Let the third side of the triangle be x.
Perimeter of the given triangle = 42 cm
18 cm + 10 cm + x = 42 x
= 14 cm
Perimeter
s =\( \frac{\text{(a + b + c)}}{2}\)
\(= \frac{42}{2} \)
= 21 cm
By Heron’s formula,
Area of a triangle \( = \sqrt{\text{s(s - a)(s - b)(s - c)}}\)
\(= \sqrt{\text{21(21 - 18)(21 - 10)(21 - 14)}}\)
\(= \sqrt{\text{21 × 3 × 11 × 7}}\)
\(= 21\sqrt{11}\) cm2
Area of the triangle \(= 21\sqrt{11}\) 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:
A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.