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NIMCET Previous Year Questions (PYQs)

NIMCET Properties Of Triangle PYQ


NIMCET PYQ
If $(4,-3)$ and $(-9,7)$ are two vertices of a triangle and $(1,4)$ is its centroid, find the area of the triangle.






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NIMCET Previous Year PYQ NIMCET NIMCET 2012 PYQ

Solution

Let the third vertex be $(x,y)$. Centroid formula: $\displaystyle \left(\frac{4 + (-9) + x}{3}, \frac{-3 + 7 + y}{3}\right) = (1,4)$

From x–coordinate: $\dfrac{4 - 9 + x}{3} = 1$ 
$x - 5 = 3$ 
$x = 8$

From y–coordinate: $\dfrac{-3 + 7 + y}{3} = 4$ 
$y + 4 = 12$ 
$y = 8$ 
So third vertex is: $(8,8)$

Vertices: $A(4,-3), B(-9,7), C(8,8)$ 

Area: $\displaystyle \text{Area} = \frac{1}{2}\left| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right|$ Substitute: 
$\displaystyle = \frac{1}{2}\left|4(7-8) + (-9)(8+3) + 8(-3-7) \right|$ 
$\displaystyle = \frac{1}{2}\left| 4(-1) - 9(11) + 8(-10) \right|$ 
$\displaystyle = \frac{1}{2}\left| -4 - 99 - 80 \right|$ 
$\displaystyle = \frac{1}{2} \times 183$ $\displaystyle = \frac{183}{2}$

NIMCET PYQ
In $\triangle ABC$, $B = 45^\circ, C = 105^\circ, c=\sqrt{2}$. Find side $a$ and $b$.





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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ

Solution

$A = 180 - (45 + 105) = 30^\circ$ Using Law of Sines: $ \displaystyle \frac{a}{\sin A} = \frac{c}{\sin C} $ $ \displaystyle a = \frac{\sin 30^\circ}{\sin 105^\circ}\cdot \sqrt{2} $ Compute: $\sin 30^\circ = \frac12$ $\sin 105^\circ = \sin(60+45)=\frac{\sqrt{6}+\sqrt{2}}{4}$ $ \displaystyle a = \sqrt{2}\cdot \frac{1/2}{(\sqrt{6}+\sqrt{2})/4} = \frac{\sqrt{2}}{(\sqrt{6}+\sqrt{2})/2} = \frac{2\sqrt{2}}{\sqrt{6}+\sqrt{2}} $ After rationalizing: $ \displaystyle a=\sqrt{3}-1 $ Similarly, $ \displaystyle b=\sqrt{2}(\sqrt{3}-1) $

NIMCET PYQ
Let $\Delta ABC$ be a triangle whose area is $10\sqrt{3}$ units with side lengths $|AB|= 8$ units and $|AC|=5$ units. Find possible values of the angle A





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NIMCET Previous Year PYQ NIMCET NIMCET 2013 PYQ

Solution


NIMCET PYQ
Triangle sides: $x^{2}+x+1,\ 2x+1,\ x^{2}-1$. Largest angle?





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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ

Solution

Largest side ≈ $x^{2}+x+1$. Use cosine rule, solve quadratic relationships → angle opposite largest side = $150^\circ$

NIMCET PYQ
In a $\Delta ABC$ , $(c+a+b)(a+b-c)=ab$ The measure of the angle C is.





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NIMCET Previous Year PYQ NIMCET NIMCET 2013 PYQ

Solution


NIMCET PYQ

If $\Delta=a^2-(b-c)^2$, where $\Delta$ is the are of the triangle ABC, then $tanA=$






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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution


NIMCET PYQ
The equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, –1). The length of the side of the triangle is.





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NIMCET Previous Year PYQ NIMCET NIMCET 2013 PYQ

Solution


NIMCET PYQ
In a triangle ABC, if the tangent of half the difference of two angles is equal to one third of the tangent of half the sum of the angles, then the ratio of the sides opposite to the angles is





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NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

Solution


NIMCET PYQ
Let $M$ be a point inside the triangle $ABC$. Then which one of the following is true?




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NIMCET Previous Year PYQ NIMCET NIMCET 2008 PYQ

Solution

By triangle inequality in $\triangle ABM$ and $\triangle ACM$, $AB>MB-AM,\ AC>MC-AM$ Adding, $AB+AC>MB+MC$ Answer: $\boxed{AB+AC>MB+MC}$

NIMCET PYQ
In an acute-angled ΔABC the least value of secA+secB+secC is 





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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution


NIMCET PYQ
If in a triangle ABC, the altitudes from the vertices A, B, C on opposite sides are in HP, then sin A, sin B, sin C are in





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NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution

If altitudes of a triangle are in HP then its side will be in AP because sides are inverse proportion to height as area is constant. a, b, c are sides of triangle.
a, b, c are in A.P.
sin A, sin B, sin C are in A.P.

NIMCET PYQ
In a triangle, if the sum of two sides is x and their product is y such that (x+z)(x-z)=y, where z is the third side of the triangle , then triangle is 





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NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ

Solution


NIMCET PYQ
In a triangle ABC, $a\cos ^2\frac{C}{2}+\, c\, \, {\cos }^2\frac{A}{2}=\frac{3b}{2}$ then the sides of the triangle are in





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NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ

Solution


NIMCET PYQ
In a triangle ABC, let angle C = π/2. If R is the inradius and R is circumradius of the triangle ABC, then 2(r + R) equals





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NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution


NIMCET PYQ
In $\triangle ABC$, $R$ is circumradius and $8R^2=a^2+b^2+c^2$. Then $\triangle ABC$ is:





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NIMCET Previous Year PYQ NIMCET NIMCET 2010 PYQ

Solution

In any triangle: $a^2+b^2+c^2 = 8R^2$ ⇔ right-angled triangle identity.

NIMCET PYQ
In a triangle ABC, angle A=90° and D is the midpoint of AC. What is the value of  equal to?






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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution


NIMCET PYQ
What is the largest area of an isosceles triangle with two edges of length 3?






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NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution


NIMCET PYQ
If A, B and C is three angles of a ΔABC, whose area is Δ. Let a, b and c be the sides opposite to the angles A, B and C respectively. Is $s=\frac{a+b+c}{2}=6$, then the product $\frac{1}{3} s^{2} (s-a)(s-b)(s-c)$ is equal to





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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ

Solution


NIMCET PYQ
The perimeter of a $\Delta ABC$ is 6 times the arithmetic mean of the sines of its angles. If the side a is 1, then the angle A is





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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Quick Solution

Given: Perimeter = 6 × Arithmetic Mean of sin A, sin B, sin C

Using Law of Sines: \( \frac{a}{\sin A} = 2R \), and a = 1 ⇒ \( \sin A = \frac{1}{2R} \)

Assume \( A = 30^\circ \Rightarrow \sin A = \frac{1}{2} \Rightarrow 2R = 2 \)

⇒ b = 2 sin B, c = 2 sin C

Perimeter = \( 1 + b + c = 1 + 2 \sin B + 2 \sin C \)

Mean = \( \frac{\sin A + \sin B + \sin C}{3} \)

Check: \( 1 + 2\sin B + 2\sin C = 6 \cdot \frac{1/2 + \sin B + \sin C}{3} \) ✅

✅ Final Answer: 30°


NIMCET PYQ
In ΔABC, if a = 2, b = 4 and ∠C = 60°, then A and B are respectively equal to





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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ

Solution


NIMCET PYQ
If the angles of a triangle are in the ratio 2 : 3 : 7, then the ratio of the sides opposite to these angles is





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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution


NIMCET PYQ
A harbour lies in a direction 60° South of West from a fort and at a distance 30 km from it, a ship sets out from the harbour at noon and sails due East at 10 km an hour. The time at which the ship will be 70 km from the fort is





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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution



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