Question

9. In the given figure, ABCD is a parallelogram in which AB = CD and BDI CD such that BD 6.8 cm, then the area of parallelogram is D с 6.8cm A 5cm B (a) 17 cm2 (b) 25 cm2 (c) 34 cm2 (d) 68 cm2​

Answer 1

Answers

Given:

Given:∠BDC=90

Given:∠BDC=90 ∠ABC=90

Let ∠BCD=x

Let ∠BCD=x Using triangle sum property in △BDC, ∠DBC=90−x

Let ∠BCD=x Using triangle sum property in △BDC, ∠DBC=90−x Also ∠ABD=x

Let ∠BCD=x Using triangle sum property in △BDC, ∠DBC=90−x Also ∠ABD=x Using triangle sum property in △ADB, ∠BAD=90−x

Let ∠BCD=x Using triangle sum property in △BDC, ∠DBC=90−x Also ∠ABD=x Using triangle sum property in △ADB, ∠BAD=90−x Now considering △BDC and △ADB

Let ∠BCD=x Using triangle sum property in △BDC, ∠DBC=90−x Also ∠ABD=x Using triangle sum property in △ADB, ∠BAD=90−x Now considering △BDC and △ADB∠BDC=∠BDA

Let ∠BCD=x Using triangle sum property in △BDC, ∠DBC=90−x Also ∠ABD=x Using triangle sum property in △ADB, ∠BAD=90−x Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD

Let ∠BCD=x Using triangle sum property in △BDC, ∠DBC=90−x Also ∠ABD=x Using triangle sum property in △ADB, ∠BAD=90−x Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DAB

Let ∠BCD=x Using triangle sum property in △BDC, ∠DBC=90−x Also ∠ABD=x Using triangle sum property in △ADB, ∠BAD=90−x Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA

Let ∠BCD=x Using triangle sum property in △BDC, ∠DBC=90−x Also ∠ABD=x Using triangle sum property in △ADB, ∠BAD=90−x Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADB

Let ∠BCD=x Using triangle sum property in △BDC, ∠DBC=90−x Also ∠ABD=x Using triangle sum property in △ADB, ∠BAD=90−x Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence CC /DB= BD /AB

AB

5.4/BB= 3.8/5.7

5.7 BC=8.1

5.7 BC=8.1Hence BC=8.1cm