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

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

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 = A₄A₅ 
• 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.

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.

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.

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.

Ans: 

Ans: 

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

Ans: In ΔABC

⇒ tan θ = tan 60°⇒ θ = 60

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


In ΔABD

     ⇒

∴

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

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.

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)]

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

In ΔADB, we have

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.

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 = B₂B₃ 

Join B₃C, draw B₂C' || B₃C.

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

ΔA'BC' is the required isosceles triangle.

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

Ans: DB = (6 - 2.54)m = 3.46 m

∴ DC = 4m

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

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

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.