MCQ
In the adjoining figure, $BC = AD$, $\text{CA}\bot\text{AB}$ and $\text{BD}\bot\text{AB}.$ The rule by which $\triangle\text{ABC}\cong\triangle\text{BAD}$ is:
- A$SAS$
- ✓$RHS$
- C$SSS$
- D$ASA$
✓
Answer
Correct option: B.
$RHS$
In $\triangle\text{ABC}$ and $\angle\text{BAC}=\angle\text{ABD}$
$BAD$, we have (Right angles)
$BC = AD $(Hypotentuses and Given)
$AB = AB$ (common in both)
Hence, $\triangle\text{ABC}\cong\triangle\text{BAD}$ by $RHS$ criterion.
$BAD$, we have (Right angles)
$BC = AD $(Hypotentuses and Given)
$AB = AB$ (common in both)
Hence, $\triangle\text{ABC}\cong\triangle\text{BAD}$ by $RHS$ criterion.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The largest sphere is cut off from a cube of side 6 cm. The volume of the sphere will be
The measure of each side of an equilateral triangle whose area is $\sqrt{3}\text{cm}^2$ is:
→The length of the longest rod that can be placed in a room of dimension $(10m × 10m × 5m)$ is:
→The mean of $100$ observations is $50$. If one of the observations $0 5$ is replaced by $150,$ the resulting mean will be:
→The positive solutums of the equation $a x+b y+c=0$ always lie in the
→Write the correct answer in the following: In Fig. if $\text{OP}||\text{RS},$ $\angle\text{OPQ}=110^\circ$ and $\angle\text{QRS}=130^\circ,$ then $\angle\text{PQR}$ is equal to,
→Area of an equilateral triangle of side $2\ cm$ is:
→The area of an isosceles right angled triangle of equal side $30\ cm,$ is given as:
→When $p(x) = x^3 - ax^2 + x$ is divided by $(x - a),$ the remainder is:
→The length of each side of an equilateral triangle having an area of $9 \sqrt{3}cm^2$ is
→