**The Triangle and Its Properties**

**Exercise 6.5**

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

In DPQR, By Pythagorean Property

QR^{2} = PQ^{2} + PR^{2}

QR^{2} = 10^{2} + 24^{2}

QR^{2} = 100 + 576

QR^{2} = 676

Taking square root on both sides

√QR^{2} = √676

QR = 26 cm

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

In DABC, By Pythagorean Property

AB^{2} = AC^{2} + BC^{2}

25^{2} = 7^{2} + BC^{2}

625 = 49 + BC^{2}

625 – 49 = BC^{2}

BC^{2 = }625 – 49

BC^{2 }= 576

Taking square root on both sides

√BC^{2} = √576

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.

Let AB be the height of the window from the ground

BC be the distance of the ladder from the wall

AC be the length of the ladder

AC = 15 m

AB = 12 m

BC = ?

In DABC, By Pythagorean Property

AC^{2} = AB^{2} + BC^{2}

15^{2} = 12^{2} + BC^{2}

225 = 144 + BC^{2}

225 – 144 = BC^{2}

BC^{2 = }225 – 144

BC^{2 }= 81

Taking square root on both sides

√BC^{2} = √81

BC = 9 m

Ans: the distance of the foot of the ladder from the wall is 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) 2.5 cm,6.5 cm, 6 cm

Greater side = 6.5 cm

6.5^{2} = 42.25

6^{2} = 36

2.5^{2} = 6.25

6^{2} + 2.5^{2} = 36 + 6.25 = 42.25

6^{2} + 2.5^{2 }= 6.5^{2}

By Pythagorean Property

Given triangle is right angled triangle.

(ii) 2 cm, 2 cm, 5 cm.

Greater side = 5 cm

5^{2} = 25

2^{2} = 4

2^{2} = 4

2^{2} + 2^{2} = 4 + 4 = 8

2^{2} + 2^{2 }≠ 5^{2}

By Pythagorean Property

Given triangle is not a right angled triangle.

(iii) 1.5 cm, 2cm, 2.5 cm.

Greater side = 2.5 cm

2.5^{2} = 6.25

2^{2} = 4

1.5^{2} = 2.25

2^{2} + 1.5^{2} = 4 + 2.25 = 6.25

2^{2} + 1.5^{2 }= 2.5^{2}

By Pythagorean Property

Given triangle is right angled triangle.

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.

Let AC be the original height of the tree and the tree is broken at B touching the ground at D

AC = ?

AB = 5 m

AD = 12 m

In DABD, By Pythagorean Property

BD^{2} = AB^{2} + AD^{2}

BD^{2} = 5^{2} + 12^{2}

BD^{2} = 25 + 144

BD^{2 } = 169

Taking square root on both sides

√BD^{2} = √169

BD = 13 m

But BC = BD

BC = 13 m

AC = 5 + BC

AC = 5 + 13

AC = 18 m

Ans: the original height of the tree is 18 m.

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

Write which of the following is true:

(i) PQ^{2} + QR^{2} = RP^{2}

(ii) PQ^{2} + RP^{2} = QR^{2}

(iii) RP^{2} + QR^{2} = PQ^{2}

Solution:

by angle sum property of a triangle

mÐP + mÐQ + mÐR = 180^{0}

mÐP + 25^{0 }+ 65^{0} = 180^{0 }

mÐP + 90^{0} = 180^{0}

* *mÐP = 180^{0 }– 90^{0}

mÐP = 90^{0}

DPQR is a right angles triangle and QR is a hypotenuse

(i) PQ^{2} + QR^{2} = RP^{2}

False

(ii) PQ^{2} + RP^{2} = QR^{2}

True

(iii) RP^{2} + QR^{2} = PQ^{2}

False

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

Let □PQRS is the given rectangle

Each angle of a rectangle is 90^{0}.

DPQR is a right angled triangle

In DPQR, By Pythagorean Property

PR^{2} = PQ^{2} + QR^{2}

41^{2} = 40^{2} + QR^{2}

1681 = 1600 + QR^{2}

1681 – 1600 = QR^{2}

QR^{2 = }1681 – 1600

QR^{2 }= 81

Taking square root on both sides

√ QR^{2} = √81

QR = 9 cm

Length = 40 cm

Breadth = 9 cm

Perimeter= 2(L+B)

= 2(40+9)

= 2(49)

= 98 cm

Ans: Perimeter of the given rectangle = 98 cm

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

Let □ABCD is a rhombus

Diagonals AC and BD intersect each other at M

AC = 16 cm

BD = 30 cm

Diagonals of rhombus bisect each other at right angles

AM = ½ AC = ½ x 16 = 8 cm

DM = ½ BD = ½ x 30 = 15 cm

DAMD is a right angled triangle

By Pythagorean Property

AD^{2} = AM^{2} + DM^{2}

AD^{2} = 8^{2} + 15^{2}

AD^{2} = 64 + 225

AD^{2} = 289

Taking square root on both sides

√ AD^{2} = √289

AD= 17 cm

All the sides of rhombus are equal

Perimeter = 4 X side

= 4 X 17

= 68 cm

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