Question

HEYA!!!♥♥♥ WELCOME TO THIS WEEK'S MIRACLE QUESTION

IF A TRIANGLE AND PARALLELOGRAM LIE ON SAME BASE AND BETWEEN

SAME PARALLELS, THENNPROVE THAT AREA OF TRIANGLE IS EQUAL TO 1/2

AREA OF PARALLELOGRAM.

ONLY BRAINLY AND MYSTERY ANSWERES BE MARKED BRAINLIEST.

Answer 1

In the figure, we have a ∆ACP and a ||gm ABCD. We have to prove that area of ∆ = 1/2 area of ||gm if they lie on same base and between the same parallel.



Method 1 :-


Drop the height of parallelogram from the point D or C. Let the height be denoted by h

So area of ||gm would be base × height

= AB × h

Now since they lie between the same parallel, the height of the traingle would be equal to the height of ||gm

So area of triangle = 1/2 × base × Height

= 1/2 × AB × h

= 1/2 × area of ||gm ABCD


Hence proved



Method 2
(Lengthy but effective)

Construct a line parallel to AP. Let the line intersect CD at Q

Now we have AB || CD (since ABCD is a ||gm)

Hence, AB || PQ (Since CD coincides with PQ)


And, BQ || AP (By construction)

So ABPQ is a ||gm (if opposite sides are parallel, then the figure is a ||gm)

And ABCD is also a ||gm.

Since both parallelogram lie between same parallel (AB || PC) and on same base (AB),

area of ABPQ = area of ABCD

Now, for ABQP, PB is the diagonal.

In ∆PAB and ∆PBQ,

PA = BQ (opposite sides of ||gm are equal)
PQ = AB(opposite sides of a ||gm are equal)
PB = PB (common)

Hence, ∆PAB is congruent to ∆PBQ (SSS)

So area of PAB = area of PBQ

(congruent figures have equal area)

And area PAB + area PBQ = area ABPQ

=> area PAB + area PAB = area ABPQ

(since, area PAB = area PBQ)

=> 2 ar(PAB) = ar(ABPQ)

=> ar(PAB) = 1/2 × ar(ABPQ)

Now ar(ABPQ) = ar(ABCD)

=> ar(PAB) = 1/2 × ar(ABCD)

Hence, if a triangle and a ||gm lie on same base and between the same parallel, then the area of triangle = 1/2 of area of parallelogram.

Answer 2

correct answer of the question....

mark me as brainliest if u like plzzz..

and thanks for asking this question...

im practicing 9th syllabus so it helped me a lot...XD