Question

In adjoining figure, ABCD is a rectangle. If BP and DQ are perpendiculars on AC from B and D respectively, then is :
(a) AB= CD. Why?
(b)∆BPA=∆DQC. Why?
(c)∆BPA =~∆DQC?
(d)BP=DQ. Why?​

Answer 1

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CBSE

Mathematics

Grade 7

Congruence of triangles

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ABCD is a rectangle in which DP and BQ are perpendiculars from D and B respectively on diagonal AC.

Show that

(i)ΔADP≅ΔCBQ

(ii)∠ADP=∠CBQ

(iii)DP=BQ

Answer

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Hint: ABCD is rectangle which means its opposite sides are equal. We can prove ΔADP≅ΔCBQ by using AAS criterion according to which if two angles and one side of a triangle are equal to the two angles and one side of another triangle then the triangles are congruent. Then by CPCT we can prove ∠ADP=∠CBQ and side DP=BQ .

Complete step-by-step answer:

Given ABCD is a rectangle then AB=CD and AD=BC

And DP and BQ are perpendiculars from D and B respectively on diagonal AC.

Then ∠P=∠Q=90∘ and two triangles ΔADP and ΔCBQ are formed.

Now we have to prove-

(i)We have to prove that ΔADP and ΔCBQare congruent.

In ΔADP and ΔCBQ ,

∠APD=∠CQB=90∘ (Given)

∠DAP=∠BCQ (Because they are alternate angles)

Since opposite sides of rectangle are equal so we can write,

Side AB=side BC

Hence by Angle-Angle-Side congruence

ΔADP≅ΔCBQ Hence Proved

(ii) We have to prove that∠ADP=∠CBQ

Since we have already proved thatΔADP≅ΔCBQ

We know that corresponding parts of two congruent triangles are always equal.

Then by CPCT (Corresponding Parts of Congruent Triangles)

∠ADP=∠CBQ Hence Proved

(iii)We have to prove that side DP=side BQ

Since we have already proved that ΔADP≅ΔCBQ

We know that corresponding parts of two congruent triangles are always equal.

Then by CPCT (Corresponding Parts of Congruent Triangles)

Side DP=side BQ Hence Proved.

Hope this helps..

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