Question

In triangle ABC, E is a point on AC such that AE:EC =5:7. ED
is drawn parallel to CB and DF is drawn parallel to BE.
Find
ar∆ADE/ar∆ADE
correct answer with explanation will be marked as the brainliest.
spammers wil get 20 answers reported ,so stay away.​

Answer 1

Answer:

In △ ADE and △ ABC, we have

∠ ADE = ∠ B

[Since DE || BC ∠ ADE = ∠ B (Corresponding angles)]

and, ∠ A = △ A [Common]

△ ADE ~ △ ABC

Therefore, (area of △ ADE / area of △ ABC) = (AD2/AB2)

3.

In the figure, PB and QA are perpendicular to segment AB. If OA = 5 cm, PO = 7cm

and area (ΔQOA) = 150 cm2, find the area of ΔPOB.

(A) 233 cm2

(B) 294 cm2

(C) 300 cm2

(D) 420 cm2

Answer: (B) 294 cm

Answer 2

∆ABC,

∠B + ∠BAC + ∠C = 180°

⇒ ∠B + 90° + ∠C = 180°

⇒ ∠B = 90° – ∠C

Similarly, In ∆ADC, ∠D AC = 90° – ∠C

In ∆ADB and ∆ADC,

∠D = ∠D = 90°

∠DBA = ∠D AC [each equal to (90° – ∠C)

∴ ∆ADB ~ ∆CDA

[by AA similarity criterion]

∴ BDAD=ADCD

⇒ BD . CD = AD2