NCERT Solutions for Class 7 Maths Exercise 10.6

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NCERT solutions for Maths Practical Geometry Download as PDF

NCERT Solutions for Class 7 Maths Practical Geometry

Class –VII Mathematics (Ex. 10.6)

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. {tex}\Delta {/tex}ABC {tex}m\angle {/tex}A = {tex}85^\circ {\text{ ;}}{/tex} {tex}m\angle {/tex}B = {tex}115^\circ {\text{ ;}}{/tex} AB = 5 cm

2. {tex}\Delta {/tex}PQR {tex}m\angle {/tex}Q = {tex}30^\circ {\text{ ;}}{/tex} {tex}m\angle {/tex}R = {tex}60^\circ {\text{ ;}}{/tex} QR = 4.7 cm

3. {tex}\Delta {/tex}ABC {tex}m\angle {/tex}A = {tex}70^\circ {\text{ ;}}{/tex} {tex}m\angle {/tex}B = {tex}50^\circ {\text{ ;}}{/tex} AC = 3 cm

4. {tex}\Delta {/tex}LMN {tex}m\angle {/tex}L = {tex}60^\circ {\text{ ;}}{/tex} {tex}m\angle {/tex}N = {tex}120^\circ {\text{ ;}}{/tex} LM = 5 cm

5. {tex}\Delta {/tex}ABC BC = 2 cm; AB = 4 cm; AC = 2 cm

6. {tex}\Delta {/tex}PQR PQ = 3.5 cm; QR = 4 cm; PR = 3.5 cm

7. {tex}\Delta {/tex}XYZ XY = 3 cm; YZ = 4 cm; XZ = 5 cm

8. {tex}\Delta {/tex}DEF DE = 4.5 cm; EF = 5.5 cm; DF = 4 cm

Answers

Miscellaneous Questions

1.In {tex}\Delta {/tex}ABC, {tex}m\angle {/tex}A = {tex}85^\circ ,m\angle {/tex}B = {tex}115^\circ ,{/tex} AB = 5 cm Construction of {tex}\Delta {/tex}ABC is not possible because {tex}m\angle {/tex}A = {tex}85^\circ + m\angle {/tex}B = {tex}200^\circ ,{/tex} and we know that the sum of angles of a triangle should be {tex}180^\circ .{/tex}

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NCERT Solutions for Class 7 Maths Exercise 10.6

2.To construct: {tex}\Delta {/tex}PQR where {tex}m\angle {/tex}Q = {tex}{30^ \circ },{/tex}{tex}m\angle {/tex}R = {tex}{60^ \circ }{/tex} and QR = 4.7 cm.

Steps of construction:

1. Draw a line segment QR = 4.7 cm.

(a) At point Q, draw {tex}\angle {/tex}XQR = {tex}{30^ \circ }{/tex} with the help of compass.

(b) At point R, draw {tex}\angle {/tex}YRQ = {tex}{60^ \circ }{/tex} with the help of compass.

(c) QX and RY intersect at point P.

It is the required triangle PQR.

3. We know that the sum of angles of a triangle is {tex}180^\circ .{/tex}

{tex}\therefore {/tex} {tex}m\angle {/tex}A + {tex}m\angle {/tex}B + {tex}m\angle {/tex}C = {tex}180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}70^\circ + 50^\circ + m\angle {/tex}C = {tex}180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}120^\circ {/tex} + {tex}m\angle {/tex}C = {tex}180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}m\angle {/tex}C = {tex}180^\circ {/tex} – {tex}120^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}m\angle {/tex}C = {tex}60^\circ {/tex}

To construct: {tex}\Delta {/tex}ABC where {tex}m\angle {/tex}A = {tex}70^\circ {/tex}, {tex}m\angle {/tex}C = {tex}60^\circ {/tex} and AC = 3 cm.

Steps of construction:

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

(b) At point C, draw {tex}\angle {/tex}YCA = {tex}{60^ \circ }.{/tex}

(c) At point A, draw {tex}\angle {/tex}XAC = {tex}70^\circ .{/tex}

(d) Rays XA and YC intersect at point B

It is the required triangle ABC.

4. In {tex}\Delta {/tex}LMN , {tex}m\angle {/tex}L = {tex}{60^ \circ },{/tex}{tex}m\angle {/tex}N = {tex}120^\circ ,{/tex} LM = 5 cm

This {tex}\Delta {/tex}LMN is not possible to construct because {tex}m\angle {/tex}L + {tex}m\angle {/tex}N = {tex}60^\circ + 120^\circ = 180^\circ {/tex} which forms a linear pair.

5. {tex}\Delta {/tex}ABC, BC = 2 cm, AB = 4 cm and AC = 2 cm

This {tex}\Delta {/tex}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.

AB < BC + AC {tex} \Rightarrow {/tex} 4 < 2 + 2 {tex} \Rightarrow {/tex} 4 = 4,

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NCERT Solutions for Class 7 Maths Exercise 10.6

6. To construct: {tex}\Delta {/tex}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 an another arc which intersects the first arc at point P.

It is the required triangle PQR.

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NCERT Solutions for Class 7 Maths Exercise 10.6

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.

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NCERT Solutions for Class 7 Maths Exercise 10.6

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.

(a) 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 for Class 7 Maths Exercise 10.6

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