Ncert Solutions Class 7 Maths Chapter 10 Exercise 10.1

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 compasses only.

Answer:

To construct: A line, parallel to given line by using ruler and compasses.

Steps of construction:

(a) Draw a line-segment AB and take a point C outside AB.

(b) Take any point D on AB and join C to D.

(c) With D as centre and take convenient radius, draw an arc cutting AB at E and CD at F.

(d) With C as centre and same radius as in step 3, draw an arc GH cutting CD at I.

(e) With the same arc EF, draw the equal arc cutting GH at J.

(f) Join JC to draw a line l.

This is the required line AB || l.

Question 2:

Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.

Answer:

To construct: A line parallel to given line when perpendicular line is also given.

Steps of construction:

(a) Draw a line l and take a point P on it.

(b) At point P, draw a perpendicular line n.

(c) Take PX = 4 cm on line n.

(d) At point X, again draw a perpendicular line m.

It is the required construction.

Question 3:

Let l be a line and P be a 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:

To construct: A pair of parallel lines intersecting other part of parallel lines.

Steps of construction:

(a) Draw a line l and take a point P outside of l.

(b) Take point Q on line l and join PQ.

(c) Make equal angle at point P such that Ð Q = Ð P.

(d) Extend line at P to get line m.

(e) Similarly, take a point R online m, at point R, draw angles such that Ð P = Ð R.

(f) Extended line at R which intersects at S online l.

(g) Draw line RS.

Thus, we get parallelogram PQRS.

Exercise 10.2

Question 1:

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

Answer:

To construct: Δ XYZ, where XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

Steps of construction:

(a) Draw a line segment YZ = 5 cm.

(b) Taking Z as centre and radius 6 cm, draw an arc.

(c) Similarly, taking Y as centre and radius 4.5 cm,

draw another arc which intersects first arc at point X.

(d) Join XY and XZ.

It is the required Δ XYZ.

Question 2:

Construct an equilateral triangle of side 5.5 cm.

Answer:

To construct: Δ ABC where AB = BC = CA = 5.5 cm

Steps of construction:

(a) Draw a line segment BC = 5.5 cm

(b) Taking points B and C as centers and radius 5.5 cm,

draw arcs which intersect at point A.

(c) Join AB and AC.

It is the required Δ ABC.

Question 3:

Draw Δ PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?

Answer:

To construction: Δ PQR, in which PQ = 4 cm, QR = 3.5 cm and PR = 4 cm.

Steps of construction:

(a) Draw a line segment QR = 3.5 cm.

(b) Taking Q as centre and radius 4 cm, draw an arc.

(c) Similarly, taking R as centre and radius 4 cm,

draw another arc which intersects first arc at P.

(d) Join PQ and PR.

It is the required isosceles Δ PQR.

Question 4:

Construct Δ ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure Ð B.

Answer:

To construct: Δ ABC in which AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm.

Steps of construction:

(a) Draw a line segment BC = 6 cm.

(b) Taking B as centre and radius 2.5 cm, draw an arc.

(c) Similarly, taking C as centre and radius 6.5 cm,

draw another arc which intersects first arc at point A.

(d) Join AB and AC.

(e) Measure angle B with the help of protractor.

It is the required Δ ABC where Ð B = 80⁰.

Exercise 10.3

Question 1:

Construct Δ DEF such that DE = 5 cm, DF = 3 cm and mÐ EDF = 90⁰.

Answer:

To construct: Δ DEF where DE = 5 cm, DF = 3 cm and mÐ EDF = 90⁰.

Steps of construction:

(a) Draw a line segment DF = 3 cm.

(b) At point D, draw an angle of 90⁰ with the help of compass i.e.,

Ð XDF = 90⁰.

(c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E.

(d) Join EF.

It is the required right angled triangle DEF.

Question 2:

Construct an isosceles triangle in which the lengths of each of its equal sides are 6.5 cm and the angle between them is 110⁰.

Answer:

To construct: An isosceles triangle PQR where PQ = RQ = 6.5 cm and Ð Q = 110⁰.

Steps of construction:

(a) Draw a line segment QR = 6.5 cm.

(b) At point Q, draw an angle of 110⁰ with the

 help of protractor, i.e., Ð YQR = 110⁰.

(c) Taking Q as centre, draw an arc with radius

 6.5 cm which cuts QY at point P.

(d) Join PR

It is the required isosceles triangle PQR.

Question 3:

Construct Δ ABC with BC = 7.5 cm, AC = 5 cm and mÐ C = 60⁰.

Answer:

To construct: Δ ABC where BC = 7.5 cm, AC = 5 cm and mÐ C = 60⁰.

Steps of construction:

(a) Draw a line segment BC = 7.5 cm.

(b) At point C, draw an angle of 60⁰

 with the help of protractor, i.e., Ð XCB = 60⁰.

(c) Taking C as centre and radius 5 cm, draw an arc, which cuts XC at the point A.

(d) Join AB

It is the required triangle ABC.

Exercise 10.4

Question 1:

Construct Δ ABC, given mÐ A = 60⁰, mÐ B = 30⁰ and AB = 5.8 cm.

Answer:

To construct: Δ ABC where mÐ A = 60⁰, mÐ B = 30⁰ and AB = 5.8 cm.

Steps of construction:

(a) Draw a line segment AB = 5.8 cm.

(b) At point A, draw an angle Ð YAB = 60⁰

 with the help of compass.

(c) At point B, draw Ð XBA = 30⁰ with the help of compass.

(d) AY and BX intersect at the point C.

It is the required triangle ABC.

Question 2:

Construct Δ PQR if PQ = 5 cm, mÐ PQR = 105⁰ and mÐ QRP = 40⁰.

Answer:

Given: mÐ PQR = 105⁰ and mÐ QRP = 40⁰

We know that sum of angles of a triangle is 180⁰.

mÐ PQR + mÐ QRP + mÐ QPR = 180⁰

105⁰ + 40⁰ + mÐ QPR = 180⁰

145⁰ + mÐ QPR = 180⁰

mÐ QPR = 180⁰ – 145⁰

mÐ QPR = 35⁰

To construct: Δ PQR where mÐ P = 35⁰, mÐ Q = 105⁰ and PQ = 5 cm.

Steps of construction:

(a) Draw a line segment PQ = 5 cm.

(b) At point P, draw Ð XPQ = 35⁰ with the help of protractor.

(c) At point Q, draw Ð YQP = 105⁰ with the help of protractor.

(d) XP and YQ intersect at point R.

It is the required triangle PQR.

Question 3:

Examine whether you can construct Δ DEF such that EF = 7.2 cm, m ÐE = 110⁰ and mÐ F = 80⁰. Justify your answer.

Answer:

Given: In Δ DEF, mÐ E = 110⁰ and mÐ F = 80⁰.

Using angle sum property of triangle, we get

        Ð𝐷 + Ð𝐸 + Ð𝐹 = 180°

=> Ð𝐷 + 110° + 80° = 180°

=> Ð𝐷 + 190° = 180°

=> Ð𝐷 = 180° − 190° = −10°

Which is not possible.

Exercise 10.5

Question 1:

Construct the right angled Δ PQR, where mÐ Q = 90⁰, QR = 8 cm and PR = 10 cm.

Answer:

To construct:

A right angled triangle PQR where mÐ Q = 90⁰, QR = 8 cm and PQ = 10 cm.

Steps of construction:

(a) Draw a line segment QR = 8 cm.

(b) At point Q, draw QX Ʇ QR.

(c) Taking R as centre, draw an arc of radius 10 cm.

(d) This arc cuts QX at point P.

(e) Join PQ.

It is the required right angled triangle PQR.

Question 2:

Construct a right angled triangle whose hypotenuse is 6 cm long and one the legs is 4 cm long.

Answer:

To construct:

A right angled triangle DEF where DF = 6 cm and EF = 4 cm

Steps of construction:

(a) Draw a line segment EF = 4 cm.

(b) At point Q, draw EX Ʇ EF.

(c) Taking F as centre and radius 6 cm, draw an arc. (Hypotenuse)

(d) This arc cuts the EX at point D.

(e) Join DF.

It is the required right angled triangle DEF.

Question 3:

Construct an isosceles right angled triangle ABC, where mÐ ACB = 90⁰ and AC = 6 cm.

Answer:

To construct:

An isosceles right angled triangle ABC where mÐ C = 90⁰, AC = BC = 6 cm.

Steps of construction:

(a) Draw a line segment AC = 6 cm.

(b) At point C, draw XC Ʇ CA.

(c) Taking C as centre and radius 6 cm, draw an arc.

(d) This arc cuts CX at point B.

(e) Join BA.

It is the required isosceles right angled triangle ABC.

Miscellaneous Questions

Questions:

Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and say why you cannot construct them.

Construct rest of the triangle.

    Triangle                                                                   Given measurements

1. Δ ABC mÐA = 85⁰;         mÐB = 115⁰;           AB = 5 cm.
2. Δ PQR mÐQ = 30⁰; mÐR = 60⁰;             QR = 4.7 cm.
3. Δ ABC mÐA = 70⁰;      mÐB = 50⁰;             AB = 3 cm.
4. Δ LMN mÐL = 60⁰; mÐN = 120⁰;           LM = 5 cm.
5. Δ ABC BC = 2 cm; AB = 4 cm;               AC = 2 cm.
6. Δ PQR PQ = 3.5 cm; QR = 4 cm;               PR = 3.5 cm.
7. Δ XYZ XY = 3 cm; YZ = 4 cm;                XY = 5 cm.
8. Δ DEF DE = 4.5 cm;         EF = 5.5 cm;            DF = 5 cm.

Answer 1:

In Δ ABC, mÐA = 85⁰, mÐB = 115⁰, AB = 5 cm.

Construction of Δ ABC is not possible because mÐ A + mÐB = 200⁰, and we know that the sum

of angles of a triangle should be 180⁰.

Answer2:

To construct: Δ PQR where mÐ Q = 30⁰, mÐ R = 60⁰ and QR = 4.7 cm.

Steps of construction:

(a) Draw a line segment QR = 4.7 cm.

(b) At point Q, draw Ð XQR = 30⁰ with the help of compass.

(c) At point R, draw Ð YRQ = 60⁰ with the help of compass.

(d) QX and RY intersect at point P.

It is the required triangle PQR.

Answer 3:

We know that the sum of angles of a triangle is 180⁰.

So, mÐ A + mÐ B + mÐ C = 180⁰

=> 70⁰ + 50⁰ + mÐ C = 180⁰

=> 120⁰ + mÐ C = 180⁰

=> mÐ C = 180⁰ – 120⁰

=> mÐ C = 60⁰

To construct: Δ ABC where mÐ A = 70⁰, mÐ C = 60⁰ and AC = 3 cm.

Steps of construction:

(a) Draw a line segment AC = 3 cm.

(b) At point C, draw Ð YCA = 60⁰.

(c) At point A, draw Ð XAC = 70⁰.

(d) Rays XA and YC intersect at point B

It is the required triangle ABC.

Answer 4:

In Δ LMN, mÐ L = 60⁰, mÐ N = 120⁰, LM = 5 cm

This Δ LMN is not possible to construct because mÐ L + mÐ N =180⁰ which forms a linear pair.

Answer 5:

In Δ ABC, BC = 2 cm, AB = 4 cm and AC = 2 cm

This Δ ABC is not possible to construct because the condition is Sum of lengths of two sides of

a triangle should be greater than the third side.

i.e. AB < BC + AC

=> 4 < 2 + 2

=> 4 = 4

Answer 6:

To construct: Δ PQR where PQ = 3.5 cm, QR = 4 cm and PR = 3.5 cm

Steps of construction:

(a) Draw a line segment QR = 4 cm.

(b) Taking Q as centre and radius 3.5 cm, draw an arc.

(c) Similarly, taking R as centre and radius 3.5 cm, draw another

arc which intersects the first arc at point P.

It is the required triangle PQR.

Answer 7:

To construct: A triangle whose sides are XY = 3 cm, YZ = 4 cm and XZ = 5 cm.

Steps of construction:

(a) Draw a line segment ZY = 4 cm.

(b) Taking Z as centre and radius 5 cm, draw an arc.

(c) Taking Y as centre and radius 3 cm, draw another arc.

(d) Both arcs intersect at point X.

It is the required triangle XYZ.

Answer 8:

To construct:

A triangle DEF whose sides are DE = 4.5 cm, EF = 5.5 cm and DF = 4 cm.

Steps of construction:

(a) Draw a line segment EF = 5.5 cm.

(b) Taking E as centre and radius 4.5 cm, draw an arc.

(c) Taking F as centre and radius 4 cm, draw an another arc

 which intersects the first arc at point D.

It is the required triangle DEF.

Ncert Solutions Class 7 Maths Chapter 10 Exercise 10.1

Source: https://www.examfear.com/cbse-ncert-solution/Class-7/Maths/Practical-Geometry/solutions.htm

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