- The circumcircle of the quadrilateral \(ABCD\) has a radius 2. \(AC, BD\) cut at \(E\) such that \(AE=EC\). If \(AB=\sqrt{2} AE, BD=2\sqrt {3}\) . Find the area of quadrilateral \(ABCD\).
- Triangle ABC is an isosceles triangle in which AB=A C. A circle is drawn passing through B and touching AC at its midpoint M. The circle cuts AB at P . Prove that BP=3AP.
- In an isosceles triangle the altitude drawn to the base is \(\frac{2}{3}\) times the radius of the circumcircle . Prove the base angle of the triangle is \(\text{arccos}\sqrt{\frac{2}{3}}\).
- The sides of a right angled triangle are all integers . Two sides are prime that differ by 50.Find the smallest value of the third side.
- In triangle ABC , D is on BC such that BD=3DC. E is on AC such that 3AE=2EC. AD and BE cut F. Area of triangle AFE=4 and area of triangle BFD=30 . Find area of triangle ABC.
- In a circle AB is a diameter . AB is produced to P such that BP=Radius of the circle . PC is tangent to the circle . The tangent at B amd AC produced cut at E . Then describe triangle CDE.

## Comments

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TopNewestQ5:

We note that \(\triangle CFE\) has the same altitude as \(\triangle AFE\), therefore their areas are in the ratio of the base lengths. That is:

\(\dfrac{[\triangle CFE]}{[\triangle AFE]} = \dfrac{EC}{AE} = \dfrac{3}{2} \quad \Rightarrow [\triangle CFE] = \dfrac{3}{2} \times [\triangle AFE] = \dfrac{3}{2} \times 4 = 6 \)

Similarly, \([\triangle CFD] = \dfrac{DC}{BD} \times [\triangle BFD] = \dfrac{1}{3} \times 30 = 10 \)

And, \([\triangle ABD] = \dfrac{BD}{DC} \times [\triangle ACD] = 3 \times ([\triangle AFE]+[\triangle CFE]+[\triangle CFD]) = 3 \times (4+6+10) = 60 \)

Therefore, \([\triangle ABC] = [\triangle ABD] + [\triangle ACD] = 60 + 20 = \boxed{80} \) – Chew-Seong Cheong · 1 year, 7 months ago

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Q3:

Let the isosceles \(\triangle ABC\) has \(AB=AC\), its base angle be \(\theta\) and the center and radius of the circumcircle be \(O\) and \(r\) respectively.

Then the extension of altitude \(AP\) to \(BC\) passes through \(O\); \(AP = \frac{2}{3}r\), \(OP = \frac{1}{3}r\) and then \(BP = \sqrt{OB^2-OP^2} = \sqrt{r^2 - \left(\frac{1}{3}r\right)^2} = \frac{2 \sqrt{2}}{3}\).

Therefore, \(\tan{\theta} = \dfrac{AP}{BP} = \dfrac{\frac{2}{3}r}{\frac{2\sqrt{2}}{3}r} = \dfrac{1}{\sqrt{2}} \quad \Rightarrow \cos \theta = \sqrt{\frac {2}{3}} \quad \Rightarrow \theta = \boxed{\cos^{-1} \left( \sqrt{\frac {2}{3}}\right)} \) – Chew-Seong Cheong · 1 year, 7 months ago

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Q:1

Q:2

– Ahmad Saad · 1 week, 4 days agoLog in to reply

Answer 4 = 60 – Saksham Rastogi · 1 year, 7 months ago

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