 Hello friends, and how are you all doing today? My name is Priyanka and the question says draw a right triangle in which the sides other than hypotenuse are of length 4 centimeter and 3 centimeter. Then construct another triangle whose sides are 5 by 3 times the corresponding sides of the given triangle. So let's start with the solution. We would be doing the construction over here and we have already written down the steps of construction for you. The first step of construction is we need to draw a right triangle, let it be the triangle ABC, right angle that B and we are given that the measurement of the sides other than hypotenuse are of length 4 centimeter and 3 centimeter. So let it be AB is equal to 4 centimeter and BC equal to 3 centimeter. So we have a right triangle where BC first of all is 3 centimeter. This is BC, right angle at B and AB is of measure 4 centimeter. So this is A, this is B and this is C to complete the triangle. We need to join AC. So this is the right triangle at B where BC is 3 centimeter and AB is 4 centimeter. Right, now the next step is to draw any ray BX making an acute angle with BC on the side opposite to the vertex A. So we have ray BX opposite to the vertex A. Now we need to locate five points as five is greater in five and three in five by in five by three, right? So we need to locate five points, let it be B1, B2, B3 and so on such that let it be B1, B2, B3, B4, B5. Such that BB1 is equal to B1, B2 is equal to B2, B3 is equal to B3, B4 and is equal to B4, B5, right? Further we need to join B3 to C as 3 is the smaller of five and three in five by three, right? So we need to join B3 to C now. So let us join B3 to C like this and now we need to extend BC and we need to draw a line which is parallel to B3C through B5 which is intersecting the extended BC at C dash. So here we have a line which is parallel to B3C. That is B5C dash This was our fourth step, which is we have drawn a line through B5 parallel to B3C intersecting the extended line segment BC at C dash. Now the next is to draw a line through C dash parallel to C A intersecting the extended line segment BA at A dash. So now we need to extend AB first and we need to now draw a line which is parallel to C A through C dash intersecting the extended BA at A dash and A dash BC dash is the required triangle. Now for justification for the construction that we have done above we have triangle A dash BC dash similar to triangle ABC that we have done by construction, right? So this implies that the corresponding sides are proportional to each other. So we have A dash B upon AB equal to A dash C dash upon AC is equal to BC dash upon BC. But we have drawn BC BC dash upon BC equivalent to BB3 upon BB5 Sorry, it's equivalent to BB5 upon BB3, which is equal to 5 upon 3. So this means that BC dash upon BC is 5 by 3. So the whole the corresponding sides are proportional to each other and the ratio is 5 is to 3. So this is the required justification for the construction that we have done above. Hope you understood the whole process. Well, do remember your steps of construction, write it neatly and have a very nice day ahead.