In the given figure, ABCD is a parallelogram.
1C is extended to a point Z such that AZ
meets diagonal DB at X and side DC at Y and
3DX = XB.
() Prove that AAXD - AZXB.
(ii) Name a triangle similar to ADXY.
(iii) Calculate the ratio of area of A AXD :
area of A ZXB.
(iv) If AB = 6 cm, AD = 4 cm, find the lengths
of BZ and DY.
Answers
Given : ABCD is a parallelogram BC is extended to point Z such that AZ meet Diagonal DB at X and DC at Y
3DX = XB
AB = 6 cm , AD = 4 cm
To Find : Show that ∠AXD = ∠ZXB
Δ Similar to ΔDXY
ratio of area of Δ AXD : area of Δ ZXB
Length of BZ and DY
Solution:
∠AXD = ∠ZXB Vertically opposite angles
ΔDXY & Δ BXA
∠DXY = ∠BXA
∠XDY = ∠XBA (alternate angles)
∠XYD = ∠XAB (alternate angles)
=> ΔDXY ≈ Δ BXA
DX/BX = DY/AB
=> DX/3DX = DY/6
=> DY = 2 cm
ΔAXD & Δ ZXB
∠AXD = ∠ZXB Vertically opposite angles
∠XDA = ∠XBZ (alternate angles)
∠XAD = ∠XZB (alternate angles)
=> ΔAXD ≈Δ ZXB
DX/XB = AD/BZ
=> DX/3DX = 4/BZ
=> 1/3 = 4/BZ
=> BZ = 12 cm
ratio of area of Δ AXD : area of Δ ZXB = (DX/XB)² = (1/3)² = 1/9
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