Question

in the given figure find PM​
please help
please

Answer 1

Let the triangle near the base be Δ1 and the other be Δ2. Let the right angles of Δ1 and Δ2 be ∠5 and ∠6 respectively. Let equal angles be ∠1 (for Δ1) and ∠2 (for Δ2), and unequal angles of both triangles be ∠3 (for Δ1) and ∠4 (for Δ2).

In the given figure, ∠1 = ∠2 (given)

Therefore, 90- ∠1 = 90- ∠2 (I)

∠1+∠3+∠5 = 180° (Sum of all angles in a triangle)

∠1+∠3+90° = 180° (given ∠5= 90°)

∠1+∠3 = 90°

∠3= 90°-∠1 (II)

Similarly, ∠4 = 90°-∠2 (III)

Substituting (II) and (III) in (I),

∠3 = ∠4 (IV)

Now, ∠3 = ∠4 (IV)

∠1 = ∠2 (given)

OP common

So, Δ1 ≅ Δ2 (ASA)

Hence, LP = MP (CPCT) (V)

In Δ2, LP²+LO²= PO² (Pythagoras Theorem)

LP²= PO²-LO²

LP²= 25- 9

LP²= 16

LP= 4 (VI)

Comparing (V) and (VI), MP= LP= 4 cm.

Therefore, length of LP= 4 cm.