
Two triangles are $...........$ if two sides and included angles of one triangle are equal to two sides and included angles of the other triangle.
$\left( a \right){\text{ congruent}}$
$\left( b \right){\text{ unequal}}$
$\left( c \right){\text{ equilateral}}$
$\left( d \right){\text{ none of these}}$
Answer
570.9k+ views
Hint:
As we know if different sides and the included point of one triangle are equivalent to the relating sides and point of another triangle. So this will be the condition of $SAS$ and through the use of this we can simply answer this question.
Complete step by step solution:
Two triangles are harmonious if different sides and included points of one triangle are equivalent to different sides and included points of the other triangle. This is the SAS condition for congruence.
Therefore, the option $\left( a \right)$ will be correct.
Additional information:
For any given set of two sides, with any particular angle between them, there is one and only one possible way to draw the third side of the triangle. Ideally, we ought to have the option to see this with a sketch or two: there is just one spot where the free finishes of the two lines can be, so there is just a single method to interface them up, which fixes the length of the third line and the size of the other two points.
If we have two triangles and we realize that two of the sides of the triangle $A$ match different sides of the triangle$B$, and the point between these sides is the equivalent in the two cases, at that point you ought to have the option to see that indeed these two triangles are precise of one another, regardless of whether one of them is a convoluted or flipped-over duplicate of the other. That is the thing that we mean when we state they are "compatible", and SAS is only an advantageous shortened form for "side-angle side": different sides, and the points between them, the equivalent.
Note:
For this type of question, we should always go through the options as it gives much more idea about the answer. And also we should know the meaning of the question stated to us and in this way we can easily find out the correct options.
As we know if different sides and the included point of one triangle are equivalent to the relating sides and point of another triangle. So this will be the condition of $SAS$ and through the use of this we can simply answer this question.
Complete step by step solution:
Two triangles are harmonious if different sides and included points of one triangle are equivalent to different sides and included points of the other triangle. This is the SAS condition for congruence.
Therefore, the option $\left( a \right)$ will be correct.
Additional information:
For any given set of two sides, with any particular angle between them, there is one and only one possible way to draw the third side of the triangle. Ideally, we ought to have the option to see this with a sketch or two: there is just one spot where the free finishes of the two lines can be, so there is just a single method to interface them up, which fixes the length of the third line and the size of the other two points.
If we have two triangles and we realize that two of the sides of the triangle $A$ match different sides of the triangle$B$, and the point between these sides is the equivalent in the two cases, at that point you ought to have the option to see that indeed these two triangles are precise of one another, regardless of whether one of them is a convoluted or flipped-over duplicate of the other. That is the thing that we mean when we state they are "compatible", and SAS is only an advantageous shortened form for "side-angle side": different sides, and the points between them, the equivalent.
Note:
For this type of question, we should always go through the options as it gives much more idea about the answer. And also we should know the meaning of the question stated to us and in this way we can easily find out the correct options.
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