Prove that the line segment joining the mid point of any two sides of a triangle is parallel to the third side and is equal to half of it . In given figure , de =3/4bc , De parallel bc and the distance between dc and bc is 3 units. If bc = 6 units , what is the area of triangle bce? The figure is is a right angle triangle at top A point and at right angle B point and another point is C . A point D at AB is join to E on AC(it seen as parallel) . E point join to point B . Plz answer me
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First I'm proving that the line segment joining the midpoints of any two lines of a triangle is parallel to the third side and its length is half of that of the third side.
Draw ΔPQR and mark the midpoints of PQ and PR as M and N respectively.
Draw MN.
Consider ΔMNP and ΔPQR,
∠P = ∠P (common angle)
PN x 2 = PR
PM x 2 = PQ
If one angle of a triangle is equal to one angle of another triangle, and the length of the sides of the angle of the first is equal to the length of the sides of the angle of the second, then both triangles will be similar.
∴ ΔMNP ~ ΔPQR.
∴ ∠N = ∠R,
∠M = ∠Q,
and MN x 2 = QR.
∴ the length of the line joining the midpoints of any two sides of a triangle is half of that of the third side.
If ∠N = ∠R and ∠M = ∠Q, MN should be parallel to QR because PR and PQ cut MN and QR.
Now let's find the area of ΔBCE.
The distance between DC and BC is also the distance between DE and BC (BD), i. e., 3 units.
Because, on considering ΔBCE and ΔBCD, DE || BC and both triangles have the same base (BC). Two triangles, which have same base and lie on the same parallel lines, have same area and same altitude.
∴ If we draw an altitude from E to BC, its length is the same as BD = 3 units.
Area of ΔBCE
= 1/2 x BC x BD
= 1/2 x 6 x 3
= 1/2 x 18 = 9 square unit.
Draw ΔPQR and mark the midpoints of PQ and PR as M and N respectively.
Draw MN.
Consider ΔMNP and ΔPQR,
∠P = ∠P (common angle)
PN x 2 = PR
PM x 2 = PQ
If one angle of a triangle is equal to one angle of another triangle, and the length of the sides of the angle of the first is equal to the length of the sides of the angle of the second, then both triangles will be similar.
∴ ΔMNP ~ ΔPQR.
∴ ∠N = ∠R,
∠M = ∠Q,
and MN x 2 = QR.
∴ the length of the line joining the midpoints of any two sides of a triangle is half of that of the third side.
If ∠N = ∠R and ∠M = ∠Q, MN should be parallel to QR because PR and PQ cut MN and QR.
Now let's find the area of ΔBCE.
The distance between DC and BC is also the distance between DE and BC (BD), i. e., 3 units.
Because, on considering ΔBCE and ΔBCD, DE || BC and both triangles have the same base (BC). Two triangles, which have same base and lie on the same parallel lines, have same area and same altitude.
∴ If we draw an altitude from E to BC, its length is the same as BD = 3 units.
Area of ΔBCE
= 1/2 x BC x BD
= 1/2 x 6 x 3
= 1/2 x 18 = 9 square unit.
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Answer:
as height is 3 units and base is 6 units of triangle BCE then area =1/2*6*3 =9 units
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