**Problem 1 :**

5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

**Solution :**

In triangle ABC,

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

5^{2} = AB^{2} + 4^{2}

AB^{2} = 25 - 16

AB^{2} = 9

AB = 3

DB = AB - AD

DB = 3 - 1.6

DB = 1.4

In triangle DBE,

ED^{2} = EB^{2} + DB^{2 }

5^{2} = EB^{2} + (1.4)^{2 }

EB^{2} = 25 - 1.96

EB = √23.04

EB = 4.8

EC = EB - BC

= 4.8 - 4

EC = 0.8 m

So, the required distance is 0.8 m.

**Problem 2 :**

The perpendicular PS on the base QR of a tiranlge PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ^{2} = 2PR^{2} + QR^{2}

**Solution :**

QR = QS + RS

QR = 3RS + RS

QR = 4RS

RS = QR/4

By applying pythagorean theorem in a triangle PQS, we get

PQ^{2} = PS^{2} + QS^{2}

PS^{2 }= PQ^{2} - QS^{2 } -----(1)

By applying pythagorean theorem in a triangle PRS, we get

PR^{2 }= PS^{2} + SR^{2}

PS^{2 }= PR^{2 }- SR^{2} -----(2)

PQ^{2} - QS^{2 }^{ }= PR^{2 }- SR^{2}

PQ^{2} - QS^{2 }^{ }= PR^{2 }- SR^{2}

PQ^{2} ^{ }= PR^{2 }- SR^{2 }+ QS^{2 }

PQ^{2} ^{ }= PR^{2 }- SR^{2 }+ (QR - RS)^{2 }

PQ^{2} ^{ }= PR^{2 }- SR^{2 }+ QR^{2} + RS^{2} - 2QR x RS

PQ^{2} ^{ }= PR^{2 }+ QR^{2} - 2QR x RS

PQ^{2} ^{ }= PR^{2 }+ QR^{2} - 2 QR x (QR/4)

PQ^{2} ^{ }= PR^{2 }+ QR^{2} - (QR^{2}/2)

PQ^{2} ^{ }= (2PR^{2 }+ 2QR^{2} - QR^{2})/2

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

**Problem 3 :**

In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE^{2} = 3AC^{2} + 5AD^{2}

Solution :

AE² = AB² + (2x)²

AE² = AB² + 4x²

In Δ ABC,

AC² = AB² + BC²

AC² = AB + (3x)²

AC² = AB² + 9x²

Now,

3AC² + 5AD² = 3(AB² + 9x²) + 5(AB² + x²)

= 8AB² + 32x²

= 8(AB² + 4x²)

= 8AE²

8AE² = 3AC² + 5AD²

Hence proved.

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