In the given figure, a circle touches all the four sides of quadrilateral ABCD with AB = 6 cm, BC = 7 cm and CD = 4 cm. The length AD is

Q5

Step 1: A circle touches all four sides of quadrilateral $ABCD$, so it is tangential. Sums of lengths of opposite sides are equal.

$AB + CD = AD + BC$

Step 2: Substitute given lengths (all in cm).

$6 + 4 = AD + 7$

Step 3: Simplify and solve for $AD$.

$10 = AD + 7$

$AD = 10 - 7 = 3$ cm

Answer: 3 cm (option 2)

3

Q6

Step 1: In an A.P., the difference between terms is $(m - k)d$, where $d$ is common difference.

Assume standard: $a_n - a_p$ where $n - p = 2$ (common in such problems, e.g., $a_4 - a_2$ or $a_3 - a_1$).

Step 2: Substitute $d = 3$ cm? No units for AP, but numerical.

$a_n - a_p = (n - p)d = 2 \times 3 = 6$

Answer: 6 (option 2)

6

Q7

Step 1: For A.P., first term $a = 1$, last term $l = 11$, sum $S_n