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.
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
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/