wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If the given figure, ABC is a right triangle right angled at A, BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y.

Show that
(i) Δ MBC Δ ABD

(ii) ar (BYXD) = 2 ar (Δ MBC)

(iii) ar (BYXD) = ar (ABMN)

(iv) Δ FCB Δ ACE

(v) ar (CYXE) = 2 ar (ΔFCB)

(vi) ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (ABMN) + ar (ACFG)

Open in App
Solution

Given:

(1) ABCD is a right angled triangle at A

(2) BCED, ACFG and ABMN are the squares on the sides of BC, CA and AB respectively.

(3) , meets BC at Y.

To prove:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Proof:

(i)

…… (1)

(ii) Triangle ABD and rectangle BYXD are on the same base BD and between the same parallels AX and BD.

Therefore

(Using (1)) …… (2)

(iii) Since ΔMBC and square MBAN are on the same base MB and between the same parallels MB and NC.

…… (3)

From (2) and (3) we get

(iv) In triangle FCB and ACE

…… (4)

(v) Since ΔACE and rectangle CYXE are on the same base CE and between the same parallels CE and AX.

…… (5)

(vi) Since ΔFCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG

…… (6)

From (5) and (6) we get

(vii) Applying Pythagoras Theorem in ΔACB, WE get


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Inequalities in Triangles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon