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Jamia Millia Islamia Previous Year Questions (PYQs)

Jamia Millia Islamia Properties Of Triangle PYQ


Jamia Millia Islamia PYQ
One vertex of an equilateral triangle with centroid at origin and one side as $x + y - 2 = 0$ is





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

Using centroid condition and perpendicular distance relation gives vertex (2, −2).

Jamia Millia Islamia PYQ
$F_1$ = 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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

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 $F_1$).

Jamia Millia Islamia PYQ
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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

Solution

For a triangle inscribed in a circle, maximum area occurs when the triangle is right-angled, since the hypotenuse = diameter = $2r$. Area $= \dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} r \times r = r^2$. $\boxed{\text{Answer: (B) Right angled triangle with side } 2r, r}$

Jamia Millia Islamia PYQ
The points $A(12,8)$, $B(-2,6)$ and $C(6,0)$ are the vertices of …





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

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 $C$.

Jamia Millia Islamia PYQ
Considering Cosine Rule of triangle ABC, possible measures of angle A include





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

Any type possible under Cosine Rule.


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