Question

1. if cdef is a parallelogram, find ( mangle fcd ).

1. if jklm is a rectangle, ( jl = 2x + 5 ), and ( mk = 7x - 40 ), find ( mk ).

1. if pqrs is a rhombus, find ( mangle pqr ).

Explanation:

Problem 7

Step1: Consecutive angles are supplementary

In a parallelogram, consecutive interior angles are supplementary, so their sum is $180^\circ$.
$$(5x-7)^\circ + (10x-23)^\circ = 180^\circ$$

Step2: Simplify and solve for $x$

Combine like terms and isolate $x$.
$$15x - 30 = 180$$
$$15x = 210$$
$$x = 14$$

Step3: Find $\angle FCD$

$\angle FCD$ is equal to $\angle FEC$ (opposite angles of parallelogram are equal), so substitute $x=14$ into $(5x-7)^\circ$.
$$m\angle FCD = 5(14) - 7 = 63$$

Step1: Diagonals of rectangle are equal

In a rectangle, the diagonals are congruent, so $JL = MK$.
$$2x + 5 = 7x - 40$$

Step2: Solve for $x$

Isolate $x$ by rearranging terms.
$$5 + 40 = 7x - 2x$$
$$45 = 5x$$
$$x = 9$$

Step3: Calculate length of $MK$

Substitute $x=9$ into the expression for $MK$.
$$MK = 7(9) - 40 = 63 - 40 = 23$$

Step1: Diagonals bisect angles in rhombus

In a rhombus, diagonals bisect the angles, so the two given angles are equal.
$$4x - 27 = 2x + 7$$

Step2: Solve for $x$

Isolate $x$ by rearranging terms.
$$4x - 2x = 7 + 27$$
$$2x = 34$$
$$x = 17$$

Step3: Find $\angle PQR$

$\angle PQR$ is twice the measure of $(2x+7)^\circ$.
$$m\angle PQR = 2(2(17)+7) = 2(34+7) = 2(41) = 82$$

Answer:

$63^\circ$

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Problem 8