RBSE Solutions for Class 7 Maths Chapter 10 Construction of Triangles In Text Exercise is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles In Text Exercise.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 7 |
| Subject | Maths |
| Chapter | Chapter 10 |
| Chapter Name | Construction of Triangles |
| Exercise | In Text Exercise |
| Number of Questions | 11 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles In Text Exercise
Do and Learn
(Page 123)
Question 1
Construct a ∆XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.
Solution:
Steps of construction:
1. Draw a line (RBSESolutions.com) segment of length XY = 4.5 cm.
2. Draw an arc of radius 6 cm from point X
3. Draw an arc of radius 5 cm from point Y, which cuts the prior arc at point Z.
4. Join XZ and YZ. Thus AXYZ is required triangle.
Question 2
Construct an equilateral (RBSESolutions.com) triangle of side 5.5 cm.
Solution:
Steps of construction:
1. Draw a line segment of length AB = 5.5 cm.
2. Draw an arc of radius 5.5 cm from A and B each which intersect each other at point C.
3. Join AC and BC.
Thus ∆ABC is required equilateral triangle.
Question 3
Draw a ∆PQR with PQ = 4 cm, QR = 3. 5 cm and PR = 4. What type (RBSESolutions.com) of triangle is this?
Solution:
Steps of construction:
1. Draw a line segment PQ = 4.
2. Draw (RBSESolutions.com) an arc of radius 4 cm from point P.
3. Draw another arc of radius 3.5 cm which intersects prior arc at point R.
4. Join PR and QR.
Thus ∆PQR is required tirangle.
also ∵ PQ = QR = 4 cm.
∴ Thus ∆PQR is an isosceles triangle.
(Page 125)
Question 1
(i) Construct a ∆DEF such (RBSESolutions.com) that DE = 5 cm, DF = 3 cm and ∠EDF = 90°.
Solution:
Steps of construction:
1. Draw a line segment DE = 5 cm.
2. Make an angle of 90° at point D.
3. Make an arc of 3 cm from point D which cuts the line of angle at point F.
4. Join EF.
Thus ∆DEF is required triangle.
Q. ii. Construct an isosceles triangle is (RBSESolutions.com) which the lengths of each of its equal side is 6.5 cm and angle between them is 110°.
Solution:
Steps of construction:
1. Draw a line segment BC = 6.5 cm
2. Draw a ray A from point C, making an angle of 110°.
3. Make an arc of 6.5 cm from point B so that it (RBSESolutions.com) intersects the ray BX at point A.
4. Join AC
Thus ΔABC is required triangle.
Q. iii. Construct ∆ABC with BC = 7.5 cm, AC = 5 cm and ∠C = 60°.
Solution:
Steps of construction:
1. Draw a line segment BC= 7.5 cm
2. Draw a ray from point C, making an angle of 60°.
3. Make an (RBSESolutions.com) arc of 5 cm from point C which inctrsects the ray CX at point A.
4. Join A to B.
Thus ∆ABC is required triangle.
(Page 126)
Question 1
Construct ∆ABC, given ∠A = 60°, ∠B = 30° and AB = 5.8.
Solution:
Steps of construction:
1. Draw a line (RBSESolutions.com) segment AB = 5.8 cm
2. Make an angle of ∠BAX = 60° at point A.
3. Make an angle of ∠ABY = 30° at point B.
4. AX and BY intersect each other at point C.
Thus ∆ABC is required triangle.
Question 2
Construct ∆PQR if PQ = 5 cm, ∠PQR = 105° and ∠QRP = 40°.
(Hint : Recall angle sum property of a triangle)
Solution:
Here PQ = 5 cm ∠Q = 105°, ∠R = 40°
∴ ∠P = ?
we know that
∠P + ∠Q + ∠R = 180°
⇒ ∠P + 105° + 40° = 180°
⇒ ∠P + 145° = 180°
⇒ ∠P = 180° – 145° = 35°
PQ = 5 cm ∠P = 35° and ∠Q= 105°
Now ∆PQR can (RBSESolutions.com) be constructed.
Steps of construction :
1. Draw a line segment PQ = 5 cm
2. Draw a ray PX making an angle of 35° at P.
3. Draw a ray QY making an angle of 105° at Q.
4. PX and QY intersect at R.
Thus ∆PQR is a required triangle.
Question 3
Examine whether you can (RBSESolutions.com) construct ∆DEF such that EF = 7.2 cm, ∠E = 110° and ∠F = 80°, justify your answer.
Solution:
Given that
∠E = 110° and ∠F=80°
∵ ∠E + ∠F = 110°+ 80°= 190°
∵ Sum of only two angles is 190° which is greater than 180° (sum of all three angles)
∴ Construction of ∆DEF is not possible.
(Page 128)
Question 1
Construct the right angled (RBSESolutions.com) triangle ∆PQR, where ∠Q = 90°, QR = 8 cm and PR = 10 cm.
Solution:
Steps of construction :
1. Draw QR = 8 cm.
2. Make ∠XQR = 90° at point Q.
3. Make an arc of 10 cm from point R which cuts QX at point P.
4. Join RP.
Thus ∆PQR is required triangle
Question 2
Construct a right-angled triangle whose (RBSESolutions.com) hypotenuse is 6 cm long and one of the legs is 4 cm long.
Solution:
Steps of construction :
1. Draw QR = 4 cm
2. Make ∠XQR = 90° at point Q.
3. Make an arc of 6 cm (RBSESolutions.com) from point Q which cuts QX at point P.
4. Join RP
Thus ∆PQR is required triangle.
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