ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm. and BP = 3cm., AQ = 1.5 cm., CQ = 4.5 cm.Prove that (area of ΔAPQ) = 1/16 (area of ΔABC).
Answers
Answer:
Step-by-step explanation:
AP = 1 cm (Given)
PB = 3cm (Given)
AQ= 1.5 cm (Given)
CQ = 4.5cm (Given)
Since, AP = 1 cm and PB = 3 cm then AB = 4 cm
As the sides are in proportion, therefore line PQ is parallel to BC
In ΔAPQ and ΔABC
∠ A = ∠ A (common)
∠APQ = ∠B (alternate interior angles)
∠AQP = ∠C (alternate interior angles)
Therefore, ΔAPQ ~ ΔABC
= (APQ)/ (ABC) = AP²/ AB²(area of two similar triangles is equal to the square of their corresponding side)
(APQ) / (ABC) = 12/ 42
(APQ) / (ABC) = 1/16
Thus, (APQ) = 1/ 16 (ABC)
Answer:
Step-by-step explanation:
Step-by-step explanation:
AP = 1 cm (Given)
PB = 3cm (Given)
AQ= 1.5 cm (Given)
CQ = 4.5cm (Given)
Since, AP = 1 cm and PB = 3 cm then AB = 4 cm
As the sides are in proportion, therefore line PQ is parallel to BC
In ΔAPQ and ΔABC
∠ A = ∠ A (common)
∠APQ = ∠B (alternate interior angles)
∠AQP = ∠C (alternate interior angles)
Therefore, ΔAPQ ~ ΔABC
= (APQ)/ (ABC) = AP²/ AB²(area of two similar triangles is equal to the square of their corresponding side)
(APQ) / (ABC) = 12/ 42
(APQ) / (ABC) = 1/16
Thus, (APQ) = 1/ 16 (ABC)