Rhombuses - PSAT Math

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.

, 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 . The diagonals are also each other's bisector, so

By the Pythagorean Theorem,

26 is the common length of the four sides of Quadrilateral , so its perimeter is .

The figure referenced is below.

The sides of a rhombus are congruent by definition, so , making isosceles. It is not equilateral, since , and an equilateral triangle must have three angles.

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

A diagonal of a rhombus bisects its angles, so

Similarly,

This makes acute.

The correct response is that is acute and isosceles, but not equilateral.

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.

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.

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.

, 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 . The diagonals are also each other's bisector, so

By the Pythagorean Theorem,

26 is the common length of the four sides of Quadrilateral , so its perimeter is .

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.

, 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 . The diagonals are also each other's bisector, so

By the Pythagorean Theorem,

26 is the common length of the four sides of Quadrilateral , so its perimeter is .

The figure referenced is below.

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

A diagonal of a rhombus bisects its angles, so

A similar argument proves that .

Since all three angles of measure , the triangle is acute. It is also equiangular, and, subsequently, equilateral.

The figure referenced is below.

The sides of a rhombus are congruent by definition, so , making isosceles. It is not equilateral, since , and an equilateral triangle must have three angles.

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

A diagonal of a rhombus bisects its angles, so

Similarly,

This makes acute.

The correct response is that is acute and isosceles, but not equilateral.

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.

The figure referenced is below.

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

A diagonal of a rhombus bisects its angles, so

A similar argument proves that .

Since all three angles of measure , the triangle is acute. It is also equiangular, and, subsequently, equilateral.

The figure referenced is below.

The sides of a rhombus are congruent by definition, so , making isosceles. It is not equilateral, since , and an equilateral triangle must have three angles.

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

A diagonal of a rhombus bisects its angles, so

Similarly,

This makes acute.

The correct response is that is acute and isosceles, but not equilateral.

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.

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.

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.

, 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 . The diagonals are also each other's bisector, so

By the Pythagorean Theorem,

26 is the common length of the four sides of Quadrilateral , so its perimeter is .

The figure referenced is below.

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

A diagonal of a rhombus bisects its angles, so

A similar argument proves that .

Since all three angles of measure , the triangle is acute. It is also equiangular, and, subsequently, equilateral.