Saved Bookmarks
| 1. |
Parallelogram ABCD. AB 5cm. AB, DC distance 4cm. P, Q, R are the point on the line DC. ADB APB AQB ARB are the triangles whose base is AB and third vertices lie on the line DC.a) What are the special features of sides AB&DC? |
|
Answer» Answer: yep Step-by-step EXPLANATION: SOLUTION : Given : In ΔABC, PQ is a line segment intersecting AB at P and AC at Q. AP = 1 cm , PB = 3cm, AQ= 1.5 cm and QC= 4.5cm. In ∆APQ and ∆ABC, ∠A = ∠A [Common] AP/AB = AQ/AC [Each equal to 1/4] ∆APQ ~ ∆ABC [By SAS similarity] We know that the ratio of the two similar triangles is equal to the ratio of the squares of their corresponding sides ar∆APQ /ar∆ABC = (AP/AB)² ar∆APQ /ar∆ABC = ( ¼)² ar∆APQ /ar∆ABC = 1/16 ar∆APQ = 1/16 × ar∆ABC pls mark as BRAINLIEST |
|