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    Ncert Solutions For Class 7 Maths 8may Chapter 6 The Triangle and Its Properties Exercise 6 5

    Uploaded by

    Shikha Behl
    The document provides solutions to exercises from NCERT Class 7 Maths Chapter 6 on triangles and their properties. It uses the Pythagorean theorem to solve problems involving right triangles, such as finding the length of the hypotenuse or another side. It also applies the theorem to find distances and perimeters in problems involving broken trees, rectangles, and rhombus shapes. Key steps involve drawing diagrams of the shapes, identifying right triangles, and setting up and solving equations based on the a2 + b2 = c2 formula.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    Ncert Solutions For Class 7 Maths 8may Chapter 6 The Triangle and Its Properties Exercise 6 5

    Uploaded by

    Shikha Behl
    32 views10 pages
    The document provides solutions to exercises from NCERT Class 7 Maths Chapter 6 on triangles and their properties. It uses the Pythagorean theorem to solve problems involving right triangles, such as finding the length of the hypotenuse or another side. It also applies the theorem to find distances and perimeters in problems involving broken trees, rectangles, and rhombus shapes. Key steps involve drawing diagrams of the shapes, identifying right triangles, and setting up and solving equations based on the a2 + b2 = c2 formula.

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    Exercise 6.5

    Original Title

    ncert-solutions-for-class-7-maths-8may-chapter-6-the-triangle-and-its-properties-exercise-6-5

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    The document provides solutions to exercises from NCERT Class 7 Maths Chapter 6 on triangles and their properties. It uses the Pythagorean theorem to solve problems involving right triangles, such as finding the length of the hypotenuse or another side. It also applies the theorem to find distances and perimeters in problems involving broken trees, rectangles, and rhombus shapes. Key steps involve drawing diagrams of the shapes, identifying right triangles, and setting up and solving equations based on the a2 + b2 = c2 formula.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    32 views10 pages

    Ncert Solutions For Class 7 Maths 8may Chapter 6 The Triangle and Its Properties Exercise 6 5

    Uploaded by

    Shikha Behl
    The document provides solutions to exercises from NCERT Class 7 Maths Chapter 6 on triangles and their properties. It uses the Pythagorean theorem to solve problems involving right triangles, such as finding the length of the hypotenuse or another side. It also applies the theorem to find distances and perimeters in problems involving broken trees, rectangles, and rhombus shapes. Key steps involve drawing diagrams of the shapes, identifying right triangles, and setting up and solving equations based on the a2 + b2 = c2 formula.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 10

    NCERT Solutions for Class 7 Maths Chapter 6 -

    The Triangle and its Properties

    Exercise 6.5 Page: 130

    1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

    Solution:-

    Let us draw a rough sketch of a right-angled triangle.

    By the rule of Pythagoras’ Theorem,

    Pythagoras’ theorem states that for any right-angled triangle, the area of the square on the hypotenuse is
    equal to the sum of the areas of squares on the legs.

    In the above figure, RQ is the hypotenuse,

    QR2 = PQ2 + PR2

    QR2 = 102 + 242

    QR2 = 100 + 576

    QR2 = 676

    QR = √676

    QR = 26 cm

    Hence, the length of the hypotenuse QR = 26 cm

    2. ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.


    NCERT Solutions for Class 7 Maths Chapter 6 -
    The Triangle and its Properties

    Solution:-

    Let us draw a rough sketch of the right-angled triangle.

    By the rule of Pythagoras’ Theorem,

    Pythagoras’ theorem states that for any right-angled triangle, the area of the square on the hypotenuse is
    equal to the sum of the areas of squares on the legs.

    In the above figure, RQ is the hypotenuse,

    AB2 = AC2 + BC2

    252 = 72 + BC2

    625 = 49 + BC2

    By transposing 49 from RHS to LHS, it becomes – 49

    BC2 = 625 – 49

    BC2 = 576

    BC = √576

    BC = 24 cm

    Hence, the length of the BC = 24 cm

    3. 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.
    NCERT Solutions for Class 7 Maths Chapter 6 -
    The Triangle and its Properties

    Solution:-

    By the rule of Pythagoras’ Theorem,

    Pythagoras’ theorem states that for any right-angled triangle, the area of the square on the hypotenuse is
    equal to the sum of the areas of squares on the legs.

    In the above figure, RQ is the hypotenuse,

    152 = 122 + a2

    225 = 144 + a2

    By transposing 144 from RHS to LHS, it becomes – 144

    a2 = 225 – 144

    a2 = 81

    a = √81

    a=9m

    Hence, the length of a = 9 m

    4. 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, 2cm, 2.5 cm

    In the case of right-angled triangles, identify the right angles.

    Solution:-

    (i) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm


    NCERT Solutions for Class 7 Maths Chapter 6 -
    The Triangle and its Properties

    Let us assume the largest value is the hypotenuse side, i.e., b = 6.5 cm.

    Then, by Pythagoras’ theorem,

    b2 = a2 + c2

    6.52 = 2.52 + 62

    42.25 = 6.25 + 36

    42.25 = 42.25

    The sum of squares of two sides of the triangle is equal to the square of the third side,

    ∴ the given triangle is a right-angled triangle.

    The right angle lies on the opposite of the greater side, 6.5 cm.

    (ii) Let a = 2 cm, b = 2 cm, c = 5 cm

    Let us assume the largest value is the hypotenuse side, i.e. c = 5 cm.

    Then, by Pythagoras’ theorem,

    c2 = a2 + b2

    52 = 22 + 22

    25 = 4 + 4

    25 ≠ 8

    The sum of squares of two sides of the triangle is not equal to the square of the third side,

    ∴ the given triangle is not a right-angled triangle.

    (iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm

    Let us assume the largest value is the hypotenuse side, i.e., b = 2.5 cm.

    Then, by Pythagoras’ theorem,

    b2 = a2 + c2

    2.52 = 1.52 + 22

    6.25 = 2.25 + 4

    6.25 = 6.25

    The sum of squares of two sides of the triangle is equal to the square of the third side,
    NCERT Solutions for Class 7 Maths Chapter 6 -
    The Triangle and its Properties

    ∴ the given triangle is a right-angled triangle.

    The right angle lies on the opposite of the greater side 2.5 cm.

    5. A tree is broken at a height of 5 m from the ground, and its top touches the ground at a distance of
    12 m from the base of the tree. Find the original height of the tree.

    Solution:-

    Let ABC is the triangle and B be the point where the tree is broken at the height of 5 m from the ground.

    Treetop touches the ground at a distance of AC = 12 m from the base of the tree,

    By observing the figure, we came to conclude that a right-angle triangle is formed at A.

    From the rule of Pythagoras’ theorem,

    BC2 = AB2 + AC2

    BC2 = 52 + 122

    BC2 = 25 + 144

    BC2 = 169

    BC = √169

    BC = 13 m

    Then, the original height of the tree = AB + BC

    = 5 + 13

    = 18 m

    6. Angles Q and R of a ΔPQR are 25o and 65o.

    Write which of the following is true:

    (i) PQ2 + QR2 = RP2


    NCERT Solutions for Class 7 Maths Chapter 6 -
    The Triangle and its Properties

    (ii) PQ2 + RP2 = QR2

    (iii) RP2 + QR2 = PQ2

    Solution:-

    Given that ∠Q = 25o, ∠R = 65o

    Then, ∠P =?

    We know that sum of the three interior angles of a triangle is equal to 180o.

    ∠PQR + ∠QRP + ∠RPQ = 180o

    25o + 65o + ∠RPQ = 180o

    90o + ∠RPQ = 180o

    ∠RPQ = 180 – 90

    ∠RPQ = 90o

    Also, we know that the side opposite to the right angle is the hypotenuse.

    ∴ QR2 = PQ2 + PR2

    Hence, (ii) is true.

    7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

    Solution:-
    NCERT Solutions for Class 7 Maths Chapter 6 -
    The Triangle and its Properties
    NCERT Solutions for Class 7 Maths Chapter 6 -
    The Triangle and its Properties

    Let ABCD be the rectangular plot.

    Then, AB = 40 cm and AC = 41 cm

    BC =?

    According to Pythagoras’ theorem,

    From the right angle triangle ABC, we have

    = AC2 = AB2 + BC2

    = 412 = 402 + BC2

    = BC2 = 412 – 402

    = BC2 = 1681 – 1600

    = BC2 = 81

    = BC = √81

    = BC = 9 cm

    Hence, the perimeter of the rectangle plot = 2 (length + breadth)

    Where, length = 40 cm, breadth = 9 cm

    Then,

    = 2(40 + 9)

    = 2 × 49

    = 98 cm

    8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

    Solution:-
    NCERT Solutions for Class 7 Maths Chapter 6 -
    The Triangle and its Properties

    Let PQRS be a rhombus, all sides of the rhombus have equal length, and its diagonal PR and SQ are
    intersecting each other at point O. Diagonals in the rhombus bisect each other at 90o.

    So, PO = (PR/2)

    = 16/2

    = 8 cm

    And, SO = (SQ/2)

    = 30/2

    = 15 cm
    NCERT Solutions for Class 7 Maths Chapter 6 -
    The Triangle and its Properties

    Then, consider the triangle POS and apply the Pythagoras theorem,

    PS2 = PO2 + SO2

    PS2 = 82 + 152

    PS2 = 64 + 225

    PS2 = 289

    PS = √289

    PS = 17 cm

    Hence, the length of the side of the rhombus is 17 cm

    Now,

    The perimeter of the rhombus = 4 × side of the rhombus

    = 4 × 17

    = 68 cm

    ∴ the perimeter of the rhombus is 68 cm.

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