ML Aggarwal Solution Class 10 Chapter 16 Costructions Exercise 16.2

 Exercise 16.2

Question 1

Draw an equilateral triangle of side 4 cm. Draw its circumcircle.













Sol :
Steps of Construction :
(i) Draw a line segment BC = 4 cm
(ii) With centres B and C, draw two arcs of radius 4 cm
which intersect each other at A.
(iii) Join AB and AC. ∆ ABC is an equilateral triangle.
(iv) Draw the right bisector of BC and AC intersecting each other at O.
(v) Join OA, OB and OC.
(vi) With centre O, and radius equal to OB or OC or OA,
draw a circle which will pass through A, B and C.
This is the required circumcircle of ∆ ABC.

Question 2

Using a ruler and a pair of compasses only, construct: (i) a triangle ABC given AB = 4cm, BC = 6 cm and ∠ABC = 90°.
(ii) a circle which passes through the points A, B and C and mark its centre as O. (2008)
Sol :
Steps of Construction:
(i) Draw a line segment AB = 4cm
(ii) At B, draw a ray BX making an angle of 90°
and cut off BC = 6 cm.
(iii) Join AC.
(iv) Draw the perpendicular bisectors of sides
AB and AC intersecting each other at O.
(v) With centre O, and radius equal to OB or OA or OC,
draw a circle which passes through A, B and C.
This is the required circle.

















Question 3

Construct a triangle with sides 3 cm, 4 cm and 5 cm. Draw its circumcircle and measure its radius.

Sol :

Steps of Construction

(i) Draw a line segment BC = 4 cm.

(ii) With centre B and radius 3 cm and with centre C

and radius 5 cm draw two arcs which intersect each other at A.

(iii) Join AB and AC.

(iv) Draw the perpendicular bisector of sides BC and AC

which intersect each other at O.

(v) Join OB.

(vi) With centre O and radius OB, draw a circle

which will pass through A, B and C.

(vii) On measuring the radius OB = 2.5cm















Question 4

Using ruler and compasses only :

(i) Construe a triangle ABC with the following data: Base AB = 6 cm, AC = 5.2 cm and ∠CAB = 60°.

(ii) In the same diagram, draw a circle which passes through the points A, B and C. and mark its centre O.

Sol :

Steps of Construction :

(i) Draw a line segment AB = 6 cm.

(ii) At A, draw a ray making an angle of 60°.

(iii) With centre B and radius 5-2 cm.

draw an arc which intersects the ray at C.

(iv) Join BC

(v) Draw the perpendicular bisector of AB and BC

intersecting each other at O.

(vi) With O as a centre and OA as a radius

draw a circle which touches the ∆ABC at A, B and C.















Question 5

Using ruler and compasses only, draw an equilateral triangle of side 5 cm and draw its inscribed circle. Measure the radius of the circle.

Sol :

Steps of Construction :

(i). Draw a line segment BC = 5 cm

(ii) With centre B and C and radius 5 cm,

draw two arcs intersecting each other at A.

(iii) Join AB and AC.

(iv) Draw the angle bisectors of ∠B and ∠C intersecting each other at I.

(v) From I, draw a perpendicular ID on BC.

(vi) With centre I and radius ID, draw a circle

which touches the sides of the triangle internally.














Question 6

(i) Conduct a triangle ABC with BC = 6.4 cm, CA = 5.8 cm and ∠ ABC = 60°. Draw its incircle. Measure and record the radius of the incircle.

(ii) Construct a ∆ABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle. (2014)

Sol :

Steps of Construction :

(i) Draw a line segment BC = 6.4 cm

(ii) Construct ∠ DBC = 60° at B.

(iii) With C as centre and radius CA = 5.8 cm.

Draw an arc cutting BD at A.

(iv) Join AC. Then ABC is the required triangle.

(v) Draw the angle bisectors of ∠B and ∠C which intersect each other at O.

(vi) Draw OE ⊥ BC, intersecting BC in E.

(vii) With O as centre and OE as radius draw the required incircle.

Measure the radius OE which is = 1.5cm

Steps of construction:
1. Draw a line BC = 6.5 cm.
2. With centre B and C draw arcs AB = 5.5 cm and AC = 5 cm
3. Join AB and AC, ABC is the required triangle.
4. Draw the angle bisetors of B and C. Let these bisectors meet at O.
5. Taking O as centre. Draw a incircle which touches all the sides of the ∆ABC.
6. From O draw a perpendicular to side BC which cut at N.
7. Measure ON which is required radius of the incircle. ON = 1.5 cm



Question 7

The bisectors of angles A and B of a scalene triangle ABC meet at O.

(i) What is the point O called?

(ii) OR and OQ are drawn a perpendicular to AB and CA respectively. What is the relation between OR and OQ ?

(iii) What is the relation between ∠ACO and ∠BCO?

Sol :

(i) The point O where the angle bisectors meet is called the incentre of the triangle.

(ii) The perpendiculars drawn from O to AB and CA are equal i.e. OR and OQ.

(iii) ∠ACO = ∠BCO

OC will bisect the ∠C








Question 8

Using ruler and compasses only, construct a triangle ABC in which BC = 4 cm, ∠ACB = 45° and the perpendicular from A on BC is 2.5 cm. Draw the circumcircle of triangle ABC and measure its radius.

Sol :

Steps of Construction

(i) Draw a line segment BC = 4 cm.

(ii) At B, draw a perpendicular and cut off BE = 2.5 cm.
















(iii) From E, draw a line EF parallel to BC.
(iv) From C, draw a ray making an angle of 45° which intersects EF at A.
(v) Join AB.
(vi) Draw the perpendicular bisectors of
sides BC and AC intersecting each other at O.
(vii) Join OB, OC and OA.
(viii) With centre 0 and radius OB or OC or OA draw a circle
which will pass through A, B and C.
This circle is the circumcircle of ∆ABC. On measuring its radius OB = 2 cm.

Question 9

Construct a regular hexagon of side 4 cm. Construct a circle circumscribing the hexagon.















Sol :
Steps of Construction
(i) Draw a line segment BC = 4 cm.
(ii) At A and B draw rays making on angle of 120° each and cut off AF = BC = 4cm.
(iii) At F and C, draw rays making angle of 120° each and cut off EF = CD = 4cm.
(iv) Join ED.
ABCDEF is the required hexagon.
(v) Draw perpendicular bisectors of sides AB and BC intersecting each other at O.
(vi) With centre O and radius equal OA or OB draw a circle
which passes through the vertices of the hexagon.
This is the required circumcircle of hexagon ABCDEF.

Question 10

Draw a regular hexagon of side 4 cm and construct its incircie.

Sol :

Steps of constructions :

















(i) Draw a regular hexagon ABCDEF of side 4 cm.

(ii) Draw the angle bisectors of ∠A and ∠B

which intersect each other at O.

(iii) Draw OL ⊥ AB.

(iv) With centre O and radius OB, draw a circle

which touches the sides of the hexagon

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs correct Solution from the given four options (1 to 7): Question 1 If log√3 27 = x, then the value of x is (a) 3 (b) 4 (c) 6 (d) 9 Sol : ⇒$\log _{\sqrt{3}} 27=x$ ⇒$(\sqrt{3})^{x}=27$ ⇒$(3)^{\frac{1}{2} x x}=3^{3}$ ⇒$3^{\frac{x}{2}}=3^{3}$ ⇒$\frac{x}{2}=3$ ⇒x=6...(c) Question 2 If log 5 (0.04) = x, then the value of x is (a) 2 (b) 4 (c) -4 (d) -2 Sol : ⇒$\log _{5}(0.04)=x$ ⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$ ∴x=-2...(d) Question 3 If log 0.5  64 = x, then the value of x is (a) -4 (b) -6 (c) 4 (d) 6 Sol : ⇒$\log _{0.5} 64=x $ ⇒$0.5^{x}=64$ ⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$ ⇒$2^{-x}=2^{6}$ ∴-x=6⇒x=-6...(b) Question 4 If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is (a) $\frac{1}{5}$ (b) $-\frac{1}{5}$ (c) -1 (d) 5 Sol : $\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$ $x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$ $x=\frac{1}{5}$ (b) Question 5 If log (3x + 1) = 2, then the value of x is (a) $\frac{1}{3}$ (b) 99 (c) 33 (d) $\frac{19}{3}$ Sol
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S
Read more