Rhombuses - PSAT Math

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Question

In Rhombus , $m \angle H = 45 ^{\circ }$ . If $\overline{RO }$ is constructed, which of the following is true about $\Delta RHO$ ?

Answer

The figure referenced is below.

The sides of a rhombus are congruent by definition, so $\overline{RH} \cong \overline{HO}$ , making $\Delta RHO$ isosceles. It is not equilateral, since $m \angle H = 45 ^{\circ }$ , and an equilateral triangle must have three $60 ^{\circ}$ angles.

Also, consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

$m \angle HRM = 180^{\circ} - m \angle H = 180^{\circ} - 45 ^{\circ} = 135^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle HRO = \frac{1}{2} m \angle HRM = \frac{1}{2} \cdot 135 ^{\circ } = 67\frac{1}{2}^{\circ }$

Similarly, $m \angle HOR = 67\frac{1}{2}^{\circ }$

This makes $\Delta RHO$ acute.

The correct response is that $\Delta RHO$ is acute and isosceles, but not equilateral.

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Question

A rhombus has a side length of 5. Which of the following is NOT a possible value for its area?

Answer

The area of a rhombus will vary as the angles made by its sides change. The "flatter" the rhombus is (with two very small angles and two very large angles, say 2, 178, 2, and 178 degrees), the smaller the area is. There is, of course, a lower bound of zero for the area, but the area can get arbitrarily small. This implies that the correct answer would be the largest choice. In fact, the largest area of a rhombus occurs when all four angles are equal, i.e. when the rhombus is a square. The area of a square of side length 5 is 25, so any value bigger than 25 is impossible to acheive.

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Question

In Rhombus , $m \angle R = 60^{\circ }$ . If $\overline{HM}$ is constructed, which of the following is true about $\Delta RHM$ ?

Answer

The figure referenced is below.

Consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

$m \angle RHO= 180^{\circ} - m \angle R = 180^{\circ} - 60^{\circ} = 120^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle RHM = \frac{1}{2} m \angle RHO= \frac{1}{2} \cdot 120^{\circ} = 60^{\circ}$

A similar argument proves that $m \angle RMH = 60^{\circ}$ .

Since all three angles of $\Delta RHM$ measure $60^{\circ}$ , the triangle is acute. It is also equiangular, and, subsequently, equilateral.

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Question

If the area of a rhombus is 24 and one diagonal length is 6, find the perimeter of the rhombus.

Answer

The area of a rhombus is found by

A = 1/2(_d_1)(_d_2)

where _d_1 and _d_2 are the lengths of the diagonals. Substituting for the given values yields

24 = 1/2(_d_1)(6)

24 = 3(_d_1)

8 = _d_1

Now, use the facts that diagonals are perpendicular in a rhombus, diagonals bisect each other in a rhombus, and the Pythagorean Theorem to determine that the two diagonals form 4 right triangles with leg lengths of 3 and 4. Since 32 + 42 = 52, each side length is 5, so the perimeter is 5(4) = 20.

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Question

Note: Figure NOT drawn to scale.

Calculate the perimeter of Quadrilateral in the above diagram if:

Answer

, so Quadrilateral is a rhombus. Its diagonals are therefore perpendicular to each other, and the four triangles they form are right triangles. Therefore, the Pythagorean theorem can be used to determine the common sidelength of Quadrilateral .

We focus on $\Delta AXB$ . The diagonals are also each other's bisector, so

$AX = \frac{1}{2} (AC) = \frac{1}{2} \cdot 48= 24$

$XB = \frac{1}{2} (BD) = \frac{1}{2} \cdot 20=10$

By the Pythagorean Theorem,

$AB = \sqrt{(AX)^{2}+ (XB)^{2}} = \sqrt{24^{2}+10^{2}} = \sqrt{576+100} = \sqrt{676} = 26$

26 is the common length of the four sides of Quadrilateral , so its perimeter is $4 \times 26 = 104$ .

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Question

In Rhombus , $m \angle H = 45 ^{\circ }$ . If $\overline{RO }$ is constructed, which of the following is true about $\Delta RHO$ ?

Answer

The figure referenced is below.

The sides of a rhombus are congruent by definition, so $\overline{RH} \cong \overline{HO}$ , making $\Delta RHO$ isosceles. It is not equilateral, since $m \angle H = 45 ^{\circ }$ , and an equilateral triangle must have three $60 ^{\circ}$ angles.

Also, consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

$m \angle HRM = 180^{\circ} - m \angle H = 180^{\circ} - 45 ^{\circ} = 135^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle HRO = \frac{1}{2} m \angle HRM = \frac{1}{2} \cdot 135 ^{\circ } = 67\frac{1}{2}^{\circ }$

Similarly, $m \angle HOR = 67\frac{1}{2}^{\circ }$

This makes $\Delta RHO$ acute.

The correct response is that $\Delta RHO$ is acute and isosceles, but not equilateral.

Compare your answer with the correct one above

Question

A rhombus has a side length of 5. Which of the following is NOT a possible value for its area?

Answer

The area of a rhombus will vary as the angles made by its sides change. The "flatter" the rhombus is (with two very small angles and two very large angles, say 2, 178, 2, and 178 degrees), the smaller the area is. There is, of course, a lower bound of zero for the area, but the area can get arbitrarily small. This implies that the correct answer would be the largest choice. In fact, the largest area of a rhombus occurs when all four angles are equal, i.e. when the rhombus is a square. The area of a square of side length 5 is 25, so any value bigger than 25 is impossible to acheive.

Compare your answer with the correct one above

Question

In Rhombus , $m \angle R = 60^{\circ }$ . If $\overline{HM}$ is constructed, which of the following is true about $\Delta RHM$ ?

Answer

The figure referenced is below.

Consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

$m \angle RHO= 180^{\circ} - m \angle R = 180^{\circ} - 60^{\circ} = 120^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle RHM = \frac{1}{2} m \angle RHO= \frac{1}{2} \cdot 120^{\circ} = 60^{\circ}$

A similar argument proves that $m \angle RMH = 60^{\circ}$ .

Since all three angles of $\Delta RHM$ measure $60^{\circ}$ , the triangle is acute. It is also equiangular, and, subsequently, equilateral.

Compare your answer with the correct one above

Question

If the area of a rhombus is 24 and one diagonal length is 6, find the perimeter of the rhombus.

Answer

The area of a rhombus is found by

A = 1/2(_d_1)(_d_2)

where _d_1 and _d_2 are the lengths of the diagonals. Substituting for the given values yields

24 = 1/2(_d_1)(6)

24 = 3(_d_1)

8 = _d_1

Now, use the facts that diagonals are perpendicular in a rhombus, diagonals bisect each other in a rhombus, and the Pythagorean Theorem to determine that the two diagonals form 4 right triangles with leg lengths of 3 and 4. Since 32 + 42 = 52, each side length is 5, so the perimeter is 5(4) = 20.

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Calculate the perimeter of Quadrilateral in the above diagram if:

Answer

, so Quadrilateral is a rhombus. Its diagonals are therefore perpendicular to each other, and the four triangles they form are right triangles. Therefore, the Pythagorean theorem can be used to determine the common sidelength of Quadrilateral .

We focus on $\Delta AXB$ . The diagonals are also each other's bisector, so

$AX = \frac{1}{2} (AC) = \frac{1}{2} \cdot 48= 24$

$XB = \frac{1}{2} (BD) = \frac{1}{2} \cdot 20=10$

By the Pythagorean Theorem,

$AB = \sqrt{(AX)^{2}+ (XB)^{2}} = \sqrt{24^{2}+10^{2}} = \sqrt{576+100} = \sqrt{676} = 26$

26 is the common length of the four sides of Quadrilateral , so its perimeter is $4 \times 26 = 104$ .

Compare your answer with the correct one above

Question

In Rhombus , $m \angle H = 45 ^{\circ }$ . If $\overline{RO }$ is constructed, which of the following is true about $\Delta RHO$ ?

Answer

The figure referenced is below.

The sides of a rhombus are congruent by definition, so $\overline{RH} \cong \overline{HO}$ , making $\Delta RHO$ isosceles. It is not equilateral, since $m \angle H = 45 ^{\circ }$ , and an equilateral triangle must have three $60 ^{\circ}$ angles.

Also, consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

$m \angle HRM = 180^{\circ} - m \angle H = 180^{\circ} - 45 ^{\circ} = 135^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle HRO = \frac{1}{2} m \angle HRM = \frac{1}{2} \cdot 135 ^{\circ } = 67\frac{1}{2}^{\circ }$

Similarly, $m \angle HOR = 67\frac{1}{2}^{\circ }$

This makes $\Delta RHO$ acute.

The correct response is that $\Delta RHO$ is acute and isosceles, but not equilateral.

Compare your answer with the correct one above

Question

A rhombus has a side length of 5. Which of the following is NOT a possible value for its area?

Answer

The area of a rhombus will vary as the angles made by its sides change. The "flatter" the rhombus is (with two very small angles and two very large angles, say 2, 178, 2, and 178 degrees), the smaller the area is. There is, of course, a lower bound of zero for the area, but the area can get arbitrarily small. This implies that the correct answer would be the largest choice. In fact, the largest area of a rhombus occurs when all four angles are equal, i.e. when the rhombus is a square. The area of a square of side length 5 is 25, so any value bigger than 25 is impossible to acheive.

Compare your answer with the correct one above

Question

In Rhombus , $m \angle R = 60^{\circ }$ . If $\overline{HM}$ is constructed, which of the following is true about $\Delta RHM$ ?

Answer

The figure referenced is below.

Consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

$m \angle RHO= 180^{\circ} - m \angle R = 180^{\circ} - 60^{\circ} = 120^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle RHM = \frac{1}{2} m \angle RHO= \frac{1}{2} \cdot 120^{\circ} = 60^{\circ}$

A similar argument proves that $m \angle RMH = 60^{\circ}$ .

Since all three angles of $\Delta RHM$ measure $60^{\circ}$ , the triangle is acute. It is also equiangular, and, subsequently, equilateral.

Compare your answer with the correct one above

Question

If the area of a rhombus is 24 and one diagonal length is 6, find the perimeter of the rhombus.

Answer

The area of a rhombus is found by

A = 1/2(_d_1)(_d_2)

where _d_1 and _d_2 are the lengths of the diagonals. Substituting for the given values yields

24 = 1/2(_d_1)(6)

24 = 3(_d_1)

8 = _d_1

Now, use the facts that diagonals are perpendicular in a rhombus, diagonals bisect each other in a rhombus, and the Pythagorean Theorem to determine that the two diagonals form 4 right triangles with leg lengths of 3 and 4. Since 32 + 42 = 52, each side length is 5, so the perimeter is 5(4) = 20.

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Calculate the perimeter of Quadrilateral in the above diagram if:

Answer

, so Quadrilateral is a rhombus. Its diagonals are therefore perpendicular to each other, and the four triangles they form are right triangles. Therefore, the Pythagorean theorem can be used to determine the common sidelength of Quadrilateral .

We focus on $\Delta AXB$ . The diagonals are also each other's bisector, so

$AX = \frac{1}{2} (AC) = \frac{1}{2} \cdot 48= 24$

$XB = \frac{1}{2} (BD) = \frac{1}{2} \cdot 20=10$

By the Pythagorean Theorem,

$AB = \sqrt{(AX)^{2}+ (XB)^{2}} = \sqrt{24^{2}+10^{2}} = \sqrt{576+100} = \sqrt{676} = 26$

26 is the common length of the four sides of Quadrilateral , so its perimeter is $4 \times 26 = 104$ .

Compare your answer with the correct one above

Question

In Rhombus , $m \angle H = 45 ^{\circ }$ . If $\overline{RO }$ is constructed, which of the following is true about $\Delta RHO$ ?

Answer

The figure referenced is below.

The sides of a rhombus are congruent by definition, so $\overline{RH} \cong \overline{HO}$ , making $\Delta RHO$ isosceles. It is not equilateral, since $m \angle H = 45 ^{\circ }$ , and an equilateral triangle must have three $60 ^{\circ}$ angles.

Also, consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

$m \angle HRM = 180^{\circ} - m \angle H = 180^{\circ} - 45 ^{\circ} = 135^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle HRO = \frac{1}{2} m \angle HRM = \frac{1}{2} \cdot 135 ^{\circ } = 67\frac{1}{2}^{\circ }$

Similarly, $m \angle HOR = 67\frac{1}{2}^{\circ }$

This makes $\Delta RHO$ acute.

The correct response is that $\Delta RHO$ is acute and isosceles, but not equilateral.

Compare your answer with the correct one above

Question

A rhombus has a side length of 5. Which of the following is NOT a possible value for its area?

Answer

The area of a rhombus will vary as the angles made by its sides change. The "flatter" the rhombus is (with two very small angles and two very large angles, say 2, 178, 2, and 178 degrees), the smaller the area is. There is, of course, a lower bound of zero for the area, but the area can get arbitrarily small. This implies that the correct answer would be the largest choice. In fact, the largest area of a rhombus occurs when all four angles are equal, i.e. when the rhombus is a square. The area of a square of side length 5 is 25, so any value bigger than 25 is impossible to acheive.

Compare your answer with the correct one above

Question

In Rhombus , $m \angle R = 60^{\circ }$ . If $\overline{HM}$ is constructed, which of the following is true about $\Delta RHM$ ?

Answer

The figure referenced is below.

Consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

$m \angle RHO= 180^{\circ} - m \angle R = 180^{\circ} - 60^{\circ} = 120^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle RHM = \frac{1}{2} m \angle RHO= \frac{1}{2} \cdot 120^{\circ} = 60^{\circ}$

A similar argument proves that $m \angle RMH = 60^{\circ}$ .

Since all three angles of $\Delta RHM$ measure $60^{\circ}$ , the triangle is acute. It is also equiangular, and, subsequently, equilateral.

Compare your answer with the correct one above

Question

If the area of a rhombus is 24 and one diagonal length is 6, find the perimeter of the rhombus.

Answer

The area of a rhombus is found by

A = 1/2(_d_1)(_d_2)

where _d_1 and _d_2 are the lengths of the diagonals. Substituting for the given values yields

24 = 1/2(_d_1)(6)

24 = 3(_d_1)

8 = _d_1

Now, use the facts that diagonals are perpendicular in a rhombus, diagonals bisect each other in a rhombus, and the Pythagorean Theorem to determine that the two diagonals form 4 right triangles with leg lengths of 3 and 4. Since 32 + 42 = 52, each side length is 5, so the perimeter is 5(4) = 20.

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Calculate the perimeter of Quadrilateral in the above diagram if:

Answer

, so Quadrilateral is a rhombus. Its diagonals are therefore perpendicular to each other, and the four triangles they form are right triangles. Therefore, the Pythagorean theorem can be used to determine the common sidelength of Quadrilateral .

We focus on $\Delta AXB$ . The diagonals are also each other's bisector, so

$AX = \frac{1}{2} (AC) = \frac{1}{2} \cdot 48= 24$

$XB = \frac{1}{2} (BD) = \frac{1}{2} \cdot 20=10$

By the Pythagorean Theorem,

$AB = \sqrt{(AX)^{2}+ (XB)^{2}} = \sqrt{24^{2}+10^{2}} = \sqrt{576+100} = \sqrt{676} = 26$

26 is the common length of the four sides of Quadrilateral , so its perimeter is $4 \times 26 = 104$ .

Compare your answer with the correct one above

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