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The parallelograms are on the same base and between the same parallels.

Area of a Parallelogram = b × h

In both parallelograms:

• The bases are equal.

• The heights are equal.

Area of parallelogram ABCD = Area of parallelogram ABGH.

Parallelograms on the same base and between the same parallels are equal in area.

**Theorem:**

Prove that the area of the triangle is equal to half the area of the parallelogram, if they are on the same base and are between the same parallels.

Given:

Δ ABP and parallelogram ABCD are on same base AB and between the same parallels AB and PC.

To Prove: ar (ΔABP) =1/2 ar ( ||gm ABCD)

Construction: Draw BQ || AP

Proof:

ar ( ||gm ABQP) = ar ( || gm ABCD (Parallelograms on the same base and between the same parallels are equal in area.)

Δ ABP ≅ Δ BQP

ar (Δ ABP) = ar (Δ BQP)

ar (Δ ABP) = (½) ar (|| gm ABQP)

ar (Δ ABP) = (½) ar (|| gm ABCD).

Recall that sum of angles of a polygon = (2n - 4) × 90°

Here number of sides, n = 9

Hence the sum of angles of the given pol...

Length = 21 cm

Breadth = 9 cm

width = 8 cm

Surface Area of tin box (excluding lid)

= 2(bh+hl) + lb...

root3/4 a suare

Given question is not clear please send us the exact question so that we can provide the solution at the earliest.

Sol:

Perimeter of triangle = ( a + b + c) = (26 + 28 + 30) cm = 84 cm

2
*s* = 84 cm

*s* = 42 cm,a =...