Given below are two statements: Statement I: Only one Rhombus ABCD can be drawn with AB = 4 cm and diagonal BD = 5 cm. Statement II: Only one parallelogram ABCD can be drawn with AB = 6 cm and diagonal BD = 8 cm. In the light of the above statements, choose the correct answer from the options given below:
Statement I is true but statement II is false
Let's analyze the given statements regarding the construction of a rhombus and a parallelogram based on specific measurements.
Statement I says that only one Rhombus ABCD can be drawn with AB = 4 cm and diagonal BD = 5 cm.
Consider the triangle ABD within the rhombus. Its sides are:
We have the lengths of all three sides of triangle ABD: 4 cm, 4 cm, and 5 cm.
According to the SSS (Side-Side-Side) criterion for triangle congruence and construction, if you know the lengths of all three sides of a triangle, you can draw only one unique triangle (up to congruence).
Since triangle ABD is uniquely determined by its side lengths (4 cm, 4 cm, 5 cm), the positions of vertices A, B, and D relative to each other are fixed. Once triangle ABD is fixed, the location of point C in the rhombus is also uniquely determined because ABCD must be a parallelogram (every rhombus is a parallelogram) and all sides must be 4 cm. Point C must be 4 cm away from B and 4 cm away from D. The intersection of arcs with radius 4 cm centered at B and D (on the opposite side of BD from A) will give the unique position for C.
Thus, only one such rhombus can be constructed with the given side and diagonal lengths.
Therefore, Statement I is true.
Statement II says that only one parallelogram ABCD can be drawn with AB = 6 cm and diagonal BD = 8 cm.
Consider the triangle ABD within the parallelogram. Its sides are:
To uniquely determine a triangle using side lengths, you need either all three side lengths (SSS) or two sides and the included angle (SAS). In triangle ABD, we only know two sides (AB = 6 cm and BD = 8 cm). The length of the third side, AD, which is the other side of the parallelogram, is not given. The angle between AB and BD (∠ABD) or the angle between AD and BD (∠ADB) or the angle at A (∠BAD) is also not given.
Since the length of the adjacent side AD can be any length (as long as it satisfies the triangle inequality for triangle ABD, i.e., $|\text{AB} - \text{BD}| < \text{AD} < \text{AB} + \text{BD}$, which means $|6-8| < \text{AD} < 6+8$, or $2 < \text{AD} < 14$), different lengths for AD will result in different possible triangles ABD. Each different triangle ABD will form a different parallelogram.
For example, if AD = 5 cm, triangle ABD has sides 6, 8, 5. This forms a parallelogram with sides 6 and 5 and diagonal 8. If AD = 7 cm, triangle ABD has sides 6, 8, 7. This forms a different parallelogram with sides 6 and 7 and diagonal 8.
Since the length of the adjacent side is not fixed, more than one parallelogram can be drawn with the given information (one side and one diagonal).
Therefore, Statement II is false.
Based on the analysis:
This matches the option which states that Statement I is true but Statement II is false.
| Figure | Given Information | Key Triangle | Triangle Information | Uniquely Determined? | Figure Uniquely Determined? | Statement Truth Value |
|---|---|---|---|---|---|---|
| Rhombus ABCD | AB=4 cm, BD=5 cm | △ABD | AB=4, AD=4, BD=5 (SSS) | Yes | Yes | True |
| Parallelogram ABCD | AB=6 cm, BD=8 cm | △ABD | AB=6, BD=8, AD=? | No (Missing AD or an angle) | No | False |
Understanding what information uniquely determines a geometric figure is crucial.
Let's review some key properties that are relevant to these constructions.
In Statement I, knowing the side length and diagonal BD allows us to form a triangle ABD with sides 4, 4, and 5. The rigidity of this triangle structure (SSS) is key to the uniqueness of the rhombus.
In Statement II, knowing side AB and diagonal BD only provides two sides of the triangle ABD. The length of the adjacent side AD is not fixed, allowing the parallelogram to be "squashed" or "stretched" along the diagonal BD while keeping AB and BD constant, thus leading to multiple possible shapes.
Match List I with List II.
| List I | List II |
|---|---|
| (A) \( \frac{14 - (x - 1)}{10} = \frac{x + 5}{6} - 3 \) | (I) 4 |
| (B) \( (x - 5)^2 - (x + 3)^2 = 48 \) | (II) \( 23^2 \) |
| (C) \( 6(x - 4) = 4(x - 3) - 3(x - 8) \) | (III) 61 |
| (D) \( (2x - 1)(2x + 3) = (2x - 7)(2x + 7) \) | (IV) -2 |
Choose the correct answer from the options given below:
Asha is twice as old as Anita. Three years ago, she was three times as old as Anita. How old is Asha now?
A cube painted green on all faces is cut into 27 small cubes of equal size. How many small cubes are painted on one face only?
Arrange the given events in ascending order of their probabilities:
A = target is hit 2 times in 20 shots
B = target is hit 175 times in 200 shots
C = target is hit 92 times in 100 shots
D = target is hit 2 times in 5 shots
E = target is hit 5 times in 13 shots
Choose the correct answer from the options given below:
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