 Hello friends and how are you all doing today? The question says construct an Isosceles triangle whose base is 8 centimeter and altitude 4 centimeter and then another triangle whose sides are one and a half times The corresponding sides of the isosceles triangle. So these steps of constructions will be the first step is to draw a triangle That is an isosceles triangle whose base is 8 centimeter and altitude 4 centimeter So this is an isosceles triangle where AB is equal to AC altitude Let's say this has AD as 4 centimeter and we have BC as 8 centimeter right now we need to Construct another triangle whose sides are one and a half times One and a half times of corresponding sides of the isosceles triangle. So the second step is To draw any vertex Or sorry any ray BX Making an acute angle with BC so we have with Bayesian array AX Sorry array BX which is opposite to the vertex A right and this is an isosceles. This is an acute angle now the third step is To locate three points Because one and a half means Three by two and the greater of three three and two in three by two is three. So we locate three points B1 B2 and B3 on BX such that B B1 is equal to B1 B2 is equal to B2 B3 So here we have B B1 equal to B1 B2 equal to B2 B3 on line Be on sorry on ray BX now what we need to do is we need to join B2 to C Let me write it down also Now the fourth step is to join B2 that is the second point as three is sorry as Two is smaller in three by two right so we will join B2 to C and draw a line through B three parallel to B2 C Intersecting the extended line segment BC at C dash okay So let us join B2 to C first and now we will extend BC and We need to now draw a line parallel through B3 Intersecting the extended BC at C dash this is with the help of the compass Now the fifth step is now to draw a line through C dash parallel to C a intersecting the extended line segment B a at a dash so Let us first extend the line B a and Now we need to draw a line which is parallel to a C through C dash intersecting the extended B a at a dash right so a dash BC dash is the required Triangle right now for justification we know that Triangle a BC is similar to triangle a dash BC dash that we have constructed So this implies that a B upon a dash B is equal to a C upon a dash C dash is Equal to BC upon BC dash isn't it? But we also know that BC upon BC dash is equal to B B to upon B B 3 which is equal to 2 by 3 or We can say that BC dash upon BC taking its reciprocal is equal to the reciprocal of 2 by 3 that is 3 by 2 and Therefore we can write that a dash B upon a B is equal to a dash C dash upon a C is Equal to BC dash upon BC is equal to 3 by 2 So this is the required justification for our construction. Hope you understood the whole process well do Make your diagram neatly and hope to see you next bye for now