Buy BOOKS at Discounted Price

Construction: Construction Of Circumcircle And Incircle Of A Triangle

Class 10th Mathematics West Bengal Board Solution
Let Us Work Out 11.1
  1. An equilateral triangle having each side of length 6 cm. Let us draw the following…
  2. An isosceles triangle whose length of the base is 5.2 cm and each of the equal sides…
  3. A right-angled triangle having two sides 4 cm and 8 cm length, containing right angle.…
  4. right-angled triangle having length of 12 cm. hypotenuse and other side is of 5 cm…
  5. A triangle of which length of one side is 6.7 cm and the two angles adjacent to this…
  6. ABC is triangle of which BC = 5 cm., ∠ABC = 100° and AB = 4 cm. Let us draw the…
  7. Given, PQ = 7.5cm, ∠QPR = 45°, ∠PQR = 75°; PQ = 7.5cm, ∠QPS = 60°, ∠PQS = 60°; Let us…
  8. Given, AB = 5cm, ∠BAC = 30°, ∠ABC = 60°, AB = 5cm, ∠BAD = 45°, ∠ABD = 45°; Let us draw…
  9. We draw a quadrilateral ABCD having AB = 4 cm, BC = 7 cm CD = 4 cm, ∠ABC = 60°, ∠BCD =…
  10. Let us draw the rectangle PQRS having PQ = 4 cm, and QR = 6 cm. Let us draw the…
  11. If any circular picture is given, then how shall we find its centre? Let us find the…
Let Us Work Out 11.2
  1. the lengths of three sides are 7 cm., 6 cm., and 5.5 cm. Let us draw the following…
  2. the length of two sides are 7.6 cm., 6 cm. and the angle included by those two sides…
  3. the length of a side is 6.2 cm. and measures of two angles adjacent to this side are…
  4. a triangle is a right-angled triangle of which two sides containing the right angle…
  5. a triangle is a right-angled triangle of which the length of hypotenuse is 9 cm. and…
  6. The triangle is an isosceles triangle having the 7.8 cm. length of base and the length…
  7. a triangle is an isosceles triangle, having the 10 cm. length of base and each of the…
  8. Let us draw an equilateral triangle having 7 cm. length of each side. By drawing the…
Let Us Work Out 11.1
Question 1.

Let us draw the following triangles. By drawing the circum-circle in each case, let us write the position of circum-centre and the length of circum-radius by measuring it.

An equilateral triangle having each side of length 6 cm.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. We name the triangle as ABC.



3. Draw perpendicular bisector of AB.



4. Draw perpendicular bisector of BC.



5. The point of intersection is the circumcenter. Name it as O.


6. With O as center and OC as radius, draw a circle.



7. The circle passes through A,B and C and is thus the circumcircle of this triangle.


8. We measure the circumradius, OC = 3.47cm.
Question 2.

Let us draw the following triangles. By drawing the circum-circle in each case, let us write the position of circum-centre and the length of circum-radius by measuring it.

An isosceles triangle whose length of the base is 5.2 cm and each of the equal sides is 7 cm.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. We name the triangle as ABC.



3. Draw perpendicular bisector of AB.



4. Draw perpendicular bisector of BC.



5. The point of intersection is the circumcenter. Name it as O.


6. With O as center and OC as radius, draw a circle.



7. The circle passes through A,B and C and is thus the circumcircle of this triangle.


8. We measure the circumradius, OC = 3.77cm.
Question 3.

Let us draw the following triangles. By drawing the circum-circle in each case, let us write the position of circum-centre and the length of circum-radius by measuring it.

A right-angled triangle having two sides 4 cm and 8 cm length, containing right angle.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. We name the triangle as ABC.



3. Draw perpendicular bisector of AB.



4. Draw perpendicular bisector of BC.



5. The point of intersection is the circumcenter. Name it as O.


6. With O as center and OC as radius, draw a circle.



7. The circle passes through A,B and C and is thus the circumcircle of this triangle.


8. We measure the circumradius, OC = 4.47cm.
Question 4.

Let us draw the following triangles. By drawing the circum-circle in each case, let us write the position of circum-centre and the length of circum-radius by measuring it.

right-angled triangle having length of 12 cm. hypotenuse and other side is of 5 cm length.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. We name the triangle as ABC.



3. Draw perpendicular bisector of AB.



4. Draw perpendicular bisector of BC.



5. The point of intersection is the circumcenter. Name it as O.


6. With O as center and OC as radius, draw a circle.



7. The circle passes through A,B and C and is thus the circumcircle of this triangle.


8. We measure the circumradius, OC = 6cm.
Question 5.

Let us draw the following triangles. By drawing the circum-circle in each case, let us write the position of circum-centre and the length of circum-radius by measuring it.

A triangle of which length of one side is 6.7 cm and the two angles adjacent to this side are 75o and 55o.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. We name the triangle as ABC.



3. Draw perpendicular bisector of AB.



4. Draw perpendicular bisector of BC.



5. The point of intersection is the circumcenter. Name it as O.


6. With O as center and OC as radius, draw a circle.



7. The circle passes through A,B and C and is thus the circumcircle of this triangle.


8. We measure the circumradius, OC = 4.37cm.
Question 6.

Let us draw the following triangles. By drawing the circum-circle in each case, let us write the position of circum-centre and the length of circum-radius by measuring it.

ABC is triangle of which BC = 5 cm., ∠ABC = 100° and AB = 4 cm.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. We name the triangle as ABC.



3. Draw perpendicular bisector of AB.



4. Draw perpendicular bisector of BC.



5. The point of intersection is the circumcenter. Name it as O.


6. With O as center and OC as radius, draw a circle.



7. The circle passes through A,B and C and is thus the circumcircle of this triangle.


8. We measure the circumradius, OC = 3.52.
Question 7.

Given, PQ = 7.5cm, ∠QPR = 45°, ∠PQR = 75°;

PQ = 7.5cm, ∠QPS = 60°, ∠PQS = 60°;

Let us draw ΔPQR and ΔPQS in such a way that the points R and S lie on the same side of PQ, let us draw the circum circle of ΔPQR and let us observe and write the position of the point S within, on and out side the circum circle. Let us find out its explanation.
Answer:

Steps of Construction:



1. Construct ΔPQR of given dimensions.


2. Construct ΔPQS of given dimensions such that R and S lie on the same side of PQ.



3. Draw perpendicular bisector of PQ.



4. Draw perpendicular bisector of QR.



5. The point of intersection is the circumcenter. Name it as O.


6. With O as center and OR as radius, draw a circle.



7. The circle passes through P,Q and R and is thus the circumcircle of this triangle.


We observe that the circumcircle also passes through S, i.e. S lies on the circle. This is because sum of the adjacent angles of the base of both the triangles are equal. That makes ∠PRQ = ∠PSQ.


We know that angle in the same segment of a circle are equal. That’s why S lies on the circle.
Question 8.

Given, AB = 5cm, ∠BAC = 30°, ∠ABC = 60°,

AB = 5cm, ∠BAD = 45°, ∠ABD = 45°;

Let us draw ΔABC and ΔABD in such a way that the point C and D lie on opposite sides of AB. Let us draw the circle circumscribing ΔABC. Let us write the position of the point D with respect to circumcircle. Let us write by understanding what other characteristics we are observing here.
Answer:

Steps of Construction:


1. Construct ΔABC of given dimensions.



2. Construct ΔABD of given dimensions such that C and D lie on the opposite side of AB.



3. Draw perpendicular bisector of AB.



4. Draw perpendicular bisector of BC.



5. The point of intersection is the circumcenter. Name it as O.


6. With O as center and OR as radius, draw a circle.



7. The circle passes through A, B and C and is thus the circumcircle of this triangle.


We observe that the circumcircle also passes through D, i.e. D lies on the circle. This is because the sum of ∠A and ∠B is 180°. We know that in a cyclic quadrilateral, sum of the opposite angles of a quadrilateral is 180°. That’s why D lies on the circle.
Question 9.

We draw a quadrilateral ABCD having AB = 4 cm, BC = 7 cm CD = 4 cm, ∠ABC = 60°, ∠BCD = 60°. Let us draw the circle circumscribing ΔABC and write what other characteristics we observe.
Answer:

Steps of Construction:


1. Construct quadrilateral ABCD of given dimensions.



2. We join A to C.


3. Draw perpendicular bisector of AB.



4. Draw perpendicular bisector of BC.



5. The point of intersection is the circumcenter. Name it as O.


6. With O as center and OC as radius, draw a circle.



7. The circle passes through A,B and C and is thus the circumcircle of this triangle.


We observe that ABCD is a cyclic quadrilateral as all the points A,B,C and D lie on the circle.
Question 10.

Let us draw the rectangle PQRS having PQ = 4 cm, and QR = 6 cm. Let us draw the diagonals of the rectangle. Let us write by calculating the position of the centre of the circum-circle of ΔPQR and the length of circum radius without drawing. By drawing circumcircle of ΔPQR, let us verify.
Answer:

In the rectangle PQRS, we join PR and QS. We see that PQR is a right-angled triangle. For a right angle triangle, the circumcenter is the midpoint of the hypotenuse and circumradius is half of the length of the hypotenuse. The hypotenuse PR = √(PQ)2 + (QR)2 = √(4)2 + (6)2 = 7.21cm.


Thus, circumradius = 7.21/2 = 3.6cm


Steps of Construction:


1. Construct rectangle PQRS of given dimensions.



2. Draw perpendicular bisector of PQ.



3. Draw perpendicular bisector of QR.



4. The point of intersection is the circumcenter. Name it as O.


5. With O as center and OR as radius, draw a circle.


6. The circle passes through P,Q and R and is thus the circumcircle of this triangle.


7. We measure the circumradius, OR = 3.6cm
Question 11.

If any circular picture is given, then how shall we find its centre? Let us find the centre of the circle in the adjoining figure.
Answer:


Steps of construction:

1. Draw a circle (Say C1), and take any three points A, B and C on its circumference



2. Join AB and BC and draw the perpendicular bisector of AB and BC which intersect each other at O.



3. O is the center of circle.
Let Us Work Out 11.2
Question 1.

Let us draw the following triangles and by drawing incircle of each circle, let us write by measuring the length of inradius in each case.

the lengths of three sides are 7 cm., 6 cm., and 5.5 cm.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. Name it as ΔABC.



3. Draw angle bisector of ∠A.



4. Draw angle bisector of ∠B.



5. They intersect at O.


6. Draw perpendicular on AB through O.



7. The perpendicular intersects with AB at P.


8. With O as center and OP as radius, draw a circle.



The inradius of the triangle, OP = 1.9cm.
Question 2.

Let us draw the following triangles and by drawing incircle of each circle, let us write by measuring the length of inradius in each case.

the length of two sides are 7.6 cm., 6 cm. and the angle included by those two sides is 75°.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. Name it as ΔABC.



3. Draw angle bisector of ∠A.



4. Draw angle bisector of ∠B.



5. They intersect at O.


6. Draw perpendicular on AB through O.



7. The perpendicular intersects with AB at P.


8. With O as center and OP as radius, draw a circle.



The inradius of the triangle, OP = 2cm.
Question 3.

Let us draw the following triangles and by drawing incircle of each circle, let us write by measuring the length of inradius in each case.

the length of a side is 6.2 cm. and measures of two angles adjacent to this side are 50° and 75°
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. Name it as ΔABC.



3. Draw angle bisector of ∠A.



4. Draw angle bisector of ∠B.



5. They intersect at O.


6. Draw perpendicular on AB through O.



7. The perpendicular intersects with AB at P.


8. With O as center and OP as radius, draw a circle.



9. The inradius of the triangle, OP = 1.8cm.
Question 4.

Let us draw the following triangles and by drawing incircle of each circle, let us write by measuring the length of inradius in each case.

a triangle is a right-angled triangle of which two sides containing the right angle have the lengths of 7 cm. and 9 cm.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. Name it as ΔABC.



3. Draw angle bisector of ∠A.



4. Draw angle bisector of ∠B.



5. They intersect at O.


6. Draw perpendicular on AB through O.



7. The perpendicular intersects with AB at P.


8. With O as center and OP as radius, draw a circle.



9. The inradius of the triangle, OP = 2.3cm.
Question 5.

Let us draw the following triangles and by drawing incircle of each circle, let us write by measuring the length of inradius in each case.

a triangle is a right-angled triangle of which the length of hypotenuse is 9 cm. and another side 6.5 cm.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. Name it as ΔABC.



3. Draw angle bisector of ∠A.



4. Draw angle bisector of ∠B.



5. They intersect at O.


6. Draw perpendicular on AB through O.



7. The perpendicular intersects with AB at P.


8. With O as center and OP as radius, draw a circle.



9. The inradius of the triangle, OP = 1.9cm.
Question 6.

Let us draw the following triangles and by drawing incircle of each circle, let us write by measuring the length of inradius in each case.

The triangle is an isosceles triangle having the 7.8 cm. length of base and the length of each equal side of 6.5 cm.
Answer:

Incircle – It is the circle inscribing the triangle and the point of intersection of angular bisectors of angles of triangle.


Steps in drawing the triangle:


● Draw a line segment of length 7.8cm and name it AB.



● Now take length of 6.5cm in protractor and make an arc from point A, and similarly from point B. Name the point of intersection as C.



Steps for incircle:


● For incentre make an arc from A on side AC and AB, From those arcs make two more arcs, and mark the point of intersection.



● Do the same from point B, and join both the points with vertices.


● Intersection of both the lines will give incentre.



● Distance from any side to incentre will be inradius.


● Inradius=2cm
Question 7.

Let us draw the following triangles and by drawing incircle of each circle, let us write by measuring the length of inradius in each case.

a triangle is an isosceles triangle, having the 10 cm. length of base and each of the equal angle is 45o
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. Name it as ΔABC.



3. Draw angle bisector of ∠A.



4. Draw angle bisector of ∠B.



5. They intersect at O.


6. Draw perpendicular on AB through O.



7. The perpendicular intersects with AB at P.


8. With O as center and OP as radius, draw a circle.



9. The inradius of the triangle, OP = 2.07cm.
Question 8.

Let us draw the following triangles and by drawing incircle of each circle, let us write by measuring the length of inradius in each case.

Let us draw an equilateral triangle having 7 cm. length of each side. By drawing the circumcircle and incircle, let us find the length of circum radius and inradius and let us write whether there is any relation between them.
Answer:

Steps of Construction:


1. Construct triangle of given dimensions.


2. Name it as ΔABC.



3. Draw angle bisector of ∠A.



4. Draw angle bisector of ∠B.



5. They intersect at O.


6. Draw perpendicular on AB through O.



7. The perpendicular intersects with AB at P.


8. With O as center and OP as radius, draw a circle.



9. The inradius of the triangle, OP = 2.02cm.


10. We know that the incenter and circumcenter of equilateral triangle is same.


11. With O as center and OC as radius, draw a circle.


12. The circumradius of triangle, OC = 4.04cm.



We observe that circumradius = 2×inradius.