Someone plz help on # 15 (idk what to do) :/
1 answer:
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Answer:
A trapezoid ABCD is show below,
AB || CD
AC and BD are diagonals intersecting at P.
(a)
In Δ ABC and ABD,
Area of triangle ABC = Area of triangle ABD
Because,
Area of triangle ABC = 1/2 × Base × Height
Base = AB
Height = h, which is same for both the triangle that's why,
Area of triangle ABC = Area of triangle ABD
(b)
As we know that,
Area of triangle ABC = Area of triangle ABD
On subtracting the area of triangle ABP from both the triangles, we get
Area of triangle ABC - area of triangle ABP = Area of triangle ABD - area of triangle ABP
⇒ area of triangle BCP = area of triangle DAP
Hence, triangles BCP and DAP have the same area.
(c)
In triangle ABP and CDP,
∠ABP = ∠CDP (opposite interior angles are equal between two parallel lines)
∠APB = ∠CPD (opposite angles are equal)
∠BAP = ∠DCP (opposite interior angles are equal between two parallel lines)
Therefore,
By AAA similarity
Triangles ABP and CDP are similar.
(d)
In triangle BCP and DAP,
Only ∠BCP = ∠APD (opposite angles are equal).
Therefore,
Triangles BCP and DAP need not be similar.
