Side of traffic signal board = a
Perimeter of the signal board = 3a = 180 cm
∴ a = 60 cm
Semi perimeter of the signal board (s) = \(\frac{3a}{2}\)
By using Heron’s formula,
Area of triangle =\(\sqrt{\text{[s(s - a)(s - b)(s - c)]}}\)
Area of given triangle
= \(\sqrt{\text{[s(s - a)(s - b)(s - c)]}}\)
=\( \sqrt{\text{[s(s - a)(s - a)(s - a)]}}\)
= \(\text{(s - a)} \sqrt{\text{[s(s - a)]}}\)
since s = \(\frac{3a}{2}\)
\((\frac{3a}{2} - a)\sqrt{\frac{3a}{2}(\frac{3a}{2} - a)}\)
\(= (\frac{a}{2}) \sqrt{\frac{3a}{2}(\frac{a}{2})}\)
\(= \frac{a}{2} × \frac{a}{2} × \sqrt3\)
=\( (\frac{\sqrt3}{4})a^2\)\( .......(1)\)
Area of the signal board = \( (\frac{\sqrt3}{4})a^2\) sq. units
perimeter = 180 cm
side of triangle = \(\frac{180}{3}\) cm
a = 60 cm
Area of the signal board = \( (\frac{\sqrt3}{4})(60)^2\)
\(= \)\( (\frac{\sqrt3}{4})(3600)\)
\(= 900\sqrt3\)
Area of the signal board \(= 900\sqrt3\) cm2
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)