in the given figure AD:DB = 1:3 and DE is parallel to BC, if the area of the triangle ABC is 432 cm² , what is the are of triangle ADE.
Answers
“In triangle ABC, AD = DB , DE is parallel to BC, and the area of triangle ABC is 40. What is the area of triangle ADE?”
If AD = DB and DE is parallel to BC then..
Tri ADE is similar to ABC.
Area of ABC = H * 1/2 * BC ; ( Area of a Triangle is the Height X 1/2 the Base)
But we are given H * 1/2* BC = 40
BD = DA, So DE must be 1/2 of BC.
(Notes: Because AD/AB is 1/2 ; then DE/BC must also be 1/2 ; Likewise H of ADE is 1/2 of H of ABC; Similar Triangles)
Area of DAE then is H*1/2 * 1/4* BC (Notes; DE is 1/2 of BC, so half of DE is 1/4 of BC ).
Juggling we get; 1/4(H*1/2*BC)
But we are given H*1/2 * BC = 40 ; so we can substitute 40 for that in our equation… 1/4 ( 40 ) = Area of Triangle DAE.
Ergo Area of Triangle DAE = 1/4 * 40 …which is 10
Answer Area of DAE = 10 Sq units.
Hope that helps.
Supplement.
Consider the Square ABCD, and in it the Triangle ABC.
40
BD = DA, So DE must be 1/2 of BC
In triangle ABC, AD = DB , DE is parallel to BC, and the area of triangle ABC is 40. What is the area of triangle ADE?
“In triangle ABC, AD = DB , DE is parallel to BC, and the area of triangle ABC is 40. What is the area of triangle ADE?”
If AD = DB and DE is parallel to BC then..
Tri ADE is similar to ABC.
Area of ABC = H * 1/2 * BC ; ( Area of a Triangle is the Height X 1/2 the Base)
But we are given H * 1/2* BC = 40
BD = DA, So DE must be 1/2 of BC.
(Notes: Because AD/AB is 1/2 ; then DE/BC must also be 1/2 ; Likewise H of ADE is 1/2 of H of ABC; Similar Triangles)
Area of DAE then is H*1/2 * 1/4* BC (Notes; DE is 1/2 of BC, so half of DE is 1/4 of BC ).
Juggling we get; 1/4(H*1/2*BC)
But we are given H*1/2 * BC = 40 ; so we can substitute 40 for that in our equation… 1/4 ( 40 ) = Area of Triangle DAE.
Ergo Area of Triangle DAE = 1/4 * 40 …which is 10
Answer Area of DAE = 10 Sq units.
Hope that helps.
Supplement.
Consider the Square ABCD, and in it the Triangle ABC.
Given AX = XB ; AZ = ZD ; XO =OY : and XY is parallel to BC, let’s illustrate that the area of
Triangle AXO is 1/4 of the area of Triangle ABC.
So XY divides the Square ABCD into halves. So the top half contains the Triangle AXO in the Square AXOZ. However AO bisects AXOZ into two equal halves, so Triangle AXO is 1/4 of the area of the Parallelogram AXYD.
Now the Triangle ABC is also 1/2 of the Square ABCD , ( Diagonal bisects square).
So by equality AXO is also 1/4 the area of Triangle ABC.
This is a Theorem in some book somewhere, if not I claim it, ha ha !!!!