Question 3 - Figure it out - Page 157-159 - Chapter 7 Class 8 - Area (Ganita Prakash II)

Transcript

Question 3 Find the area of ΔSUB, given that it is isosceles, SE is perpendicular to UB, and the area of ΔSEB is 24 sq. units.We use the property: In an Isosceles triangle, altitude bisects the side So, we can write UE = BE = 𝟏/𝟐 UB We can prove this using Congruent triangles We need to find Area ∆ SUB In ∆ SUB Base = UB Height = SE Thus, Area of ∆ SUB = 1/2 × Base × Height = 𝟏/𝟐 × UB × SE Now, Given that area of ΔSEB is 24 sq. units. In ∆ SEB Base = EB = 𝟏/𝟐 UB Height = SE Thus, Area of ∆ SEB = 1/2 × Base × Height 24 = 𝟏/𝟐 × (𝟏/𝟐 UB) × SE 24 × 2 = 1/2 × UB × SE 48 = 𝟏/𝟐 × UB × SE Putting Area of ∆ SUB = 1/2 × UB × SE 48 = Area of ∆ SUB Area of ∆ SUB = 48 square units We could have directly written this as well Direct answer An Isosceles triangle is perfectly symmetrical down the middle. Because line SE drops straight down to form a right angle, it cuts the large triangle into two identical right-angled triangles. If the area of the right half ∆ SEB is 24 sq units, then the left half ∆ SIB must also be 24 sq units. Total Area = 24 + 24 = 48 sq units.