Question

triangle pqr is right angle right angled at q such that QR = b and a = area of triangle pqr if QN and is perpendicular PR then show that 2a×b/√b⁴ + 4a²

Answer 1

Answer:

Step-by-step explanation:

In the picture attached,

ΔPQR is a right angle triangle in which ∠Q = 90°, therefore, height of the triangle is PQ and base QR = b

Now it has been given that area of the triangle PQR = a

From the formula,

Area of the triangle = $\frac{1}{2}\times (\text{Base})\times (\text Height)$

a = $\frac{1}{2}(PQ)(b)$

PQ = $\frac{2a}{b}$

Now PR = $\sqrt{(\frac{2a}{b})^{2}+(b)^{2}}$

PR = $\sqrt{(\frac{4a^{2}+b^{4}}{b^{2}})}$

Another height of the triangle is QN and Base is PR.

Therefore, Area of the triangle = $\frac{1}{2}(QN)(PR)$

a = $\frac{1}{2}(QN)(\sqrt{(\frac{4a^{2}+b^{4}}{b^{2}})})$

QN = $\frac{2a}{\sqrt{(\frac{4a^{2}+b^{4}}{b^{2}})}}$

QN = $\frac{2ab}{\sqrt{4a^{2}+b^{4}}}$

Learn more about the right angle triangles from https://brainly.in/question/6344829