In △PQR, ∠Q = 90° and QM is perpendicular to PR. Prove that : (i) PQ 2 = PM × PR (ii) QR 2 = PR × MR (iii) PQ 2 + QR 2 = PR 2

△PQR is shown in the figure below: (i) In △PQR and △PMQ, ⇒ ∠PMQ = ∠PQR [Both = 90°] ⇒ ∠QPM = ∠RPQ [Common] ∴ △PQR ~ △PMQ [By AA] Since, corresponding sides of similar triangles are proportional we have : ⇒ ⇒ PQ 2 = PM × PR Hence, proved that PQ 2 = PM × PR. (ii) In △QRM and △PRQ, ⇒ ∠QMR = ∠PQR [Both = 90°] ⇒ ∠QRM = ∠QRP [Common] ∴ △QRM ~ △PRQ [By AA] Since, corresponding sides of similar triangles are proportional we have : ⇒ ⇒ QR 2 = PR × MR Hence, proved that QR 2 = PR × MR. (iii) Adding equations from (i) and (ii) we get, ⇒ PQ 2 + QR 2 = PM × PR + PR × MR .........(1) ⇒ PQ 2 + QR 2 = PR(PM + MR) From figure, PM + MR = PR ⇒ PQ 2 + QR 2 = PR 2 . Hence, proved that PQ 2 + QR 2 = PR 2 .

In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm, L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. Find the lengths of PN and RM.

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In △ABC, ∠B = 90° and BD ⊥ AC.
(i) If CD = 10 cm and BD = 8 cm; find AD.
(ii) If AC = 18 cm and AD = 6 cm; find BD.
(iii) If AC = 9 cm and AB = 7 cm; find AD.

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In the given figure, ΔABC and ΔAMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove that : ∆ABC ~ ∆AMP.
(ii) Find AB and BC.

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D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that: CA² = CB x CD.

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Mathematics

In △PQR, ∠Q = 90° and QM is perpendicular to PR. Prove that :
(i) PQ² = PM × PR
(ii) QR² = PR × MR
(iii) PQ² + QR² = PR²

Similarity

Related Questions

In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm, L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. Find the lengths of PN and RM.

In △ABC, ∠B = 90° and BD ⊥ AC.
(i) If CD = 10 cm and BD = 8 cm; find AD.
(ii) If AC = 18 cm and AD = 6 cm; find BD.
(iii) If AC = 9 cm and AB = 7 cm; find AD.

In the given figure, ΔABC and ΔAMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove that : ∆ABC ~ ∆AMP.
(ii) Find AB and BC.

D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that: CA² = CB x CD.

Mathematics

In △PQR, ∠Q = 90° and QM is perpendicular to PR. Prove that :(i) PQ2 = PM × PR(ii) QR2 = PR × MR(iii) PQ2 + QR2 = PR2

Similarity

Related Questions

In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm, L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. Find the lengths of PN and RM.

In △ABC, ∠B = 90° and BD ⊥ AC.(i) If CD = 10 cm and BD = 8 cm; find AD.(ii) If AC = 18 cm and AD = 6 cm; find BD.(iii) If AC = 9 cm and AB = 7 cm; find AD.

In the given figure, ΔABC and ΔAMP are right angled at B and M respectively.Given AC = 10 cm, AP = 15 cm and PM = 12 cm.(i) Prove that : ∆ABC ~ ∆AMP.(ii) Find AB and BC.

D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that: CA2 = CB x CD.

In △PQR, ∠Q = 90° and QM is perpendicular to PR. Prove that :
(i) PQ² = PM × PR
(ii) QR² = PR × MR
(iii) PQ² + QR² = PR²

In △ABC, ∠B = 90° and BD ⊥ AC.
(i) If CD = 10 cm and BD = 8 cm; find AD.
(ii) If AC = 18 cm and AD = 6 cm; find BD.
(iii) If AC = 9 cm and AB = 7 cm; find AD.

In the given figure, ΔABC and ΔAMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove that : ∆ABC ~ ∆AMP.
(ii) Find AB and BC.

D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that: CA² = CB x CD.