# Selina Solutions Concise Maths Class 7 Chapter 15: Triangles Exercise 15C

Selina Solutions Concise Maths Class 7 Chapter 15 Triangles Exercise 15C has stepwise answers, which help in the construction of triangles. The solutions contain explanations in simple language, based on the understanding abilities of students. Important formulas, shortcut tricks and steps are covered in the PDF to boost the exam preparation of students. The solutions created are 100% accurate, as per the ICSE exam pattern. In order to improve the academic performance, Selina Solutions Concise Maths Class 7 Chapter 15 Triangles Exercise 15C, PDF can be downloaded from the links which are provided here.

## Selina Solutions Concise Maths Class 7 Chapter 15: Triangles Exercise 15C Download PDF

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### Access Selina Solutions Concise Maths Class 7 Chapter 15: Triangles Exercise 15C

#### Exercise 15C page: 185

1. Construct a âˆ†ABC such that:
(i) AB = 6 cm, BC = 4 cm and CA = 5.5 cm
(ii) CB = 6.5 cm, CA = 4.2 cm and BA = 51 cm
(iii) BC = 4 cm, AC = 5 cm and AB = 3.5 cm

Solution:

(i) Steps of Construction

1. Construct a line segment BC = 4 cm.

2. Taking B as centre and 6 cm as radius construct an arc.

3. Taking C as centre and 5.5 cm as radius construct another arc which intersects the first arc at the point A.

4. Now join AB and AC.

Therefore, âˆ†ABC is the required triangle.

(ii) Steps of Construction

1. Construct a line segment CB = 6.5 cm

2. Taking C as centre and 4.2 cm as radius construct an arc.

3. Taking B as centre and 5.1 cm as radius construct another arc which intersects the first arc at the point A.

4. Now join AC and AB.

Therefore, âˆ†ABC is the required triangle.

(iii) Steps of Construction

1. Construct a line segment BC = 4 cm

2. Taking B as centre and 3.5 cm as radius construct an arc.

3. Taking C as centre and 5 cm as radius construct another arc which intersects the first arc at the point A.

4. Now join AB and AC.

Therefore, âˆ†ABC is the required triangle.

2. Construct a âˆ† ABC such that:
(i) AB = 7 cm, BC = 5 cm and âˆ ABC = 60Â°
(ii) BC = 6 cm, AC = 5.7 cm and âˆ ACB = 75Â°
(iii) AB = 6.5 cm, AC = 5.8 cm and âˆ A = 45Â°

Solution:

(i) Steps of Construction

1. Construct a line segment AB = 7 cm.

2. At the point B construct a ray which makes an angle 600 and cut off BC = 5cm.

3. Now join AC.

Therefore, âˆ†ABC is the required triangle.

(ii) Steps of Construction

1. Construct a line segment BC = 6 cm.

2. At the point C construct a ray which makes an angle 750 and cut off CA = 5.7 cm.

3. Now join AB.

Therefore, âˆ†ABC is the required triangle.

(iii) Steps of Construction

1. Construct a line segment AB = 6.5 cm.

2. At the point A construct a ray which makes an angle 450 and cut off AC = 5.8 cm.

3. Now join CB.

Therefore, âˆ†ABC is the required triangle.

3. Construct a âˆ† PQR such that :
(i) PQ = 6 cm, âˆ Q = 60Â° and âˆ P = 45Â°. Measure âˆ R.
(ii) QR = 4.4 cm, âˆ R = 30Â° and âˆ Q = 75Â°. Measure PQ and PR.
(iii) PR = 5.8 cm, âˆ P = 60Â° and âˆ R = 45Â°.
Measure âˆ Q and verify it by calculations

Solution:

(i) Steps of Construction

1. Construct a line segment PQ = 6 cm.

2. At point P construct a ray which makes an angle 450.

3. At point Q construct another ray which makes an angle 600 which intersect the first ray at point R.

Therefore, âˆ† PQR is the required triangle.

By measuring âˆ R = 750.

(ii) Steps of Construction

1. Construct a line segment QR = 4.4 cm.

2. At point Q construct a ray which makes an angle 750.

3. At point R construct another ray which makes an angle 300 which intersect the first ray at point R.

Therefore, âˆ† PQR is the required triangle.

By measuring the length, PQ = 2.1 cm and PR = 4.4 cm.

(iii) Steps of Construction

1. Construct a line segment PR = 5.8 cm.

2. At point P construct a ray which makes an angle 600.

3. At point R construct another ray which makes an angle 450 which intersect the first ray at point Q.

Therefore, âˆ† PQR is the required triangle.

By measuring âˆ Q = 750.

Verification â€“

âˆ P + âˆ Q + âˆ R = 1800

Substituting the values

600Â + âˆ Q + 450 = 1800

By further calculation

âˆ Q = 180 â€“ 105 = 750

4. Construct an isosceles âˆ† ABC such that:
(i) base BC = 4 cm and base angle = 30Â°
(ii) base AB = 6.2 cm and base angle = 45Â°
(iii) base AC = 5 cm and base angle = 75Â°.
Measure the other two sides of the triangle.

Solution:

(i) Steps of Construction

In an isosceles triangle the base angles are equal

1. Construct a line segment BC = 4 cm.

2. At the points B and C construct rays which makes an angle 300 intersecting each other at the point A.

Therefore, âˆ† ABC is the required triangle.

By measuring the equal sides, each is 2.5 cm in length approximately.

(ii) Steps of Construction

In an isosceles triangle the base angles are equal

1. Construct a line segment AB = 6.2 cm.

2. At the points A and B construct rays which makes an angle 450 intersecting each other at the point C.

Therefore, âˆ† ABC is the required triangle.

By measuring the equal sides, each is 4.3 cm in length approximately.

(iii) Steps of Construction

In an isosceles triangle the base angles are equal

1. Construct a line segment AC = 5 cm.

2. At the points A and C construct rays which makes an angle 750 intersecting each other at the point B.

Therefore, âˆ† ABC is the required triangle.

By measuring the equal sides, each is 9.3 cm in length approximately.

5. Construct an isosceles âˆ†ABC such that:
(i) AB = AC = 6.5 cm and âˆ A = 60Â°
(ii) One of the equal sides = 6 cm and vertex angle = 45Â°. Measure the base angles.
(iii) BC = AB = 5-8 cm and ZB = 30Â°. Measure âˆ A and âˆ C.

Solution:

(i) Steps of Construction

1. Construct a line segment AB = 6.5 cm.

2. At point A construct a ray which makes an angle 600.

3. Now cut off AC = 6.5cm

4. Join BC.

Therefore, âˆ† ABC is the required triangle.

(ii) Steps of Construction

1. Construct a line segment AB = 6 cm.

2. At point A construct a ray which makes an angle 450.

3. Now cut off AC = 6cm

4. Join BC.

Therefore, âˆ† ABC is the required triangle.

By measuring âˆ B and âˆ C, both are equal toÂ 67 Â½ 0.

(iii) Steps of Construction

1. Construct a line segment BC = 5.8 cm.

2. At point B construct a ray which makes an angle 300.

3. Now cut off BA = 5.8cm

4. Join AC.

Therefore, âˆ† ABC is the required triangle.

By measuring âˆ C and âˆ A is equal to 750.

6. Construct an equilateral triangle ABC such that:
(i) AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB and PC.
(ii) Each side is 6 cm.

Solution:

(i) Steps of Construction

1. Construct a line segment AB = 5cm.

2. Taking A and B as centres and 5 cm radius, construct two arcs which intersect each other at the point C.

3. Now join AC and BC where âˆ† ABC is the required triangle.

4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point p.

5. Join PA, PB and PC.

By measuring each is 2.8 cm.

(ii) Steps of Construction

1. Construct a line segment AB = 6cm.

2. Taking A and B as centres and 6 cm radius, construct two arcs which intersect each other at the point C.

3. Now join AC and BC

Therefore, âˆ† ABC is the required triangle.

7. (i) Construct a âˆ† ABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct a circumcircle of this triangle.
(ii) Construct an isosceles âˆ†PQR such that PQ = PR = 6.5 cm and âˆ PQR = 75Â°. Using ruler and compasses only construct a circumcircle to this triangle.
(iii) Construct an equilateral triangle ABC such that its one side = 5.5 cm.
Construct a circumcircle to this triangle.

Solution:

(i) Steps of Construction

1. Construct a line segment BC = 4.5 cm.

2. Taking B as centre and 6 cm radius construct an arc.

3. Taking C as centre and 5.5 cm radius construct another arc which intersects the first arc at point A.

4. Now join AB and AC

Therefore, âˆ† ABC is the required triangle.

5. Construct a perpendicular bisector of AB and AC which intersect each other at the point O.

6. Now join OB, OC and OA.

7. Taking O as centre and radius OA construct a cirlce which passes through the points A, B and C.

This is the required circumcircle of âˆ† ABC.

(ii) Steps of Construction

1. Construct a line segment PQ = 6.5 cm.

2. At point Q, construct an arc which makes an angle 750.

3. Taking P as centre and radius 6.5 cm construct an arc which intersects the angle at point R.

4. Join PR.

âˆ† PQR is the required triangle.

4. Construct the perpendicular bisector of sides PQ and PR which intersects each other at the point O.

5. Join OP, OQ and OR.

6. Taking O as centre and radius equal to OP or OQ or OR construct a circle which passes through P, Q and R.

This is the required circumcircle of âˆ† PQR.

(iii) Steps of Construction

1. Construct a line segment AB = 5.5 cm.

2. Taking A and B as centres and radius 5.5 cm construct two arcs which intersect each other at point C.

3. Now join AC and BC.

âˆ† ABC is the required triangle.

4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point O.

5. Now join OA, OB and OC.

6. Taking O as centre and OA or OB or OC as radius construct a circle which passes through A, B and C.

This is the required circumcircle.

8. (i) Construct a âˆ†ABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle and measure its radius.
(ii) Construct an isosceles âˆ† MNP such that base MN = 5.8 cm, base angle MNP = 30Â°. Construct an incircle to this triangle and measure its radius.
(iii) Construct an equilateral âˆ†DEF whose one side is 5.5 cm. Construct an incircle to this triangle.
(iv) Construct a âˆ† PQR such that PQ = 6 cm, âˆ QPR = 45Â° and angle PQR = 60Â°. Locate its incentre and then draw its incircle.

Solution:

(i) Steps of Construction

1. Construct a line segment AB = 6 cm.

2. Taking A as centre and 6.5 cm as radius and B as centre and 5.6 cm as radius construct arcs which intersect each other at point C.

3. Now join AC and BC.

4. Construct the angle bisector of âˆ A and âˆ B which intersect each other at point I.

5. From the point I construct IL which is perpendicular to AB.

6. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ†ABC internally.

By measuring the required incircle the radius is 1.6 cm.

(ii) Steps of Construction

1. Construct a line segment MN = 5.8 cm.

2. At points M and N construct two rays which make an angle 300 each intersecting each other at point P.

3. Construct the angle bisectors of âˆ M and âˆ N which intersect each other at point I.

4. From the point I draw perpendicular IL on MN.

5. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ† PMB internally.

By measuring the required incircle the radius is 0.6 cm.

(iii) Steps of Construction

1. Construct a line segment BC = 5.5 cm.

2. Taking B and C as centres and 5.5 cm radius construct two arcs which intersect each other at point A.

3. Now join AB and AC.

4. Construct the perpendicular bisectors of âˆ B and âˆ C which intersect each other at the point I.

5. From the point I construct IL which is perpendicular to BC.

6. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ†ABC internally.

This is the required incircle.

(iv) Steps of Construction

1. Construct a line segment PQ = 6 cm.

2. At the point P construct rays which make an angle of 450 and at point Q which makes an angle 600Â thats intersects each other at point R.

3. Construct the bisectors of âˆ P and âˆ Q which intersect each other at point I.

4. From the point I construct IL which is perpendicular to PQ.

5. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ†PQR internally.

This is the required incircle where the point I is incentre.