Question

Answer soon please! Question and answers are in the photograph attached.

Answer 1

Answer:

Δ MLP  ~  Δ HKG

By S-A-S similarity Postulate.

Step-by-step explanation:

Given:

In Δ MLP

MP = 20 ,

LP = 15

∠ P = 42°

In Δ HKG

HG = 16 ,

KG = 12

∠G = 42°

To Prove:

Δ MLP ~Δ HKG

Proof:

In Δ MLP and Δ HKG

$\dfrac{MP}{HG}= \dfrac{20}{16}=\dfrac{5}{4}$ ..........( 1 )

$\dfrac{LP}{KG}= \dfrac{15}{12}=\dfrac{5}{4}$  ..............( 2 )

∴ $\dfrac{MP}{HG}=\dfrac{LP}{KG}$ ...From 1 and 2

       ∠P ≅ ∠ G = 42°          ...............Given

∴ Δ MLP ~Δ HKG ......{ By S-A-S similarity test}............Proved