Given: AB = 25 cm, AC = 7 cm Let BC be x cm. In right angled triangle ACB, (Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem] ⇒ (AB)² = (AC)² + (BC)² ⇒ (25)² = (7)² ⇒ 625 = 49 + x² ⇒ x² = 625 – 49 = 576 ⇒ x² = √576 = 24 cm Thus, the length of BC is 24 cm. Class 7 Maths Chapter 6 ExerRead more
Given: AB = 25 cm, AC = 7 cm
Let BC be x cm.
In right angled triangle ACB,
(Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem]
⇒ (AB)² = (AC)² + (BC)²
⇒ (25)² = (7)²
⇒ 625 = 49 + x²
⇒ x² = 625 – 49 = 576
⇒ x² = √576 = 24 cm
Thus, the length of BC is 24 cm.
Given: PQ = 10 cm, PR = 24 cm Let QR be x cm. In right angled triangle QPR, (Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem] ⇒ (QR)² = (PQ)² + (PR)² ⇒ x² = (10)² + (24)² ⇒ x² = 100 + 576 = 676 ⇒ x = √676 = 26 cm Thus, the length of QR is 26 cm. Class 7 Maths Chapter 6 Exercise 6.5Read more
Given: PQ = 10 cm, PR = 24 cm
Let QR be x cm.
In right angled triangle QPR,
(Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem]
⇒ (QR)² = (PQ)² + (PR)²
⇒ x² = (10)² + (24)²
⇒ x² = 100 + 576 = 676
⇒ x = √676 = 26 cm
Thus, the length of QR is 26 cm.
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side. It is given that two sides of triangle are 12 cm and 15 cm. Therefore, the third side should be less than 12 + 15 = 27 cm. And also the third side cannot be less than the difference of the two sRead more
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
It is given that two sides of triangle are 12 cm and 15 cm.
Therefore, the third side should be less than 12 + 15 = 27 cm.
And also the third side cannot be less than the difference of the two sides.
Therefore, the third side has to be more than 15 – 12 = 3 cm.
Hence, the third side could be the length more than 3 cm and less than 27 cm.
Therefore, In AOB, AB < OA + OB ……….(i) In ∆BOC, BC < OB + OC ……….(ii) In ∆COD, CD < OC + OD ……….(iii) In ∆AOD, DA < OD + OA ……….(iv) Adding equations (i), (ii), (iii) and (iv), we get AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA ⇒ AB + BC + CD + DA < 2OA + 2OB + 2OCRead more
Therefore, In AOB, AB < OA + OB ……….(i)
In ∆BOC, BC < OB + OC ……….(ii)
In ∆COD, CD < OC + OD ……….(iii)
In ∆AOD, DA < OD + OA ……….(iv)
Adding equations (i), (ii), (iii) and (iv), we get
AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA
⇒ AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD
⇒ AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]
⇒ AB + BC + CD + DA < 2(AC + BD)
Hence, it is proved.
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side. Therefore, In ∆ABC, AB + BC > AC ……….(i) In ∆ADC, AD + DC > AC ……….(ii) In ∆DCB, DC + CB > DB ……….(iii) In ∆ADB, AD + AB > DB ……….(iv) Adding equations (i), (ii), (iii) and (iv), weRead more
Since, the sum of lengths of any two sides in a triangle should be greater than the length
of third side.
Therefore, In ∆ABC, AB + BC > AC ……….(i)
In ∆ADC, AD + DC > AC ……….(ii)
In ∆DCB, DC + CB > DB ……….(iii)
In ∆ADB, AD + AB > DB ……….(iv)
Adding equations (i), (ii), (iii) and (iv), we get
AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB
⇒ (AB + AB) + (BC + BC) + (AD + AD) + (DC + DC) > 2AC + 2DB
⇒ 2AB + 2BC + 2AD + 2DC > 2(AC + DB)
⇒ 2(AB + BC + AD + DC) > 2(AC + DB)
⇒ AB + BC + AD + DC > AC + DB
⇒ AB + BC + CD + DA > AC + DB
Hence, it is true.
ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Given: AB = 25 cm, AC = 7 cm Let BC be x cm. In right angled triangle ACB, (Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem] ⇒ (AB)² = (AC)² + (BC)² ⇒ (25)² = (7)² ⇒ 625 = 49 + x² ⇒ x² = 625 – 49 = 576 ⇒ x² = √576 = 24 cm Thus, the length of BC is 24 cm. Class 7 Maths Chapter 6 ExerRead more
Given: AB = 25 cm, AC = 7 cm
Let BC be x cm.
In right angled triangle ACB,
(Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem]
⇒ (AB)² = (AC)² + (BC)²
⇒ (25)² = (7)²
⇒ 625 = 49 + x²
⇒ x² = 625 – 49 = 576
⇒ x² = √576 = 24 cm
Thus, the length of BC is 24 cm.
Class 7 Maths Chapter 6 Exercise 6.5
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Given: PQ = 10 cm, PR = 24 cm Let QR be x cm. In right angled triangle QPR, (Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem] ⇒ (QR)² = (PQ)² + (PR)² ⇒ x² = (10)² + (24)² ⇒ x² = 100 + 576 = 676 ⇒ x = √676 = 26 cm Thus, the length of QR is 26 cm. Class 7 Maths Chapter 6 Exercise 6.5Read more
Given: PQ = 10 cm, PR = 24 cm
Let QR be x cm.
In right angled triangle QPR,
(Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem]
⇒ (QR)² = (PQ)² + (PR)²
⇒ x² = (10)² + (24)²
⇒ x² = 100 + 576 = 676
⇒ x = √676 = 26 cm
Thus, the length of QR is 26 cm.
Class 7 Maths Chapter 6 Exercise 6.5
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side. It is given that two sides of triangle are 12 cm and 15 cm. Therefore, the third side should be less than 12 + 15 = 27 cm. And also the third side cannot be less than the difference of the two sRead more
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
It is given that two sides of triangle are 12 cm and 15 cm.
Therefore, the third side should be less than 12 + 15 = 27 cm.
And also the third side cannot be less than the difference of the two sides.
Therefore, the third side has to be more than 15 – 12 = 3 cm.
Hence, the third side could be the length more than 3 cm and less than 27 cm.
Class 7 Maths Chapter 6 Exercise 6.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?
Therefore, In AOB, AB < OA + OB ……….(i) In ∆BOC, BC < OB + OC ……….(ii) In ∆COD, CD < OC + OD ……….(iii) In ∆AOD, DA < OD + OA ……….(iv) Adding equations (i), (ii), (iii) and (iv), we get AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA ⇒ AB + BC + CD + DA < 2OA + 2OB + 2OCRead more
Therefore, In AOB, AB < OA + OB ……….(i)
In ∆BOC, BC < OB + OC ……….(ii)
In ∆COD, CD < OC + OD ……….(iii)
In ∆AOD, DA < OD + OA ……….(iv)
Adding equations (i), (ii), (iii) and (iv), we get
AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA
⇒ AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD
⇒ AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]
⇒ AB + BC + CD + DA < 2(AC + BD)
Hence, it is proved.
Class 7 Maths Chapter 6 Exercise 6.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side. Therefore, In ∆ABC, AB + BC > AC ……….(i) In ∆ADC, AD + DC > AC ……….(ii) In ∆DCB, DC + CB > DB ……….(iii) In ∆ADB, AD + AB > DB ……….(iv) Adding equations (i), (ii), (iii) and (iv), weRead more
Since, the sum of lengths of any two sides in a triangle should be greater than the length
of third side.
Therefore, In ∆ABC, AB + BC > AC ……….(i)
In ∆ADC, AD + DC > AC ……….(ii)
In ∆DCB, DC + CB > DB ……….(iii)
In ∆ADB, AD + AB > DB ……….(iv)
Adding equations (i), (ii), (iii) and (iv), we get
AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB
⇒ (AB + AB) + (BC + BC) + (AD + AD) + (DC + DC) > 2AC + 2DB
⇒ 2AB + 2BC + 2AD + 2DC > 2(AC + DB)
⇒ 2(AB + BC + AD + DC) > 2(AC + DB)
⇒ AB + BC + AD + DC > AC + DB
⇒ AB + BC + CD + DA > AC + DB
Hence, it is true.
Class 7 Maths Chapter 6 Exercise 6.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/