Question

Find the area of the red triangle.​

Answer 1

Concept used :-

• Area of a triangle = (1/2) × (Base × Height).
• Area of a parallelogram = (Base × Height).
• Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
• In other words, area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height).

From image we can see That :-

→ ∆EIB & ∆ADE are on same Base as AB, and between same parallel Lines AB & DC.

→ Both ∆'s have Same Height as Equal to the Height of ||gm.

So ,

→ Ar.[∆EIB] + Ar.[∆ADE]

→ (1/2 * EB * h) + (1/2 * AE * h)

→ (1/2 *h)[ EB + AE ]

→ (1/2 *h) * AB

→ (1/2)* (h * AB)

→ (1/2) * Area of llgm

So ,

we can say That,

→ Ar.[∆EIB] + Ar.[∆ADE] = (1/2) * Area of llgm .

→ (72+d+8) + (x + b) = (1/2) * Area of llgm ------------ Equation (1) .

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Now,

∆AFD is on Base AD and b/w Two parallel lines AD & BC..

So,

→ Ar.[AFD] = (1/2) * Area of llgm

→ (b + 79 + d + 10) = (1/2) * Area of llgm ----------------- Equation (2).

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From Equation (1) & (2) now, we get,

→ (72+d+8) + (x + b) = (b + 79 + d + 10)

→ 80 + x + b + d = 89 + b + d

→ x + b + d - b - d = 89 - 80

→ x = 9 unit². (Ans).

Answer 2

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