Sec-3 Sec-A
Sec-3 Sec-A
Sec-3 Sec-A
Solution:
4
2. If , a, and 2 are three consecutive terms of an A.P., then find the value of a.
5
Solution:
7
Thus, the value of a is .
5
3. If the areas of two similar triangles are in the ratio 25:64, write the ratio of their
corresponding sides.
CBSE X Mathematics 2009 Solution (SET 3)
Solution:
We know that the ratio of the areas of two similar triangles is equal to the square
of the ratio of their corresponding sides.
It is given that the areas of two similar triangles are in the ratio 25:64.
2
25 5
64 8
Thus, the ratio of the corresponding sides of the two similar triangles is 5:8.
Solution:
AQ + QC = AC
QC = AC – AQ
Using the values AQ = 4 cm and AC = 11 cm,
QC = 11 cm – 4 cm
QC = 7 cm
BC = BP + CP
On using equations (1) and (2), we obtain
BC = 3 cm + 7 cm
CBSE X Mathematics 2009 Solution (SET 3)
BC = 10 cm
5. Two coins are tossed simultaneously. Find the probability of getting exactly one
head.
Solution:
If two coins are tossed simultaneously, then the possible outcomes are
S = {HT, TH, TT, HH}
Thus, the total number of possible outcomes is 4.
Out of all the four outcomes, {HT} and {TH} are cases of exactly one head.
Favourable outcomes 2 1
Required probability =
Total possible outcomes 4 2
6. Find the [HCF × LCM] for the numbers 100 and 190.
Solution:
7. If 1 is a zero of the polynomial p(x) = ax2 – 3(a – 1) x – 1, then find the value of a.
Solution:
Solution:
Diameter = 14 cm
Diameter 14 cm
Radius = 7 cm
2 2
22
Length of the semicircular part = πr (7) 22 cm
7
10. Find the number of solutions of the following pair of linear equations:
x + 2y – 8 = 0
2x + 4y = 16
Solution:
a1 b1 c1 1
a2 b2 c2 2
Hence, the given pair of linear equations has infinitely many solutions.