Chapter 6: The Triangle and Its Properties

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

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

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 = 9 m

Hence, the length of a = 9 m

Solution:-

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

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,

∴ the given triangle is a right-angled triangle.

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

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

Solution:-

Given that ∠Q = 25^o, ∠R = 65^o

Then, ∠P =?

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

∠PQR + ∠QRP + ∠RPQ = 180^o

25^o + 65^o + ∠RPQ = 180^o

90^o + ∠RPQ = 180^o

∠RPQ = 180 – 90

∠RPQ = 90^o

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

∴ QR² = PQ² + PR²

Hence, (ii) is true.

Solution:-

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

Solution:-

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 90^o.

So, PO = (PR/2)

= 16/2

= 8 cm

And, SO = (SQ/2)

= 30/2

= 15 cm

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.