# in the pictures answer question 4,7,8,10,11,18,21,22,23,24,25,26

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(4) You can tile the plane with regular hexagons since the sum of the interior angles that meet at a point is 360 degrees. (Each interior angle is 120 degrees -- the sum of the interior angles is 180(6-2)=720; dividing by 6 yields that each interior angle is 120 degrees. Alternatively, the exterior angle of a regular convex polygon can be found by `360/n ` where n is the number of sides; here 360/6=60 so the exterior angles are all 60, and since the interior angles are supplements the interior angles are 180-60=120 degrees.)

(7) Given that each interior angle of a regular polygon is 140 degrees: `140=(180(n-2))/n `

Then `140n=180n-360 ==> 40n=360 ==> n=9 ` so the polygon has 9 sides (a nonagon.)

Since the exterior angles are supplements, each exterior angle will be 180-140=40 degrees. Since there are 9 angles, the sum of the exterior angles is 9(40)=360 degrees.

(8) Each exterior angle of a regular octagon has measure `360/8=45^@ ` . Since the interior angles are supplements each interior angle has measure 135; since there are 8 angles the interior sum is 8(135)=1080 degrees. (Check with the formula: `S=180(n-2)=180(6)=1080^@ `

(10) The interior angles of a regular pentagon are 108 degrees. ` `( `(180(3))/5=108^@ ` ) Since `Delta PLO, Delta MNO ` are isosceles, `m/_PLO=m/_MLN=36 ` (`108+2m/_PLO=180 ==> m/_PLO=36 ` )

` Delta PLO cong Delta MNL` by SAS, so `bar(LO) cong bar(LN) ` by corresponding parts of congruent triangles are congruent (CPCTC) Then `Delta LON ` is isosceles by definition.

(11) Sandy should have used `(180(10-2))/10=144^@ ` .

(21) `Delta PRQ cong Delta YSU ` by ASA so `/_P cong /_Y,/_R cong /_S,/_Q cong /_U, bar(PR) cong bar(YS),bar(RQ) cong bar(SU), bar(PQ) cong bar(YU) `

(22) Yes: `Delta ABC cong Delta CDA ` by SSS (Since AC=AC by the reflexive property)

(23) `m/_PQR=180-(m/_QRP+m/_RPQ) `

`m/_PQT=m/_QSR ` sine alternate interior angles formed by parallel lines and a transversal are congruent.

Then ` m/_RQT=m/_PQR-m/_PQT `

(25) Let the shared side of the polygons be `bar(BD) ` .

`m/_ABD=108,m/_CBD=120 ` ==> ` m/_ABC=132 `

`Delta ABC ` is isosceles, so `m/_A=m/_C=24 `

(26) ` `Let the parallelogram be ABCD, with O the intersection of the diagonals.

Then `Delta ABO cong Delta CDO, Delta AOD cong Delta COB ` by SSS since the opposite sides of a parallelogram are congruent and the diagonals bisect each other. (If you do not have this yet, you can use ASA since the diagonals will be transversals creating pairs of alternate interior angles.)

**Sources:**

24) It is best to give angle z a reasonable number so I'll say <z =110.

Use <z to find <y because the two angles are supplementary. So <z + <y = 180 then solve for <y. So /180-<z = <y, 180 - 110 = 70. So <y = 70 degrees.

Since Triangle ABC is isosceles, that means that side AB=BC and also <b=2<a (since there are 2 a's for the one angle.) Now look at the triangle BCD to find <a because the total angle measure of a triangle is 180 degrees. So use this to find <a. So 90+<y+2<a=180 degrees and

solve for <a=180-90-70

<a = 20/2 = 10 degrees

Then use the same formula to find <v

<v + <y +<a = 180

<v = 180 - <y - <a , 180-70-10=100, <v = 100 degrees.

Then you can use <v to find <u because they are supplementary angles.

<v + <u =180

<u = 180 - <v, 180 - 100 = 80. <u=80 degrees

Then repeat to find <w

<w+<u=180

<w= 180-<u, 180-80=100, <w = 100 degrees

You know the angle measures for <b and <w, so use the small triangle to find out <x.

<b+<w+<x = 180,

<x=180 -<b-<w, 180-20-100=60, <x=60 degrees.

Then <x and <s are supplementary so subtract <x from 180. So <s=120 degrees.

Then do the same to find <t using <w as the supplementary angles. So <t=80 degrees.

There you have all of them with just using one given angle measure.

What about 18 and 24 ?