
By, Heron's formula
Area = \(\sqrt{\text{s(s - a)(s - b)(s - c)}}\)
The sides of the triangular walls are a = 11 m, b = 6 m and c = 15 m.
Semi Perimeter:
\(s =\)\( \frac{\text{(a + b + c)}}{2}\)
\(= \frac{\text{(11 + 6 + 15)}}{2}\)
\(= \frac{32}{2}\)
= 16 m
By using Heron’s formula,
Area of triangular wall = \(\sqrt{\text{s(s - a)(s - b)(s - c)}}\)
Substituting the given values in formula,
\(= \sqrt{\text{16(16 -11)(16 - 6)(16 -15)}}\)
\(= \sqrt{\text{16 × 5 × 10 × 1}}\)
\(= \)\(\sqrt{800}\) m²
\(= 20\sqrt2\) m²
Area of the wall of the park to be painted in colour \(= 20\sqrt2\) m2.
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Let \( A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix} \) and \( P = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \theta > 0. \) If \( B = P A P^T \), \( C = P^T B P \), and the sum of the diagonal elements of \( C \) is \( \frac{m}{n} \), where gcd(m, n) = 1, then \( m + n \) is:
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ACP = ∠ QCD

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∠∆BAD
(iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)