#### If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

#### In a right triangle ABC, right angled at B, AC = 13 cm, BC = 5 cm. The length of AB is ____.

AB^{2} = AC^{2} – BC^{2} (Pythagoras theorem) ⇒ AB^{2} = 13^{2 }− 5^{2} ⇒ AB^{2} = 169 – 25 ⇒ AB^{2} = 144 ∴ AB = 12 cm Hence, the length of AB is 12 cm

#### If P and Q are the mid points of the sides CA and CB respectively of a Δ ABC right angled at C, then AQ^{2 } in terms of AC and BC is _____.

Construction: Join AQ, BP and PQ. In a Δ ACQ AQ^{2} = AC^{2} + QC^{2} = AC^{2} + (1 /2 BC)^{2} = AC^{2} + BC^{2} /4 ∴ 4AQ^{2} = 4AC^{2} + BC^{2}

#### In a right triangle XYZ, right angled at Y, XY = 6 cm, XZ = 10 cm. The length of YZ is ____.

YZ^{2} = XZ^{2 }– XY^{2} (By Pythagoras theorem) ⇒ YZ^{2} = 10^{2} - 6^{2 } ⇒ YZ^{2 }= 100 – 36 ⇒ YZ^{2} = 64 ∴ YZ = 8cm Hence, the length of YZ is 8 cm.

#### The hypotenuse of a right triangle is 25 units and one of its perpendicular sides is 7 units. The its perimeter is ____.

By pythagoras theorem, AC^{2} = AB^{2 }+ BC^{2} ⇒ 625 = AB^{2} + 49 ⇒ AB^{2} = 576 ⇒ AB = 24 Perimeter = 25 + 7 + 24 = 56 Hence, the perimeter of the triangle is 56 units.

#### The triangle whose sides are 6 cm, 8 cm and 10 cm, is a right triangle.

In △ABC, a = 6 cm, b = 8 cm and c = 10 cm a^{2} +b^{2 }= c^{2} (By Pythagoras theorem) ⇒ 6^{2} + 8^{2} = 10^{2} ⇒ 100 = 100 Hence, the given triangle is a right triangle.

#### Two triangles are said to be similar, if their corresponding angles are equal and the ratio of their corresponding sides is equal.

Two triangles are said to be similar, if their corresponding angles are equal and the ratio of their corresponding sides is equal.

#### A man goes 150 m due east and then 200 m due north. The distance from the starting point is ____.

Let A be the starting point Let AB = 150 m, BC = 200 m and ∠B = 90° In △ABC, By Pythagoras theorem, AC^{2} = AB^{2 }+ BC^{2 } = (150)^{2} + (200)^{2} = 22500 + 40000 AC^{2} = 62500 ∴ AC = 250 m Hence, he is 250 m from the starting point.

#### In a trapezium ABCD, AB∥DC and AB = 2 CD. If area ΔAOB = 84 cm^{2}, then the area of ΔCOD = ____.

Given that AB∥ CD and AB = 2CD In ΔAOB and ΔCOD, ∠AOB = ∠COD (vertically opposite angles) ∠OAB = ∠OCD (Alternate angles) ∠OBA = ∠OCD (Alternate angles) ⇒ ΔAOB ~ ΔCOD ⇒ Area(ΔAOB) / Area(ΔCOD) = AB^{2} /CD^{2 } = (2CD)^{2} /CD^{2} = 4 = Area(ΔCOD) = Area(ΔAOB) /4 = 84 /4 ∴ Area (ΔCOD) = 21 cm^{2}

#### A ladder is placed against a wall such that its foot is at a distance of 7 m and its top reaches a window. If the length of the ladder is 25 m, then the distance between the window and the ground is ____.

Let BC be the height of the wall with window at C and AC is the length of the ladder ⇒AB = 7 m and AC = 25 m By Pythagoras theorem, AC^{2} = AB^{2} + BC^{2} ⇒ BC^{2} = AC^{2 }– AB^{2} = 25^{2} - 7^{2} = 625 – 49 ⇒ BC^{2} = 576 ⇒ BC = 24 Hence, the the distance between the window and the ground is 24 m.