✔ 最佳答案
23.
(2n - 4) x 90o = 156o x n
180n - 360 = 156n
24n = 360
n = 15
22.
x
= (Exterior of a pentagon) + (Exterior of a square)
= (360o/5) + (360o/4)
= 72o + 90o
= 162o
21.
BAE + ABE + AEB = 180o ( sum of D)
BAE + 30o + 30o = 180o
BAE = 120o
ABD = AED = 30o (given)
BD = ED (given)
AB = AE (opp. equal in D)
DABD DAED (SAS)
BAC = EAC (corr. s of cong. D)
But BAC + EAC = BAE = 120o
Hence, BAC = 60o
BAC + ABC + ACB = 180o ( sum of D)
60o + (30o + 30o) + ACB = 180o
ACB = 60o
BAC = ABC = ACB = 60o
Hence, ABC is an equilateral D.
20.
BCF = ACF (given)
But BCF + ACF = 60o (int. of equil. D)
Hence, BCF = ACF = 30o
AEF + DCF = 180o (int. s, EA//CD)
AEF + (90o + 30o) = 180o
AEF = 60o
19.
(a)
A = C (given)
BC = AB (opp. s equal in D)
3x + 4 = 5x
2x = 4
x = 2
(b)
Perimeter
= {5(2) + [3(2) + 4] + 3(2)} cm
= 26 cm
18.
Draw a st. line HCK which // AB and //DE.
(H on the left of C, while K on the right.)
HCB + ABC = 180o (int. s, HCK // AB)
HCB + 120o = 180o
HCB = 60o
KCD + EDC = 180o (int. s, HCK // DE)
KCD + 130o = 180o
KCD = 50o
HCB + BCD + KCD = 180o (adj. s on a st. line)
60o + c + 50o = 180o
c = 70o
17.
BAE + EAH = 180o (adj. s on a st. line)
BAE + (2x - 10o) = 180o
BAE = 190o - 2x
DEA = FEG (vert. opp. s)
DEA = x + 60o
Sum of int. s = (2 x 5 - 4) x 90o
(190o - 2x) + 2x + 2x + (3x - 40o) + (x + 60o) = 540o
6x + 210o = 540o
x = 55o
16.
Let each of the two vertically opposite s be y.
Consider the upper (isos.) D:
(2x + 15o) + (2x + 15o) + y = 180o
y = 180o - 2(2x + 15o)
y = 150o - 4x
Consider the lower D:
The exterior is equal to the sum of the two int. opp. s.
y + 72o = 126o
(150o - 4x) + 72o = 126o
4x = 96o
x = 24o
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