Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions

Extra Questions for Class 10 Maths Constructions with Answers

Extra Questions for Class 10 Maths Chapter 11 Constructions. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

You can also download Maths Class 10 to help you to revise complete syllabus and score more marks in your examinations.

Constructions Class 10 Extra Questions Very Short Answer Type

Question 1.
In figure, P divides AB internally in what ratio?

Answer:
∵ APC ~ ∆ BPC’
∴ \(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AC}}{\mathrm{BC}^{\prime}}\)
or = 4:5

Question 2.
To find a point P on a line segment AB such that \(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{7}\), the segment AB is to be divided in what ratio?
Answer:

Question 3.
If a similar triangle whose sides are \(\frac{3}{5}\) times the corresponding sides of a given triangle is to be constructed, the point on the side of a given triangle divides the side in what ratio?

Answer:
∵ BC || B’C’ (By construction)
∴ ∆AB’C’ ~ ∆ABC
∴ \(\frac{C^{\prime} \mathrm{A}}{\mathrm{AC}}=\frac{\mathrm{C}^{\prime} \mathrm{B}^{\prime}}{\mathrm{BC}}\)
∴ \(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{3}{5}\)
∴ The point on the side of a given triangle divides the side in ratio 3 :5.

Question 4.
Write steps of construction of a tangent to a circle at a given point.
Answer:
Steps of construction:
1. With centre O and any radius draw a circle.
2. Take any point say P on its circumference.
3. Join that point with centre O.
4. At P construct a right angle i.e., of 90°. Thus, a required tangent is constructed.

Constructions Class 10 Extra Questions Short Answer Type-1

Question 1.
Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point? cm AB such that \(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{5}\).
Answer:

Steps of construction:
1. Draw AB = 7 cm.
2. Draw an acute angle ∠BAX.
3. Along AX locate points A₁, A₂, A₃ at equal distances such that AA₁ = A1A2 = A2A3.
4. Join A₃B.
5. Draw A₂P || A₃B which meets AB in P.
Hence, P is the required point.
∴ \(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{3}{2}\)

Question 2.
Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°.
Answer:
Steps of construction:
1. Draw a circle of radius 4.5 cm.

2. Join OA. (Take A any point on the circle)
3. Construct ∠AOB = 135°
4. Draw tangent at A and B.
5. Then ∠APB = 45°.

Justification:
∠AOB + ∠P = 180°
i.e., 135°+ 45° = 180°
180° = 180°

Question 3.
Draw a ∆ABC with sides AB = 6 cm, BC = 7.5 cm and AC = 6.6 cm. Construct another triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of ∆ABC. [CBSE 2017]
Answer:
Steps of construction:
1. Draw AB = 6 cm.
2. With A and B as centres taking 6.6 cm and 7.5 cm as radii, draw two arcs intersecting each other at C.

3. Join ∆ABC as the given triangle.
4. Draw ∠BAX an acute angle.
5. Along AX draw points A₁, A₂, A₃, A₄ at equal distance such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄.
6. Join BA₄.
7. Draw A₃B’ || A₄B which intersects AB at B’.
8. Draw B’C’ 11 BC which intersects AC at C’.
Hence, ∆AB’C’ is the required triangle.

Constructions Class 10 Extra Questions Short Answer Type-2

Question 1.
Draw a circle of radii 4 cm. From the point 7 cm away from its centre, construct the pair of tangents to the circle.
Or
Draw a line segment of length 8 cm and divide it in the ratio 2:3. [CBSE SQP 2019 (Basic)]
Answer:
Steps of construction:
(i) Draw a circle of radius 4 cm and mark its centre as O.

(ii) Take a point P at a distance OP = 7 cm from centre O.
(iii) Draw a perpendicular bisector of OP intersecting OP at C.
(iv) With C as centre and OC = CP as radius draw a dotted circle intersecting the given circle at S and T.
(v) Join PT and PS.
Thus PT and PS are required tangents.

Or

Steps of construction:
(i) Draw the line segment AB = 8 cm.
(ii) Draw any ray AX, making an acute angle ∠BAX.
(iii) Locate 2 + 3 = 5 (given ratio is 2:3) points A₁, A₂, A₃, A₄, A₅, on ray AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.

(iv) Join BA₅.
(v) Through point A₂ draw a line A₂C || A₅B at A₂ intersect in AB at C.
Then, AC: CB = 2:3.

Question 2.
Construct a ∆ABC in which CA = 6 cm, AB = 5 cm and ∠ABC = 45°. Then construct a triangle whose sides are \(\frac{3}{5}\) of the corresponding sides of ∆ABC. [CBSE 2019]
Answer:
Steps of construction:
(i) Draw AC = 6 cm.
(ii) At A draw an angle ∠CAX = 45°.
(iii) With A as centre draw an arc of radius 5 cm to cut AX at B. So that AB = 5 cm.
(iv) Join CB to get ∆ABC.
(v) On the other side of AC draw an acute angle ∠CAY.
(vi) Mark 5 points namely B₁, B₂, B₃, B₄, B₅ at equal distances such that AB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
(vii) Join B₅C.
(viii) Take B₃ and draw B₃C’ parallel to B₅C to cut AC at C’.
(ix) Draw C’B’ || CB to get required ∆AC’B’.
∴ AC’B’ is required triangle.

Question 3.
Draw two concentric circle of radii 3 cm and 5 cm. Construct a tangent to similar circle from a point on the larger circle. Also measure its length. [CBSE Delhi 2016]
Answer:
Following are the steps of constructions:
Step 1: Draw a circle of 3 cm radius with centre O on the given plane.
Step 2: Draw a circle of 5 cm radius, taking O as its centre. Locate a point P on this circle and join OP.
Step 3: Bisect OP. Let M be the midpoint of PO.
Step 4: Taking M as its centre and MO as its radius, draw a circle. Let it intersect the given circle at points Q and R.
Step 5: Join PQ and PR, PQ and PR are the required tangents.
It can be observed that PQ and PR are of length 4 cm each.

In ∆PQO
Since, PQ is a tangent
∠PQO = 90°
PO = 5 cm
QO = 3 cm
Applying Pythagoras theorem in ∆PQO,
we obtain
PQ² + QO² = PQ²
PQ² + (3)² = (5)²
⇒ PQ = \(\sqrt{25-9}\) = 4 cm

Question 4.
Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other. [CBSE Outside Delhi 2016]
Answer:
Steps of constructions:

(i) Draw a circle of radius 4 cm and mark its centre as O.
(ii) Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.
(iii) Draw a radius OB making an angle of 120° (180° – 60°) with OA.
(iv) Draw a perpendicular to OB at B. Let both the perpendicular meet at point T.
Then, TA and TB are the required tangents at an angle of 60°.

Question 5.
Construct a circle of radius 3.5 cm from a point on the concentric circle of radius 6.5 cm. Draw a tangent to the first circle and measure the length of the tangent drawn. Also find the length of the tangent by actual calculation.
Answer:
Steps of construction:
1. Draw concentric circles with centre O and radii 3.5 cm and 6.5 cm.

2. Take a point P on the outer circle and join OP.
3. Locate the mid-point M at OP.
4. Taking M as centre and radius = MP draw an arc of circle which intersects the inner circle at Q (say). See figure.
5. Join PQ. Here PQ is the required tangent (One more tangent can be drawn on the other side of PO and of length = PQ)
6. We measure PQ = 5.5 cm (approx.).
7. By Pythagoras theorem,

Question 6.
Draw a pair of tangents to a circle of radius 4 cm, which are inclined to each other at an angle of 60°.
Answer:
Steps of construction:

1. Take a point O on a sheet of paper and draw a circle of radius OA = OP = 4 cm.
2. Produce OA to B such that OA = AB = 4 cm.
3. Taking A as centre draw a circle of radius OA = AB = 4 cm.
4. It intersects the circle at P and Q.
5. Join BP and BQ to get the required tangents.

Justification:
OA = OP = 4 cm (radii)
AP = OA = 4 cm (radii)
So, ∆OAP is an equilateral ∆.
In ∆OAP we have
i.e, ∠PAO = 60°
∠BAP = 180° – ∠AOP =180° – 60° = 120°
In ∆ BAP we have
AB = AP (= 4 cm)
∴ ∠ABP = ∠APB = 30°
(by angle sum property)

Constructions Class 10 Extra Questions Long Answer Type 1

Question 1.
Construct an isosceles triangle with base 8 cm and altitude 4 cm. Construct another triangle whose sides are \(\frac{2}{3}\) times the corresponding sides of the isosceles triangle. [CBSE Delhi 2017]
Answer:

∆ABC is triangle with base 8 cm & altitude 4 cm.
∆AB’C’ is triangle whose sides are \(\frac{2}{3}\) times the corresponding sides of ∆ABC.

Question 2.
Given a rhombus ABCD in which AB = 4 cm and ∠ABC = 60°, divide it into two triangles say, ABC and ADC. Construct the triangle AB’C’ similar to ∆ABC with scale factor \(\frac{2}{3}\). Draw a line segment C’D’ parallel to CD where D’ lies on AD. Is AB’C’D’ a rhombus? Give reasons.
Answer:
Given: For rhombus ABCD, AB = 4 cm and ∠ABC = 60°.

Required: To construct ∆AB’C’ similar to ∆ABC with scale factor \(\frac{2}{3}\).

Steps of construction:
1. Draw AB = 4 cm.
2. At B, draw an ∠ABM = 60° and cut off BC = 4 cm on BM.
3. At C, draw an ∠BCN = 180° – 60° = 120°.
4. Cut off CD = 4 cm on CN.
5. Join AD to complete the rhombus ABCD.
6. Draw a ray AX below the side AB, such that ∠BAX = 30°.
7. Mark three points P₁, P₂, P₃ on ray AX such that AP₁ = P₁P₂ = P₂P₃.
8. Join P₃B.
9. Draw a line P₂B’ || P₃B intersecting AB at B’.
10. Draw a line through B’C’ 11 BC intersecting AC at C’.
Thus, ∆AB’C’ is similar to ∆ABC with scale factor \(\frac{2}{3}\).
11. Finally draw segment C’D’ 11 CD.
Yes, AB’C’D’ is a rhombus, we show it as given below

Question 3.
Draw a circle of diameter 8 cm. From a point P, 7 cm away from its centre construct a pair of tangents to the circle.
Answer:
Steps of construction:
1. Draw a line segment AB = 8 cm
2. Draw a circle (C₁) taking C as centre which is the perpendicular bisector of AB i.e., radius AC = CB = 4 cm.

3. Produce AB and cut off CP = 7 cm.
4. Draw a perpendicular bisector of CP.
5. Taking Q as centre and radius (= PQ) draw a circle C₂ which intersects the circle C₁ at M and N.
6. Join PM and PN.
Now, PM and PN are the required two tangents.

Constructions Class 10 Extra Questions HOTS

Multiple Choice Questions

Choose the correct option out of four given in each of the following:

Question 1.
Which of the following is false?
(a) It is possible to divide a line segment in the ratio √3 : \(\frac{1}{\sqrt{3}}\) by geometrical construction.
(b) By geometrical construction, it is possible to divide a line segment in the ratio √2 : \(\frac{3}{\sqrt{2}}\)
(c) By geometrical construction, it is possible to divide a line segment in the ratio \(\frac{1}{3}: \frac{1}{2}\).
(d) By geometrical construction, it is possible to divide a line segment in the ratio (1 + √2):(1 – √2).
Answer:
(d) By geometrical construction, it is possible to divide a line segment in the ratio (1 + √2):(1 – √2).

Question 2.
To divide a line segment AB in the ratio m:n(m,n are positive integers), draw a ray AX so that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points are
(a) greater of m and n
(b) smaller of m and n
(c) m + n
(d) m + n – 1
Answer:
(c) m + n

Question 3.
To divide a line segment in the ratio 3:5, first a ray AX is drawn so that ∠BAX is 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:
(a) 8

Question 4.
To divide a line segment AB in the ratio 3:8, draw a ray AX such that ∠BAX is an acute angle and then points A₁, A₂, A₃, …… are located at equal distances on the ray AX and the point B is joined to
(a) A₅
(b) A₁₀
(c) A₁₁
(d) A₉
Answer:
(c) A₁₁

Question 5.
To divide a line segment AB in the ratio 5:6, draw a ray AX such that ∠BAX is an acute angle. Then draw a ray BY parallel to AX and the points A₁, A₂, A₃, …………. and B₁, B₂, B₃, …………….. are located at equal distances on AX and BY respectively then the points joined are
(a) A₅ and B₆
(b) A₆ and B₅
(c) A₄ and B₅
(d) A₄ and B₄
Answer:
(a) A₅ and B₆

Question 6.
To construct a triangle similar to given ∆ABC with its sides \(\frac{3}{7}\) of the corresponding sides of ∆ABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite sides of A w.r.t. BC. Then locate points B₁, B₂, B₃, …………….. on BX at equal distances and 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

Question 7.
To construct a triangle similar to a given ∆ABC 8 with its sides ~ of the corresponding sides of ∆ABC draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A w.r.t. 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

Question 8.
To draw a pair of tangents to a circle which are inclined to each other at an angle of 45°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be
(a) 135°
(b) 90°
(c) 60°
(d) 120°
Answer:
(a) 135°

Question 9.
Which of the following is not true ?
(a) A pair of tangents can be constructed to a circle inclined at an angle of 170°.
(b) A pair of tangents can be constructed from point P to a circle of radius 4 cm situated at a distance of 3.9 cm from the centre.^~
(c) A quadrilateral ABCD with its sides \(\left(\frac{5}{7}\right)\) the corresponding sides of quad. PQRS can be constructed.
(d) None of these.
Answer:
(b) A pair of tangents can be constructed from point P to a circle of radius 4 cm situated at a distance of 3.9 cm from the centre.^~

Question 10.
Which of the following is not a step in the construction of tangent from an external point P to a circle of centre ‘O’ ?
(a) Join OP.
(b) Bisect OP at M.
(c) With M as centre and radius equal to MP draw a circle intersecting given circle at T₁ and T₂.
(d) Join OT₁, OT₂ to get required tangents.
Answer:
(d) Join OT₁, OT₂ to get required tangents.

Fill in the Blanks

Question 1.
______________ tangent can be drawn from a point lying inside a circle.
Answer:
No

Question 2.
Number of tangents that can be drawn from a point lying in exterior of circle is ______________ .
Answer:
two

Question 3.
To construct a tangent at a point P on the circle with centre O, first join ______________ and then construct a ______________ angle at P.
Answer:
OP, right

Question 4.
______________ theorem forms the basis for dividing a line segment in the given ratio.
Answer:
Basic Proportionality

Question 5.
To locate centre of a circle we draw any two ______________ and draw their ______________ .
Answer:
non-parallel chords, perpendicular bisectors

Question 6.
To draw a pair of tangents to a circle which are inclined to each other at an angle of 30°, it is required to draw tangents at the end point of two radii of the circle, the angle between which is ______________ .
Answer:
150°

Question 7.
To divide a given line segment PQ in the ratio 6 : 5, draw any ray PX making any ______________ angle with PQ and then mark a minimum of ______________ points on ray PX.
Answer:
acute, elven

Question 8.
To divide a line segment AB in the ratio 3:5, draw a ray AX such that ∠BAX is an acute angle, then draw BY parallel to AX and and mark the points A₁, A₂, …….. and B₁, B₂, ………….. at equal distances on ray AX and BY respectively, then the points _____________ and _____________ are joined.
Answer:
A₃, A₅

Question 9.
Length of tangents drawn from external point to a circle are _____________ .
Answer:
equal

Question 10.
To construct a triangle similar to ∆ABC with scale factor \(\frac{m}{n}\) > 1. We construct ∠CBX < 90° and then n mark _____________ points at equal distances on ray BX.
Answer:
m.

Extra Questions for Class 10 Maths

NCERT Solutions for Class 10 Maths

The post Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions appeared first on Learn CBSE.

source https://www.learncbse.in/constructions-cbse-class-10-maths-chapter-11-extra-questions/