Exercise 136 Page 525

a Consider the given diagram.

The figure is a quadrilateral with a pair of parallel opposite sides and of equal length. By the Opposite Sides Parallel and Congruent Theorem this quadrilateral is a parallelogram. Now let's recall the Parallelogram Opposite Angles Theorem.

Parallelogram Opposite Angles Theorem
If a quadrilateral is a parallelogram, then its opposite angles are congruent.

We can use this theorem to equate the measures of the given angles. 3x-60^(∘) = 2x + 50^(∘) Let's solve this equation for x.

3x-60^(∘) = 2x + 50^(∘)

SubEqn

LHS-2x=RHS-2x

x - 60^(∘) = 50^(∘)

AddEqn

LHS+60^(∘)=RHS+60^(∘)

x = 110^(∘)

b We want to find the value of t in the given diagram.

Since the perimeter of a polygon is the sum of all of its side lengths, we can equate the side lengths in the diagram to the given perimeter value. (5t+1)+(2t+5)+(3t-2)+t = 202 Let's solve the equation for t.

(5t+1)+(2t+5)+(3t-2)+t = 202

AddSubTerms

Add and subtract terms

11t+4=202

SubEqn

LHS-4=RHS-4

11t = 198

DivEqn

.LHS /11.=.RHS /11.

t = 18

c Like in previous parts, let's first take a look at the given diagram.

We are told that the quadrilateral CARD is a rhombus, so let's recall the Rhombus Diagonals Theorem.

Rhombus Diagonals Theorem
A parallelogram is a rhombus if and only if its diagonals are perpendicular.

Since CARD is a rhombus, its diagonals are perpendicular. This means that the measure from the diagram must equal 90^(∘). 4x-2^(∘) = 90^(∘) Let's solve the equation above for x.

4x-2^(∘) = 90^(∘)

AddEqn

LHS+2^(∘)=RHS+2^(∘)

4x = 92^(∘)

DivEqn

.LHS /4.=.RHS /4.

x = 23^(∘)

d Let's take a look at the final given diagram.

The diagram shows a trapezoid with a pair of angles of equal measure near one of the bases. Let's recall the Converse Isosceles Trapezoid Base Angles Theorem.

If a trapezoid has a pair of congruent base angles, then it is an isosceles trapezoid.

By this theorem we know that our trapezoid is isosceles, which means that the legs of this trapezoid are congruent. Let's equate the expressions for their lengths. 13m - 9 = 7m+15 Now let's solve the equation for m.

13m - 9 = 7m+15

SubEqn

LHS-7m=RHS-7m

6m-9 = 15

AddEqn

LHS+9=RHS+9

6m = 24

DivEqn

.LHS /6.=.RHS /6.

m = 4

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Exercise 136 Page 525

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