Question

ABC is a triangle in which ∠A= 90°,AN⊥BC , BC = 12 cm and AC = 5 cm. Find the ratio of the areas of ΔANC and ΔABC .

Answer 1

The ratio of the areas of ΔANC and ΔABC is $\frac{25}{144}$.

Step-by-step explanation:

Given :

∠A= 90°,AN⊥BC , BC = 12 cm and AC = 5 cm

In ΔCAB and ΔCAN

∠ACN = ∠ACB  (∠C is common)

∠CAB = ∠CNA = 90° (AN ⊥BC)

By applying the Angle - Angle Similarity theorem,

we get ΔABC ≈ ΔANC.

According to the Area of Similar Triangle theorem, the ratio of area of similar triangles is proportional to  square of  ratio of the corresponding sides.

$\frac{\text{ Area of triangle ANC}}{\text{Area of triangle ABC}} = \frac{\textt{AC}^2}{\text{BC}^2}$

$\frac{\text{ Area of triangle ANC}}{\text{Area of triangle ABC}} = \frac{5^2}{12^2}$                                            

$\frac{\text{ Area of triangle ANC}}{\text{Area of triangle ABC}} = \frac{25}{144}$                                                    

To Learn More......

1. Proof of the theorem: ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

brainly.in/question/2605465

2. Question 9 In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that: (i) ΔABC ∼ ΔAMP

brainly.in/question/1345158