Class 9 Exam > Class 9 Notes > Theory - To show that the Area of Rhombus is half the Product of its Diagonals, Math, Class 9
Theory - To show that the Area of Rhombus is half the Product of its Diagonals, Math, Class 9 PDF Download
As performed in the real lab:
Materials Required:
Coloured papers, sketch pens, geometry box, a pair of scissors, fevicol and eraser.
Procedure:
1.Draw □ ABCD with length d2 and breadth d1 units on a coloured paper.
2. Mark points E, F, G and H as midpoints of the sides AD, DC, CB and BA respectively of sides of the □ ABCD.
3. Join HF and EG. Mark their intersection as point O. Fold the rectangle ABCD along EG and HF dividing the □ ABCD into four congruent rectangles, namely OEAH, OEDF, OFCG and OGBH.
4. Divide each of the four rectangles into two congruent triangles by drawing their respective diagonals.
As performed in the simulator:
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Provide dimensions of a rectangle i.e. length and breadth in the input box to draw the rectangle.
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Mark midpoint of each side of the rectangle by clicking on the midpoint of each side.
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Connect midpoints of opposite sides.
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Connect midpoints of adjacent sides to form a rhombus.
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Cut four triangles that are in the corner of rectangle ABCD using "Cut Traingle" button.
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Place each triangle to the position mentioned in "Instructions".
Observation:
1. As the small rectangles are congruent, their diagonals EH, HG, GF, FE are equal. Thus EHGF is a rhombus.
2. In the rectangle AHOE, triangles AHE and EHO are congruent and hence equal in area.
3. Thus area of the right triangle EOH is half the area of the rectangle AEOH and similarly, the area of right triangles HOG, GOF, FOE are half the area of the rectangles HBGO, OGCF and FOED respectively.
4. Thus the area of rhombus EFGH = (1/2 X Area of rectangle AEOH) + (1/2 X Area of rectangle HBGO) + (1/2 X Area of rectangle OGCF) + (1/2 X Area of rectangle FOED)
= 1/2 X ( Area of rectangle AEOH + Area of rectangle HBGO + Area of rectangle OGCF + Area of rectangle FOED)
= 1/2 x Area of rectangle ABCD
= 1/2 x d1 x d2
= 1/2 X HF X EG
= 1/2 x product of diagonals of rhombus EFGH
Result:
Area of rhombus is half the product of its diagonals.
FAQs on Theory - To show that the Area of Rhombus is half the Product of its Diagonals, Math, Class 9
| 1. How can I prove that the area of a rhombus is half the product of its diagonals? |  |
| 2. What are the diagonals of a rhombus? | |
Ans. In a rhombus, the diagonals are line segments that connect opposite vertices. There are two diagonals in a rhombus, and they intersect each other at a 90-degree angle. The diagonals of a rhombus bisect each other, meaning they divide each other into two equal parts.
| 3. How do you find the length of the diagonals of a rhombus? | |
Ans. To find the length of the diagonals of a rhombus, you can use the Pythagorean theorem. If you have the lengths of the sides of the rhombus, you can divide one side by 2 to find the length of the shorter diagonal. The longer diagonal can be found by multiplying the shorter diagonal by the square root of 2.
| 4. Can a rhombus have diagonals of different lengths? | |
Ans. No, a rhombus cannot have diagonals of different lengths. The diagonals of a rhombus are always perpendicular to each other and bisect each other. This means that they divide each other into two equal parts, resulting in both diagonals having the same length.
| 5. How is the area of a rhombus related to the length of its diagonals? | |
Ans. The area of a rhombus is directly related to the length of its diagonals. The formula for the area of a rhombus is (d1 * d2) / 2, where d1 and d2 represent the lengths of the diagonals. This formula shows that the area is half the product of the diagonals. Thus, if the length of the diagonals increases, the area of the rhombus will also increase. Similarly, if the length of the diagonals decreases, the area of the rhombus will decrease.
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