Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 6 Answers Solutions Chapter 1 Geometrical Constructions.

## Geometrical Constructions Class 7 Practice Set 6 Answers Solutions Chapter 1

Question 1.

Write the names of pairs of congruent line segments. (Use a divider to find them.)

i. ___

ii. ___

iii. ___

iv. ___

Solution:

i. seg BG ≅ seg CG

ii. seg NG ≅ seg MG ≅ seg EG ≅ seg RG

Question 2.

On the line below, the distance between any two adjoining points shown on it is equal. Hence, fill in the blanks.

i. seg AB ≅ seg ___

ii. seg AP ≅ seg ___

iii. seg AC ≅ seg ___

iv. seg ___ ≅ seg BY

v. seg __ ≅ seg YQ

vi. seg BW ≅ seg ___

Solution:

i. BC

ii. QW

iii. QZ

iv. AZ

v. AY

vi. AC

Note: The above problem has many solutions. Students may write solutions other than the ones given.

**Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 6 Intext Questions and Activities**

Question 1.

Try to draw triangles with the following data. Can you draw these triangles. If not, look for the reason why you could not draw so. (Textbook pg. no. 7)

i. ∆ABC in which m∠A = 85°, m∠B = 115°, l(AB) = 5cm.

Solution:

m∠A + m∠B = 85° + 115°

= 200°>180°

But the sum of the measures of the angles of a triangle is 180°

Hence, ∆ABC cannot be drawn.

ii. ∆PQR in which l(QR) = 2cm, l(PQ) = 4cm, l(PR) = 2cm.

Solution:

l(QR) + l(PR) = 2 cm + 2cm

= 4 cm

= l(PQ)

But in a triangle, the sum of the length of any two sides of a triangle is always greater than the length of the third side.

Hence, ∆PQR cannot be drawn.

Question 2.

Draw ∆ABC such that l(BC) = 8 cm, l(CA) = 6 cm, m∠ABC = 40°.

Draw a ray to make an angle of 40° with the base BC, l(BC) = 8 cm. We have to obtain point ‘A’ on the ray. With ‘C’ as the centre, draw an arc of radius 6 cm to do so. What do we observe? The arc intersects the ray in two different points. Thus, we get two triangles of two different shapes having the given measures. (Textbook pg. no. 7)

Solution:

Here ∠B is an acute angle. ∠C can be an acute angle or an obtuse angle.

Hence we get two triangles of two different shapes.

Question 3.

Can a triangle be drawn if the three angles are given, but not any side? How many such triangles can be drawn? (Textbook pg. no. 7)

Solution:

Yes a triangle can be drawn.

Since the length of side is not given, any length of side can be selected and then triangle can be constructed. We will get different triangles for different length of sides.

Question 4.

Using the ruler, measure the lengths of seg AB and seg PQ. Are they of same length? Trace the seg AB on a sheet of transparent paper. Now place this new segment on PQ verify that if point A is placed on point P, then B falls on Q. (Textbook pg. no. 7)

l(AB) = ___

l(PQ) = ___

Solution:

l(AB) = 4 cm

l(PQ) = 4 cm

Since the length of two segments is the same, if placed on one another, they will coincide.

Question 5.

From the shape shown below, write the names of the pairs of congruent line segments. (Textbook pg. no. 8)

i. seg AB ≅ seg DC

ii. seg AE ≅ seg BH

iii. seg EF ≅ seg ___

iv. seg DF ≅ seg ___

Solution:

seg EF ≅ seg AD ≅ seg BC ≅ seg HG

seg DF ≅ seg CG ≅ seg AE ≅ seg BH

Question 6.

Take a rectangular paper. Place two opposite sides upon each Other. What do you observe? (Textbook pg. no. 7)

Solution:

Opposite sides of the rectangular paper coincide and hence are congruent.