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On January 25, 2024, 11:35 AM

Exercise 10.2 - Chapter 10 Practical Geometry class 7 ncert solutions Maths - SaraNextGen [2024]


Question 1:

Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

Answer:

The rough figure of this triangle is­ as follows.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1448/Chapter%2010_html_m53a16978.jpg

The required triangle is constructed as follows.

(i) Draw a line segment YZ of length 5 cm.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1448/Chapter%2010_html_64e9d8c4.jpg

(ii) Point X is at a distance of 4.5 cm from point Y. Therefore, taking point Y as centre, draw an arc of 4.5 cm radius.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1448/Chapter%2010_html_m57fec990.jpg

(iii) Point X is at a distance of 6 cm from point Z. Therefore, taking point Z as centre, draw an arc of 6 cm radius. Mark the point of intersection of the arcs as X. Join XY and XZ.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1448/Chapter%2010_html_m7c03097d.jpg

XYZ is the required triangle.

Question 2:

Construct an equilateral triangle of side 5.5 cm.

Answer:

An equilateral triangle of side 5.5 cm has to be constructed. We know that all sides of an equilateral triangle are of equal length. Therefore, a triangle ABC has to be constructed with AB = BC = CA = 5.5 cm.

The steps of construction are as follows.

(i) Draw a line segment BC of length 5.5 cm.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1449/Chapter%2010_html_m133259fb.jpg

(ii) Taking point B as centre, draw an arc of 5.5 cm radius.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1449/Chapter%2010_html_48902b57.jpg

(iii) Taking point C as centre, draw an arc of 5.5 cm radius to meet the previous arc at point A.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1449/Chapter%2010_html_m75e8ae02.jpg

(iv) Join A to B and C.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1449/Chapter%2010_html_m4f0be8cb.jpg

ABC is the required equilateral triangle.

Question 3:

Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of

triangle is this?

Answer:

The steps of construction are as follows.

(i) Draw a line segment QR of length 3.5 cm.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1458/Chapter%2010_html_m7a30b2a3.jpg

(ii) Taking point Q as centre, draw an arc of 4 cm radius.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1458/Chapter%2010_html_75b5b4f2.jpg

(iii) Taking point R as centre, draw an arc of 4 cm radius to intersect the previous arc at point P.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1458/Chapter%2010_html_7b85f23.jpg

(iv) Join P to Q and R.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1458/Chapter%2010_html_m56d2f71f.jpg

ΔPQR is the required triangle. As the two sides of this triangle are of the same length (PQ = PR), therefore, ΔPQR is an isosceles triangle.

Question 4:

Construct ΔABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.

Answer:

The steps of construction are as follows.

(i) Draw a line segment BC of length 6 cm.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1461/Chapter%2010_html_191fc863.jpg

(ii) Taking point C as centre, draw an arc of 6.5 cm radius.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1461/Chapter%2010_html_m6621f186.jpg

(iii) Taking point B as centre, draw an arc of radius 2.5 cm to meet the previous arc at point A.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1461/Chapter%2010_html_39ead55b.jpg

(iv) Join A to B and C.

https://img-nm.mnimgs.com/img/study_content/curr/1/7/3/40/1461/Chapter%2010_html_35f97070.jpg

ΔABC is the required triangle. ∠B can be measured with the help of protractor. It comes to 90º.

Also Read : Exercise-10.3-Chapter-10-Practical-Geometry-class-7-ncert-solutions-Maths

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