Practical Geometry| Worksheet Solutions| Class 7

Worksheet solutions for chapter: Practical Geometry for class 7 are provided below.


(1) Radha wants to construct a parallel line to a line AB of 4 cm length. But, her friend challenges Radha to draw it with compass. Can Radha draw it if yes how?


Yes, Radha can draw a parallel line. Steps of construction a parallel line is showing below-

(i)Draw a line AB of length 4 cm. Take a point C outside it and take a point P on it.

(ii)Join point C to point P, with P as centre and a convenient radius; draw an arc     cutting AB at D and PC at E.

(iii)Now with same radius and C as centre, draw an arc FG cutting PC at H.

(iv)Place the tip of the compasses at D and adjust the opening so that the pencil tip is at E.

It does not need drawing.

(v)With the same opening and with H as centre, draw an arc cutting the arc FG at I.

(vi)Join A and I to draw a parallel line.


(2) A gardener wants to create a triangle area in a park. He wants to make a triangle with lengths 7 m, 6 m and 1 m. Can he make a triangle?


Condition for making a triangle – The sum of the lengths of any two sides would be greater than the length of the third side.

Is 7 + 6 > 1?  Yes

Is 7 + 1 > 6?  Yes

Is 6 + 1 > 7? No

Hence, the gardener can not make the triangle with given lengths.


(3)Draw a right-angled triangle with hypotenuse of 6 cm and one arm is 4 cm.


(i) Draw QR of length 4 cm.


(ii)At Q draw QX ⊥ QR.

(iii)With R as center, draw an arc of radius 6 cm.

(iv)P has to be on the perpendicular line QX as well as on the arc drawn with centre R.

Therefore, P is the meeting point of these two.

ΔPQR is now obtained.


(4)Draw a triangle which has an angle of 600 and lengths of two sides are 4 cm and 4.8 cm.


(i) Draw a line segment BC of length 4.8 cm.


(ii)At C, draw CX making 600 with BC.


(iii)With C as center, draw an arc of radius 4 cm. It cuts CX at the point A.


(iv)Join AB, ΔABC is now obtained.


(5)Draw a triangle whose angles are 600, 750 and 1100.


(6)If two sides of a triangle and angle between them is given.The criterion is known as

(a)SAS criterion

(b)SSS criterion

(c)ASA criterion


(a) SAS criterion


(7) In a RHS criterion triangle, an angle must be

(a)obtuse angle

(b)right- angle

(c)acute angle




(8) Construct a triangle of whose two angles are 900 and 450.One leg between these angles is 4cm.


(i)Draw PQ of length 4 cm.

(ii)At point P, draw an angle of 900.

(iii)At point Q, draw a ray YQ making an angle of 450 with line PQ.

(iv)R has to lie on both the rays XP and YQ. So, the point of intersection of the two rays is R.

PQR is required triangle.


(9) Can a triangle be construct whose lengths are 2 cm, 2 cm and 4 cm?


Condition for making a triangle – The sum of the lengths of any two sides would be greater than the length of the third side.

Is 2 + 2 > 4?  No

Is 2 + 4 > 2?  No

Is 4 + 2 > 2? No

Hence, a triangle can not make with given lengths.


(10) Match the followings-

Column I                                                   Column II

(i)3 cm, 5c m, 6 cm                                   ASA criterion

(ii)300, 600, 6 cm                                      SAS criterion

(iii)8 cm, 450, 5 cm                                   SSS criterion


Column I                                                Column II

(i)3 cm, 5 cm, 6 cm                                    SSS criterion

(ii)300, 600, 6 cm                                      ASA criterion

(iii)8 cm, 450, 5 cm                                   SAS criterion


Helping Topics

Practical Geometry

NCERT Solutions Class 7