Question

In the figure drawn algonside, ∆XYZ is a right triangle, right angled at Y such that YZ = b

and A(∆XYZ) = a.

If YP ⊥ XZ, then show that

YP =

4 2

2ab

b 4a +

​

Answer 1

Answer:

BC=√3DE

Step-by-step explanation:

Explanation:

In triangle ABC, seg DE || SEG BC. if twice area of triangle ADE = area of quadrilateral DBCE, find AB:AD. SHOW THAT BC = √3 d

DE || BC

hence ADE & ABC are similar

BC/DE = AB / AD = AC / AE = k

area of ABC = k^2 Area of ADE

Area of DBCE = Area of ABC - ARea of ADE

2(Area of ADE) = Area of DBCE

2(Area of ADE) = Area of ABC - ARea of ADE

3(Area of ADE) = Area of ABC

3(Area of ADE) = k^2 Area of ADE

k^2 = 3

k = √3

BC/DE = k = √3

BC = √3DE