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Q9(i):

In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that: (i) $\triangle ABC \sim \triangle AMP$

Solution :

Given:

Two triangles, $\triangle ABC$ and $\triangle AMP$.

$\angle ABC = 90^\circ$ (Given that $\triangle ABC$ is right-angled at $B$).

$\angle AMP = 90^\circ$ (Given that $\triangle AMP$ is right-angled at $M$).

To Prove:

$\triangle ABC \sim \triangle AMP$

Visual Representation:

B C A M P

Detailed Steps:

Step 1: Identify the triangles to be compared.

We are considering $\triangle ABC$ and $\triangle AMP$.

Step 2: Compare the angles of the two triangles.

In $\triangle ABC$ and $\triangle AMP$:

(i) $\angle ABC = \angle AMP = 90^\circ$ [Given that both are right-angled triangles at $B$ and $M$ respectively].

(ii) $\angle BAC = \angle MAP$ [This is the common angle shared by both triangles, as vertex $A$ is common to both].

Step 3: Apply the Similarity Criterion.

According to the AA (Angle-Angle) Similarity Criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Since we have established that:

1. $\angle ABC = \angle AMP$ ($90^\circ$ each)

2. $\angle BAC = \angle MAP$ (Common angle)

Therefore, by the AA similarity criterion, we conclude that:

$\triangle ABC \sim \triangle AMP$

Final Answer:

Since two corresponding angles are equal, $\triangle ABC \sim \triangle AMP$ is proved by the AA similarity criterion.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.3


CBSE Solutions for Class 10 Mathematics Triangles


Chapters in CBSE - Class 10 Mathematics


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