Class 7 Maths

# Properties of Triangles

## Exercise 6.5

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

QR^2 = PQ^2 + PR^2

Or, QR^2 = 10^2 + 24^2 = 100 + 576 = 676

Or, QR^2 = 2 xx 2 xx 13 xx 13

Or, QR = 2 xx 13 = 26 cm

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

AB^2 = AC^2 + BC^2

Or, 25^2 = 7^2 + BC^2

Or, 625 = 49 + BC^2

Or, BC^2 = 625 – 49 = 576

Or, BC^2 = 2 xx 2 xx 2 xx 2 xx 2 xx 2 xx 3 xx 3 = 2^2 xx 2^2 xx 2^2 xx 3^2

Or, BC = 2 xx 2 xx 2 xx 3 = 24

Question 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.

Answer: The ladder is making hypotenuse while the wall is making one of the legs of the triangle.

Using Pythagoras rule, h^2 = p^2 + b^2

Or, b^2 = h^2 – p^2 = 15^2 – 12^2

Or, b^2 = 225 – 144 = 81

Or, b^2 = 3 xx 3 xx 3 xx 3 = 3^2 xx 3^2

Or, b = 3 xx 3 = 9

Question 4: Which of the following can be the sides of a right triangle?

(a) 2.5 cm, 6.5 cm, 6 cm

Answer: Square of the longest side = 6.5^2 = 42.25

Now, 2.5^2 = 6.25 and 6^2 = 36

Adding above two, we get; 6.25 + 36 = 42.25 which is equal to the square of the longest side.

Hence, it is a right angled triangle.

(b) 2 cm, 2 cm, 5 cm

Answer: Square of the longest side = 5^2 = 25

Now, 2^2 = 4

Adding the squares of remaining two sides, we get, 4 + 4 = 16

This figure is not equal to the square of the longest side.

Hence, this is not a right angled triangle.

(c) 1.5 cm, 2 cm, 2.5 cm

Answer: Square of the longest side = 2.5^2 = 6.25

Now, 1.5^2 = 2.25 and 2^2 = 4

Adding the above two, we get, 2.25 + 4 = 6.25

This figure is equal to the square of the longest side.

Hence, it is a right angled triangle.

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

Answer: In options ‘a’ and ‘c’, right angle is made by the smaller sides or legs of the triangle.

Question 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.

Answer: We have; two legs of right angled triangle = 5 m and 12 m

Hypotenuse can be calculated as follows:

h^2 = p^2 + b^2

Or, h^2 = 5^2 +12^2 = 25 + 144 = 169

Or, h^2 = 13 xx 13

Or, h = 13 m

So, original height of tree = 12 + 13 = 25 m

Question 6: Angles Q and R of a ΔPQR are 25° and 65°. Write which of the following is true.

(a) PQ^2 + QR^2 = RP^2

(b) PQ^2 + RP^2 = QR^2

(c) RP^2 + QR^2 = PQ^2

Answer: Here, Sum of angles Q and R = 25° + 65° = 90°

So, angle P = 90°

This means, h = QR while other sides are PQ and PR.

So, QR^2 = PQ^2 + PR^2

Hence, option ‘b’ is correct.

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

Answer: Breadth of the rectangle can be calculated by using Pythagoras rule because length, breadth and diagonal would make a right angled triangle.

So, h^2 = p^2 + b^2

Or, 41^2 = 40^2 + b^2

Or, 1681 = 1600 + b^2

Or, b^2 = 1681 – 1600 = 81

Or, b^2 = 9 xx 9

Or, b = 9 cm.

Perimeter can be calculated as follows:

= 2(40 + 9) = 2 × 49 = 98 cm

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

Answer: Halves of diagonals of a rhombus make the legs of a right angle triangle while hypotenuse is made by a side of the rhombus. So, side of the rhombus can be calculated by using Pythagoras rule;

We know, h^2 = p^2 + b^2

Or, h^2 = 8^2 + 15^2 = 64 + 225 = 289

Or, h^2 = 17 xx 17

Or, h = 17 cm

Hence, Perimeter = 17 × 4 = 68 cm