Q26 (b) ABCD is a parallelogram BP and DQ are the perpendiculars from B and D on the - Maths - Areas of Parallelograms and Triangles

Question

Q26
(b) ABCD is a parallelogram BP and DQ are the perpendiculars from B
and D on the diagonal AC
(i) Show that BP = DQ
(ii) Show that ar ( △ ADX) = ar ( △
ABX), where X is any point in AC

Ankita Agarwal · Accepted Answer

Dear Student,

Consider the triangles , ∆BCA and ∆DACAB=CD              opposite sides of parallelogramAD=BC              opposite sides of parallelogramAC=AC              CommonHence ∆BCA ≅∆DAC   By SSSSo,i The area of the corresponding triangles are also equal ar∆BCA =ar∆DAC 12×DQ×AC=12×BP×ACHence BP=DQ   provediiNow consider the triangle ∆ADX and∆ABXAX is the base common to both the triangles  ∆ADX and∆ABXAs, Area of the triangle ∆ADX=12×AX×DQ     ∵ DQ is the altitudeSimilarly , Area of the triangle ∆ABX=12×AX×BP     ∵ BP is the altitudeSince BP=DQ from part iwe have, 12×AX×DQ=12×AX×BP So, ar∆ADX=ar∆ABX  Proved
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Q26. (b) ABCD is a parallelogram. BP and DQ are the perpendiculars from B and D on the diagonal AC. (i) Show that BP = DQ. (ii) Show that ar ( △ ADX) = ar ( △ ABX), where X is any point in AC.

Q26.
(b) ABCD is a parallelogram. BP and DQ are the perpendiculars from B and D on the diagonal AC.
(i) Show that BP = DQ.
(ii) Show that ar ( $△$ ADX) = ar ( $△$ ABX), where X is any point in AC.