Previous Year Questions: Constructions

# Class 10 Maths Previous Year Questions - Constructions (Old Syllabus)

 Table of contents Previous Year Questions 2019 Previous Year Questions 2018 Previous Year Questions 2017 Previous Year Questions 2016

## Previous Year Questions 2019

Q1: Draw two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and construct a pair of tangents PA and PB to the smaller circle. Measure PA.    [CBSE, Allahabad 2019]

Ans: Steps of Construction:

• Draw concentric circles of radius OQ = 2 cm and OP = 5 cm having same centre O.
• Mark these circles as C and C'.
• Points O, Q and P lie on the same line.
• Draw perpendicular bisector of OP, which intersects OP at O'.
• Take O' as centre, draw a circle of radius OO' which intersects circle C at points A and B.
• Join PA and PB, these are the required tangents.
• Length of these tangents are approx. 4.6 cm.

Justification:
Join OA and OB
OA ⊥ PA   [Radius ⊥ to tangent]
In right angled ΔOAP

A pair of equal tangents can be drawn to a circle from an external common point outside the circle.
∴ PA = PB

Q2: Construct a ΔABC in which CA = 6 cm, AB = 5 cm and ΔBAC = 45°. Then construct a triangle whose sides are 3/5 of the corresponding sides of ΔABC.   [Delhi 2019]

Ans: Steps of construction:

• A line segment AC = 6 cm is drawn.
• ∠CAB = 45° is constructed at A.
• An arc of 5 cm radius to be drawn with A as centre, cutting AY at B.
• B and C are joined. Then, ΔABC is constructed.
• An acute angle CAX is drawn below AC.
• Points A1, A2, A3, A4 and A5 are taken on AX, such that AA1 = A3A2 = A2A3 = A3A4 = A4A
• A5 and C are joined.
• A3C' is drawn parallel to A5C, meeting AC at C' .
• C'B' is drawn parallel to CB, meeting AB at B'.
• AB'C' is the required triangle similar to ΔABC whose sides a 3/5 of the corresponding sides of ΔABC.

Q3: Two poles of equal heights are standing opposite to each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.     [NCERT; CBSE 2019 (30/1/2)]

Ans: Let AB and CD be two poles of equal height & metre and let P be any point between the poles, such that
∠APB = 60° and ∠DPC = 30°
The distance between two poles is 80m. (Given)
Let AP = x m, then PC = (80 - x) m
Now, in ΔAPB, we have

Now, putting the value of x in equation (i), we have

Hence, the height of the pole is 20 √3 m and the distance of the point from first pole is 20 m and that of the second pole is 60 m.

Q4: A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of 30°. A girl standing on the roof of 20 metre high building, finds the angle of elevation of the same bird to be 45°. Both the boy and the girl are on opposite sides of the bird. Find the distance o f bird from the girl. [Given √2 = 1.414]     [CBSE 2019 (30/5/1)]

Ans: Let B be the position of bird. O and P be the positions of boy and girl respectively and PQ be the building
We have, ∠AOB = 30° and, ∠BPM = 45°
In ΔAOB, we have

Hence, distance of bird from girl is 30 √2 m.

## Previous Year Questions 2018

Q5: Construct a triangle with sides 6 cm, 8 cm and 10 cm. Construct another triangle whose sides are 3/5 of the corresponding sides of the original triangle.   [CBSE 2018 (C)]

Ans: Steps of construction:

• Draw ΔABC, such that
AB = 8 cm, BC = 6 cm
AC = 10 cm
• Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
• Draw 5 equal marks, C1, C2, C3, C4, C5 such that BC1 = C1C2 = C2C3 = C3C4 = C4C5
• Join CC5. Draw C3C' || C5C, where C' is any point on BC
• Draw C'A' || AC
• ΔA'BC' is the required triangle.

## Previous Year Questions 2017

Q6: A ladder 15 m long makes an angle of 60° with the wall. Find the height of the point where the ladder touches the wall.    [CBSE(F) 2017]

Ans:

Q7: The ratio of the height of a tower and the length of its shadow on the ground is  What is the angle of elevation of the sun?    [CBSE Delhi 2017]

Ans:

Q8: Draw a line segment of length 8 cm and divide it internally in the ratio 4 : 5.    [Delhi 2017]

Ans: Steps of construction:

• Draw a line segement AB of length 8 cm.
• Draw any ray AX making an acute angle with AB.
• Locate 9(i.e. 4+5) points A1, A2, A3, ... A9 on AX so that AA1 = A1A2 = ... A8A9
• Join A9B.
• Through the point A4 draw a line parallel to A9B by making an angle equal to ∠AA9 B at A4 intersecting AB at point P.

Then AP : PR = 4 : 5

Q9: If a tower 30 m high, casts a shadow 10 √3 m long on the ground, then what is the angle of elevation of the sun?     [CBSE (Al) 2017]

Ans: In ΔABC

⇒ tan θ = tan 60°⇒ θ = 60

Q10: A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h.     [CBSE Delhi 2017]

Ans: Let the speed of boat be x m/min
∴ CD = 2x
In ΔABC

In ΔABD

Q11: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.     [NCERT, CBSE Delhi 2017]
OR
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower. [Use √3 = 1.732]     [CBSE (F) 2017]

Ans: Let PQ be the building of height 7 metres and AB be the cable tower. Now it is given that the angle of elevation of the top A of the tower observed from the top P of building is 60° and the angle of depression of the base B of the tower observed from P is 45°. (Fig. 11.38)
So, ∠APR = 60° and ∠QBP = 45°
Let QB = x m, AR = h m then, PR = x m
Now, in ΔAPR, we have

...(i)
Again, in ΔPBQ we have

Putting the value of x in equation (i), we have

So, the height of tower

Q12: A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.   [NCERT, CBSE Delhi 2017 (C)]

Ans: Let OA be the tower of height h, and P be the initial position of the car when the angle of depression is 30°.
After 6 seconds, the car reaches to Q such that the angle of depression at Q is 60°. Let the speed of the car be v metre per second. Then,
PQ = 6v     (∴ Distance = speed x time)
and let the car take i seconds to reach the tower OA from Q (Fig. 11.41). Then, OQ = vt metres.
Now, in ΔAQO, we have

Now, substituting the value of h from (i) into (ii), we have

Hence, the car will reach the tower from Q in 3 seconds.

Q13: From the top of a tower, 100 m high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression 30° and 45°. Find the distance between the cars.  [Take √3= 1.732]      [CBSE (AI) 2017]

Ans: Let AO be the tower of height 100 m. Car B and Car C are in opposite direction and at distance of x m and y m respectively.
In ΔABO

⇒ x = 100   ...(i)
In ΔACO,
...(ii)
Distance between the cars - x + y
[From equation (i) and (ii)]

Q14: The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the sun is 60°. Find the angle of elevation of the sun at the time of the longer shadow.     [CBSE(F) 2017]

Ans: Let AB be the flagstaff and BC be the length of its shadow when the Sun rays meet the ground at an angle of 60°. Let 0 be the angle between the Sun rays and the ground when the length of the shadow of the flagstaff is BD. Let h be the height of the flagstaff (Fig. 11.52).
Let BC = x  ∴ BD = 3x and CD = 2x
In ΔABC, we have

Q15: Construct a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then construct another triangle whose sides are 3/4 times the corresponding sides of the ΔABC.    [AI 2017]

Ans: In ΔABC,

Steps of Construction:

• Draw a line segment BC = 7 cm. At point B, draw ∠B = 45° and at C, draw ∠C = 30° and get ΔABC.
• Draw an acute CBX on the base BC at point B (In the downward direction). Mark the ray BX with four equal points B1, B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4.
• Join B4 to C. Draw B3C'||B4C.
• At C', draw C'A'||AC.
• ΔABC' is the required triangle.

Q16: Draw an isosceles triangle ABC in which the base BC is 8 cm long and its altitude AD through A is 4 cm long. Then draw another triangle whose sides are 2/3 of the corresponding sides of the ΔABC.    [AI 2017 (C)]

Ans: Steps of construction:

Draw an isosceles ΔABC in which BC = 8 cm, AD = 4 cm and AB = AC

Draw an acute angle CBX below BC at point B.

Draw three equal marks B1, B2, B3 on BX, such that
BB1 = B1B2 = B2B

Join B3C, draw B2C' || B3C.

At point C', draw C'A' || AC.

ΔA'BC' is the required isosceles triangle.

Q17: From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 45° and 30° respectively. Find the height of the hill.     [CBSE (E) 2017]

Ans: Let the height of the hill be h m, C and D are two consecutive stones having a distance of 1000 m between them and 4C = x m.
In ΔABC,

⇒ x = h   ...(i)
In ΔABD,

Hence, the height of the hill

## Previous Year Questions 2016

Q18: In Fig. 11.13, AB is a 6 m high pole and CD is a ladder inclined at an angle of 60° to the horizontal and reaches up to a point D of pole. If AD = 2.54 m, find the length of the ladder. (Use √3 =1.73)     [CBSE Delhi 2016]

Ans: DB = (6 - 2.54)m = 3.46 m

∴ DC = 4m

Q19: An observer, 1.7 m tall, is 20 √3 m away from a tower. The angle of elevation from the eye of the observer to the top of the tower is 30°. Find the height of the tower.     [CBSE (F) 2016]

Ans: Let AB be the height of tower and DE be the height of observer.

Q20: A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 5 m. From a point on the ground the angles of elevation of the top and bottom of the flagstaff are 60° and 30° respectively. Find the height of the tower and the distance of the point from the tower. (Take √3 = 1.732)     [CBSE (F) 2016]

Ans: Let height of tower be x m and distance of point from tower be y m.

⇒ x + 5 = 3x ⇒ x = 5/2 = 2.5
Height of tower = 2.5 m
Distance of point from tower = y = √3x
= (2.5 X 1.732) or 4.33 m

Q21: Two men on either side of a 75 m high building and in line with base of building observe the angle of elevation of the top of the building as 30° and 60°. Find the distance between the two men. (Use √3 = 1.73)    [CBSE (F) 2016]

Ans:

ln ΔABM1 ,
Let AB be the building having height 75 m and the angles of elevation are 30° and 60° from the point M1 and M2 respectively;

In ΔABM2,

∴ Distance between two men = 173 m.

The document Class 10 Maths Previous Year Questions - Constructions (Old Syllabus) is a part of the SSS 1 Course Mathematics for SSS 1.
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## FAQs on Class 10 Maths Previous Year Questions - Constructions (Old Syllabus)

 1. What are the basic tools required for constructions?
Ans. The basic tools required for constructions include a measuring tape, a compass, a ruler, a pencil, a set square, and a protractor.
 2. How can we construct a perpendicular bisector of a line segment?
Ans. To construct a perpendicular bisector of a line segment, follow these steps: 1. Use a compass to draw arcs on both sides of the line segment. 2. Draw two arcs of the same radius from two points on the line segment. 3. Draw a straight line connecting the two points where the arcs intersect. This line is the perpendicular bisector.
 3. What is the significance of constructions in mathematics?
Ans. Constructions in mathematics help in visualizing geometric concepts, understanding the properties of shapes, and solving mathematical problems using precise constructions.
 4. How can we construct an angle bisector of a given angle?
Ans. To construct an angle bisector of a given angle, follow these steps: 1. Use a compass to draw an arc from the vertex of the angle. 2. Without changing the compass width, draw two more arcs intersecting the sides of the angle. 3. Draw a line connecting the vertex to the point where the arcs intersect. This line is the angle bisector.
 5. What are some common errors to avoid while making constructions?
Ans. Some common errors to avoid while making constructions include inaccurate measurements, using a blunt pencil, not drawing arcs accurately with a compass, and not using a ruler for straight lines.

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