Constructions Class 10th Mathematics CBSE Solution

Class 10th Mathematics CBSE Solution

Exercise 11.1
Question 1.

Draw a line segment of length 7.6 cm and divide it in the ratio 5: 8. Measure the two parts.


Answer:

Step 1: Draw line segment AB = 7.6 cm



Step 2: Draw a ray AC making an acute angle with line segment AB .



Step 3: Along AC, divide (5+8) 13 equals points


A1,A2,A3,……………….,A13 on AX such that


AA1=AA2=A2A3 and so on .



Step 4: Join BA13



Step 5: Through the point A5, draw a line parallel to BA13 (by making an angle equal to ∠AA13B) at A5 intersecting AB at point D.


C is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8.


The lengths of AD and DB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.



Justification


The construction can be justified by proving that



By construction, we have A5D∥A13B. By applying Basic proportionality theorem for



From the figure, it can be observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments respectively.



On comparing equations (1) and (2), we obtain



This justifies the construction.


Question 2.

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are  of the corresponding sides of the first triangle.


Answer:

Step 1: Draw a line segment AB of 5 cm.



Step 2: Taking A and B as Center, draw arcs of 6 cm and 5 cm radius respectively. Let these arcs intersect each other at point C. △ABC is the required triangle having length of sides as 5 cm , 6 cm , 7 cm respectively.



Step 3: Draw a ray AX making acute angle with line AB on the opposite side of vertex C.



Step 4: Locate 7 points, A1,A2,A3,A4, A5,A6,A7( as 7 is greater between 5 and 7), on line AX such that AA1=A1 A2=A2 A3=A3 A4=A4 A5=A5 A6=A6 A7



Step 5: Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B′.



Step 6: Draw a line through B′ parallel to BC intersecting the extended line segment AC at C′.△AB’C’ is the required triangle.



Justification


∆AB' C'~∆ABCby AAA Similarity Condition



And 



Question 3.

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are  of the corresponding sides of the first triangle.


Answer: 1. Now in order to make a triangle, draw a line segment AB = 5 cm.From point A, draw an arc at 6 cm. Draw an arc at 7 cm from point B intersecting the previous arc.

2. Join the point of intersection from A and B.

Hence, this gives the required triangle ABC

3. Dividing the base, draw a ray AX which is at an acute angle from AB


4. Plot seven points on AX such that:

AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7.

5. Join A5 to B.

6. Draw a line from point A7that is parallel to A5B and joins AB’ (AB extended to AB’).

7. Draw a line B’C’ || BC.

Hence, triangle AB’C’ is the required triangle.


Question 4.

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are  times the corresponding sides of the isosceles triangle.


Answer:

△A′BC′ whose sides  of the corresponding sides of can be drawn as follows.


Step 1 Draw a △ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = .



Step 2 Draw a ray BX making an acute angle with BC on the opposite side of vertex A.



Step 3 Locate 4 points (as 4 is greater in 3 and 4), B1,B2,B3,B4 on line segment BX.


Step 4 Join B4C and draw a line through B3 , parallel to B4C intersecting BC at C′



Step 5 Draw a line through C' parallel to AC intersecting AB at A'.△A′BC′ is the required triangle.



Justification


∠A=60° (Common)


∠C=∠C'


 By AA similarity condition



Also, 


AB’=AB ,B’C’=BC and AC’=AC


Question 5.

Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ ABC = 60°. Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.


Answer:

Given in Δ ABC,

  • Length of side BC = 6 cm.
  • Length of side AB = 5 cm.
  • ∠ ABC = 60°.

Steps of Construction:

  1. Draw a line segment BC of length 6 cm.

2. With B as center, draw a line which makes an angle of 60° with BC.

Construction of 60° angle at B:
a. With B as centre and with some convenient radius draw an arc which cuts the line BC at D.


b. With D as radius and with same radius (in step a), draw another arc which cuts the previous arc at E.


c. Join BE. The line BE makes an angle 60° with BC.

  1. Again with B as centre and with radius of 5 cm, draw an arc which intersects the line BE at point A.


4. Join AC. This is the required triangle.

  1. Now, from B, draw a ray BX which makes an acute angle on the opposite side of the vertex A.

6. With B as center, mark four points B1, B2, B3 and B4 on BX such that they are equidistant. i.e. BB1 = B1B2 = B2B3 = B3B4.


7. Join B4C and then draw a line from B3 parallel to B4C which meets the line BC at P.


8. From P, draw a line parallel to AC and meets the line AB at Q. Thus ΔBPQ is the required triangle.

 
Question 6.

Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are times the corresponding sides of Δ ABC.


Answer:

Steps of construction:


1. Draw a line segment BC = 7 cm.



2. Draw ∠ABC = 45° and ∠ACB = 30° i.e. ∠BAC = 105°.



We obtain ΔABC.


3. Draw a ray BX making an acute angle with BC. Mark four points B1, B2, B3,B4 at equal distances.



4. Through B3 draw B3C and through B4 draw B4C1 parallel to B3C.


Then draw A1C1 parallel to AC.



∴ A1BC1 is the required triangle.


Question 7.

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are  times the corresponding sides of the given triangle.


Answer:

Steps of construction:

  1. Now in order to make a triangle, draw a line segment AB = 3 cm.


2. Make a right angle at point A and draw AC = 4 cm from this point.

3. Join points A and B to get the right triangle ABC.


4. Now, Dividing the base, draw a ray AX such at it forms an acute angle from AB.


5. Then, plot 5 points on AX such that:

AG = GH = HI = IJ = JK.

6. Join I to point line AB and Draw a line from K which is parallel to IB such that it meets AB at point M.

7. Draw MN || CB.


This is the required construction, thus forming AMN which have all the sides 5/3 times the sides of ABC

Triangle AMN is the required triangle.



Exercise 11.2
Question 1.

Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.


Answer:

Step1: Draw circle of radius 6cm with center A, mark point C at 10 cm from center



Step 2: find perpendicular bisector of AC



Step3: Take this point as center and draw a circle through A and C



Step4:Mark the point where this circle intersects our circle and draw tangents through C



Length of tangents = 8cm


AE is perpendicular to CE (tangent and radius relation)


In ΔACE


AC becomes hypotenuse


AC2 = CE2 + AE2


102 = CE2 + 62


CE2 = 100-36


CE2 = 64


CE = 8cm


Question 2.

Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.


Answer:

Steps of construction:

i. Draw two concentric circles with radii 4 cm and 6 cm respectively.


ii. Now, draw the radius OP of the larger circle.


iii. Construct a perpendicular bisector of OP intersecting OP at point O’.


iv. Considering O’P as radius, draw another circle.


v. From point P, draw tangents PQ and PR(can see in the figure)



Justification: By applying Pythagoras theorem, we have;


PQ2 = OP2 – OQ2


= 62 – 42


= 36 – 16 = 20


Or, PQ = 2√5 cm


Question 3.

Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.


Answer:

Steps of construction:

i. At first draw a circle with radius 3 cm.

ii. Now, extend the diameter to A and B on both the sides.


iii. Then, draw the perpendicular bisectors of OA and OB.


Such that:


Perpendicular bisector of OA intersects it at point O’.


And,


Perpendicular bisector of OB intersects it at point O”.


iv. Considering O’A as radius, construct another circle.


v. considering O”B as radius, construct the third circle.


vi. From point A, draw tangents AP and AQ.


vii. From point B, draw tangents BR and BS.


 
Question 4.

Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.


Answer: Steps of construction:
1) Draw a circle of radius 5 cm, and draw a radius OA anywhere in the circle.

2) Taking OA as base, draw an angle AOB such that ∠AOB = 120°.

3) At A, Draw a line AX such that AX ⊥ OA.

4) At B, Draw a line BY such that BY ⊥ OB.

5) AX and BY intersect at P; AP and BP are required tangents.


Justification:
1) Clearly, AP and BP are tangents since tangent at a point on the circle is perpendicular to the radius through point of contact.
2) In Quadrilateral OAPB, we have
∠OAP + ∠APB + ∠OBP + ∠AOB = 360° [By Angle Sum Property]
⇒ ∠OAP + 90° + 90° + 120° = 360°
⇒ ∠OAP = 60°
 
Question 5.

Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.


Answer:

Step1: Draw a line PQ=8cm. taking P and Q as a center draw circle of 3cm and 4cm.


Step2: Now bisect PQ. We get midpoint of PQ be T.


Now take T as a center , draw a circle of PT radius , this will intersect the circle at point A,B,C,D. Join PB,PD,AQ,QC.