Question

if BD:DE:BC =2:5:3 and area of triangle ABC is 10 sq units,then find area of triangle ABC

Answer 1

Answer:

9/6 is the answer

Step-by-step explanation:

● Given = DE /BC = 3/5

●Solution =

Since, triangle's ( ADE ~ ABC )

Then, area (triangle ADE) /area ( triangle ABC ) = DE^2 /BC^2

= DE^2/BC^2 = 3^2/5^2 = 9/25

Let area of triangle ADE = 9 square units

& area of triangle ABC = 25 square units

Therefore, area of trap. BCED = area of triangle ABC - area of triangle ADE

= ( 25 - 9 )x square units

= 16 x square units

Therefore, area ( triangle ADE ) / area ( trap. BCED ) = 9x /16x

= 9 / 16

Hope it helps ✌

Answer 2

Answer: 2

Step-by-step explanation:

So the height of all the three triangles is the same.

As given the ratio of the lengths are 2:5:3.

Lets take the unknown value as x

lets take the height as h

so the area of each triangle is

ABD - 1/2 x 2xh

ADE - 1/2 x 5xh

AEC- 1/2 x 3xh

adding all the areas we get the total area which is 10

2xh/2 + 5xh/2 + 3xh/2 = 10

(2xh + 5xh + 3xh) / 2 = 10

10xh = 20

xh = 2

Hence the area of ABD is  1/2 x 2xh

                                           = 1/2 x 2 x 2

                                          = 2

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