QUESTION IMAGE
Question
1 n.b: you need a compass and a straightedge.
a) draw segment ab.
b) draw perpendicular bisector of segment ab.
c) take point p on the perpendicular bisector of ab. to which is the point p closer to?
2 you need compass, straightedge and paper.
a) draw segment ab.
b) draw the perpendicular bisector of ab.
c) on the perpendicular line, mark - off point p such that the distance from p to segment ab is greater than 0.5ab and less than segment ab.
d) what is name of the triangle abp?
a) what can you say about side ap and bp?
2 look at the given figure.
a) what is the value of x?
b) what is name of triangle klm? why do you say so?
Step1: Solve for x in the figure
Since the two line - segments are equal (assuming they are congruent based on the context of the circles), we set up the equation $3x + 2=x + 6$.
Subtract x from both sides: $3x - x+2=x - x + 6$, which simplifies to $2x+2 = 6$.
Subtract 2 from both sides: $2x+2 - 2=6 - 2$, so $2x=4$.
Divide both sides by 2: $\frac{2x}{2}=\frac{4}{2}$, and we get $x = 2$.
Step2: Determine the type of triangle KLM
In triangle KLM, since the two line - segments (assumed to be radii of the circles) are equal (from the equation we just solved), two sides of the triangle are equal. A triangle with two equal sides is an isosceles triangle.
Step3: Answer part 1c
A point P on the perpendicular bisector of segment AB is equidistant from A and B. So the point P is neither closer to A nor closer to B.
Step4: Answer part 2d
In triangle ABP, since P is on the perpendicular bisector of AB, AP = BP. So triangle ABP is an isosceles triangle.
Step5: Answer part 2a
Since P is on the perpendicular bisector of AB, by the property of the perpendicular bisector, the side AP and BP are equal.
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1c) Neither A nor B.
2d) Isosceles triangle.
2a) AP and BP are equal.
3a) $x = 2$
3b) Isosceles triangle because two of its sides are equal.