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