To construct: An isosceles right angled triangle ABC where m∠C = 90° , AC = BC = 6 cm. Steps of construction: (a) Draw a line segment AC = 6 cm. (b) At point C, draw XC ⊥ 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 isoscelRead more
To construct:
An isosceles right angled triangle ABC where m∠C = 90° , AC = BC = 6 cm.
Steps of construction:
(a) Draw a line segment AC = 6 cm.
(b) At point C, draw XC ⊥ 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.
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 ⊥ 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 rightRead more
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 ⊥ 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.
To construct: A right angled triangle PQR where m∠Q =90°, QR = 8 cm and PQ = 10 cm. Steps of construction: (a) Draw a line segment QR = 8 cm. (b) At point Q, draw QX ⊥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 angleRead more
To construct:
A right angled triangle PQR where m∠Q =90°, QR = 8 cm and PQ = 10 cm.
Steps of construction:
(a) Draw a line segment QR = 8 cm.
(b) At point Q, draw QX ⊥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.
Given: In ∆DEF, m∠E = 110° and m∠F = 80°. Using angle sum property of triangle ∠𝐷 + ∠𝐸 + ∠𝐹 = 180° ⟹ ∠𝐷 + 110° + 80° = 180° ⟹ ∠𝐷 + 190° = 180° ⟹ ∠𝐷 = 180° − 190° = −10° Which is not possible. Class 7 Maths Chapter 10 Exercise 10.4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutioRead more
Given: In ∆DEF, m∠E = 110° and m∠F = 80°.
Using angle sum property of triangle
∠𝐷 + ∠𝐸 + ∠𝐹 = 180°
⟹ ∠𝐷 + 110° + 80° = 180°
⟹ ∠𝐷 + 190° = 180°
⟹ ∠𝐷 = 180° − 190° = −10°
Which is not possible.
Given: m∠PQR = 105° and m∠QRP = 40° We know that sum of angles of a triangle is 180°. ⇒ m∠PQR + m∠QRP + m∠QPR = 180° ⇒ 105° + 40° + m∠QPR = 180° ⇒ 145° + m∠QPR = 180° ⇒ m∠QPR = 180° – 145° ⇒ m∠QPR = 35° To construct: ∆PQR where m∠P = 35° , m∠Q = 105° and PQ = 5 cm. Steps of construction: (a) Draw aRead more
Given: m∠PQR = 105° and m∠QRP = 40°
We know that sum of angles of a triangle is 180°.
⇒ m∠PQR + m∠QRP + m∠QPR = 180°
⇒ 105° + 40° + m∠QPR = 180°
⇒ 145° + m∠QPR = 180°
⇒ m∠QPR = 180° – 145°
⇒ m∠QPR = 35°
To construct: ∆PQR where m∠P = 35° , m∠Q = 105° and PQ = 5 cm.
Steps of construction:
(a) Draw a line segment PQ = 5 cm.
(b) At point P, draw ∠XPQ = 35° with the help of protractor.
(c) At point Q, draw ∠YQP = 105° with the help of protractor.
(d) XP and YQ intersect at point R.
It is the required triangle PQR.
Construct an isosceles right angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
To construct: An isosceles right angled triangle ABC where m∠C = 90° , AC = BC = 6 cm. Steps of construction: (a) Draw a line segment AC = 6 cm. (b) At point C, draw XC ⊥ 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 isoscelRead more
To construct:
An isosceles right angled triangle ABC where m∠C = 90° , AC = BC = 6 cm.
Steps of construction:
(a) Draw a line segment AC = 6 cm.
(b) At point C, draw XC ⊥ 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.
Class 7 Maths Chapter 10 Exercise 10.5
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/
Construct a right angled triangle whose hypotenuse is 6 cm long and one the legs is 4 cm long.
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 ⊥ 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 rightRead more
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 ⊥ 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.
Class 7 Maths Chapter 10 Exercise 10.5
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/
Construct the right angled ∆PQR, where m∠Q = 90°, QR = 8 cm and PR = 10 cm.
To construct: A right angled triangle PQR where m∠Q =90°, QR = 8 cm and PQ = 10 cm. Steps of construction: (a) Draw a line segment QR = 8 cm. (b) At point Q, draw QX ⊥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 angleRead more
To construct:
A right angled triangle PQR where m∠Q =90°, QR = 8 cm and PQ = 10 cm.
Steps of construction:
(a) Draw a line segment QR = 8 cm.
(b) At point Q, draw QX ⊥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.
Class 7 Maths Chapter 10 Exercise 10.5
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/
Examine whether you can construct ∆DEF such that EF = 7.2 cm, m∠E = 110° and m∠F = 80°. Justify your answer.
Given: In ∆DEF, m∠E = 110° and m∠F = 80°. Using angle sum property of triangle ∠𝐷 + ∠𝐸 + ∠𝐹 = 180° ⟹ ∠𝐷 + 110° + 80° = 180° ⟹ ∠𝐷 + 190° = 180° ⟹ ∠𝐷 = 180° − 190° = −10° Which is not possible. Class 7 Maths Chapter 10 Exercise 10.4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutioRead more
Given: In ∆DEF, m∠E = 110° and m∠F = 80°.
Using angle sum property of triangle
∠𝐷 + ∠𝐸 + ∠𝐹 = 180°
⟹ ∠𝐷 + 110° + 80° = 180°
⟹ ∠𝐷 + 190° = 180°
⟹ ∠𝐷 = 180° − 190° = −10°
Which is not possible.
Class 7 Maths Chapter 10 Exercise 10.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/
Construct ∆PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°.
Given: m∠PQR = 105° and m∠QRP = 40° We know that sum of angles of a triangle is 180°. ⇒ m∠PQR + m∠QRP + m∠QPR = 180° ⇒ 105° + 40° + m∠QPR = 180° ⇒ 145° + m∠QPR = 180° ⇒ m∠QPR = 180° – 145° ⇒ m∠QPR = 35° To construct: ∆PQR where m∠P = 35° , m∠Q = 105° and PQ = 5 cm. Steps of construction: (a) Draw aRead more
Given: m∠PQR = 105° and m∠QRP = 40°
We know that sum of angles of a triangle is 180°.
⇒ m∠PQR + m∠QRP + m∠QPR = 180°
⇒ 105° + 40° + m∠QPR = 180°
⇒ 145° + m∠QPR = 180°
⇒ m∠QPR = 180° – 145°
⇒ m∠QPR = 35°
To construct: ∆PQR where m∠P = 35° , m∠Q = 105° and PQ = 5 cm.
Steps of construction:
(a) Draw a line segment PQ = 5 cm.
(b) At point P, draw ∠XPQ = 35° with the help of protractor.
(c) At point Q, draw ∠YQP = 105° with the help of protractor.
(d) XP and YQ intersect at point R.
It is the required triangle PQR.
Class 7 Maths Chapter 10 Exercise 10.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/