The document Previous Year Questions - Constructions Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.

All you need of Class 10 at this link: Class 10

**Short Answer Type Questions**

**Q.1. 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:**

(i) Draw concentric circles of radius OQ = 2 cm and OP = 5 cm having same centre O.

(ii) Mark these circles as C and C'.

(iii) Points O, Q and P lie on the same line.

(iv) Draw perpendicular bisector of OP, which intersects OP at O'.

(v) Take O' as centre, draw a circle of radius OO' which intersects circle C at points A and B.

(vi) Join PA and PB, these are the required tangents.

(vii) Length of these tangents are approx. 4.6 cm.

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

Ans.

(a) Draw ΔABC, such that

AB = 8 cm, BC = 6 cm

AC = 10 cm

(b) Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

(c) Draw 5 equal marks, C

(d) Join CC

(e) Draw C'A' || AC

(f) ΔA'BC' is the required triangle.

Ans.

Ans.

Ans. Steps of construction:

(i) Draw a line segement AB of length 8 cm.

(ii) Draw any ray AX making an acute angle with AB.

(iii) Locate 9(i.e. 4+5) points A

(iv) Join A

(v) Through the point A

Then AP : PR = 4 : 5

Ans.

⇒ tan θ = tan 60°

⇒ θ = 60

Ans.

∴ CD = 2x

In ΔABC

In ΔABD

⇒

∴

Ans.

∴ DC = 4m**Q.9. An observer, 1.7 m tall, is 20 √3 m away from a tower. The angle of elevation from the eye of observer to the top of tower is 30°. Find the height of tower. [CBSE (F) 2016] Ans.** Let AB be the height of tower and DE be the height of observer.

Ans.

⇒ 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

Ans.

Let AB be the building having height 75 m and the angles of elevation are 30° and 60° from the point M

In ΔABM

∴ Distance between two men — 173 m.

**Long Answer Type Questions**

**Q.1. 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:

(i) A line segment AC = 6 cm is drawn.

(ii) ∠CAB = 45° is constructed at A.

(iii) An arc of 5 cm radius to be drawn with A as centre, cutting AY at B.

(iv) B and C are joined. Then, ΔABC is constructed.

(v) An acute angle CAX is drawn below AC.

(vi) Points A

(vii) A

(viii) A

(ix) C'B' is drawn parallel to CB, meeting AB at B'.

(x) AB'C' is the required triangle similar to ΔABC whose sides a 3/5 of the corresponding sides of ΔABC.

Ans.

∠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.

**Q.3. 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.

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.

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

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 die tower from Q in 3 seconds.

Ans.

In ΔABO

⇒ x = 100 ...(i)

In ΔACO,

⇒ ...(ii)

Distance between the cars - x + y

[From equation (i) and (ii)]

Let BC = x ∴ BD = 3x and CD = 2x

In ΔABC, we have

In ΔADB, we have

Ans.

1. Draw a line segment BC = 7 cm. At the point B, draw ∠B = 45° and at C, draw ∠C = 30° and get ΔABC.

2. Draw an acute CBX on the base BC at the point B (In downward direction). Mark the ray BX with four equal points B

3. Join B

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

5. ΔABC' is the required triangle.

Ans.

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

2. Draw an acute angle CBX below BC at point B.

3. Draw three equal marks B

BB

4. Join B

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

6. ΔA'BC' is the required isosceles triangle.

Ans.

In ΔABC,

⇒ x = h ...(i)

In ΔABD,

⇒

⇒

⇒

Hence, the height of the hill

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

51 videos|346 docs|103 tests