1
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\mathrm{A} \equiv(0,0), \mathrm{B}(3,0), \mathrm{C}(0,-4)$ are vertices of $\triangle A B C$, then the co-ordinates of incentre of $\triangle \mathrm{ABC}$ is

A
$\left(\frac{45}{14}, \frac{3}{14}\right)$
B
$\left(\frac{45}{14}, \frac{-3}{14}\right)$
C
$\left(\frac{3}{14}, \frac{45}{14}\right)$
D
$\left(\frac{-3}{14}, \frac{45}{14}\right)$
2
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+2
-0
In a triangle ABC with usual notations if $b \sin C(b \cos C+c \cos B)=42$, then area of triangle $\mathrm{ABC}=$
A
42 sq. units
B
21 sq. units
C
24 sq. units
D
12 sq. units
3
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+2
-0

In a triangle with one of the angles $120^{\circ}$, the lengths of the sides form an A.P. If length of the greatest side is 7 m , then the area of the triangle is

A
$\frac{15 \sqrt{3}}{4} \mathrm{~m}^2$
B
$\frac{15 \sqrt{3}}{2} \mathrm{~m}^2$
C
$\frac{15}{2} \mathrm{~m}^2$
D
$\frac{15}{4} \mathrm{~m}^2$
4
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+2
-0

In $\triangle \mathrm{ABC}$, with usual notations, if $\cos \frac{B}{2}=\sqrt{\frac{c+a}{2 a}}$, then $a^2=$

A
$\mathrm{b}^2-\mathrm{c}^2$
B
$\mathrm{b}+\mathrm{c}$
C
$\mathrm{b}^2+\mathrm{c}^2$
D
$\mathrm{b}-\mathrm{c}$
MHT CET Subjects
EXAM MAP