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Question

Match the two columns for the figure shown below.



Column - IColumn - II(A) Area of DABC(p) 41(B) Area of ΔADC(q) 126(C) Area ofΔABC(r) 306(D) AC(s) 180


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Solution

Using Pythagorus theorem in ΔABC

AC=AB2+BC2

=92+402

=81+1600=41 m

(A) Area of \(\square DABC = area~(\Delta ABC) + area~(\Delta ADC)\

area (ΔABC)=12(a)(40)=180 m2forΔADC

S=28+15+412=42 m

area (ΔADC)=42(4228)(4215)(4241)

=42(14)(27)

=2 × 3 × 7 × 2 × 7 × 3 × 3 ×3

= 2 x 3 x 3 x 7 = 126 m2

areaDABC=180+126=306 m2

(B) Area of ΔABC=126 m2

(C) Area of ΔABC=180 m2

(D) AC=41m

(A)r;(B)q;(C)s;(D)p


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