Area of Rhombus Formula Proof, Maths Activity, Project, TLM

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Area of Rhombus Formula Proof, Maths Activity, Project, TLM
area of rhombus formula
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Пікірлер: 12
@nitinvalani6792
@nitinvalani6792 25 күн бұрын
Thank You
@vbabatechnicalguru0143
@vbabatechnicalguru0143 Жыл бұрын
Very nice 👍👍🙂
@somnathkhan3961
@somnathkhan3961 Жыл бұрын
Thanks........... Mam... ❤
@minakshithakur5419
@minakshithakur5419 3 жыл бұрын
Thanku mam
@mkmandy120
@mkmandy120 3 жыл бұрын
👌👌👌👌
@502akshayar9
@502akshayar9 3 жыл бұрын
Very nice
@LearningNotebook
@LearningNotebook 3 жыл бұрын
Thanks dear. We have many art integrated projects. You can see the list of all such activities, projects and models here on our website as well: learningnotebookyoutube.blogspot.com/p/maths-art-integrated-projects.html learningnotebookyoutube.blogspot.com/p/cbse-art-integrated-projects.html
@vaishnavisingh3459
@vaishnavisingh3459 2 жыл бұрын
Thnx mam this help full for me
@jasimharoon5335
@jasimharoon5335 Жыл бұрын
I am not clear how D1 = D2 as you take 13 inch for D1 and 11 inch for D2
@laislacollins2299
@laislacollins2299 Жыл бұрын
I agree that the actual proof of a rhombus is missing from this video. First, you do not need the diagonals to be equal. The diagonals are NOT equal in the actual proof of a rhombus. It helps to label all of the points on both construction paper rhombuses and then superimpose the blue triangles onto the red rhombus as you carry out each step in this proof, which was given to me by “Experience Math’s” video on the proof of a rhombus. -------- 1. Let ABCD be a rhombus with four congruent sides and two perpendicular bisectors called diagonal d1 (AC) and diagonal d2 (BD). 2. Let AC = d1 units. 3. Let BD = d2 units. 4. Let O be the intersection of the diagonals AC and BD. 5. The area of the rhombus is the area of triangle ADC and the area of triangle ABC. 6. Thus, the area of rhombus ABCD = [the area of ADC] + [the area of ABC] = [1/2 x AC x OD] + [1/2 x AC x OB]. 7. Factoring obtains the area of rhombus ABCD as = 1/2 x AC x (OD + OB). 8. OD + OB is the diagonal BD. 9. Apply the substitution property of equality to replace (OD + OB) with (BD) …and [1/2 x AC x (OD + OB)] becomes [1/2 x AC x BD]. 10. Recall that AC = d1 and BD = d2 and turn [area of ABCD = 1/2 x AC x BD] into [area of ABCD = 1/2 x d1 x d2]. Hope that helps!
@jasimharoon5335
@jasimharoon5335 Жыл бұрын
@@laislacollins2299 Thank you
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