Qus : 1
NIMCET PYQ
3
If $(4,-3)$ and $(-9,7)$ are two vertices of a triangle and $(1,4)$ is its centroid, find the area of the triangle.
1
$\dfrac{138}{2}$ 2
$\dfrac{319}{2}$ 3
$\dfrac{183}{2}$ 4
$\dfrac{381}{2}$ Go to Discussion
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:
$\displaystyle
= \frac{1}{2}\left|4(7-8) + (-9)(8+3) + 8(-3-7) \right|$
$\displaystyle
= \frac{1}{2}\left| 4(-1) - 9(11) + 8(-10) \right|$
$\displaystyle
= \frac{1}{2}\left| -4 - 99 - 80 \right|$
$\displaystyle
= \frac{1}{2} \times 183$
$\displaystyle
= \frac{183}{2}$
Qus : 2
NIMCET PYQ
1
In $\triangle ABC$, $B = 45^\circ, C = 105^\circ, c=\sqrt{2}$.
Find side $a$ and $b$.
1
$A=30^\circ, a=\sqrt{3}-1, b=\sqrt{2}(\sqrt{3}-1)$ 2
$A=30^\circ, a=\sqrt{3}+1, b=\sqrt{2}(\sqrt{3}-1)$ 3
$A=30^\circ, a=1-\sqrt{3}, b=\sqrt{2}(\sqrt{3}+1)$ 4
$A=30^\circ, a=\sqrt{3}-1,; b=\sqrt{2}(\sqrt{3}+1)$ Go to Discussion
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) $
Qus : 9
NIMCET PYQ
2
Let $M$ be a point inside the triangle $ABC$. Then which one of the following is true?
1
$AB+AC2
$AB+AC>MB+MC$ 3
$AB+AC\le MB+MC$ 4
None of these Go to Discussion
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: $\boxed{AB+AC>MB+MC}$
Qus : 11
NIMCET PYQ
3
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
1
HP 2
AGP 3
AP 4
GP Go to Discussion
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.
Qus : 18
NIMCET PYQ
2
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
1
$2 \Delta$ 2
$2 \Delta^{2}$ 3
$\sqrt{2} \Delta$ 4
$\Delta^{2}$ Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2014 PYQ
Solution
Qus : 19
NIMCET PYQ
1
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
1
$\frac{\pi}{6}$ 2
$\frac{\pi}{3}$ 3
$\frac{\pi}{2}$ 4
$\pi$ Go to Discussion
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°
Qus : 21
NIMCET PYQ
1
If the angles of a triangle are in the ratio 2 : 3 : 7, then the ratio of the sides opposite to these
angles is
1
$$\sqrt{2} : 2 : \sqrt{3}+1$$ 2
$$2 : \sqrt{2} : \sqrt{3}+1$$ 3
$$2 : \sqrt{2} : \frac{\sqrt{2}}{\sqrt{3}-1}$$ 4
$$\frac{1}{\sqrt{2}} : 2 : \frac{\sqrt{3}+1}{2}$$ Go to Discussion
NIMCET Previous Year PYQ
NIMCET NIMCET 2015 PYQ
Solution
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