CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard
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CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard
Last updated at April 16, 2024 by Teachoo
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Given PA = 10 cm We need to find perimeter of ∆PCD i.e. we need to find PC, PD and CD From Point A We know that Tangent from external point are equal ∴ PA = PB = 10 cm From Point C Since Tangent from external point are equal ∴ CA = CE From Point C Since Tangent from external point are equal ∴ DB = DE Now, Perimeter of ∆PCD = PC + PD + CD Putting CD = CE + ED = PC + PD + (CE + ED) Putting CE = AC, DE = DB = PC + PD + (AC + DB) = (PC + AC) + (PD + DB) = PA + PB = 10 + 10 = 20 cm
