Given: Diagonals AC = 30 cm and DB = 16 cm. Since the diagonals of the rhombus bisect at right angle to each other. Therefore, OD = DB/2 = 16/2 = 8cm And OC = AC/2 = 30/2 = 15 cm Now, In right angle triangle DOC, (DC)² = (OD)² + (OC)² [By Pythagoras theorem] ⇒ (DC)² = (8)² + (15)² ⇒ (DC)² = 64 + 225Read more
Given: Diagonals AC = 30 cm and DB = 16 cm.
Since the diagonals of the rhombus bisect at right angle to each other.
Therefore, OD = DB/2 = 16/2 = 8cm
And OC = AC/2 = 30/2 = 15 cm
Now, In right angle triangle DOC,
(DC)² = (OD)² + (OC)² [By Pythagoras theorem]
⇒ (DC)² = (8)² + (15)²
⇒ (DC)² = 64 + 225 = 289
⇒ DC = √289 = 17
Perimeter of rhombus = 4 x side = 4 x 17 = 68 cm
Thus, the perimeter of rhombus is 68 cm.
Given diagonal (PR) = 41 cm, length (PQ) = 40 cm Let breadth (QR) be x cm. Now, in right angled triangle PQR, (PR)² = (RQ)² + (PQ)² [By Pythagoras theorem] ⇒ (41)² = x² + 1600 ⇒ 1681 = x² + 1600 ⇒ x² = 1681 – 1600 ⇒ x² = 81 ⇒ x= √81 = 9 cm Therefore the breadth of the rectangle is 9 cm. Perimeter ofRead more
Given diagonal (PR) = 41 cm, length (PQ) = 40 cm
Let breadth (QR) be x cm.
Now, in right angled triangle PQR,
(PR)² = (RQ)² + (PQ)² [By Pythagoras theorem]
⇒ (41)² = x² + 1600
⇒ 1681 = x² + 1600
⇒ x² = 1681 – 1600
⇒ x² = 81
⇒ x= √81 = 9 cm
Therefore the breadth of the rectangle is 9 cm.
Perimeter of rectangle = 2(length + breadth)
= 2 (9 + 49)
= 2 x 49 = 98 cm
Hence, the perimeter of the rectangle is 98 cm.
Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem, (Hypotenuse)² = (Base)² + (Perpendicular)² (i) 2.5 cm, 6.5 cm, 6 cm In ∆ABC, (AC)² = (AB)² + (BC)² L.H.S. = (6.5)² = 42.25 cm R.H.S. = (6)² + (2.5)² = 36 + 6.25 = 42.25 cm Since, L.H.S. = R.H.S. Therefore, the giveRead more
Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)²
(i) 2.5 cm, 6.5 cm, 6 cm
In ∆ABC, (AC)² = (AB)² + (BC)²
L.H.S. = (6.5)² = 42.25 cm
R.H.S. = (6)² + (2.5)² = 36 + 6.25 = 42.25 cm
Since, L.H.S. = R.H.S.
Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.
(ii) 2 cm, 2 cm, 5 cm
In the given triangle, (5)² = (2)² + (2)²
L.H.S. = (5)² = 25
R.H.S. = (2)² + (2)² = 4+4=8
Since, L.H.S. ≠ R.H.S.
Therefore, the given sides are not of the right angled triangle.
(iii) 1.5 cm, 2 cm, 2.5 cm
In ∆ PQR, (PR)² = (PQ)² + (RQ)²
L.H.S. = (2.5)² = 6.25
R.H.S. = (1.5)² + (2)² = 2.25 + 4 = 6.25 cm
Since, L.H.S. = R.H.S.
Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q.
Let AC be the ladder and A be the window. Given: AC = 15 m, AB = 12 m, CB = a m In right angled triangle ACB, (Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem] ⇒ (AC)² = (CB)² + (AB)² ⇒ (15)² = (a)² + (12)² ⇒ 225 = a² + 144 ⇒ a² = 225 – 144 = 81 ⇒ a √81 = 9 cm Thus, the distance ofRead more
Let AC be the ladder and A be the window.
Given: AC = 15 m, AB = 12 m, CB = a m
In right angled triangle ACB,
(Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem]
⇒ (AC)² = (CB)² + (AB)²
⇒ (15)² = (a)² + (12)²
⇒ 225 = a² + 144
⇒ a² = 225 – 144 = 81
⇒ a √81 = 9 cm
Thus, the distance of the foot of the ladder from the wall is 9 m.
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.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Given: Diagonals AC = 30 cm and DB = 16 cm. Since the diagonals of the rhombus bisect at right angle to each other. Therefore, OD = DB/2 = 16/2 = 8cm And OC = AC/2 = 30/2 = 15 cm Now, In right angle triangle DOC, (DC)² = (OD)² + (OC)² [By Pythagoras theorem] ⇒ (DC)² = (8)² + (15)² ⇒ (DC)² = 64 + 225Read more
Given: Diagonals AC = 30 cm and DB = 16 cm.
Since the diagonals of the rhombus bisect at right angle to each other.
Therefore, OD = DB/2 = 16/2 = 8cm
And OC = AC/2 = 30/2 = 15 cm
Now, In right angle triangle DOC,
(DC)² = (OD)² + (OC)² [By Pythagoras theorem]
⇒ (DC)² = (8)² + (15)²
⇒ (DC)² = 64 + 225 = 289
⇒ DC = √289 = 17
Perimeter of rhombus = 4 x side = 4 x 17 = 68 cm
Thus, the perimeter of rhombus is 68 cm.
Class 7 Maths Chapter 6 Exercise 6.5
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Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Given diagonal (PR) = 41 cm, length (PQ) = 40 cm Let breadth (QR) be x cm. Now, in right angled triangle PQR, (PR)² = (RQ)² + (PQ)² [By Pythagoras theorem] ⇒ (41)² = x² + 1600 ⇒ 1681 = x² + 1600 ⇒ x² = 1681 – 1600 ⇒ x² = 81 ⇒ x= √81 = 9 cm Therefore the breadth of the rectangle is 9 cm. Perimeter ofRead more
Given diagonal (PR) = 41 cm, length (PQ) = 40 cm
Let breadth (QR) be x cm.
Now, in right angled triangle PQR,
(PR)² = (RQ)² + (PQ)² [By Pythagoras theorem]
⇒ (41)² = x² + 1600
⇒ 1681 = x² + 1600
⇒ x² = 1681 – 1600
⇒ x² = 81
⇒ x= √81 = 9 cm
Therefore the breadth of the rectangle is 9 cm.
Perimeter of rectangle = 2(length + breadth)
= 2 (9 + 49)
= 2 x 49 = 98 cm
Hence, the perimeter of the rectangle is 98 cm.
Class 7 Maths Chapter 6 Exercise 6.5
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Angles Q and R of a ∆PQR are 25° and 65°. Write which of the following is true: (i) PQ² + QR² = RP² (ii) PQ² + RP² = QR² (iii) RP² + QR² = PQ²
In ∆ PQR, ∠PQR + ∠QRP + ∠RPQ = 180° [By Angle sum property of a ∆ ] ⇒ 25° + 65° + ∠RPQ = 180 ⇒ 90° ∠RPQ=180° ⇒ ∠RPQ = 180° - 90° = 90° Thus, ∆ PQR is a right angled triangle, right angled at P. ∴ (Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem] ⇒ (QR)² = (PR)² + (QP)² Hence, OptioRead more
In ∆ PQR,
∠PQR + ∠QRP + ∠RPQ = 180° [By Angle sum property of a ∆ ]
⇒ 25° + 65° + ∠RPQ = 180
⇒ 90° ∠RPQ=180°
⇒ ∠RPQ = 180° – 90° = 90°
Thus, ∆ PQR is a right angled triangle, right angled at P.
∴ (Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem]
⇒ (QR)² = (PR)² + (QP)²
Hence, Option (ii) is correct.
Class 7 Maths Chapter 6 Exercise 6.5
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Which of the following can be the sides of a right triangle? (i) 2.5 cm, 6.5 cm, 6 cm (ii) 2 cm, 2 cm, 5 cm (iii) 1.5 cm, 2 cm, 2.5 cm In the case of right angled triangles, identify the right angles.
Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem, (Hypotenuse)² = (Base)² + (Perpendicular)² (i) 2.5 cm, 6.5 cm, 6 cm In ∆ABC, (AC)² = (AB)² + (BC)² L.H.S. = (6.5)² = 42.25 cm R.H.S. = (6)² + (2.5)² = 36 + 6.25 = 42.25 cm Since, L.H.S. = R.H.S. Therefore, the giveRead more
Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)²
(i) 2.5 cm, 6.5 cm, 6 cm
In ∆ABC, (AC)² = (AB)² + (BC)²
L.H.S. = (6.5)² = 42.25 cm
R.H.S. = (6)² + (2.5)² = 36 + 6.25 = 42.25 cm
Since, L.H.S. = R.H.S.
Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.
(ii) 2 cm, 2 cm, 5 cm
In the given triangle, (5)² = (2)² + (2)²
L.H.S. = (5)² = 25
R.H.S. = (2)² + (2)² = 4+4=8
Since, L.H.S. ≠ R.H.S.
Therefore, the given sides are not of the right angled triangle.
(iii) 1.5 cm, 2 cm, 2.5 cm
In ∆ PQR, (PR)² = (PQ)² + (RQ)²
L.H.S. = (2.5)² = 6.25
R.H.S. = (1.5)² + (2)² = 2.25 + 4 = 6.25 cm
Since, L.H.S. = R.H.S.
Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q.
Class 7 Maths Chapter 6 Exercise 6.5
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A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a Find the distance of the foot of the ladder from the wall
Let AC be the ladder and A be the window. Given: AC = 15 m, AB = 12 m, CB = a m In right angled triangle ACB, (Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem] ⇒ (AC)² = (CB)² + (AB)² ⇒ (15)² = (a)² + (12)² ⇒ 225 = a² + 144 ⇒ a² = 225 – 144 = 81 ⇒ a √81 = 9 cm Thus, the distance ofRead more
Let AC be the ladder and A be the window.
Given: AC = 15 m, AB = 12 m, CB = a m
In right angled triangle ACB,
(Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras theorem]
⇒ (AC)² = (CB)² + (AB)²
⇒ (15)² = (a)² + (12)²
⇒ 225 = a² + 144
⇒ a² = 225 – 144 = 81
⇒ a √81 = 9 cm
Thus, the distance of the foot of the ladder from the wall is 9 m.
Class 7 Maths Chapter 6 Exercise 6.5
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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
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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
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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
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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
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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
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