Question

ABCD is a trapezium, segAB I l seg DC,

seg DE side AB, seg CF I side AB.

Find: (i) DE and CF (ii). BF (iii). AB.​

Answer 1

Answer:

Construction: Draw seg DE ⊥ seg AB, A – E – B  and seg CF ⊥ seg AB, A – F- B. In ∆ ACB, ∠ACB = 90° [Given]  ∴ AB2 = AC2 + BC2 [Pythagoras theorem]  ∴ 252 = AC2 + 152  ∴ AC2 = 625 – 225 = 40 ∴ AC = √400 [Taking square root of both sides]  = 20 units  Now, A(∆ABC) = 1/2 × BC × AC  Also, A(∆ABC) = 1/2 × AB × CF ∴ BC × AC = AB × CF  ∴ 15 × 20 = 25 × CF  ∴ CF = (15 x 20)/25 = 12 units  In ∆CFB, ∠CFB 90° [Construction]  ∴ BC2 = CF2 + FB2 [Pythagoras theorem]  ∴ 152 = 122 + FB2  ∴ FB2 = 225 – 144  ∴ FB2 = 81Read more on Sarthaks.com - https://www.sarthaks.com/851222/in-a-trapezium-abcd-seg-ab-seg-dc-seg-bd-seg-ad-seg-ac-seg-bc-if-ad-15-bc-15-and-ab-25-find-a-abcd

Answer 2

Answer:

Step-by-step explanation:

in reality the question asked is different