Q4. - Geometry

Question

Step-by-Step Solution

Answer

The line $JM$ is an altitude of the triangle $JAM$ from the vertex $J$.

Step 1. Given information:

   JAM is a right-angled triangle.

Step 2. Concept used:

 We use basic geometric and angle concept.                                  

Step 3. Applying the concept:

line $JM$ is perpendicular to the side $MA.$

Therefore, the line $JM$ is an altitude of the triangle $JAM$from the vertex $J$.

The line $AM$ is also an altitude of the triangle $JAM$ from the vertex $J$.

Step 1. Given information:

   JAM is a right-angled triangle.

Step 2. Concept used:

 We use basic geometric and angle concept.                                  

Step 3. Applying the concept:

By observing the given figure:

The line $AM$ is an altitude of the triangle from the vertex $A$.

The line $JM$ is an altitude of the triangle $JAM$ from the vertex $J$.

Step 1. Given information:

 $JAM$ is a right-angled triangle.

Step 2. Concept used:

 We use basic geometric and angle concept.                                  

Step 3. Applying the concept:

From the point $M$ draw two arcs on the line $JA$ with equal lengths.

From $X$ and $Y$ draw other two arcs and mark their intersection as a point $Z$ and join  $MZ$. $MXZ$ and $JA$ intersect at a point $D.$

  $MD$ is an altitude of triangle $JAM$ that contains the vertex $M.$ .

Step 1. Given information:

  $JAM$   is a right-angled triangle.

Step 2. Concept used:

 We use basic geometric and angle concept.                                  

Step 3. Applying the concept:

The perpendicular bisectors of a right-angled trianglemeet at a point on the hypotenuse of this triangle as shown below:

The final answer is: yes, it supports.

Step 1. Given information:

  $JAM$   is a right-angled triangle.

Step 2. Concept used:

 We use basic geometric and angle concept.                                  

Step 3. Applying the concept:

In the given figure,

$\begin{array}{l}\overline{XJ}=\overline{XA}\\ \overline{XJ}=\overline{XM}\\ \overline{XA}=\overline{XM}\\ =\overline{XJ}\end{array}$

$X$ is equidistant from the vertices $M$, $J$, $A$. hence, the theorem $10-2$ supports $\left(d\right)$.