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# ABC is a right-angled triangle with ∠ABC=90∘. D is any point on AB and DE is perpendicular to AC. Prove that:  [4 MARKS] i) ΔADE∼ΔACB ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm, find DE and AD. iii) Find, area of ΔADE : area of quadrilateral BCED Solution

## Applying theorems: 2 Marks Calculation: 2 Marks i) Given:   ∠ABC=90∘ DE⊥AC ⇒ar(ΔAED)ΔABC=19 ⇒ar(ΔABC)ΔAED=91 ⇒ar(ΔAED)+ar(quad.BCED)ar(ΔAED)=91 ⇒1+ar(quad.BCED)ar(ΔAED)=9 ⇒ar quad. BCEDar(ΔAED)=8 ⇒ar(ΔAED):ar(quad. BCED)=1:8 To prove: ΔADE∼ΔACB Proof: Considering ΔADE and ΔACB, ∠A=∠A     [Common] ∠AED=∠ABC   [90∘] ∴ΔAED∼ΔABC   [By A.A] ii) Given: AC = 1.3 cm BC = 5 cm AE = 4 cm We have, ΔAED∼ΔABC    [Proved above] ⇒AEAB=EDBC=ADAC ⇒4AB=ED5=AD4  .... (1) In ΔABC, By pythagoras theorem, AB=√AC2−BC2 =√132−52=√169−25 =√144=12 cm From (1), 412=ED5=AD4 13=ED5 ⇒ED=53=123 cm Also, 13=AD4 ⇒AD=43=113 cm iii) As ΔAED∼ΔABC ⇒ar(ΔAED)ar(ΔABC)=AE2AB2      [Ratio of areas of two similar Δ′s is equal to the ratio of the squares of corresponding sides] ⇒ar(ΔAED)ar(ΔABC)=42122  Suggest corrections   