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.

Answer: Using Pythagoras rule;
`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.

Answer: Using Pythagoras rule;
`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:

Perimeter = 2(length + breadth)
`= 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



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