CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard
You are here
CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard
Last updated at Dec. 13, 2024 by Teachoo
Since opposite sides of parallelogram are equal AB = CD Let AB = CD = x Point P divides AB in ratio 2 : 3 Thus, π¨π·/π©π·=π/π Adding 1 both sides π΄π/π΅π+1=2/3+1 (π΄π + π΅π)/π΅π=(2 + 3)/3 π΄π΅/π΅π=5/3 Since AB = x π₯/π΅π=5/3 BP = π/π π Also, AP = π/π π Point P divides AB in ratio 2 : 3 Thus, π¨π·/π©π·=π/π Adding 1 both sides π΄π/π΅π+1=2/3+1 (π΄π + π΅π)/π΅π=(2 + 3)/3 π΄π΅/π΅π=5/3Point Q divides DC in ratio 4 : 1 Thus, π«πΈ/πΈπͺ=π/π Adding 1 both sides π·π/ππΆ+1=4+1 (π·π + ππΆ)/ππΆ=5 π·πΆ/ππΆ=5 In ΞAOP & ΞCOQ β AOP = β COQ β APO = β CQO β΄ ΞAOP ~ ΞCOQ Since sides of similar triangle are proportional Thus, π¨π·/πͺπΈ=πΆπ¨/πΆπͺ Putting AP = π/π π and QC = π/π π (π/π π)/(π/π π)=πΆπ¨/πΆπͺ 2/1=ππ΄/ππΆ ππ=π/π ππ β΄ OC is half of OA Hence proved
