WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 7 Construction Of Circumcircle And Incircle Of A Triangle

WBBSE Solutions For Class 10 Maths Solid Geometry Chapter 7 Construction Of Circumcircle And Incircle Of A Triangle

Circumcircle, circumcentre And Circumradius Of A Triangle

Circumcircle of a triangle

Definition: The circle passing through the three vertices of a triangle, is called the circumcircle of the triangle.

The circle having O as the centre and passing through the vertices A, B and C is the circumcircle of the triangle ΔABC.

Circumcentre of a triangle

Definition: The centre of the circumcircle of a triangle is called the circumcentre of the triangle.

∴ O is the circumcentre of the ΔABC. Again, ∵ O is the circumcentre of Δ ABC,

∴ the distances of all the points on the circumcircle, from O are equal. ∵ OA = OB = OC

Read and Learn More WBBSE Solutions for Class 10 Maths

i.e., ΔBOC, ΔCOA and ΔAOB are all isosceles triangles.

We know that the perpendicular drawn from any vertex of an isosceles triangle to its base bisects the base.

WBBSE Solutions for Class 10 History	WBBSE Solutions for Class 10 Geography and Environment
WBBSE Class 10 History Long Answer Questions	WBBSE Solutions for Class 10 Life Science And Environment
WBBSE Class 10 History Short Answer Questions	WBBSE Solutions for Class 10 Maths
WBBSE Class 10 History Very Short Answer Questions	WBBSE Solutions for Class 10 Physical Science and Environment
WBBSE Class 10 History Multiple Choice Questions

 

∴ The perpendiculars OD, OE and OF from O to the sides BC, CA and AB respectively of the ΔABC have bisected the corresponding bases and these perpendicular bisectors intersect each other at a point O.

Therefore, we can say that the perpendicular bisectors of the sides of a triangle are concurrent and the point at which the perpendicular bisectors intersect each other, is the circumcentre of the triangle.

Thus, we take the point at which any two perpendicular bisectors meet one another as the circumcentre of the triangle.

If the triangle be

1. an acute-angled triangle, then the circumcentre of the triangle lie inside the triangle.

2. a right-angled triangle, then the circumcentre of the triangle lie on the hypotenuse of the triangle and it bisects the hypotenuse, i.e., the mid-point of the hypotenuse of a right-angled triangle is its circumcentre and circumradius = \(\frac{1}{2}\) x length of the hypotenuse.

3. an obtuse-angled triangle, then the circumcentre of its circum- circle lie on the outside of the triangle.

Circumradius of a triangle

Definition: The radius of the circumcircle of a triangle is called the circumradius of the triangle.

i.e. the distance of any vertex of a triangle from the point at which the perpendicular bisectors of the sides of a triangle meet each other is called the circumradius of the triangle.

OA or OB or OC is called the circumradius of the ΔABC.

Construction of the circumcircle of a given acute triangle.

Let ΔABC be an acute triangle. We have to construct a circumcircle of this triangle.

Principle: To construct the circumcircle we have to take the point of intersection of the two perpendicular bisectors of any two sides of ΔABC as the centre and the distance of any vertex of the triangle from that centre has to be taken as the radius and then the circumcircle is drawn.

Method of construction:

1. Let us draw the perpendicular bisector PQ of the side BC of ΔABC.
2. Let us draw the perpendicular bisector RS of the side AB of ΔABC.
3. Let PQ and RS intersect each other at O.
4. Let us draw the circle with centre at O and radius equal to OA or OB or OC.

Then the circle with centre at O and radius OA or OB or OC is the required circumcircle of ΔABC.

Proof:

Let us join the points O, A; O, B and O, C.

Now, O lie on the perpendicular bisector of AB.

∴ the point O is equidistant from points A and B.

∴ OA = OB

Similarly, it can be proved that OB = OC.

∴ OA = OB = OC.

∴ the circle with a centre at O and radius O A must pass through the vertices A, B and C of ΔABC.

Hence that very circle is the required circumcircle of ΔABC.

From the above construction, we see that the circumcircle of any acute triangle lie within the triangle.

We shall now consider the case when the triangle is an obtuse angle.

“WBBSE Class 10 circumcircle and incircle solved examples”

Construction of the circumcircle of a given obtuse triangle.

Let ΔABC be an obtuse angle of which ∠A = obtuse. We have to construct the circumcircle of ΔABC.

Principle: The circle, drawn with centre at the point of intersection of two perpendicular bisectors of any two sides of ΔABC and the distance of any one of the vertices of ΔABC from that centre being taken as the radius, is the required circumcircle of the obtuse triangle.

Method of construction:

1. Let us draw the perpendicular bisector PQ of the side AC of ΔABC.
2. Let us draw the perpendicular bisector RS of the side AB of ΔABC.
3. Let PQ and RS intersect at point O.
4. Let us draw the circle with the centre at O and a radius equal to OA or OB or OC.

Then that very circle is the required circumcircle of the obtuse triangle.

Proof:

Let us join O, A; O, B and O, C.

Since O lie on the perpendicular bisector of AB, ∴ O is equidistant from the points A and B.

∴ OA = OB.

Similarly, it can be proved that OB = OC.

∴ OA = OB = OC.

∴ each of A, B and C is equidistant from the point O and the circle passes through A, B and C.

Hence the circle with a centre at O and radius equal to OA or OB or OC is the required circumcircle of ΔABC.

From the above construction, we see that the circumcentre of the ΔABC lie outside the triangle.

Hence we can say that the circumcentre of any obtuse triangle lie outside the triangle.

Now, if the triangle be a right-angled triangle, then where the circumcentre of the triangle lie we shall examine that in the following construction.

Construction of the circumcircle of a given right-angled triangle.

Let ΔABC be a right-angled triangle of which ∠A = right angle. We have to construct a circumcircle of ΔABC.

Method of construction:

1. Let us draw the perpendicular bisector EF of the side AB of ΔABC.
2. Let us draw the perpendicular bisector PQ of the side AC.
3. Let EF and PQ intersect each other at point O on the Side BC.
4. Let us draw a circle with a centre at O and radius equal to OA or OB or OC.

Then the circle with centre at O and a radius to OA or OB or OC is the required circumcircle of ΔABC.

Proof:

Let us join O, A.

O lies on the perpendicular bisector of AB.

the points A and B are equidistant from point O. ∴ OA = OB.

Similarly, it can be proved that OB = OC.

OA = OB = OC.

“Construction of circumcircle of a triangle for Class 10 Maths”

∴ the three vertices A, B and C of the AABC are equidistant from the point O, i.e. the circle passes through the vertices A, B and C of ΔABC.

Hence the circle with centre at O and a radius equal to OA or OB or OC is the required circumcircle of the ΔABC.

From the above construction, we see that the circumcentre of the right-angled triangle AABC lies on the hypotenuse BC of ΔABC, Again, since OB = OC,

∴ O is the mid-point of BC.

∴ We can say that the circumcentre of any right-angled triangle lie on the mid-point of its hypotenuse.

In the following examples we have discussed how to construct circumcircles of different types of triangles using the above constructions.

Solid Geometry Chapter 7 Construction Of Circumcircle And Incircle Of A Triangle Examples

“Chapter 7 circumcircle and incircle exercises WBBSE solutions”

Example 1. Construct the following triangles as directed and then construct the circumcircle in each case stating the location of the circumcentres. Also determine the length of the circumradius in each case.

Solution:

1. An equilateral triangle of sides 6 cm each.

we see that the circum-centre of the triangle lies inside the triangle and the length of the circumradius = 3.4 cm.

2. An isosceles triangle, the base of which is 5.2 cm and the length of each of the equal sides is 7cm.

It is clearly seen that the circumcentre of the triangle lies inside the triangle and the length of its circumradius = 3.75 cm (approx.).

3. An right-angled triangle, the lengths of whose two adjacent sides of right angle are 4 cm and 8 cm respectively.

We see that the circumcentre of the triangle lie on the mid-point of the hypotenuse and the circumradius = 4.5 cm (approx.)

4. A right-angled triangle, the hypotenuse of which is 12 cm and the length of any other side is 5 cm.

We see that the circumcentre of the right-angled triangle lies on the mid-point of its hypotenuse and the length of its circumradius = 6 cm.

“Class 10 Maths steps to draw circumcircle and incircle”

5. A triangle, the length of one of whose sides is 6.7 cm and the magnitudes of its two adjacent angles are 75° and 55° respectively.

We see that the circumcentre of the triangle lies inside the triangle and the length of its circumradius = 3.7 cm (approx.)

6. A triangle ABC of which BC = 5 cm, ∠ABC = 100° and AB = 4 cm.

We see that the circumcentre of the triangle lies outside the triangle and the length of the circumradius = 3.6 cm (approx.).

“Understanding circumcircle and incircle in triangles for Class 10”

Example 2. Given that PQ = 7.5 cm, ∠QPR = 45°, ∠PQR = 75°, ∠QPS = 60°, ∠PQS = 60°. Construct the triangle PQR and PQS in such a way that the points R and S lie on the same side of PQ. Then by drawing the circumcircle ofΕPQR, write the position of the point S within, on and outside the circumcircle. Also, explain your answer.

Solution:

Given:

PQ = 7.5 cm, ∠QPR = 45°, ∠PQR = 75°, ∠QPS = 60°, ∠PQS = 60°.

It is seen that the position of point S is on the circumcircle of ΔPQR, i.e., the points S and R are concyclic.

Explanation:

∠S = 180° – (∠SPQ + ∠QPS) = 180° – (60° + 60°)
= 180° – 120° = 60°

and ∠R = 180° – (∠RPQ + ∠RQP) = 180° – (45° + 75°)
= 180° – 120° = 60°.

∴ ∠S = ∠R.

i.e. ∠S and ∠R are the same angles in circle produced by the same arc PQ.

Hence S and R are concyclic.

Example 3.Given that AB = 5 cm, ∠BAC = 30°, ∠ABC = 60°, ∠BAD = 45°, ∠ABD = 45°. Construct two triangles ΔABC and ΔABD in such a way that point C and D lie on opposite sides of AB. By drawing the circumcircle of ΔABC write the position of the point D with respect to the circumcircle. Also state what other characteristics you are observing here.

Solution:

Given:

AB = 5 cm, ∠BAC = 30°, ∠ABC = 60°, ∠BAD = 45°, ∠ABD = 45°.

We see that the position of the point D is on the circumcircle of ΔABC. i.e., points C and D are concyclic.

Because, here ∠BAC = 30°, ∠ABC = 60°.

∠ACB =180° – (∠BAC + ∠ABC)

= 180° – (30° + 60°) = 180° – 90° = 90°.

∴ in ΔABC, ∠C = 90°, i.e., ∠C is a semicircular angle.

Similarly, ∠ABD = 45°, ∠BAD = 45°

∴ ∠ADB = 180° – (∠ABD + ∠BAD)

= 180° – (45° + 45°) = 180° – 90° = 90°

∴ ∠D = 90°, i.e., ∠D is a semicircular angle.

∴ ∠C + ∠D = 90° + 90° = 180°,

i.e., the sum of two opposite angles of the quadrilateral ABCD is 180°.

Hence the points C and D are concyclic.

“Step-by-step solutions for circumcircle and incircle Class 10”

Example 4. Given that AB = 4 cm, BC = 7 cm, CD = 4 cm, ∠ABC = 60°, ∠BCD = 60°. Construct the quadrilateral ABCD. Then construct the circumcircle of ΔABC and also state what other characteristics you observe.

Solution:

Given:

AB = 4 cm, BC = 7 cm, CD = 4 cm, ∠ABC = 60°, ∠BCD = 60°.

We see that point D lie on the circumcircle of ΔABC.

Also the quadrilateral ABCD is an isosceles trapezium of which AD II BC and AB = CD = 4 cm.

Example 5. Given that PQ = 4 cm, QR = 6 cm. Construct the rectangle PQRS. Also draw the diagonals of the rectangle and without drawing the circumcircle write the position of the centre of the circumcircle of ΔPQR and find the length of circumradius. After all by drawing the circumcircle of ΔPQR verify your answer.

Solution:

Given:

PQ = 4 cm, QR = 6 cm.

The centre of the circumcircle of APQR will be the point of intersection of the diagonals of the rectangle PQRS.

The length of the circumradius = \(\frac{\sqrt{4^2+6^2}}{2}\) cm = \(\frac{\sqrt{52}}{2}\) cm = \(\frac{\sqrt{4 \times 13}}{2}\) cm = \(\frac{2 \sqrt{13}}{2}\) cm = √13 cm

Example 6. If any circular picture is given, then how shall you find its centre? Find the centre of the circle in the adjoining figure.

Solution: We can find the centre of any given circular figure in the following manner:

1. At first let us draw any two chords of any length of the given circle. Let PQ and RS be two such chords of the circle.
2. Let us then draw the perpendicular bisectors of the two chords PQ and RS. Let the bisectors are AB and CD respectively.
3. The point of intersection at which the two perpendicular bisectors AB ad CD of the chords PQ and RS respectively intersect will be the required centre of the circle. Let the perpendicular bisectors AB and CD intersect each other at the point O.

Hence O is the centre of the circle.

Example 7. By drawing the following triangles construct their circumcircles

Solution:

1. The lengths of any two sides of the triangle are 5 cm and 6 cm and their internal angle is 60°.

2. The length of one of the sides of the triangle is 5.4 cm and the two adjacent angles of that side are 60° and 45°.

3. The length of one of the sides of a right-angled triangle is 8 cm and the length of its hypotnuse is 10 cm.

4. The lengths of three sides of the triangle are 5.5 cm, 6.6 cm and 7.7 cm.

“WBBSE Mensuration Chapter 7 practice questions on circles”

Example 8. By constructing the circumcircle of an equilateral triangle of sides 6 cm each, determine the position of the circumcentre and the length of circumradius.

Solution:

We see that the circumcentre of the ΔABC entirely lies inside the triangle and the circumradius = 3.5 cm.

Example 9. Construct an angle of measure 120° without any help of the protractor. Then draw the triangle PQR, where ∠P = 120°. PQ = 4 cm and PR = 3 cm. Also construct the circumcircle of ΔPQR.

Solution: ∠ABC = 120° which has been drawn without any help of a protractor.

We see that the circumcentre of the triangle lies outside the triangle and the circumradius = 3.5 cm.

Example 10. Construct a triangle of sides 5 cm, 7.5 cm and 4 cm. By constructing the circumcircle of the triangle determine the circumradius of the triangle.

Solution: we see that the circumradius = 4 cm.

Example 11. Construct a right-angled triangle of which the two adjacent sides of the right angle are 11 cm and 4.5 cm respectively. Also by drawing the circumcircle of this triangle determine the circumradius.

Solution:

We see that the circumcentre of the ΔABC lies on the mid-point of its hypotenuse AC and the circumradius.

= \(\frac{\sqrt{4.5^2+11^2}}{2}\) cm = 5.94 cm (approx)

Solid Geometry Chapter 7 Construction Of Circumcircle And Incircle Of A Triangle Incircle, Incentre And Inradius Of A Triangle

“Examples of triangle circumcircle and incircle constructions for Class 10”

Incircle

Definition: The circle which completely lie inside any triangle and its circumference just touches all three sides of the triangle is called the incircle of the triangle.

The incentre of the incircle of any triangle is usually denoted by I. In the given figure beside, the circle with the centre I.

1. lie completely inside the ΔABC and
2. its circumference have just touched the sides BC, CA and AB at points D, E and F respectively.

∴ the circle DEFD with centre at I is a incircle of the ΔABC.

Incentre

Definition: The centre of the incircle of any triangle is called the incentre of the triangle.

I is the incentre of the ΔABC.

Now, the circle have touched BC at D, ∴ ID ⊥ BC.

Similarly, IE ⊥ CA and IF ⊥ AB.

Since the distances of any point on the circumference of a circle from its centre are all equal, we have, ID = IE = IF ……(1)

Then, in the triangles ΔAEI and ΔAFI, IE = IF  [∵ by (1)]

∠AEI = ∠AFI [∵ each is a right angle] and AI is common to both.]

∴ ΔAEI ≅ ΔAFI [∵ by the condition of R-H-S congruency ]

∴ ∠ IAE = ∠LAF [∵ they are the similar angles of congruent triangles ]

∴ AI, is the bisector of ∠A.

Similarly, it can be proved that BI and Cl are the bisectors of ∠B and ∠C. respectively.

Therefore, the incentre of the incircle of any ΔABC lie on the bisectors of its angles and they are concurrent.

Hence, the point at which the bisectors of the angles of a triangle intersect is called the incentre of the triangle.

The incentre of any type of triangle lie inside it.

Since the incentre of any triangle lie on the bisectors of its angles,

∴ To find the incentre of any triangle we take the point as incentre at which any two bisectors of its angles intersect and to draw the incircle we take the perpendicular distance of any of its side from this incentre as its radius.

Inradius

Definition: The radius of the incircle of any triangle is called its inradius, i.e., the perpendicular distance of any of the three sides of a triangle from the point at which the bisectors of its angles intersect is called the incentre of the triangle.

ID, IE and IF are all the inradii of ΔABC. Clearly, ID = IE = IF and ID ⊥BC, IE ⊥ CA and IF ⊥ AB.

Ex-circle, Ex-centre and Ex-radius of a triangle

1. Ex-circle: Taking the point of intersection of the bisectors of any two exterior angles of any triangle and the bisector of the third interior angle, as centre and the perpendicular distance of any of its sides from that point as the radius we can draw a circle, which is called the ex-circle of the triangle.

Given beside,the circle DHGD with centre E is a excircle of the ΔABC.

2. Ex-centre: The centre of the ex-circle of any triangle is called its ex-centre.

In besides, E is the ex-centre of the ΔABC.

The perpendicular distance of any side of a triangle from the point of intersection at which the external bisectors of any two angles of the triangle and the internal bisector of the third angle intersect is called the ex-radius of the triangle.

Besides, ED or EG or EH are the ex-radius of the ΔABC. Clearly, ED = EG = EH.

From the above discussion, we see that the incircle and incentre of any type of triangle always lie inside the triangle.

We shall now discuss how the incircle of a triangle is constructed.

Construction of incircle of a given acute triangle.

Let ΔABC be an acute triangle. We have to construct the incircle of this triangle.

Principle: The circle with centre at the point of intersection of the two internal bisectors of any two angles of the triangle and radius as the perpendicular distance a of any side from that centre will be the incircle of the triangle.

Method of construction:

1. Let us draw the triangle as per the given measurement.
2. Now, let us draw the internal bisectors of the angles of ΔABC. Let the bisectors of the angles intersect each other at the point I.
3. A perpendicular ID is drawn from point I to the side BC which intersects BC at point D.
4. Now, let us draw a circle with centre at I and a radius equal to ID, which touches the sides BC, CA and AB of the ΔABC at the points D, E and F respectively.

Hence the circle with centre I and radius ID will be required to incircle of the triangle.

∴ If the triangle be an obtuse or a right-angled triangle.

Construction of incircle of a given obtuse triangle.

Let ΔABC be an obtuse angle of which ∠B is obtuse. We have to construct the incircle of this triangle.

Method of construction:

1. Let us construct the triangle as per the given measurement.
2. Let us draw the internal bisectors of ∠BAC and ∠ACB. Let the bisectors intersect each other at the point I.
3. Let us draw ID perpendicular to BC from the point I, which intersects the side BC at D.
4. Now, let us draw a circle with the centre at I and with radius equal to ID.

Hence that very circle is the required incircle of the ΔABC.

Now we shall discuss how the incircle of a given right-angled triangle is constructed.

Construction of incircle of a given right-angled triangle.

Let ΔABC be a right-angled triangle of which ∠B = right angle. We have to construct an incircle of the ΔABC.

Method of construction:

1. Let us construct the triangle ABC of which ∠B = right angle.
2. Let us draw the internal bisectors of ∠BAC and ∠ACB which intersect each other at the point I.
3. Now, let us draw a perpendicular ID on the side AC from the point I, which intersects AC at point D.
4. Let us then draw a circle with centre at I and with radius equal to ID.

Hence that very circle is the required incircle of the ΔABC.

In the following examples how incircles of different triangles are constructed is discussed.

Solid Geometry Chapter 7 Construction Of Circumcircle And Incircle Of A Triangle Incircle, Incentre And Inradius Of A Triangle Examples

“WBBSE Class 10 Maths solved problems on circumcircle and incircle”

Example 1. The lengths of three sides of a triangle are 7 cm, 6 cm and 5.5 cm. Construct the triangle and determine the length of the inradius of that triangle.

Solution:

Given:

The lengths of three sides of a triangle are 7 cm, 6 cm and 5.5 cm.

Let the lengths of the sides AB, BC and CA of ΔABC are 7 cm, 6 cm and 5.5 cm respectively. We have to construct the triangle.

 

The circle with centre at I is the required incircle of ΔABC and the length of its inradius = 1.7 cm.

Example 2. The length of two sides of a triangle are 7.6 cm and 6 cm and the internal angle between these two sides is 75°. Construct the triangle and then by drawing the incircle of this triangle, determine its inradius of it.

Solution:

Given:

The length of two sides of a triangle are 7.6 cm and 6 cm and the internal angle between these two sides is 75°.

The circle with centre I is the required incircle of the ΔABC and the inradius of this triangle = 2 cm.

The inradius of this triangle = 2 cm.

Example 3. The lengths of two adjacent sides of the right angle of a right-angled triangle are 7 cm and 9 cm. Construct the right-angled triangle. Also construct the incircle of this triangle.

Solution:

Given:

The lengths of two adjacent sides of the right angle of a right-angled triangle are 7 cm and 9 cm.

The circle with centre at I is the required incircle of the ΔABC and the inradius of it = 2.3 cm (approx.)

Example 4. Construct a right-angled triangle, the hypotenuse of which is 11.4 cm and the length of another side is 9 cm. Then by constructing the incircle of this triangle find its inradius of it.

Solution:

Let ΔABC be a right-angled triangle of which ∠A = right angle, hypotenuse BC = 11.4 cm and another side AC = 9 cm.

We have to construct the triangle.

The circle with centre I is the required incircle of the ΔABC and the inradius = 2.35 cm (approx.)

Example 5. The base of an isosceles triangle is 10 cm and each of the equal angles is 45°. By drawing this triangle, construct the incircle of it and hence determine its inradius.

Solution:

Given:

The base of an isosceles triangle is 10 cm and each of the equal angles is 45°.

The circle with centre at I is the required incircle of the ΔABC and the inradius of it = 2 cm (approx.)

Example 6. Construct an equilateral triangle of sides 7 cm each. By constructing its circumcircle and incircle, determine whether there is any relation between their circumradius and inradius or not.

Solution:

The circle with centre at O and passing through the vertices A, B and C of ΔABC is the required circumcircle of the ΔABC and its circumradius = 4 cm.

Again, the circle with centre at O and passing through the points D, E and F is the required incircle of the ΔABC and its inradius = 2 cm.

Hence the circumradius of ΔABC is double of its inradius, which is the required relation.

Example 7. The base of an isosceles triangle is 8.4 cm and the length of each of its equal sides is 9 cm. Construct this triangle and then by drawing its incircle, determine its inradius.

Solution:

Given:

The base of an isosceles triangle is 8.4 cm and the length of each of its equal sides is 9 cm.

The circle with centre I is the required incircle of ΔABC and its inradius = 2.5 cm

Example 8. The two adjacent sides of the right angle of a right-angled triangle are 3 cm and 4 cm. Construct the triangle and then by drawing its incircle, determine the inradius of this triangle.

Solution:

Given:

The two adjacent sides of the right angle of a right-angled triangle are 3 cm and 4 cm.

Let ΔABC be a right-angled triangle of which ∠A = right angle. AB and AC are two adjacent sides of its right angle ∠A, the lengths of which one 3 cm and 4 cm respectively. We have to construct this triangle.

The circle with centre I is the required incircle of the ΔABC and its inradius = 1 cm (approx.)

Example 9. The length of the hypotenuse of a right-angled triangle is 13 cm and its another side is 5 cm. Construct the triangle and then by drawing the incircle of this triangle determine its in radius.

Solution:

Given:

The length of the hypotenuse of a right-angled triangle is 13 cm and its another side is 5 cm.

Let the hypotenuse AC = 13 cm of the ΔABC and the other side BC = 5 cm. We have to construct this triangle.

The circle with centre I is the required incircle of the ΔABC and its inradius = 2 cm (approx.)

“Circumradius and inradius calculations for triangles in Class 10 Maths”

Example 10. Three sides of a triangle are 5 cm, 6 cm and 9 cm respectively. Construct the triangle and then by drawing its incircle, determine the inradius of this triangle.

Solution:

Given:

Three sides of a triangle are 5 cm, 6 cm and 9 cm respectively.

Let in the ΔABC, AB = 5 cm, AC = 6 cm and BC = 9 cm. We have to construct the triangle.

The circle with the centre at I is the required incircle of the ΔABC and its radius = 1.5 cm