NCERT Solutions for Class 7 Maths Exercise 6.3

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NCERT solutions for Maths Triangles and its Properties Download as PDF

NCERT Solutions for Class 7 Maths Triangles and its Properties

Class –VII Mathematics (Ex. 6.3)

Question 1.Find the value of unknown {tex}x{/tex} in the following diagrams:

Answer:

(i) In {tex}\Delta {\text{ABC}},{/tex}

{tex}\angle {/tex}BAC + {tex}\angle {/tex}ACB + {tex}\angle {/tex}ABC = {tex}180^\circ {/tex} [By angle sum property of a triangle]

{tex} \Rightarrow {/tex} {tex}x + 50^\circ + 60^\circ = 180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}x + 110^\circ = 180^\circ {/tex} {tex} \Rightarrow {/tex} {tex}x = 180^\circ – 110^\circ = 70^\circ {/tex}

(ii) In {tex}\Delta {/tex}PQR,

{tex}\angle {/tex}RPQ + {tex}\angle {/tex}PQR + {tex}\angle {/tex}RPQ = {tex}180^\circ {/tex} [By angle sum property of a triangle]

{tex} \Rightarrow {/tex} {tex}{90^ \circ } + {30^ \circ } + x = 180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}x + 120^\circ = 180^\circ {/tex} {tex} \Rightarrow {/tex} {tex}x = 180^\circ – 120^\circ = 60^\circ {/tex}

(iii) In {tex}\Delta {/tex}XYZ,

{tex}\angle {/tex}ZXY + {tex}\angle {/tex}XYZ + {tex}\angle {/tex}YZX = {tex}180^\circ {/tex} [By angle sum property of a triangle]

{tex} \Rightarrow {/tex} {tex}{30^ \circ } + {110^ \circ } + x = 180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}x + 140^\circ = 180^\circ {/tex} {tex} \Rightarrow {/tex} {tex}x = 180^\circ – 140^\circ = 40^\circ {/tex}

(iv) In the given isosceles triangle,

{tex}x + x + 50^\circ = 180^\circ {/tex} [By angle sum property of a triangle]

{tex} \Rightarrow {/tex} {tex}2x + 50^\circ = 180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}2x = 180^\circ – 50^\circ {/tex} {tex} \Rightarrow {/tex} {tex}2x = 130^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}x = \frac{{130^\circ }}{2} = 65^\circ {/tex}

(v) In the given equilateral triangle,

{tex}x + x + x = 180^\circ {/tex} [By angle sum property of a triangle]

{tex} \Rightarrow {/tex} {tex}3x = 180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}x = \frac{{180^\circ }}{3} = 60^\circ {/tex}

(vi) In the given right angled triangle,

{tex}x + 2x + 90^\circ = 180^\circ {/tex} [By angle sum property of a triangle]

{tex} \Rightarrow {/tex} {tex}3x + 90^\circ = 180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}3x = 180^\circ – 90^\circ {/tex} {tex} \Rightarrow {/tex} {tex}3x = 90^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}x = \frac{{90^\circ }}{3} = 30^\circ {/tex}

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NCERT Solutions for Class 7 Maths Exercise 6.3

Question 2.Find the values of the unknowns {tex}x{/tex} and {tex}y{/tex} in the following diagrams:

Answer:

(i) {tex}50^\circ + x = 120^\circ {/tex} [Exterior angle property of a {tex}\Delta {/tex} ]

{tex} \Rightarrow {/tex} {tex}x = 120^\circ – 50^\circ = 70^\circ {/tex}

Now, {tex}50^\circ + x + y = 180^\circ {/tex} [Angle sum property of a {tex}\Delta {/tex} ]

{tex} \Rightarrow {/tex} {tex}50^\circ + 70^\circ + y = 180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}120^\circ + y = 180^\circ {/tex} {tex} \Rightarrow {/tex} {tex}y = 180^\circ – 120^\circ = 60^\circ {/tex}

(ii) {tex}y = 80^\circ {/tex} ……….(i) [Vertically opposite angle]

Now, {tex}50^\circ + x + y = 180^\circ {/tex} [Angle sum property of a {tex}\Delta {/tex} ]

{tex} \Rightarrow {/tex} {tex}50^\circ + 80^\circ + y = 180^\circ {/tex}

[From eq. (i)]

{tex} \Rightarrow {/tex} {tex}130^\circ + y = 180^\circ {/tex} {tex} \Rightarrow {/tex} {tex}y = 180^\circ – 130^\circ = 50^\circ {/tex}

(iii) {tex}50^\circ + 60^\circ = x{/tex} [Exterior angle property of a {tex}\Delta {/tex} ]

{tex} \Rightarrow {/tex} {tex}x = 110^\circ {/tex}

Now {tex}50^\circ + 60^\circ + y = 180^\circ {/tex} [Angle sum property of a {tex}\Delta {/tex} ]

{tex} \Rightarrow {/tex} {tex}110^\circ + y = 180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}y = 180^\circ – 110^\circ {/tex} {tex} \Rightarrow {/tex} {tex}y = 70^\circ {/tex}

(iv) {tex}x = {60^ \circ }{/tex} ……….(i) [Vertically opposite angle]

Now, {tex}30^\circ + x + y = 180^\circ {/tex} [Angle sum property of a {tex}\Delta {/tex} ]

{tex} \Rightarrow {/tex} {tex}50^\circ + 60^\circ + y = 180^\circ {/tex} [From eq. (i)]

{tex} \Rightarrow {/tex} {tex}90^\circ + y = 180^\circ {/tex} {tex} \Rightarrow {/tex} {tex}y = 180^\circ – 90^\circ = 90^\circ {/tex}

(v) {tex}y = {90^ \circ }{/tex} ……….(i) [Vertically opposite angle]

Now, {tex}y + x + x = 180^\circ {/tex} [Angle sum property of a {tex}\Delta {/tex} ]

{tex} \Rightarrow {/tex} {tex}90^\circ + 2x = 180^\circ {/tex} [From eq. (i)]

{tex} \Rightarrow {/tex} {tex}2x = 180^\circ – 90^\circ {/tex} {tex} \Rightarrow {/tex} {tex}2x = 90^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}x = \frac{{{{90}^ \circ }}}{2} = {45^ \circ }{/tex}

(vi) {tex}x = y{/tex} ……….(i) [Vertically opposite angle]

Now, {tex}x + x + y = 180^\circ {/tex} [Angle sum property of a {tex}\Delta {/tex} ]

{tex} \Rightarrow {/tex} {tex}2x + x = 180^\circ {/tex} [From eq. (i)]

{tex} \Rightarrow {/tex} {tex}3x = 180^\circ {/tex} {tex} \Rightarrow {/tex} {tex}x = \frac{{180^\circ }}{3} = 60^\circ {/tex}

NCERT Solutions for Class 7 Maths Exercise 6.3

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