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### Course: ತರಗತಿ 9 ಗಣಿತ (ಭಾರತ) > Unit 7

Lesson 3: Proofs: Rhombus# Proof: Rhombus area

Sal proves that we can find the area of a rhombus by taking half the product of the lengths of the diagonals. ಸಾಲ್ ಖಾನ್ ರವರು ರಚಿಸಿದ್ದಾರೆ.

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## ವೀಡಿಯೊ ಪ್ರತಿಲಿಪಿ

So quadrilateral ABCD, they're
telling us it is a rhombus, and what we need to
do, we need to prove that the area of this rhombus is
equal to 1/2 times AC times BD. So we're essentially proving
that the area of a rhombus is 1/2 times the product of
the lengths of its diagonals. So let's see what
we can do over here. So there's a bunch of
things we know about rhombi and all rhombi are
parallelograms, so there's tons of things that
we know about parallelograms. First of all, if it's a rhombus,
we know that all of the sides are congruent. So that side length is
equal to that side length is equal to that side length is
equal to that side length. Because it's a
parallelogram, we know the diagonals bisect each other. So we know that
this length-- let me call this point
over here B, let's call this E. We know that BE
is going to be equal to ED. So that's BE, we know that's
going to be equal to ED. And we know that
AE is equal to EC. We also know, because
this is a rhombus, and we proved this
in the last video, that the diagonals, not only
do they bisect each other, but they are also perpendicular. So we know that this
is a right angle, this is a right angle,
that is a right angle, and then this is a right angle. So the easiest way
to think about it is if we can show that
this triangle ADC is congruent to triangle ABC,
and if we can figure out the area of one of them,
we can just double it. So the first part is
pretty straightforward. So we can see that
triangle ADC is going to be congruent
to triangle ABC, and we know that by
side-side-side congruency. This side is congruent
to that side. This side is congruent to
that side, and they both share a C right over here. So this is by side-side-side. And so we can say that the
area-- so because of that, we know that the
area of ABCD is just going to be equal to
2 times the area of, we could pick
either one of these. We could say 2 times
the area of ABC. Because area of ABCD-- actually
let me write it this way. The area of ABCD is equal to
the area of ADC plus the area of ABC. But since they're
congruent, these two are going to be the
same thing, so it's just going to be 2 times
the area of ABC. Now what is the area of ABC? Well area of a triangle is
just 1/2 base times height. So area of ABC is just
equal to 1/2 times the base of that triangle times its
height, which is equal to 1/2. What is the length of the base? Well the length
of the base is AC. So it's 1/2-- I'll
color code it. The base is AC. And then what is the height? What is the height here? Well we know that this
diagonal right over here, that it's a
perpendicular bisector. So the height is just
the distance from BE. So it's AC times BE,
that is the height. This is an altitude. It intersects this base
at a 90-degree angle. Or we could say BE is the
same thing as 1/2 times BD. So this is-- let me write it. This is equal to, so it's
equal to 1/2 times AC, that's our base. And then our height is
BE, which we're saying is the same thing
as 1/2 times BD. So that's the area
of just ABC, that's just the area of this broader
triangle right up there, or that larger triangle
right up there, that half of the rhombus. But we decided that the
area of the whole thing is two times that. So if we go back, if we
use both this information and this information right over
here, we have the area of ABCD is going to be equal to
2 times the area of ABC, where the area of ABC is
this thing right over here. So 2 times the area
of ABC, area of ABC is that right over there. So 1/2 times 1/2 is
1/4 times AC times BD. And then you see
where this is going. 2 times 1/4 fourth is
1/2 times AC times BD. Fairly straightforward,
which is a neat result. And actually, I haven't
done this in a video. I'll do it in the next video. There are other ways of finding
the areas of parallelograms, generally. It's essentially base times
height, but for a rhombus we could do that because
it is a parallelogram, but we also have this
other neat little result that we proved in this video. That if we know the
lengths of the diagonals, the area of the rhombus
is 1/2 times the products of the lengths of the diagonals,
which is kind of a neat result.