Question

Solve any two sub question.
1) Point P is on the perpendicular bisector of Seg AB .such that, distance AB is more than 7cm than distance AP. If perimeter of triangleABP is 38cm then find the sides of
triangleABp



​

Answer 1

Answer:

MARK AS A BRAINLIST

Step-by-step explanation:

 l (BP) = l (AP) = y [By perpendicular bisector theorem]

As per first condition,

y = x + 7

∴ – x + y = 7 .......eq. no. (1)

As per the second condition,

Perimeter of ∆ABP = 38

∴ x + y + y = 38

∴  x + 2y = 38 ......(2)

Adding (1) and (2),

- x + y = 7

x + 2y = 38

   3y = 45

∴ y = 45/3

∴ y = 15

Substituting y = 15 in equation (2)

∴ x + 2y = 38

∴ x + 2(15) = 38

∴ x + 30 = 38

∴ x = 38 – 30

∴ x = 8

∴l (AB) = 8 cm, l (BP) = l (AP) = 15 cm.

The length of sides of ∆ ABP are 8 cm, 15 cm and 15 cm.

Answer 2

Answer:

Let the length of AB =x cm

P ison the perpedicular bisector,

AP=x+7

BP=x+7(since P is on the⊥bisector;AP=BP)

x+x+7+x+7=38

3x=24

x=8

x+7=15cm

So the three sides are8,15,15cm