Quadrilaterals

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PSAT Math › Quadrilaterals

Questions 1 - 10

Thingy_3

Refer to the above diagram. .

Give the area of Quadrilateral .

Explanation

$\angle C \cong \angle BDX$ , since both are right; by the Corresponding Angles Theorem, $\overline{AC} || \overline{BD}$ , and Quadrilateral is a trapezoid.

By the Angle-Angle Similarity Postulate, since

$\angle C \cong \angle BDX$

and

$\angle X \cong \angle X$ (by reflexivity),

$\Delta ACX \sim \Delta BDX$ ,

and since corresponding sides of similar triangles are in proportion,

$\frac{AC}{CX} = \frac{BD}{DX}$

$\frac{AC}{24} = \frac{6}{9}$

$\frac{AC}{24} \times 24 = \frac{6}{9} \times 24$

, the larger base of the trapozoid;

The smaller base is .

, the height of the trapezoid.

The area of the trapezoid is

$A = \frac{1}{2} (b + B)h$

$A = \frac{1}{2} (BD + AC) \cdot CD$

$A = \frac{1}{2} (6+ 16) \cdot 15 = \frac{1}{2} (22) \cdot 15 = 165$

Thingy_3

Refer to the above diagram. .

Give the area of Quadrilateral .

Explanation

$\angle C \cong \angle BDX$ , since both are right; by the Corresponding Angles Theorem, $\overline{AC} || \overline{BD}$ , and Quadrilateral is a trapezoid.

By the Angle-Angle Similarity Postulate, since

$\angle C \cong \angle BDX$

and

$\angle X \cong \angle X$ (by reflexivity),

$\Delta ACX \sim \Delta BDX$ ,

and since corresponding sides of similar triangles are in proportion,

$\frac{AC}{CX} = \frac{BD}{DX}$

$\frac{AC}{24} = \frac{6}{9}$

$\frac{AC}{24} \times 24 = \frac{6}{9} \times 24$

, the larger base of the trapozoid;

The smaller base is .

, the height of the trapezoid.

The area of the trapezoid is

$A = \frac{1}{2} (b + B)h$

$A = \frac{1}{2} (BD + AC) \cdot CD$

$A = \frac{1}{2} (6+ 16) \cdot 15 = \frac{1}{2} (22) \cdot 15 = 165$

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

$\Delta RHM$ is acute and equilateral

$\Delta RHM$ is acute and isosceles, but not equilateral

$\Delta RHM$ is obtuse and isosceles, but not equilateral

$\Delta RHM$ is acute and scalene

$\Delta RHM$ obtuse and scalene

Explanation

The figure referenced is below.

Rhombus

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.

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

$\Delta RHM$ is acute and equilateral

$\Delta RHM$ is acute and isosceles, but not equilateral

$\Delta RHM$ is obtuse and isosceles, but not equilateral

$\Delta RHM$ is acute and scalene

$\Delta RHM$ obtuse and scalene

Explanation

The figure referenced is below.

Rhombus

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.

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

$\Delta RHM$ is acute and equilateral

$\Delta RHM$ is acute and isosceles, but not equilateral

$\Delta RHM$ is obtuse and isosceles, but not equilateral

$\Delta RHM$ is acute and scalene

$\Delta RHM$ obtuse and scalene

Explanation

The figure referenced is below.

Rhombus

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.

Thingy_3

Refer to the above diagram. .

Give the area of Quadrilateral .

Explanation

$\angle C \cong \angle BDX$ , since both are right; by the Corresponding Angles Theorem, $\overline{AC} || \overline{BD}$ , and Quadrilateral is a trapezoid.

By the Angle-Angle Similarity Postulate, since

$\angle C \cong \angle BDX$

and

$\angle X \cong \angle X$ (by reflexivity),

$\Delta ACX \sim \Delta BDX$ ,

and since corresponding sides of similar triangles are in proportion,

$\frac{AC}{CX} = \frac{BD}{DX}$

$\frac{AC}{24} = \frac{6}{9}$

$\frac{AC}{24} \times 24 = \frac{6}{9} \times 24$

, the larger base of the trapozoid;

The smaller base is .

, the height of the trapezoid.

The area of the trapezoid is

$A = \frac{1}{2} (b + B)h$

$A = \frac{1}{2} (BD + AC) \cdot CD$

$A = \frac{1}{2} (6+ 16) \cdot 15 = \frac{1}{2} (22) \cdot 15 = 165$

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

Explanation

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.

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

Explanation

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

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

Explanation

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

The two rectangles shown below are similar. What is the length of EF?

Sat_mah_166_02

Explanation

When two polygons are similar, the lengths of their corresponding sides are proportional to each other. In this diagram, AC and EG are corresponding sides and AB and EF are corresponding sides.

To solve this question, you can therefore write a proportion:

AC/EG = AB/EF ≥ 3/6 = 5/EF

From this proportion, we know that side EF is equal to 10.

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