ML Aggarwal Solutions for Class 7 Maths Chapter 13 Practical Geometry consist of answers designed by subject experts at BYJU’S. The solutions clear the doubts of students instantly, which arise while practising textbook questions. Continuous practice of ML Aggarwal Solutions boosts confidence in solving complex problems with ease. Students can cross-check their answers while revising exercise problems before the final examination. This helps them analyse and understand the concepts they have covered. For more details, students are suggested to follow ML Aggarwal Solutions for Class 7 Maths Chapter 13 Practical Geometry PDF from the link available below.
Chapter 13 deals with the construction of parallel lines to a line, through a point not on the line and various other related topics on Practical Geometry. These solutions help students to improve their problem-solving and time-management skills, which are vital for exam preparation.
ML Aggarwal Solutions for Class 7 Maths Chapter 13 Practical Geometry
Access ML Aggarwal Solutions for Class 7 Maths Chapter 13 Practical Geometry
1. Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. If this line meets l at S, then what shape do the two sets of parallel lines inclose?
Solution:
Steps of Construction:
(i) Draw a line l and P be a point not on the line l
(ii) Take a point A on line l and join PA
(iii) Draw a line m which is parallel to line l on point P
(iv) Take a point Q on line l and join PQ
(v) From a point R on line m, draw a line parallel to PQ, which meets l at point S
Here,
We can observe that PQRS is a parallelogram.
2. Construct a triangle ABC, given that
(i) AB = 5 cm, BC = 6 cm and AC = 7 cm
(ii) AB = 4.5 cm, BC = 5 cm and AC = 6 cm
Solution:
(i) Steps of Construction:
1. Draw a line segment BC, such that BC = 6 cm
2. Taking B as the centre and radius 5 cm, and taking C as the centre and radius 7 cm, draw arcs which intersect each other at point A.
3. Now, join AB and AC.
Therefore,
△ABC is the required triangle.
(ii) Steps of Construction:
1. Draw a line segment BC, such that BC = 5 cm
2. Taking B as the centre and radius 4.5 cm, and taking C as the centre and radius 6 cm, draw arcs which intersect each other at point A.
3. Now, join AB and AC.
Therefore,
△ABC is the required triangle.
3. Construct a triangle PQR given that PQ = 5.4 cm, QR = PR = 4.7 cm. Name the triangle.
Solution:
Steps of Construction:
1. Draw a line segment PQ of length 5.4 cm
2. Taking P and Q as centres and radius 4.7 cm, draw two arcs intersecting each other at point R.
3. Join PR and QR.
Here, the two sides are equal in length, 4.7 cm.
Hence, the required △RPQ is an isosceles triangle.
4. Construct a triangle LMN such that the length of each side is 5.4 cm. Name the triangle.
Solution:
Steps of Construction:
1. Draw a line segment MN, such that MN = 5.4 cm
2. Taking M and N as centres and radius 5.4 cm, draw two arcs intersecting each other at point L.
Here, we can observe that all the sides of the triangle = 5.4 cm
Therefore,
The required △LMN is an equilateral triangle.
5. Construct a triangle ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠ABC and name the triangle.
Solution:
Steps of Construction:
1. Draw a line segment such that BC = 6 cm
2. Taking B as the centre and radius 2.5 cm, and taking C as the centre and radius 6.5 cm, draw two arcs intersecting each other at point A.
3. Join AB and AC.
Now, the △ABC is the required triangle.
On measuring, we get ∠ABC = 900
Hence, △ABC is a right-angled triangle.
6. Construct a triangle PQR, given that PQ = 3 cm, QR = 5.5 cm and ∠PQR = 600
Solution:
Steps of Construction:
1. Draw a line segment QR of length 5.5 cm.
2. Taking Q as the centre, draw a ray QX making an angel of 600
3. Now, cut off PQ = 3 cm
4. Join PR
Therefore,
△PQR is the required triangle.
7. Construct △DEF such that DE = 5 cm, DF = 3 cm and ∠EDF = 900
Solution:
Steps of Construction:
1. Draw a line segment DE of length 5 cm
2. Taking D as the centre, draw a ray DX making an angle of 900
3. Now, cut off DF = 3 cm
4. Join FE
△FDE is the required triangle.
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