Question

Prove that triangles on the same base and between the same parallels are equal in area.

Answer 1

Let, ABC and DBC are two Δs on the same base.

Given: Δs ABC and DBC are on the same base BC and between the same line l and BC.

Construct: Through B, draw BP|| CD, intersecting line l in P and through C,CQ||BA, intersecting line l in Q

To prove: ar(ΔABC)=ar(ΔDBC)

Proof:

Quad. BCQA is a ||gm [by construction]

Quad. BCDP is a ||gm [by construction]

∴ar(||gm BCQA)= ar(||gm BCDP) [||gms on the same base between the same parrallels.]

Also, ar(ΔABC) = ar(ΔAQC)

                        = 1/2 ar(||gm BCQA) [Diag. AC bisects ||gm ABCQ]

ar(ΔDBC)=1/2 ar(||BCQA) [Diag. AC bisects ||gm ABCQ]

Hence, ar(ΔABC)=ar(ΔDBC) (is proved)

Answer 2

Answer:

Step-by-step explanation:

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

area of parallelogram

, ABCD is a parallelogram and AC is one of its diagonals. A line ‘AN’ is drawn which is perpendicular to the line ‘DC’. We know that a diagonal divides a parallelogram into two congruent triangles. Hence,

∆ ADC ≅ ∆ CBA

Also, we know that congruent figures have equal areas. So,

Area (ADC) = Area (CBA)

In simple words, the two triangles of the same (or equal) base and lying between the same parallels are equal in area. Taking the equation further, we can write

Area (ADC) = 1/2 [Area (ABCD)]

Now, the area of parallelogram is the length of the base multiplied by height. Hence, we have

Area (ADC) = 1/2 (DC x AN)

Or, Area of ∆ ADC = 1/2 (base DC × corresponding altitude AN)

Area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height).