CBSE Class 7

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.5

Question 1.
Construct the right-angled ∆PQR, where m∠Q = 90° QR = 8 cm and PR = 10 cm. Ans: Steps of Construction
(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line segment QR = 8 cm.
(iii) At Q, Construct ∠RQX = 90° .

(iv) With centre at R and radius 10 cm, draw an arc to cut QX at P.
(v) Join PR.
Thus, PQR is the required right-angled triangle.

Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Answer:
Steps of Construction


(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line segment QR = 4 cm.
(iii) At Q, construct ∠RQX = 90
(iv) With R as centre and radius 6 cm draw an arc to cut QX at R
(v) Join PR.
Thus, PQR is the required right angled triangle.

Question 3.
Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
Answer:
Steps of Construction

(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line segment AC = 6 cm.
(iii) At ‘C’, construct ∠ACX – 90°.
(iv) With centre ‘C’ and radius 6 cm, draw an arc to cut CX at B.
(v) Join AB.
Thus, ∆ABC is the required isosceles right-angled triangle.

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.4

Question 1.
Construct ∆ABC given m∠A = 60° , m ∠B = 30 and AB = 5.8 cm.
Answer:
Steps of Construction
(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line segment AB = 5.8 cm.
(iii) Construct ∠BAX = 60° at A.
(iv) At B, construct ∠ABY = 30° .

(v) Let the rays AX and BY intersect at C. Thus, ∆ABC is the required triangle.

Question 2.
Construct ∆PQR if PQ = 5 cm, m∠PQR = 105° and m ∠QRP = 40°.
(Hint : Recall angle-sum property of a triangle)
Answer:
Here the line segment PQ is given. We can construct the triangle, if the measure of ∠QPR is known.
∠QPR = 180° – [∠PQR + ∠QRP]
= 180°-(105°+ 40°)
= 180° – 145°
∠QPR = 35°
Steps of Construction


(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line, segment PQ = 5 cm.
(iii) At P, construct ∠QPX = 35 .
(iv) At Q, construct ∠PQY = 105 .
(v) Let the rays \(\overline{\mathrm{PX}}\) and \(\overline{\mathrm{QY}}\) intersect at R.
Thus, ∆PQR is the required triangle.

Question 3.
Examine whether you can construct ∆DEF such that EF = 7.2 cm, m∠E = 110 and m∠F = 80°. Justify your answer.
Answer:
Here m∠E =110 and m∠F = 80°
i. e. m∠E + m∠F= 110°+ 80°= 190°>180°
Since, sum of the three angles of a triangle is 180°.
∆DEF is not possible, having sum of two angles > 180°
Thus, ∆DEF cannot be constructed.

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.3

Question 1.
Construct a ADEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90 .
Answer:
Steps of Construction

(i) First we draw a rough sketch with measures marked on it.
(ii) Draw a line segment DE = 5 cm.
(iii) At D, construct ∠EDX = 90
(iv) From the ray DX, cut off DF = 3 cm.
(v) Join EF.
Then, ∆DEF is the required triangle.

Question 2.
Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110.
Answer:
Steps of Construction

(i) First we draw a rough sketch with measures marked on it.
(ii) Draw a line segment AB = 6.5 cm.
(iii) At B, using protractor construct ∠ABX = 110°.
(iv) From BX, cut off BC = 6.5 cm.
(v) Join AC.
Thus, ∆ABC is the required isosceles triangle.

Question 3.
Construct ∆ABC with BC = 7.5 cm AC = 5 cm and m∠C – 60
Answer:
Steps of Construction
(i) First we draw a rough sketch with measures marked on it.
(ii) Draw a line segment BC = 7.5 cm.
(iii) At ‘C’ using protractor construct ∠BCX = 60°.

(iv) From the ray CX, cut off CA = 5 cm.
(v) Join AB.
Thus, ∆ABC is the required triangle.

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°.

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.1

Question 1.
Draw a line say AB, take a point ‘C’ outside it. Through C, draw a line parallel to AB using ruler and compass only.
Answer:
Steps of Construction

(i) Draw a line AB.
(ii) Take a point C outside it.
(iii) Take any point ‘D’ on AB.
(iv) Join C to D.
(v) With D as centre and a convenient radius draw an arc cutting AB at F and CD at E.
(vi) Now with C as centre and the same radius as in step 5, draw an arc GH cutting CD at I.
(vii) Place the pointed tip of the compass at F and adjust the opening so that the pencil tip is at E.
(viii) With the same opening as in step 7 and with I as centre, draw an arc cutting the arc GH at J.
(ix) Now Join CJ to draw a line ‘KL’ Then KL is the required line.

Question 2.
Draw a line 1. Draw a perpendicular to T at any point on T. On this perpendicular Choose a point X, 4 cm away from l. Through X, draw a line ‘m’ parallel to T
Answer:
Steps of Construction

(i) Draw a line ‘l’.
(ii) Take any point A on line l.
(iii) Construct an angle of 90° at point ‘A’ on line ‘l’ and draw a line ‘AL’ perpendicular to line ‘l’.
(iv) Make a point ‘X’ on AL such that AX = 4 cm.
(v) At X, construct an angle of 90 and draw a line XC perpendicular to line AL.
(vi) Then line XC (line m) is the required line through X such that m || l.

Question 3.
Let ‘l’ be a line and ‘P’ be 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. Let this meet ‘l’ at ‘S’. What shape do the two sets of parallel lines enclose?
Answer:
Steps of Construction
(i) Draw a line ‘l’ and take a point ‘P’ not on it.
(ii) Take any point ‘Q’ on ‘l’.
(iii) Join ‘Q’ to ‘P’.
(iv) Draw a line ‘m’ parallel to the line T as shown in figure.
Then line ‘m’ || line ‘l’.
(v) Join ‘P’ to any point ‘Q’ on ‘l’.
(vi) Choose any point ‘R’ on ‘m’.

(vii) Join R to Q
(viii) Through R, draw a line ‘n’ parallel to the line PQ.
(ix) Let the line ‘n’ meet the line ‘l’ at ‘S’.
(x) Then the shape enclosed by the two sets of parallel lines is a ‘Parallelogram’.

These NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Exercise 5.2

Question 1.
State the property that is used in each of the following statements?
(i) If a || b, then ∠1 = ∠5.
(ii) If ∠4 = ∠6, then a || b.
(iii) If ∠4 + ∠5 = 180°, then a || b.

Answer:
(i) If two parallel lines are intersected by a transversal, then corresponding angles are equal.
(ii) If two given lines are intersected by a transversal such that alternate interior angles are equal, then the lines are parallel.
(iii) If two parallel lines are cut by a transversal such that the pairs of a interior angles on the same side of the transversal are supplementary, then the lines are parallel.

Question 2.
In the figure given below, identify
(i) the pairs of corresponding angles.
(ii) the pairs of alternate interior angles.
(iii) the pairs of interior angles on the same side of the transversal.
(iv) the vertically opposite angles.

Answer:
(i) The pairs of corresponding angles
are (∠1, ∠5); (∠2, ∠6); (∠3, ∠7); and (∠4, ∠8)
(ii) The pairs of alternate interior angles are (∠2, ∠8) and (∠3, ∠5)
(iii) The pairs of interior angles on the same side of the transversal are
(∠2, ∠5) and (∠3, ∠8)
(iv) The vertically opposite angles are (∠1, ∠3); (∠2, ∠4); (∠5, ∠7); and (∠6, ∠8)

Question 3.
In the figure given below, p || q. Find the unknown angles.
Answer:
Since, ∠e + 125° = 180°… (linear pair)
∠e = 180° – 125°
∠e = 55°
∠e = ∠f
(vertically opposite angles)
∴ ∠f = 55°
∴ ∠a = ∠e
(corresponding angles)
∠a = 55°
Again ∠b = 125°
(alternate exterior angles)
since, ∠b and ∠c form a linear pair
∴ ∠b+ ∠c= 180°
or 125° + ∠c = 180°
∠c = 180° – 125°
= 55°

Now, ∠b and ∠d are vertically opposite angles.
∴ ∠d = ∠b = 125°
[∵ ∠b = 125°]
Thus, the required measures are as follows: ∠a = 55°, ∠b = 125°, ∠c = 55°,∠d = 125°, ∠e = 55°, ∠f = 55°

Question 4.
Find the value of x in each of the following figures if 1 || m.

Answer:
(i) ∠x = ∠P
(alternate angles are equal)
But ∠P + 110° = 180° (linear pair)
∠P = 180°- 110°
= 70°
∴ ∠x = 70°

(ii) 1 and m are parallel and ‘a’ is a transversal.
∴ ∠x = 100°
(corresponding angles are equal)

Question 5.
In the given figure, the arms of two angles are parallel.
If ∠ABC = 70°, then find
(i) ∠DGC
(ii) ∠DEF

Answer:
We have AB || ED and BC || EF
(i) BC is a transversal
∴ ∠DGC – ∠ABC
(corresponding angles)
But∠ABC – 70°
∴ ∠DGC = 70°

(ii) ED is a transversal to BC || EF
∴ ∠DEF = ∠DGC
(corresponding angles)
But∠DGC = 70°
∴ ∠DEF = 70°

Question 6.
In the given figures below, decide whether 1 is parallel to m.

Answer:
(i) 44° + 126° = 170° ≠ 180°
Sum of the interior angles on the same side of the transversal is not 180°.
∴ 1 and m are not parallel.

(ii) n is a transversal to 1 and m and ∠P = 75°(vertically opposite angles)
also ∠P + 75°= 75° + 75° = 150° ≠ 180°
∴ 1 and m are not parallel as sum of interior angles on same side of transversal is not 180°.

(iii) ∠P + 123°= 180° (linear pair)
∠P = 180° – 123° = 57°
∠P = 57°(linear pair)
i.e., Corresponding angles are equal.
∴ 1 and m are parallel (corresponding angles are not equal)

(iv) ∠1 + ∠2 = 180°
∠1 + 98° = 180°
∠1 = 180°- 98°= 82°
∠3 = 72°
∠1 ≠∠3
∴1 and m are not parallel.