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in triangle abc, P,Q,R are mid points. Area of triangle PQR is equal to 16 cm square then find the area of ABC

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Answered by Anonymous
3

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Answered by pasajereynoel
1

Answer:

The area of triangle ABC is 64 cm².

Step-by-step explanation:

In triangle ABC, P,Q,R are mid points. Area of triangle PQR is equal to 16 cm squared, then find the area of ABC.

By drawing the figure based on the given, we can apply the Midpoint Theorem of the triangle which states that the line segment that connects the midpoints of two sides of the triangle is parallel and congruent to one-half of the third side.

Let us prove that ΔPQR ≅ ΔCRQ.

Statement                           Reason

1. PR ║ QC               1. Midpoint Theorem of the Triangle

2. PR ≅ QC              2. Midpoint Theorem of the Triangle

3. ∠PRQ ≅ ∠CQR    3. Alternate interior angles are congruent if two parallel

                                     lines are cut by transversal

4. QR ≅ QR              4. Reflexive Property

5. ΔPQR ≅ ΔCRQ    5. SAS Congruence

Let us prove that ΔPQR ≅ ΔQPA.

Statement                           Reason

1. PR ║ AQ               1. Midpoint Theorem of the Triangle

2. PR ≅ AQ              2. Midpoint Theorem of the Triangle

3. ∠QPR ≅ ∠PQA    3. Alternate interior angles are congruent if two parallel

                                     lines are cut by transversal

4. PQ ≅ PQ              4. Reflexive Property

5. ΔPQR ≅ ΔQPA    5. SAS Congruence

Let us prove that ΔPQR ≅ ΔRBP.

Statement                           Reason

1. PQ ║ BR               1. Midpoint Theorem of the Triangle

2. PQ ≅ BR              2. Midpoint Theorem of the Triangle

3. ∠QPR ≅ ∠BRQ    3. Alternate interior angles are congruent if two parallel

                                     lines are cut by transversal

4. PR ≅ PR              4. Reflexive Property

5. ΔPQR ≅ ΔRBP   5. SAS Congruence

Thus,

ΔPQR ≅ ΔCRQ ≅ ΔQPA = ΔRBP

Area of ΔPQR = 16 cm²

Area of ΔCRQ = 16 cm²

Area of ΔQPA = 16 cm²

Area of ΔRBP = 16 cm²

Area of ΔABC = Area of ΔPQR + Area of ΔCRQ + Area of ΔQPA + Area of ΔRBP

Area of ΔABC = 16 cm² + 16 cm² + 16 cm² + 16 cm²

Area of ΔABC = 64 cm²

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