Jamia Millia Islamia MCA Properties Of Triangle Previous Year Questions (PYQs)
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Showing 5 questions
One vertex of an equilateral triangle with centroid at origin and one side as $x + y - 2 = 0$ is
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Solution
Using centroid condition and perpendicular distance relation gives vertex (2, −2).
$F_1$ = set of parallelograms, $F_2$ = rectangles, $F_3$ = rhombuses, $F_4$ = squares, $F_5$ = trapeziums.
$F_1$ may be equal to:
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Solution
$F_4\subseteq F_2$ and $F_4\subseteq F_3$, and $F_2\subseteq F_1,\ F_3\subseteq F_1$.
Thus $F_1\cup F_2\cup F_3\cup F_4=F_1$ (since the others are subsets of $F_1$).
If a triangle is inscribed in a circle of radius $r$, then which of the following triangles can have maximum area?
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Solution
For a triangle inscribed in a circle, maximum area occurs when the triangle is right-angled,
since the hypotenuse = diameter = $2r$.
Area $= \dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} r \times r = r^2$.
$\boxed{\text{Answer: (B) Right angled triangle with side } 2r, r}$
The points $A(12,8)$, $B(-2,6)$ and $C(6,0)$ are the vertices of …
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Solution
$AB^{2}=14^{2}+2^{2}=200,\ BC^{2}=(-8)^{2}+6^{2}=100,\ CA^{2}=(-6)^{2}+(-8)^{2}=100$.
Since $BC=CA$ the triangle is isosceles, and $BC^{2}+CA^{2}=AB^{2}$ ⇒ right-angled at $C$.
Considering Cosine Rule of triangle ABC, possible measures of angle A include
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Solution
Any type possible under Cosine Rule.
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