CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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) ΔADEΔACB
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
DEAC

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:
ΔADEΔACB
Proof: Considering ΔADE and ΔACB,
A=A     [Common]
AED=ABC   [90]
ΔAEDΔABC   [By A.A]

ii) Given: AC = 1.3 cm BC = 5 cm AE = 4 cm
We have, ΔAEDΔABC    [Proved above]
AEAB=EDBC=ADAC
4AB=ED5=AD4  .... (1)
In ΔABC, By pythagoras theorem,
AB=AC2BC2
=13252=16925
=144=12 cm
From (1),
412=ED5=AD4
13=ED5
ED=53=123 cm
Also, 13=AD4
AD=43=113 cm

iii) As ΔAEDΔABC
ar(ΔAED)ar(ΔABC)=AE2AB2      [Ratio of areas of two similar Δs is equal to the ratio of the squares of corresponding sides]
ar(ΔAED)ar(ΔABC)=42122

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image