Download App

Triangle And Its Angles — Maths STD 9 — Question

CBSE BoardEnglish MediumSTD 9MathsTriangle And Its Angles1 Mark
MCQ
If $A B C$ and $D E F$ are two triangles such that $\triangle A B C \cong \triangle F D E$ and $A B=5 cm, \angle B=40^{\circ}$ and $\angle A=80^{\circ}$. Then, which of the following is true?
  • A
    $D F=5 cm, \angle F=60^{\circ}$
  • B
    $D E=5 cm, \angle E=60^{\circ}$
  • ✓
    $D F=5 cm, \angle E=60^{\circ}$
  • D
    $D E=5 cm, \angle D=40^{\circ}$
✓

Answer

Correct option: C.
$D F=5 cm, \angle E=60^{\circ}$
c

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q" from Triangle And Its Angles→Triangle And Its Angles — all question types→Maths STD 9 question papers→CBSE Board STD 9 question papers→CBSE Board question paper generator→

Similar questions

The perimeter of an equilateral triangle is $60m.$ The area is:
→
If $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, then $x^2+x y+y^2=$
→
Write the correct answer in the following: The graph of the linear equation $2x + 3y = 6 $ is a line which meets the $X-$axis at the point.
→
If the base radius and the height of a right circular cone are increased by $20\%,$ then the percentage increase in volume is approximately:
→
In Fig. $ABCD$ is a cyclic quadrilateral. If $\angle\text{BAC}=50^\circ$ and $\angle\text{DBC}=60^\circ$ then find $\angle\text{BCD}=\ ?$
→
If $(3, 2)$ is the solution $3x - ky = 5$, then $k$ equals of the equation.
→
Write the correct answer in the following: The probability that bulbs selected randomly from the lot has life less than $900$ hours is:
→
Two point on a circle makes the:
→
The number $1.\overline{27}$ in the form $\frac{\text{p}}{\text{q}},$ where $p$ and $q$ are integers and $\text{q}\neq0,$ is:
→
$ABCD$ is a parallelogram and $E$ and $F$ are the centroids of triangles $ABD$ and $BCD$ respectively, then $EF =$
→