In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate: (i) EF (ii) AC

(i) In ∆PCD and ∆PEF, ∠CPD = ∠EPF [Vertically opposite angles are equal] ∠DCP = ∠FEP [Alternate angles are equal] ∴ ∆PCD ~ ∆PEF [By AA] Since corresponding sides of similar triangle are proportional to each other. Hence, EF = 13.5 cm. (ii) In ∆CEF and ∆CAB ∠FCE = ∠BCA [Common angles] ∠CFE = ∠CBA [As FE || BA, corresponding angles are equal] ∴ ∆CEF ~ ∆CAB [By AA] Since corresponding sides of similar triangle are proportional to each other. Hence, AC = 37.5 cm.

View Answer Bookmark Now

In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.
(i) Prove that ΔACD is similar to ΔBCA.
(ii) Find BC and CD.
(iii) Find the area of ΔACD : area of ΔABC.

View Answer Bookmark Now

In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB,
(ii) AG : GD = 2 : 1

View Answer Bookmark Now

Mathematics

In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.
Calculate: (i) EF (ii) AC

Similarity

Related Questions

In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.
(i) Prove that ΔACD is similar to ΔBCA.
(ii) Find BC and CD.
(iii) Find the area of ΔACD : area of ΔABC.

In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB,
(ii) AG : GD = 2 : 1

Mathematics

In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.Calculate: (i) EF (ii) AC

Similarity

Related Questions

In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.(i) Prove that ΔACD is similar to ΔBCA.(ii) Find BC and CD.(iii) Find the area of ΔACD : area of ΔABC.

In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that:(i) EF = FB,(ii) AG : GD = 2 : 1

In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.
Calculate: (i) EF (ii) AC

In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.
(i) Prove that ΔACD is similar to ΔBCA.
(ii) Find BC and CD.
(iii) Find the area of ΔACD : area of ΔABC.

In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB,
(ii) AG : GD = 2 : 1