## Maharashtra State Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4

Question 1.

Select the correct alternative for each of the following questions.

i. The number of tangents that can be drawn to a circle at a point on the circle is ______

(A) 3

(B) 2

(C) 1

(D) 0

Answer:

(C)

ii. The maximum number of tangents that can be drawn to a circle from a point outside it is ______

(A) 2

(B) 1

(C) one and only one

(D) 0

Answer:

(A)

iii. If ∆ABC ~ ∆PQR and \(\frac { AB }{ PQ } \) = \(\frac { 7 }{ 5 } \), then ______

(A) AABC is bigger.

(B) APQR is bigger.

(C) both triangles will be equal

(D) can not be decided

Answer:

(A)

Question 2.

Draw a circle with centre O and radius 3.5 cm. Take point P at a distance 5.7 cm from the centre. Draw tangents to the circle from point P.

Solution:

Analysis:

As shown in the figure, let P be a point in the exterior of circle at a distance of 5.7 cm.

Let PQ and PR be the tangents to the circle at points Q and R respectively.

∴ seg OQ ⊥ tangent PQ …[Tangent is perpendicular to radius]

∴ ∠OQP = 90°

∴ point Q is on the circle having OP as diameter. …[Angle inscribed in a semicircle is a right angle]

Similarly, point R also lies on the circle having OP as diameter.

∴ Points Q and R lie on the circle with OP as diameter.

On drawing a circle with OP as diameter, the points where it intersects the circle with centre O, will be the positions of points Q and R respectively.

Question 3.

Draw any circle. Take any point A on it and construct tangent at A without using the centre of the circle.

Solution:

Analysis:

As shown in the figure, line l is a tangent to the circle at point A.

seg BA is a chord of the circle and ∠BCA is an inscribed angle.

By tangent secant angle theorem,

∠BCA = ∠BAR

By converse of tangent secant angle theorem,

If we draw ∠BAR such that ∠BAR = ∠BCA, then ray AR (i.e. line l) is a tangent at point A.

Question 4.

Draw a circle of diameter 6.4 cm. Take a point R at a distance equal to its diameter from the centre. Draw tangents from point R.

Solution:

Diameter of circle = 6.4 cm 6.4

Radius of circle = \(\frac { 6.4 }{ 2 } \) = 3.2 cm

Analysis:

As shown in the figure, let R be a point in the exterior of circle at a distance of 6.4 cm.

Let RQ and RS be the tangents to the circle at points Q and S respectively.

∴ seg PQ ⊥tangent RQ …[Tangent is perpendicular to radius]

∴ ∠PQR = 90°

∴ point Q is on the circle having PR as diameter. …[Angle inscribed in a semicircle is a right angle]

Similarly,

Point S also lies on the circle having PR as diameter.

∴ Points Q and S lie on the circle with PR as diameter.

On drawing a circle with PR as diameter, the points where it intersects the circle with centre P, will be the positions of points Q and S respectively.

Question 5.

Draw a circle with centre P. Draw an arc AB of 100° measure. Draw tangents to the circle at point A and point B.

Solution:

m(arc AB) = ∠APB = 100°

Analysis:

seg PA ⊥ line l

seg PB ⊥line m … [Tangent is perpendicular to radius]

The perpendicular to seg PA and seg PB at points A and B respectively will give the required tangents at A and B.

Steps of construction:

i. With centre P, draw a circle of any radius and take any point A on it.

ii. Draw ray PA.

iii. Draw ray PB such that ∠APB = 100°.

iv. Draw line l ⊥ray PA at point A.

v. Draw line m ⊥ ray PB at point B.

Lines l and m are tangents at points A and B to the circle.

Question 6.

Draw a circle of radius 3.4 cm and centre E. Take a point F on the circle. Take another point A such that E – F – A and FA = 4.1 cm. Draw tangents to the circle from point A.

Solution:

Analysis:

Draw a circle of radius 3.4 cm

As shown in the figure, let A be a point in the exterior of circle at a distance of (3.4 + 4.1) = 7.5 cm.

Let AP and AQ be the tangents to the circle at points P and Q respectively.

∴ seg EP ⊥ tangent PA … [Tangent is perpendicular to radius]

∴ ∠EPA = 90°

∴ point P is on the circle having EA as diameter. …[Angle inscribed in a semicircle is a right angle]

Similarly, point Q also lies on the circle having EA as diameter.

∴ Points P and Q lie on the circle with EA as diameter.

On drawing a circle with EA as diameter, the points where it intersects the circle with centre E, will be the positions of points P and Q respectively.

Question 7.

∆ABC ~ ∆LBN. In ∆ABC, AB = 5.1 cm, ∠B = 40°, BC = 4.8 cm, \(\frac { AC }{ LN } \) = \(\frac { 4 }{ 7 } \). Construct ∆ABC and ∆LBN.

Solution:

Analysis:

As shown in the figure,

Let B – C – N and B – A – L.

∆ABC ~ ∆LBN …[Given]

∴ ∠ABC ≅ ∠LBN …[Corresponding angles of similar triangles]

\(\frac { AB }{ LB } \) = \(\frac { BC }{ BN } \) = \(\frac { AC }{ LN } \) …(i)[Corresponding sides of similar triangles]

But. \(\frac { AC }{ LN } \) = \(\frac { 4 }{ 7 } \) …(ii)[Given]

∴ \(\frac { AB }{ LB } \) = \(\frac { BC }{ BN } \) = \(\frac { AC }{ LN } \) = \(\frac { 4 }{ 7 } \) …[From(i)and(ii)]

∴ sides of ∆LBN are longer than corresponding sides of ∆ABC.

∴ If seg BC is divided into 4 equal parts, then seg BN will be 7 times each part of seg BC.

So, if we construct ∆ABC, point N will be on side BC, at a distance equal to 7 parts from B.

Now, point L is the point of intersection of ray BA and a line through N, parallel to AC.

∆LBN is the required triangle similar to ∆ABC.

Steps of construction:

i. Draw ∆ABC of given measure. Draw ray BD making an acute angle with side BC.

ii. Taking convenient distance on compass, mark 7 points B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6} and B_{7} such that

BB_{1} = B_{1}B_{2} = B_{2}B_{3 }B_{3}= B4_{4} = B_{4}B_{5} = B_{5}B_{6} = B_{6}B_{7}.

iii. Join B_{4}C. Draw line parallel to B_{4}C through B_{7} to intersects ray BC at N.

iv. Draw a line parallel to side AC through N. Name the point of intersection of this line and ray BA as L.

∆LBN is the required triangle similar to ∆ABC.

Question 8.

Construct ∆PYQ such that, PY = 6.3 cm, YQ = 7.2 cm, PQ = 5.8 cm. If = \(\frac { YZ }{ YQ } \) = \(\frac { 6 }{ 5 } \) then construct ∆XYZ similar to ∆PYQ.

Solution:

Analysis:

As shown in the figure,

Let Y – Q – Z and Y – P – X.

∆XYZ ~ ∆PYQ …[Given]

∴ ∠XYZ ≅ ∠PYQ …[Corresponding angles of similar triangles]

\(\frac { XY }{ PY } \) = \(\frac { YZ }{ YQ } \) = \(\frac { XZ }{ PQ } \) …(i)[Corresponding sides of similar triangles]

But, \(\frac { YZ }{ YQ } \) = \(\frac { 6 }{ 5 } \) ,..(ii)[Given]

∴ \(\frac { XY }{ PY } \) = \(\frac { YZ }{ YQ } \) = \(\frac { XZ }{ PQ } \) = \(\frac { 6 }{ 5 } \) …[From (i) and (ii)]

∴ sides of ∆XYZ are longer than corresponding sides of ∆PYQ.

∴ If seg YQ is divided into 5 equal parts, then seg YZ will be 6 times each part of seg YQ.

So, if we construct ∆PYQ, point Z will be on side YQ, at a distance equal to 6 parts from Y.

Now, point X is the point of intersection of ray YP and a line through Z, parallel to PQ.

∆XYZ is the required triangle similar to ∆PYQ.

Steps of construction:

i. Draw ∆ PYQ of given measure. Draw ray YT making an acute angle with side YQ.

ii. Taking convenient distance on compass, mark 6 points Y_{1}, Y_{2}, Y_{3}, Y_{4}, Y_{5} and Y_{6} such that

YY_{1} = Y_{1}Y_{2} = Y_{2}Y_{3} = Y_{3}Y_{4} = Y_{4}Y_{5} = Y_{5}Y_{6}.

iii. Join Y_{5}Q. Draw line parallel to Y_{5}Q through Y_{6} to intersects ray YQ at Z.

iv. Draw a line parallel to side PQ through Z. Name the point of intersection of this line and ray YP as X.

∆XYZ is the required triangle similar to ∆PYQ.