way.
Hence, the given statement is false.

Question:3

Write True or False and give reasons for your answer in each of the following:
A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of 3 cm from the centre

Answer [False]
Solution
Radius, r = 3.5 cm
Point distance from centre = 3 cm
But, $r>d\Rightarrow 3.5cm>3cm$
So, the point P lies inside the circle.
So, a pair of tangents cannot be drawn to point P to a circle.

Question:4

Write True or False and give reasons for your answer in each of the following:
A pair of tangents can be constructed to a circle inclined at an angle of ${170}^7{0}^{\circ }$.

Answer [True]
Solution
Tangent: - It is a straight line that touches the curve but does not cross it. A pair of tangents can be constructed to a circle inclined at an angle greater than $0^{\circ }$ but less than ${180}^{\circ }$. Here, the inclination angle is ${170}^{\circ }$. Hence, it is possible.

Question:1

Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio $3:5$.

Solution
Given: AB = 7cm
The required ratio is $3:5$
Let m = 3, n = 5
$m+n=3+5=8$

Steps of construction
1. Draw a line segment AB = 7cm.
2. Draw a ray AX making an acute angle with AB
3. Locate 8 points $A_1,A_2,A_3\cdots A_8$ on AX on equidistant. (because $m+n=8$ )
4. Join $B{A}_8$
5. Through the point $A_3$ draw a line parallel to $B{A}_8$ which intersect line AB at P..
Here triangle AA₃P is similar to triangle $A{A}_{8}B$
$A{A}_3/A{A}_8=AP/AB=3/5$ (by construction)
Therefore $AP:BP=3:5$

Question:2

Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and $\mathrm{\angle }B={90}^{\circ }$.Construct a triangle similar to it and of scale factor $\frac{2}{3}$. Is the new triangle also a right triangle

Solution
Given BC = 12cm, AB = 5 cm, $\mathrm{\angle }B={90}^{\circ }$

Steps of construction
1. Draw a line $BC=12\mathrm{cm}$
2. From $B$ draw $AB=5\mathrm{cm}$ which makes an angle of ${90}^{\circ }$ at $B$.
3. Join AC
4. Make an acute angle at B as $<CBX$
5. On BX mark 3 points at equal distance $X_1,X_2,X_3$.
6. Join $X_{3}C$
7. From $X_2$ draw $X_2{C}^{\prime }\Vert X_{3}C$ intersect AB at $C^{\prime }$
8. From point $C^{\prime }$ draw $C^{\prime }{A}^{\prime }\Vert CA$ intersect $\mathrm{AB}{A}^{\prime }$

Now $\mathrm{△}{A}^{\prime }B{C}^{\prime }$ is the required triangle and $\mathrm{△}{A}^{\prime }B{C}^{\prime }$ is also a right triangle.

Question:3

Draw a triangle ABC in which BC = 6 cm, CA = 5 cm, and AB = 4cm. Construct a triangle similar to it and of scale factor$\frac{5}{3}$

Solution
Given : BC = 6cm, CA = 5cm, AB = 4cm
Scale factor = 5/3
Let m = 5 and n = 3

Steps of construction

1. Draw a line BC = 6c cm

2. Taking B and C as centre mark arcs of length 4cm and 5cm respectively,y intersects each other at point A.

3. Join BA and CA

4 . Draw a ray BX making a g acute angle with BC.

5.Locate 5 Points on BX in equidistant as B₁, B₂, B₃, B₄, B₅.

6.Join B₃ to C and draw a line through B5 parallel to B₃C to intersect at BC extended at $C^{\prime }$.

7. Draw a line through $C^{\prime }$ parallel to AC intersect AB extended at $A^{\prime }$ .
Now $A^{\prime }B{C}^{\prime }$ is required triangle.

Question:4

Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre.

Solution
Given : Radius = 4cm

Steps of construction
1. Draw a circle at radius r = 4 cm at point O.
2. Take a point P at a distance of 6cm from point O and join PO
3. Draw a perpendicular bisector of line PO. M is the midpoint of PO.
4. Taking M as the centre, we draw another circle of radius equal to MO, and it intersects the given circle at points Q and R
5. Now, join PQ and PR.
Then PQ and PR one the required two tangents.

Question:1

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,y such that $AP=\frac{3}{4}AB$ and $AQ=\frac{1}{4}AB$. Join P and Q and measure the length of 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
$AP=\frac{3}{4}AB$
$AP=\frac{3}{4}\times 5=\frac{15}{4}[\because AB=5cm]$
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 a line segment AB = 5 cm
2. Now draw ray AO which makes an angle,
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, you join PQ, and its measurement is 3.25 cm.

Question:2

Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm, and ABC=60^ 60^{\circ}$,anddivideitintotrianglesBCDandABDbythediagonalBD.ConstructthetriangleB{D}^{\prime }{C}^{\prime }similartoDBDCwithscalefactor$\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 $\mathrm{\angle }ABP={60}^{\circ }$
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 $\mathrm{\angle }{Q}^{\prime }AX={60}^{\circ }$
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 po, int B, i.e., $\mathrm{\angle }CBQ$ is the 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 $B_4,C^{\prime }$ parallel to $B_{3}C$ which intersect at point $C^{\prime }$
11. From point $C^{\prime }$ draw line $C^{\prime }{D}^{\prime }$ which is parallel to CD
12. Now draw a line segment $D^{\prime }{A}^{\prime }$ parallel to DA
Note : Here $A^{\prime },C^{\prime }$ and $D^{\prime }$ are the extended sides.
13. $A^{\prime }B{C}^{\prime }{D}^{\prime }$ is a parallelogram in which $A^{\prime }{D}^{\prime }=6\cdot 5cm$ and $A^{\prime }B=4cm$ and $A^{\prime }B{D}^{\prime }={60}^{\circ }$ divide it into triangles $B{C}^{\prime }{D}^{\prime }$ and $A^{\prime }{BD}^{\prime }$ by the diagonal ${BD}^{\prime }$

Question:3

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 cijoin, 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. Thu, PQ and PR are required tangents
On measuring PQ and PR, we find that PQ = PR = 4 cm
Calculations
$△OQP,\mathrm{\angle }OQP={90}^{\circ }$
[usiPythagoras' theorem]
${\left(5\right)}^2={\left(3\right)}^2+{\left(PQ\right)}^2$
$25-9=P{Q}^2$
$16=P{Q}^2$
$\sqrt{16}=PQ$
$4cm=PQ$
H, the length of both tangents is 4 cm.

Question: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. a Draw a 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 $CB{B}^{\prime }$
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
$\because B_{4}D\parallel B_{3}C$
$\therefore \frac{BC}{CD}=\frac{3}{1}\Rightarrow \frac{CD}{BC}=\frac{1}{3}$
$Now\therefore \frac{BD}{BC}=\frac{BC+CD}{BC}=1+\frac{CD}{BC}=1+\frac{1}{3}=\frac{4}{3}$
$AlsoDE\parallel CA$
$\therefore △ABC\sim △DBE$
$\frac{EB}{AB}=\frac{DE}{CA}=\frac{BD}{BC}=\frac{4}{3}$

Question:5

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 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 $A{A}^{\prime }$ with acute angle $\mathrm{\angle }BA{A}^{\prime }$

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 $B_{5}X\parallel B_{7}BBY$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, $XY\parallel BC$
$\therefore △AXY\sim △ABC$
$\frac{AX}{AB}=\frac{AY}{AC}=\frac{YX}{BC}=\frac{5}{7}$

Question:6

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 = 8 =8cm
4. Now, taking A as the centre, we draw the circle of radius AP = OA = 4 cm
5Which intersectsst the circle at x and y
6. Join PX and PY
7. PX and PY are the tangents to the circle
Justification
In $△OAX$ we have
$OA=OP=4cm$ (Radius)
$AX=4cm$ (Radius of circle with centre A)
$\therefore OAX$ is equilateral triangle
$\mathrm{\angle }OAX={60}^{\circ }$
$\Rightarrow \mathrm{\angle }XAP={120}^{\circ }$
In $△PAX$ we have
$PA=AX=4cm$
$\mathrm{\angle }XAP={120}^{\circ }$
$\mathrm{\angle }APX={30}^{\circ }$
$\Rightarrow APY={30}^{\circ }$
Hence $\mathrm{\angle }XPY={60}^{\circ }$

Question:7

Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to $△ABC$ with scale factor $\frac{3}{2}$ . Justify the construction. Are the two triangles congruent? Note that all 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 centres, draw an arc of radius AB 4cm and AC = 9 cm

3. Join AB and AC

4 . Triangle ABC is a required triangle. From Bd raw ray BM with acute angle $\mathrm{\angle }XBM$

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 $△BXY$ is the required triangle whose sides are equal to$\frac{3}{2}$ of the $△ABC$
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 $XY\parallel CA$
$△ABC\sim △YBX$
$\therefore \frac{YB}{AB}=\frac{YX}{AC}=\frac{BX}{BC}=\frac{3}{2}$.

Here are three angles that are the same, but the three sides are not the same.
$\therefore$The two triangles are not congruent because, if two triangles are congruent, then they have the same shape and size.