MP 7 Maths

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.2

Question 1.
Construct ΔXYZ in which XY = 4.5 cm YZ = 5 cm and ZX = 6 cm.
Answer:
Steps of Construction
(i) First we draw a rough sketch with given measure.


(ii) Draw a line segment YZ = 5 cm
(iii) With centre Y and radius 4.5 cm, draw an arc.
(iv) With centre Z and radius 6 cm, draw another arc to cut the previous arc at X.
(v) Join XY and XZ.
Thus, ΔXYZ is the required triangle.

Question 2.
Construct an equilateral triangle of side 5.5 cm.
Answer:
Steps of Construction

(i) First we draw a rough sketch with given measure.
(ii) Draw a line segment, BC = 5.5 cm.
(iii) With B as centre, 5.5 cm as radius draw an arc.
(iv) With C as centre, 5.5 cm as radius draw an arc to cut the previous arc at A.
(v) Join AB and AC.
Thus, ΔABC 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:
Steps of Construction
(i) First we draw a rough sketch with given measure.
(ii) Draw a line segment QR = 3.5 cm.

(iii) With Q as centre, 4 cm as radius, draw an arc.
(iv) With R as centre, 4 cm as radius, draw an arc to cut the previous arc at P.
(v) Join PQ and PR.
Thus, ΔPQR is the required triangle since PQ = PR = 4 cm.
.’. ΔPQR is an isosceles triangle.

Question 4.
Construct AABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Answer:
Steps of Construction

(i) First we draw a rough sketch with the given measure.
(ii) Draw a line segment BC = 6 cm.
(iii) With B as centre and 2.5 cm as radius, draw an arc.
(iv) With C as centre and 6.5 cm as radius, draw an arc to cut the previous arc at A.
(v) Join AB and AC.
Thus, ΔABC is the required triangle. On measuring, we find that ∠B = 90°.

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