01.12.2014 Views

Skrivnosti Å¡tevil in oblik 7

Skrivnosti Å¡tevil in oblik 7

Skrivnosti Å¡tevil in oblik 7

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

RE©ITVE<br />

OBSEGI IN PLO©»INE<br />

6.1 OBSEG IN PLO©»INA VE»KOTNIKOV<br />

1. o = 30 cm, p = 39 cm 2<br />

2. a = 5 cm, p = 25 cm 2<br />

3. a) o = 22 cm, p = 10 cm 2<br />

b) o = 22 cm, p = 18 cm 2<br />

c) o = 22 cm, p = 24 cm 2<br />

Vsi pravokotniki imajo enake<br />

obsege <strong>in</strong> razliËne ploπË<strong>in</strong>e.<br />

4. a) 10 · 6 = 60 dm 2 ; DA b) Ostanka je 0,4 m 2 .<br />

5. Bazen je pravokotnik s stranicama 6 <strong>in</strong> 9 m. Folija je pravokotnik<br />

s stranicama 6 m <strong>in</strong> 9 m. VodopivËevi morajo kupiti 54 m 2 folije.<br />

6. a) 10,5 · 5 — (2 · (1,5 · 1,2) + 3 · 1 · 1 + 2 · 2,2) = 41,5<br />

Pleskar mora prepleskati 41,5 m 2 .<br />

b) 3 · (4 · 1) + 2 · (2 · 1,5 + 2 · 1,2) + 2,2 + 2 + 2,2 = 29,2<br />

Pleskar bo potreboval 29,2 m lepilnega traku.<br />

6.2 OBSEG IN PLO©»INA PARALELOGRAMA<br />

1. a) o = 2 · 4 + 2 · 3 = 14; o = 14 cm; p = 4 · 2 = 8; p = 8 cm 2<br />

b) o = 2 · 2 + 2 · 5,4 = 14,8; o = 14,8 cm;<br />

p = 5,4 · 1,5 = 8,1; p = 8,1 cm 2<br />

2. a) o = 42 cm; p = 66 cm 2<br />

b) o = 3,4 dm; p = 0,3 dm 2 1 2<br />

c) o = 2 m; p = m 2<br />

2 3<br />

3. a) p = 13 · 12 = 156; p = 156 cm 2<br />

b) a = 30 : 5 = 6; a = 6 cm<br />

c) b = (60 — 30) : 2 = 15; b = 15 cm<br />

4. a) o = 20 dm; p = 20 dm 2<br />

b) a = 12,5 cm; p = 100 cm 2<br />

c) v a = 16 m; o = 36 m<br />

5. a) D a C<br />

1 cm<br />

b<br />

v a<br />

α<br />

b<br />

D<br />

C<br />

A a B<br />

b = 2,6 cm<br />

o = 17,2 cm<br />

p = 15 cm 2<br />

b) D<br />

b<br />

A<br />

v b<br />

a<br />

a<br />

a = 2,3 cm<br />

o = 14,6 cm<br />

p = 10 cm 2<br />

C<br />

b<br />

β<br />

B<br />

A<br />

1 cm<br />

v a<br />

a<br />

D<br />

v<br />

b<br />

b<br />

B<br />

C<br />

b) PloπË<strong>in</strong>e paralelogramov so enake, ker imajo vsi enako dolgo<br />

stranico a <strong>in</strong> enako dolgo viπ<strong>in</strong>o na stranico a.<br />

9. a) Poti je 32 m 2 .<br />

b) Travnika je 400 m 2 32 2<br />

. c) = 432 27<br />

10. En parkirni avto zaseda 15 m 2 .<br />

6.3 OBSEG IN PLO©»INA TRIKOTNIKA<br />

1. a) a = 3,9 cm; b = 6,2 cm; c = 6,6 cm; v c = 3,6 cm; o = 16,7 cm;<br />

p = 11,88 cm 2<br />

b) a = 4 cm; b = 3,4 cm; c = 3,3 cm; v a = 2,7 cm; o = 10,7 cm;<br />

p = 5,4 cm 2<br />

c) a = 1,8 cm; b = 6,2 cm; c = 6,5 cm; o = 14,5 cm; p = 5,58 cm 2<br />

Toleranca za stranico je ±2 mm.<br />

2. a) o = 90 dm; p = 300 dm 2 b) o = 30 cm; p = 30 cm 2<br />

3. a) o = 44 mm; p = 66 mm 2 2<br />

; v a = 10 13 mm; v b = 6,6 mm<br />

1<br />

b) v c = 8 3 mm<br />

4. p = pΔABD + pΔBCD<br />

C<br />

p = 6 cm 2 + 6 cm 2 = 12 cm 2 2 cm<br />

5.<br />

C<br />

2<br />

D 0,9<br />

1,5<br />

A<br />

4<br />

p = pΔABC + pΔCDA<br />

p =<br />

4 $ 15 , 2 $ 09 ,<br />

+ = 3,9 cm<br />

2 2<br />

6. Pravilni so odgovori: Ë, d <strong>in</strong> e.<br />

7. D<br />

C<br />

3<br />

1,5<br />

pΔATD =<br />

3 $ 3<br />

= 4,5<br />

2<br />

pΔATD = 4,5 cm 2<br />

T p ABCD = 9 cm 2<br />

1,5<br />

D<br />

2 cm<br />

A<br />

6 cm<br />

B<br />

T<br />

6 cm<br />

R<br />

B<br />

c) D a<br />

b<br />

α v f<br />

b<br />

A a<br />

a = 2,7 cm<br />

b = 3,7 cm<br />

o = 12,8 cm<br />

p = 7,4 cm 2<br />

B<br />

C<br />

b<br />

Toleranca za stranico je ± 2mm.<br />

A<br />

6. Paralelogram pod c), ker je p = 5 · 1,2 = 6 cm 2 .<br />

7. IzraËunamo lahko obseg, ploπË<strong>in</strong>o <strong>in</strong> viπ<strong>in</strong>o na stranico b.<br />

o = 18 cm; p = 10 cm 2 ; v b = 2,5 cm<br />

8. a) Najmanjπi obseg ima lik A, ker ima najkrajπo dolæ<strong>in</strong>o stranice b<br />

(zaradi pravega kota), vsi paralologrami pa imajo enake dolæ<strong>in</strong>e a.<br />

a<br />

1 cm<br />

D<br />

β<br />

B<br />

v<br />

b<br />

A<br />

C<br />

β<br />

a<br />

b<br />

B<br />

A<br />

8.<br />

D<br />

3<br />

B<br />

PloπË<strong>in</strong>a trikotnika je polovica ploπË<strong>in</strong>e kvadrata. Kvadratu smo<br />

odrezali dvakrat po 4<br />

1 lika, kar je ravno 2<br />

1 lika, zato je druga<br />

polovica lika ostala (obavano).<br />

pΔ 1 = pΔ 3 =<br />

7 $ 2<br />

= 7<br />

2<br />

Δ 3<br />

pΔ 1 = pΔ 3 = 7 dm 2<br />

Δ 4<br />

4<br />

Δ 2<br />

4 $ 35 ,<br />

Δ pΔ 2 = pΔ 4 = = 7<br />

1 2<br />

A<br />

7<br />

B pΔ 2 = pΔ 4 = 7 dm 2<br />

PloπË<strong>in</strong>e vseh πtirih trikotnikov so med seboj enake.<br />

oΔ 1 = oΔ 3 = 2 · 4 dm + 7 dm = 15 dm<br />

oΔ 2 = oΔ 4 = 2 · 4 dm + 4 dm = 12 dm<br />

Po dva nasproti leæeËa trikotnika imata enake obsege.<br />

C<br />

Prelom SSIO 7 20.<strong>in</strong>dd 161<br />

25

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!