Application 1: Finding Missing Lengths [of Area of Triangle] - Applications of Area of Triangle Formula

part 2 - Application 1: Finding Missing Lengths (The "Two Bases" Trick) - Applications of Area of Triangle Formula - Chapter 7 Class 8 - Area (Ganita Prakash II) - Class 8 (Ganita Prakash - 1, 2 & Old NCERT)
part 3 - Application 1: Finding Missing Lengths (The "Two Bases" Trick) - Applications of Area of Triangle Formula - Chapter 7 Class 8 - Area (Ganita Prakash II) - Class 8 (Ganita Prakash - 1, 2 & Old NCERT) part 4 - Application 1: Finding Missing Lengths (The "Two Bases" Trick) - Applications of Area of Triangle Formula - Chapter 7 Class 8 - Area (Ganita Prakash II) - Class 8 (Ganita Prakash - 1, 2 & Old NCERT)

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Application 1: Finding Missing Lengths (The "Two Bases" Trick)A triangle has three sides, which means it actually has three different base-and-height combinations. Let’s use this to find a missing measurement Find BY. We can find Area of Triangle ∆ ABC using Height AX, Base BC Or Height BY, Base AC Finding Area of ∆ ABC using AX & BC Now, Base = BC = 5 Height = AX = 3 Thus, Area of ∆ ABC = 1/2 × Base × Height = 𝟏/𝟐 × BC × AX = 1/2 × 5 × 3 = 𝟏𝟓/𝟐 square units Finding Area of ∆ using BY & AC Now, Base = AC = 4 Height = BY = ? Area of ∆ ABC = 𝟏𝟓/𝟐 Thus, Area of ∆ ABC = 1/2 × Base × Height Area of ∆ ABC = 𝟏/𝟐 × AC × BY 15/2 = 1/2 × 4 × BY 15/2 = 2 × BY 15/2 ×1/2 = BY 15/4 = BY BY = 𝟏𝟓/𝟒 BY = 3.75 units

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 16 years. He also provides Accounts Tax GST Training in Delhi, Kerala and online.